JEE Fluid Mechanics Questions

When the ball reaches terminal velocity, the net force acting on it is zero. The forces involved are:

1. Weight of the ball acting downward: $$W = Mg$$

2. Buoyant force acting upward: This equals the weight of the glycerine displaced by the ball. Let the volume of the ball be $$V$$. The buoyant force is given by $$F_b = \rho_g V g$$, where $$\rho_g$$ is the density of glycerine.

3. Viscous force acting upward: Denoted as $$F_v$$.

Given that the density of glycerine is half the density of the ball, let the density of the ball be $$\rho_b$$. Then:

$$\rho_g = \frac{1}{2} \rho_b$$

The mass of the ball is related to its density and volume by:

$$M = \rho_b V$$

Solving for volume:

$$V = \frac{M}{\rho_b}$$

Substitute this into the buoyant force formula:

$$F_b = \rho_g V g = \left(\frac{1}{2} \rho_b\right) \times \left(\frac{M}{\rho_b}\right) g = \frac{1}{2} M g$$

At terminal velocity, the net force is zero, so upward forces equal downward forces:

$$F_b + F_v = W$$

Substitute the expressions:

$$\frac{1}{2} M g + F_v = M g$$

Solve for the viscous force $$F_v$$:

$$F_v = M g - \frac{1}{2} M g = \frac{1}{2} M g$$

Therefore, the viscous force acting on the ball is $$\frac{Mg}{2}$$.

The correct option is D. $$\frac{Mg}{2}$$

The tank is completely filled, so the vertical distance between the top free surface and the hole is the full column above the hole.

Depth of hole below the top surface:
$$h = 1.6\,\text{m} - 0.9\,\text{m} = 0.7\,\text{m}$$

1. Pressure at the hole due to the water column alone (hydrostatic law $$P = \rho g h$$):
Density of water $$\rho = 1000\,\text{kg/m}^3$$, acceleration due to gravity $$g = 10\,\text{m/s}^2$$.
$$P_{\text{water}} = \rho g h = 1000 \times 10 \times 0.7 = 7000\,\text{Pa}$$

2. Extra pressure caused by the 50 kg load placed on the water surface:
Area of the tank’s cross-section $$A = 0.5\,\text{m}^2$$.
Force of the load $$F = mg = 50 \times 10 = 500\,\text{N}$$.
Extra pressure $$P_{\text{load}} = \dfrac{F}{A} = \dfrac{500}{0.5} = 1000\,\text{Pa}$$

3. Total gauge pressure at the level of the hole:
$$P_{\text{total}} = P_{\text{water}} + P_{\text{load}} = 7000 + 1000 = 8000\,\text{Pa}$$

4. Speed of efflux (Torricelli’s theorem with an added surface pressure):
For a small orifice the velocity is obtained from energy balance
$$v = \sqrt{\dfrac{2 P_{\text{total}}}{\rho}}$$

Substituting values:
$$v = \sqrt{\dfrac{2 \times 8000}{1000}} = \sqrt{16} = 4\,\text{m/s}$$

Therefore, the velocity of water emerging from the hole is 4 m/s.
Option D.

Since the vessels have equal cross-sectional areas and are connected at the bottom, water will flow until the heights are equal. The total volume is conserved.

$$h_f = \frac{h_1 + h_2}{2} = \frac{10 + 6}{2} = 8 \text{ m}$$

The potential energy of a liquid column of mass $$M$$ and height $$h$$ is $$U = Mg\frac{h}{2}$$ (where $$h/2$$ is the center of mass). In terms of density ($$\rho$$) and area ($$A$$), this is $$U = \frac{1}{2} \rho Ag h^2$$.

$$U_i = \frac{1}{2} \rho Ag (h_1^2 + h_2^2)$$

$$U_i = \frac{1}{2} \times 10^3 \times 2 \times 10 \times (10^2 + 6^2) = 10^4 \times (100 + 36) = 136 \times 10^4 \text{ J}$$

$$U_f = \frac{1}{2} \rho Ag (h_f^2 + h_f^2) = \rho Ag h_f^2$$

$$U_f = 10^3 \times 2 \times 10 \times (8^2)$$

$$U_f = 2 \times 10^4 \times 64 = 128 \times 10^4 \text{ J}$$

$$W_g = U_i - U_f$$ (work done by gravity is the negative change in potential energy)

$$W_g = (136 - 128) \times 10^4 = 8 \times 10^4 \text{ J}$$

Water flows in a horizontal pipe. When valve is closed, pressure is $$P_1$$. When opened, pressure falls to $$P_2$$. We need to find the proportionality of speed.

Apply Bernoulli's equation:

When the valve is closed, velocity = 0, pressure = $$P_1$$.

When the valve is opened, velocity = v, pressure = $$P_2$$.

$$P_1 + \frac{1}{2}\rho(0)^2 = P_2 + \frac{1}{2}\rho v^2$$

$$P_1 - P_2 = \frac{1}{2}\rho v^2$$

$$v = \sqrt{\frac{2(P_1 - P_2)}{\rho}}$$

Therefore, $$v \propto \sqrt{P_1 - P_2}$$.

The correct answer is Option 4: $$\sqrt{P_1 - P_2}$$.

Mass of the cube is given as $$m = 400\text{ g} = 0.4\text{ kg}$$ and edge length is $$a = 10\text{ cm} = 0.1\text{ m}$$.

Total volume of the cube is$$V_{\text{total}} = a^3 = (0.1)^3 = 1.0\times 10^{-3}\text{ m}^3\;. $$

For floating equilibrium, weight of cube = buoyant force. We state the formula:

Buoyant force = $$\rho_{\text{water}}\;V_{\text{submerged}}\;g$$
Weight = $$m\,g$$
Equating, we get$$\rho_{\text{water}}\;V_{\text{submerged}}\;g = m\,g\quad-(1)$$

Cancel $$g$$ on both sides of $$(1)$$ to obtain the submerged volume:

$$V_{\text{submerged}} = \frac{m}{\rho_{\text{water}}} \quad-(2)$$

Substitute $$m = 0.4\text{ kg}$$ and $$\rho_{\text{water}} = 1000\text{ kg/m}^3$$ into $$(2)$$:

$$V_{\text{submerged}} = \frac{0.4}{1000} = 4.0\times 10^{-4}\text{ m}^3\;. $$

Convert $$V_{\text{submerged}}$$ to cubic centimetres using $$1\text{ m}^3 = 10^6\text{ cm}^3$$:

$$V_{\text{submerged}} = 4.0\times 10^{-4}\times 10^6 = 400\text{ cm}^3\;. $$

Therefore, the volume of the cube outside the water is

$$V_{\text{outside}} = V_{\text{total}} - V_{\text{submerged}} = 1000\text{ cm}^3 - 400\text{ cm}^3 = 600\text{ cm}^3\;. $$

Final Answer: Option B, $$600\text{ cm}^3\;.$$

For a small sphere moving slowly through a viscous liquid, Stokes’ law gives the viscous drag force as $$F_{v} = 6 \pi \eta r v$$, where $$\eta$$ is the viscosity, $$r$$ the radius of the sphere and $$v$$ its speed.

At terminal velocity, the sphere moves with constant speed, so the net force on it is zero. The downward forces (weight) are balanced by the upward forces (buoyant force + viscous drag):

$$\text{Weight}\;-\;\text{Buoyant force}\;=\;\text{Viscous drag}$$
$$\bigl(\rho_{s} V g - \rho_{o} V g\bigr)\;=\;6 \pi \eta r v_{t}$$

Here
$$\rho_{s} = 7825\;\text{kg m}^{-3}$$ (density of steel)
$$\rho_{o} = 925\;\text{kg m}^{-3}$$ (density of oil)
$$V = \tfrac{4}{3}\pi r^{3}$$ is the volume of the sphere
$$v_{t} = 2.45 \times 10^{-2}\;\text{m s}^{-1}$$ is the terminal velocity.

Cancelling the common factor $$V g$$ and simplifying:

$$(\rho_{s}-\rho_{o})\;\tfrac{4}{3}\pi r^{3} g = 6 \pi \eta r v_{t}$$

Divide both sides by $$\pi r$$ to isolate $$\eta$$:

$$\eta = \frac{2 r^{2} g\,(\rho_{s}-\rho_{o})}{9 v_{t}}\quad -(1)$$

Now substitute the numerical data.
Diameter of ball $$= 3.6\;\text{mm} \Rightarrow r = 1.8\;\text{mm} = 1.8 \times 10^{-3}\;\text{m}$$

Density difference:
$$(\rho_{s}-\rho_{o}) = 7825 - 925 = 6900\;\text{kg m}^{-3}$$

Using $$g = 9.8\;\text{m s}^{-2}$$ and $$v_{t} = 2.45 \times 10^{-2}\;\text{m s}^{-1}$$ in equation $$(1)$$:

$$\eta = \frac{2\,(1.8 \times 10^{-3})^{2}\,(9.8)\,(6900)}{9\,(2.45 \times 10^{-2})}$$

Step-wise calculation:
$$r^{2} = (1.8 \times 10^{-3})^{2} = 3.24 \times 10^{-6}\;\text{m}^{2}$$
$$2 r^{2} = 2 \times 3.24 \times 10^{-6} = 6.48 \times 10^{-6}$$
$$6.48 \times 10^{-6} \times 6900 = 4.471 \times 10^{-2}$$
$$4.471 \times 10^{-2} \times 9.8 = 0.438\;(\text{approximately})$$
Denominator $$= 9 \times 2.45 \times 10^{-2} = 0.2205$$

Finally,
$$\eta = \frac{0.438}{0.2205} \approx 1.99\;\text{Pa·s}$$

The viscosity of the oil is therefore $$\mathbf{1.99\;Pa\cdot s}$$.

Hence, the correct option is Option D.

We analyze each statement about the experiment measuring viscosity $$\eta$$ using a falling ball in a liquid.

Statement A: Graph between terminal velocity V and R is a parabola.

The terminal velocity formula from Stokes' law is: $$V_t = \frac{2R^2(\rho_s - \rho_l)g}{9\eta}$$

Since $$V_t \propto R^2$$, the graph of $$V_t$$ vs $$R$$ is indeed a parabola. Statement A is CORRECT.

Statement B: Terminal velocities of different diameter balls are constant for a given liquid.

Since $$V_t \propto R^2$$, different diameter balls have different terminal velocities. Statement B is INCORRECT.

Statement C: Measurement of terminal velocity depends on temperature.

Viscosity $$\eta$$ is temperature-dependent. As temperature changes, the terminal velocity changes. Statement C is CORRECT.

Statement D: This experiment can assess the density of a given liquid.

From the terminal velocity equation $$V_t = \frac{2R^2(\rho_s - \rho_l)g}{9\eta}$$, if we know $$V_t$$, $$R$$, $$\rho_s$$, and $$\eta$$, we can determine $$\rho_l$$. Statement D is CORRECT.

Statement E: If balls are dropped with some initial speed, the value of $$\eta$$ will change.

Viscosity is an intrinsic property of the liquid. Regardless of the initial speed of the ball, it will eventually reach the same terminal velocity. The measured $$\eta$$ remains unchanged. Statement E is INCORRECT.

The correct statements are A, C, and D. The answer is Option C) A, C and D Only.

Work to break a drop of radius R into 27 drops is 10 J and we need the work to break it into 64 drops.

Since the work is given by $$W = T \times \Delta A = T(n \times 4\pi r^2 - 4\pi R^2),$$

and volume conservation implies $$n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow r = R/n^{1/3},$$

substituting in yields $$W = 4\pi TR^2(n^{1/3}-1).$$

For 27 drops $$W_1 = 4\pi TR^2(27^{1/3}-1) = 4\pi TR^2(3-1) = 8\pi TR^2 = 10$$ J

For 64 drops $$W_2 = 4\pi TR^2(64^{1/3}-1) = 4\pi TR^2(4-1) = 12\pi TR^2$$

This gives $$\frac{W_2}{W_1} = \frac{12\pi TR^2}{8\pi TR^2} = \frac{3}{2},$$

therefore $$W_2 = \frac{3}{2} \times 10 = 15$$ J

An air bubble of radius $$r = 0.1$$ cm = $$0.001$$ m at depth $$h = 20$$ cm = $$0.20$$ m has excess pressure of 2100 N/m² above atmospheric. We need the surface tension.

The pressure inside the bubble exceeds atmospheric pressure by both the hydrostatic pressure and the surface tension pressure:

$$ P_{inside} - P_{atm} = \rho g h + \frac{2T}{r} $$

Note: For a bubble in a liquid (not a soap bubble), there is only one surface, so the excess pressure due to surface tension is $$\frac{2T}{r}$$.

$$ 2100 = 1000 \times 10 \times 0.20 + \frac{2T}{0.001} $$

$$ 2100 = 2000 + 2000T $$

$$ 100 = 2000T $$

$$ T = \frac{100}{2000} = 0.05 \text{ N/m} $$

The correct answer is Option (2): 0.05.

For a tube with varying cross-sectional areas, the product of the area ($$A$$) and the velocity ($$v$$) is constant.

$$A_1 v_1 = A_2 v_2$$

$$A = \pi r^2$$

$$\pi r_1^2 v_1 = \pi r_2^2 v_2$$

$$r_1^2 v_1 = r_2^2 v_2$$

$$(2)^2 \cdot 2 = (1)^2 \cdot v_2$$

$$4 \cdot 2 = 1 \cdot v_2$$

$$v_2 = 8 \text{ m/s}$$

The window has area $$A = 100\ \text{cm}^2$$.
Since $$1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$$, we get
$$A = 100 \times 10^{-4}\ \text{m}^2 = 0.01\ \text{m}^2$$.

Depth of the window below the liquid surfaces is $$h = 3\ \text{m}$$.

Pressure at depth $$h$$ inside a liquid of density $$\rho$$ is given by
$$P = \rho g h\quad$$(hydrostatic pressure formula).

Side-1 is filled with water: $$\rho_1 = 1000\ \text{kg m}^{-3}$$.
Pressure on this side:
$$P_1 = \rho_1 g h = 1000 \times 10 \times 3 = 3.0 \times 10^{4}\ \text{Pa}$$.

Side-2 is filled with the heavier liquid: $$\rho_2 = 1.5 \times 10^{3}\ \text{kg m}^{-3}$$.
Pressure on this side:
$$P_2 = \rho_2 g h = 1.5 \times 10^{3} \times 10 \times 3 = 4.5 \times 10^{4}\ \text{Pa}$$.

The window experiences a pressure difference
$$\Delta P = P_2 - P_1 = 4.5 \times 10^{4} - 3.0 \times 10^{4} = 1.5 \times 10^{4}\ \text{Pa}$$.

Resultant force on the window due to this pressure difference is
$$F = \Delta P \times A = 1.5 \times 10^{4} \times 0.01 = 150\ \text{N}$$.

Therefore, one must apply a force of $$150\ \text{N}$$ on the hinged door to keep it from opening.

Given a hydraulic lift with input piston area $$A_{1}=6\;\text{cm}^{2}$$ and output piston area $$A_{2}=1500\;\text{cm}^{2}$$. A force $$F_{1}=100\;\text{N}$$ is applied on the input piston to raise the output piston by $$h_{2}=20\;\text{cm}$$. We need to find the work done by the input force in kJ.

According to Pascal’s law, the pressure transmitted is the same throughout the fluid: $$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}\quad-(1)$$ From $$(1)$$, the output force is $$F_{2}=\frac{A_{2}}{A_{1}}\;F_{1}=\frac{1500}{6}\times100=25000\;\text{N}.$$

By conservation of fluid volume, the volume displaced by the input piston equals the volume risen by the output piston: $$A_{1}\,h_{1}=A_{2}\,h_{2}\quad-(2)$$ Substituting values in $$(2)$$: $$6\;\text{cm}^{2}\times h_{1}=1500\;\text{cm}^{2}\times20\;\text{cm}$$ $$h_{1}=\frac{1500\times20}{6}=5000\;\text{cm}=50\;\text{m}.$$

The work done by the input force is $$W=F_{1}\,h_{1}=100\;\text{N}\times50\;\text{m}=5000\;\text{J}=5\;\text{kJ}.$$

Thus, the required work done is $$5\;\text{kJ}$$.

The bulk modulus $$B$$ of a liquid is defined as
$$B = -\,\frac{\Delta P}{\dfrac{\Delta V}{V}}$$
where $$\Delta P$$ is the change in pressure, $$\Delta V$$ is the change in volume (negative for compression), and $$V$$ is the initial volume.

We are given:

• Initial pressure $$P_1 = 1$$ atm, final pressure $$P_2 = 5$$ atm.
Hence the pressure rise is
$$\Delta P = P_2 - P_1 = 5 - 1 = 4$$ atm.

• Take $$1$$ atm $$= 10^{5}$$ Pa, so
$$\Delta P = 4 \times 10^{5}\ \text{Pa}$$.

• Bulk modulus $$B = 2$$ GPa $$= 2 \times 10^{9}$$ Pa.

• Magnitude of volume decrease $$|\Delta V| = 0.8\ \text{cm}^3 = 0.8 \times 10^{-6}\ \text{m}^3$$.

Using the definition of $$B$$ and taking magnitudes (since we know the sign corresponds to compression),
$$B = \frac{\Delta P \, V}{|\Delta V|} \quad\Longrightarrow\quad V = \frac{B\,|\Delta V|}{\Delta P}\,.$$

Substituting the numerical values:
$$V = \frac{2 \times 10^{9}\ \text{Pa} \;\times\; 0.8 \times 10^{-6}\ \text{m}^3}{4 \times 10^{5}\ \text{Pa}}$$

Step-wise calculation:
$$2 \times 0.8 = 1.6$$
$$10^{9} \times 10^{-6} = 10^{3}$$
Numerator $$= 1.6 \times 10^{3} = 1600$$
Denominator $$= 4 \times 10^{5}$$
Therefore,
$$V = \frac{1600}{4 \times 10^{5}} = 0.4 \times 10^{-2}\ \text{m}^3 = 0.004\ \text{m}^3.$$

Conversion to litres (as $$1\ \text{m}^3 = 1000\ \text{L}$$):
$$V = 0.004\ \text{m}^3 \times 1000\ \frac{\text{L}}{\text{m}^3} = 4\ \text{L}.$$

Hence, the initial volume of the liquid was $$4$$ litres.

We need to find the radius of curvature of the common surface when two soap bubbles of radii 2 cm and 4 cm are in contact.

When two soap bubbles of radii $$r_1$$ and $$r_2$$ (where $$r_1 < r_2$$) are in contact, the radius of curvature of the common surface is given by:

$$\frac{1}{R} = \frac{1}{r_1} - \frac{1}{r_2}$$

This is because the excess pressure inside the smaller bubble is greater, and the common surface bulges into the larger bubble.

$$r_1 = 2$$ cm, $$r_2 = 4$$ cm

$$\frac{1}{R} = \frac{1}{2} - \frac{1}{4} = \frac{2-1}{4} = \frac{1}{4}$$

$$R = 4$$ cm

The radius of curvature of the common surface is 4 cm.

An air bubble of radius 1.0 mm at depth 20 cm in a liquid with surface tension 0.095 J/m² and density 10³ kg/m³.

We first calculate the pressure inside the bubble using

$$P_{inside} = P_{atm} + \rho g h + \frac{2T}{r}$$

Then the difference between this pressure and the atmospheric pressure is given by

$$P_{inside} - P_{atm} = \rho g h + \frac{2T}{r}$$

Substituting the numerical values gives

$$= 10^3 \times 10 \times 0.20 + \frac{2 \times 0.095}{1.0 \times 10^{-3}}$$

$$= 2000 + 190$$

$$= 2190 \text{ N/m}^2$$

The answer is 2190.

We need to find $$\alpha$$ where the angular velocity is $$\sqrt{\frac{F}{\alpha M}}$$ for a tube of length 1 m filled with liquid of mass 2M, rotated about one end.

Set up the problem

A tube of length $$L = 1$$ m is filled with liquid of mass $$2M$$. It rotates about one end with angular velocity $$\omega$$. We need the force at the other end (the far end).

Find the centripetal force

Linear mass density: $$\lambda = \frac{2M}{L} = 2M$$ (since $$L = 1$$).

Consider an element at distance $$r$$ from the axis: $$dm = \lambda \, dr = 2M \, dr$$.

The centrifugal force on this element: $$dF = dm \cdot \omega^2 r = 2M\omega^2 r \, dr$$.

The force at the far end ($$r = L = 1$$) is due to the liquid beyond some point — but since it's the end, the force is exerted by the liquid on the closed end.

Actually, the force at the far end is the integral of centripetal acceleration of all the liquid:

$$F = \int_0^L \lambda \omega^2 r \, dr = 2M\omega^2 \int_0^1 r \, dr = 2M\omega^2 \times \frac{1}{2} = M\omega^2$$

$$\omega^2 = \frac{F}{M} \Rightarrow \omega = \sqrt{\frac{F}{M}}$$

So $$\alpha = 1$$.

Conclusion

$$\alpha = 1$$.

Therefore, the answer is 1.

The cube is attached at one end of a uniform rigid rod of total length $$27\text{ cm}$$. A wedge placed under the rod acts as a fulcrum. Given distance from the fulcrum to the $$200\text{ g}$$ mass is $$25\text{ cm}$$, the distance from the fulcrum to the cube is

$$27\text{ cm}-25\text{ cm}=2\text{ cm}$$.

Let the actual (true) mass of the cube be $$M\;(\text{kg})$$. Before the cube touches water, rotational equilibrium would require

$$Mg(2\text{ cm}) = 0.2g(25\text{ cm}) \; \Longrightarrow \; M=2.5\text{ kg}.$$

Since the rod is not balanced initially, the true mass must be greater than $$2.5\text{ kg}$$ (otherwise the $$200\text{ g}$$ side could never dominate).

A beaker is now placed beneath the cube and water is added until exactly half of the cube is submerged. By Archimedes’ principle, the upward buoyant force equals the weight of the displaced water.

Side of cube $$=10\text{ cm}=0.10\text{ m}$$. Total volume of cube: $$V = (0.10\text{ m})^{3}=1.0\times10^{-3}\text{ m}^{3}$$. Half the volume is submerged, so

$$V_{\text{disp}}=\frac{V}{2}=0.5\times10^{-3}\text{ m}^{3}.$$

With water density $$\rho_{w}=1000\text{ kg m}^{-3}$$, the buoyant force is

$$F_{B} = \rho_{w}gV_{\text{disp}} = 1000g(0.5\times10^{-3}) = 0.5g\;\text{N}.$$

This buoyant force is equivalent to the weight of a $$0.5\text{ kg}$$ mass. Hence the cube’s effective downward (apparent) weight becomes

$$\bigl(M-0.5\bigr)g.$$

At the moment of balance, clockwise and counter-clockwise moments about the fulcrum are equal:

$$\bigl(M-0.5\bigr)g(2\text{ cm}) = 0.2g(25\text{ cm}).$$

Cancel $$g$$ and substitute the lever arms (convert to the same units is optional since they cancel too):

$$\bigl(M-0.5\bigr)\times2 = 0.2\times25$$ $$\Longrightarrow \bigl(M-0.5\bigr)\times2 = 5$$ $$\Longrightarrow M-0.5 = \frac{5}{2} = 2.5$$ $$\Longrightarrow M = 2.5 + 0.5 = 3.0\text{ kg}.$$

Therefore, the mass of the cube is $$\mathbf{3\;kg}$$.

For the hydraulic press to be in equilibrium, the pressure in both arms must be equal (Pascal's Law)

$$P_1 = P_2 \implies \frac{F_1}{A_1} = \frac{F_2}{A_2}$$

Since the cross-sectional area $$A$$ is proportional to the square of the diameter ($$A = \pi \frac{d^2}{4}$$), 

$$F_2 = F_1 \times \left( \frac{d_2}{d_1} \right)^2$$

$$F_2 = 10 \times \left( \frac{14}{1.4} \right)^2$$

$$F_2 = 10 \times (10)^2$$

$$F_2 = 10 \times 100 = 1000\text{ N}$$

We apply Bernoulli’s principle for incompressible, non‐viscous flow in horizontal streamlines: $$p + \frac{1}{2}\rho v^2 = \text{constant} \quad-(1)$$

From $$(1)$$ for upper and lower surfaces, we have:
$$p_{\text{lower}} + \frac{1}{2}\rho v_{\text{lower}}^2 = p_{\text{upper}} + \frac{1}{2}\rho v_{\text{upper}}^2$$

Rearranging gives the pressure difference (lift per unit area):
$$p_{\text{lower}} - p_{\text{upper}} = \frac{1}{2}\,\rho\,(v_{\text{upper}}^2 - v_{\text{lower}}^2)\quad-(2)$$

Convert speeds from km h$$^{-1}$$ to m s$$^{-1}$$:
$$180\;\text{km h}^{-1} = 180 \times \frac{5}{18} = 50\;\text{m s}^{-1}\quad-(3)$$
$$252\;\text{km h}^{-1} = 252 \times \frac{5}{18} = 70\;\text{m s}^{-1}\quad-(4)$$

Substitute $$\rho = 1\;\text{kg m}^{-3}$$, $$v_{\text{upper}}=70\;\text{m s}^{-1}$$, $$v_{\text{lower}}=50\;\text{m s}^{-1}$$ into $$(2)$$:
$$\Delta p = \frac{1}{2} \times 1 \times (70^2 - 50^2) = \frac{1}{2}\,(4900 - 2500) = 1200\;\text{Pa}\quad-(5)$$

Each wing has area $$40\;\text{m}^2$$, so total lifting area is
$$A_{\text{total}} = 2 \times 40 = 80\;\text{m}^2\quad-(6)$$

Lift generated is pressure difference times area:
$$L = \Delta p \times A_{\text{total}} = 1200 \times 80 = 96000\;\text{N}\quad-(7)$$

In level flight at constant speed, lift equals weight: $$L = mg\quad-(8)$$
Using $$g = 10\;\text{m s}^{-2}$$:
$$m = \frac{L}{g} = \frac{96000}{10} = 9600\;\text{kg}\quad-(9)$$

Final Answer: The mass of the plane is $$9600\;\text{kg}$$.

Lift = $$\frac{1}{2}\rho(v_1^2 - v_2^2) \times A = \frac{1}{2}(1.2)(70^2 - 65^2)(2) = 1.2(4900-4225) = 1.2 \times 675 = 810$$ N.

The answer is $$\boxed{810}$$.

Mercury fills a tube of radius 2 cm to height 30 cm. Find the force on the bottom.

The total pressure at the bottom is atmospheric pressure plus the pressure due to the mercury column: $$P = P_0 + \rho g h$$. Substituting gives $$P = 10^5 + (1.36 \times 10^4)(10)(0.3) = 10^5 + 40800 = 140800 \text{ N/m}^2$$.

Using $$r = 2$$ cm = 0.02 m, the area of the bottom is $$A = \pi r^2 = \frac{22}{7} \times (0.02)^2 = \frac{22}{7} \times 4 \times 10^{-4} = \frac{88}{7} \times 10^{-4} \text{ m}^2$$.

The force on the bottom is $$F = P \times A = 140800 \times \frac{88}{7} \times 10^{-4}$$, which simplifies as $$F = 140800 \times \frac{88}{70000} = 140800 \times 1.2571 \times 10^{-3}$$, and further $$F = \frac{140800 \times 88}{70000} = \frac{12390400}{70000} = 177.006 \approx 177 \text{ N}$$.

The correct answer is 177 N.

We are given that small water droplets with radius $$r = 0.01$$ mm fall with terminal velocity $$v_t = 10$$ cm/s. When eight such droplets coalesce to form a larger drop, we need to find the new terminal velocity.

Recall that for a small sphere falling through a viscous medium, Stokes’ law gives the terminal velocity as $$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$$, where $$r$$ is the radius, $$\rho$$ is the density of the sphere, $$\sigma$$ is the density of the medium, $$g$$ is the acceleration due to gravity, and $$\eta$$ is the viscosity of the medium. From this expression, it follows that $$v_t \propto r^2$$.

When eight small droplets merge, volume conservation implies $$\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3$$, which simplifies to $$R^3 = 8r^3$$ and hence $$R = 2r$$. This shows that the radius of the larger drop is twice that of each small droplet.

Since the terminal velocity varies as the square of the radius, the ratio of the new terminal velocity to the original one is $$\frac{v'}{v_t} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$$. From this it follows that $$v' = 4 \times v_t = 4 \times 10 = 40\text{ cm/s}$$.

The answer is 40 cm/s.

Using Bernoulli's equation (the pipe is horizontal and both points are at the same height):

$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$

When the valve is closed, $$v_1 = 0$$, $$P_1 = 4.5 \times 10^4$$ N/m$$^2$$.

When opened, $$P_2 = 2.0 \times 10^4$$ N/m$$^2$$, $$v_2 = v$$.

$$4.5 \times 10^4 = 2.0 \times 10^4 + \frac{1}{2}(1000)v^2$$

$$2.5 \times 10^4 = 500v^2$$

$$v^2 = 50$$

$$v = \sqrt{50}$$ m/s

So $$V = 50$$.

The answer is $$\boxed{50}$$.

Let the cross-sectional area of each tank be $$A$$ (large) and the cross-sectional area of the small hole (and of the pipe) be $$a$$.

At any instant let $$y$$ be the height of water column above the hole inside the tank (so $$y$$ decreases from $$h$$ to $$0$$ as the tank empties).

Case 1: Tank 1 (hole open to atmosphere at height $$H$$ above ground)
The exit point of water is the hole itself. The instantaneous speed of efflux is given by Torricelli’s theorem: $$v_1 = \sqrt{2 g y}$$

Volume flow rate: $$a v_1$$. Fall of water level in the tank: $$A \, dy = -\,a v_1\, dt$$.

Hence $$A \frac{dy}{dt} = -\,a \sqrt{2 g y}$$ $$\Rightarrow dt = -\frac{A}{a\sqrt{2 g}} \frac{dy}{\sqrt{y}}$$

Total time $$t_1$$: integrate $$y$$ from $$h$$ to $$0$$, changing sign to keep time positive. $$t_1 = \frac{A}{a\sqrt{2 g}}\int_{0}^{h} \frac{dy}{\sqrt{y}} = \frac{A}{a\sqrt{2 g}}\left[\,2\sqrt{y}\,\right]_{0}^{h} = \frac{2A}{a\sqrt{2 g}}\sqrt{h}$$

Case 2: Tank 2 (hole connected to a vertical pipe ending at ground level)
The exit point is now at ground (height $$0$$). The free surface of water is at height $$H+y$$ above ground. Applying Bernoulli between the free surface and the ground exit (both open to atmosphere): $$v_2 = \sqrt{2 g\,(H+y)}$$

Again, $$A \, dy = -\,a v_2\, dt$$ gives $$dt = -\frac{A}{a\sqrt{2 g}} \frac{dy}{\sqrt{H+y}}$$

Integrate from $$y=h$$ to $$0$$: $$t_2 = \frac{A}{a\sqrt{2 g}}\int_{0}^{h} \frac{dy}{\sqrt{H+y}}$$

Put $$s = H+y \; \Rightarrow \; ds = dy,$$ and when $$y=0,\,s=H$$; when $$y=h,\,s=H+h$$. $$t_2 = \frac{A}{a\sqrt{2 g}}\int_{H}^{H+h} \frac{ds}{\sqrt{s}} = \frac{A}{a\sqrt{2 g}}\left[\,2\sqrt{s}\,\right]_{H}^{H+h} = \frac{2A}{a\sqrt{2 g}}\left(\sqrt{H+h}-\sqrt{H}\right)$$

Ratio of times
$$\frac{t_1}{t_2} = \frac{\sqrt{h}}{\sqrt{H+h}-\sqrt{H}}$$

Given $$H = \frac{16}{9}\,h$$.

Compute the square roots:
$$\sqrt{H} = \sqrt{\tfrac{16}{9}h} = \tfrac{4}{3}\sqrt{h},$$
$$\sqrt{H+h} = \sqrt{\left(\tfrac{16}{9}+1\right)h} = \sqrt{\tfrac{25}{9}h} = \tfrac{5}{3}\sqrt{h}.$$

Denominator:
$$\sqrt{H+h}-\sqrt{H} = \left(\tfrac{5}{3}-\tfrac{4}{3}\right)\sqrt{h} = \tfrac{1}{3}\sqrt{h}.$$

Therefore $$\frac{t_1}{t_2} = \frac{\sqrt{h}}{\tfrac{1}{3}\sqrt{h}} = 3.$$

Hence the required ratio is 3.

Final Answer: 3

According to Newton's second law, the net force on the ball is: $$ma = mg - F_b - 6\pi\eta rv$$

Initially, when the ball is dropped ($$v=0$$), the acceleration is at its maximum. As the velocity increases, the upward viscous drag also increases, which reduces the net force and causes the acceleration ($$a = \frac{dv}{dt}$$) to decrease over time.

Eventually, the upward forces balance the weight, and the acceleration becomes zero. At this point, the ball moves with a constant terminal velocity.

Among the given options, graph (B) accurately represents this behavior.

The ball moves in two stages:
(i) free fall through air for a height $$h$$ (ignore air drag),
(ii) motion in water with constant speed (terminal velocity).

Stage (ii): terminal velocity in water
For a sphere of radius $$r$$ moving slowly through a viscous liquid, Stokes’ law gives the viscous force
$$F_{\text{viscous}} = 6\pi \eta r v$$.
At terminal velocity all forces balance:
weight $$= \text{buoyancy} + \text{viscous force}$$

Weight: $$W = \rho_s \, V \, g$$, Buoyant force: $$B = \rho_w \, V \, g$$,
where $$V = \frac{4}{3}\pi r^{3}$$ is the volume.
Hence
$$\rho_s V g = \rho_w V g + 6\pi \eta r v$$

Simplify (take $$V$$ common):
$$(\rho_s - \rho_w)V g = 6\pi \eta r v$$

Substitute $$V = \frac{4}{3}\pi r^{3}$$:
$$(\rho_s - \rho_w)\left(\frac{4}{3}\pi r^{3}\right)g = 6\pi \eta r v$$

Cancel $$\pi$$ and rearrange for $$v$$:
$$v = \frac{2}{9}\,\frac{(\rho_s - \rho_w) g r^{2}}{\eta}$$ $$-(1)$$

Insert the given data:
$$\rho_s = 1.0\times10^{5}\,\text{kg m}^{-3}$$,
$$\rho_w = 1.0\times10^{3}\,\text{kg m}^{-3}$$,
$$r = 1.0\times10^{-4}\,\text{m}$$,
$$\eta = 9.8\times10^{-6}\,\text{N s m}^{-2}$$,
$$g = 9.8\,\text{m s}^{-2}$$.

Compute the difference in densities:
$$(\rho_s - \rho_w) = 1.0\times10^{5} - 1.0\times10^{3} = 9.9\times10^{4}\,\text{kg m}^{-3}$$

Now evaluate $$v$$ from $$(1)$$:
$$v = \frac{2}{9}\,\frac{(9.9\times10^{4})(9.8)(1.0\times10^{-4})^{2}}{9.8\times10^{-6}}$$

$$r^{2} = (1.0\times10^{-4})^{2} = 1.0\times10^{-8}$$,
so
$$v = \frac{2}{9}\,\frac{(9.9\times10^{4})(9.8)(1.0\times10^{-8})}{9.8\times10^{-6}}$$

The factor $$9.8$$ in numerator and denominator cancels, giving
$$v = \frac{2}{9}\,\frac{9.9\times10^{4}\times1.0\times10^{-8}}{1.0\times10^{-6}}$$

$$9.9\times10^{4}\times1.0\times10^{-8}=9.9\times10^{-4}$$;
dividing by $$1.0\times10^{-6}$$ multiplies by $$10^{6}$$:
$$\frac{9.9\times10^{-4}}{1.0\times10^{-6}} = 9.9\times10^{2}=990$$

Therefore
$$v = \frac{2}{9}\times 990 \approx 220\,\text{m s}^{-1}$$

Stage (i): free fall in air
For free fall without air resistance
$$v^{2} = 2 g h \;\Rightarrow\; h = \frac{v^{2}}{2g}$$ $$-(2)$$

Insert $$v \approx 220\,\text{m s}^{-1}$$ and $$g = 9.8\,\text{m s}^{-2}$$ into $$(2)$$:
$$h = \frac{(220)^{2}}{2\times9.8} = \frac{48400}{19.6} \approx 2.47\times10^{3}\,\text{m}$$

The nearest value among the options is $$\boxed{2518\ \text{m}}$$ (Option B).

Let $$V_w$$ be the volume of ice in water and $$V_k$$ be the volume in kerosene. 

The total volume of the ice is $$V = V_w + V_k$$.

$$\text{Weight of Ice} = \text{Buoyant force (Water)} + \text{Buoyant force (Kerosene)}$$

$$\rho_i (V_w + V_k)g = \rho_w V_w g + \rho_k V_k g$$

Using specific gravities ($$\rho_w = 1$$, $$\rho_i = 0.9$$, and $$\rho_k = 0.8$$):

$$0.9(V_w + V_k) = 1(V_w) + 0.8(V_k)$$

$$0.9V_w + 0.9V_k = 1V_w + 0.8V_k$$

$$0.1V_k = 0.1V_w$$

$$\frac{V_w}{V_k} = \frac{0.1}{0.1} = \frac{1}{1}$$

The ratio of the volume immersed in water to that in kerosene is $$1 : 1$$.

We need to find the terminal velocity of a ball of radius $$2r$$ with the same mass as a ball of radius $$r$$.

According to Stokes' law, the terminal velocity is given by $$v_t = \frac{2r^2(\rho_s - \rho_l)g}{9\eta}$$. Since the medium has negligible density ($$\rho_l \approx 0$$), this reduces to $$v_t = \frac{2r^2\rho_s g}{9\eta}$$.

Because the spheres have the same mass, we have $$\frac{4}{3}\pi r_1^3 \rho_1 = \frac{4}{3}\pi r_2^3 \rho_2$$ with $$r_1 = r$$ and $$r_2 = 2r$$. Substituting these radii gives $$\rho_2 = \rho_1 \left(\frac{r_1}{r_2}\right)^3 = \rho_1 \left(\frac{r}{2r}\right)^3 = \frac{\rho_1}{8}$$.

Then the ratio of their terminal velocities is $$\frac{v_2}{v_1} = \frac{r_2^2 \rho_2}{r_1^2 \rho_1} = \frac{(2r)^2 \times \rho_1/8}{r^2 \times \rho_1} = \frac{4r^2 \times \rho_1/8}{r^2 \times \rho_1} = \frac{4}{8} = \frac{1}{2}$$. Hence $$v_2 = \frac{v}{2}$$.

The correct answer is Option 1: $$\frac{v}{2}$$.

A sphere of relative density $$\sigma$$, outer diameter $$D$$, with a concentric cavity of diameter $$d$$, just floats on water.

"Just floats" means the sphere is fully submerged with no net force (weight = buoyant force).

Mass of the sphere = $$\sigma \rho_w \times \frac{\pi}{6}(D^3 - d^3)$$ (volume of material times density of material, where density = $$\sigma \rho_w$$).

Weight = $$\sigma \rho_w g \cdot \frac{\pi}{6}(D^3 - d^3)$$.

Buoyant force = $$\rho_w g \cdot \frac{\pi}{6}D^3$$ (displaced water volume = total sphere volume).

Setting equal:

$$\sigma(D^3 - d^3) = D^3$$

$$\sigma D^3 - \sigma d^3 = D^3$$

$$D^3(\sigma - 1) = \sigma d^3$$

$$\frac{D^3}{d^3} = \frac{\sigma}{\sigma - 1}$$

$$\frac{D}{d} = \left(\frac{\sigma}{\sigma - 1}\right)^{1/3}$$

The correct answer is Option 2: $$\left(\frac{\sigma}{\sigma - 1}\right)^{1/3}$$.

Viscosity of gases is not greater than that of liquids; in fact, liquids generally exhibit much higher viscosities than gases. For example, the viscosity of water at 20 °C is approximately $$1 \times 10^{-3}$$ Pa·s, while the viscosity of air at 20 °C is approximately $$1.8 \times 10^{-5}$$ Pa·s, which is about fifty times smaller. This difference arises because in liquids, viscosity is governed by strong intermolecular cohesive forces, whereas in gases it results from momentum transfer between molecular layers; the cohesive forces in liquids are far stronger, leading to higher viscosity.

Insoluble impurities such as dust particles, powders, or oils that spread on the surface of a liquid reduce the cohesive forces between the liquid molecules at the interface, thereby lowering its surface tension. For instance, camphor particles on water demonstrate this effect by decreasing the water’s surface tension.

The correct answer is Option (2): Statement I is incorrect but Statement II is correct.

At terminal velocity: Weight = Buoyant force + Viscous force.

$$mg = \frac{m\rho_0}{\rho}g + F_v$$ (buoyant force = $$V\rho_0 g = \frac{m}{\rho}\rho_0 g$$).

$$F_v = mg - \frac{m\rho_0 g}{\rho} = mg\left(1 - \frac{\rho_0}{\rho}\right)$$.

The correct answer is Option (4): $$mg\left(1 - \frac{\rho_0}{\rho}\right)$$.

We need to identify the correct form of Bernoulli's equation.

Bernoulli's equation applies to steady, incompressible, non-viscous fluid flow along a streamline. It is derived from the work-energy theorem (or equivalently, conservation of energy per unit volume).

The three terms represent pressure energy, gravitational potential energy, and kinetic energy — all per unit volume:

$$P + \rho g h + \frac{1}{2}\rho v^2 = \text{constant}$$

where $$P$$ is the fluid pressure, $$\rho$$ is the density, $$g$$ is gravitational acceleration, $$h$$ is the height, and $$v$$ is the fluid velocity.

Option (1) uses mass $$m$$ instead of density $$\rho$$ — incorrect (that would give energy, not energy per unit volume).

Option (2): $$P + \rho gh + \frac{1}{2}\rho v^2 = \text{constant}$$ — this is the correct form.

Option (3) is missing the $$\frac{1}{2}$$ in the kinetic term — incorrect.

Option (4) has an extra $$\frac{1}{2}$$ in the gravitational term — incorrect.

The correct answer is Option (2).

We have the same pressure $$P$$ applied to 1 litre of water and 2 litres of a liquid. Water gets compressed by 0.01% and the liquid by 0.03%.

The bulk modulus is defined as:

$$B = \frac{P}{\frac{\Delta V}{V}}$$

Note that the fractional compression $$\frac{\Delta V}{V}$$ is independent of the initial volume, so the volume of liquid (2 litres vs 1 litre) does not affect the bulk modulus calculation.

For water: $$B_w = \frac{P}{0.01\%} = \frac{P}{0.0001}$$

For the liquid: $$B_l = \frac{P}{0.03\%} = \frac{P}{0.0003}$$

Now the ratio of bulk moduli is:

$$\frac{B_w}{B_l} = \frac{P/0.0001}{P/0.0003} = \frac{0.0003}{0.0001} = 3 = \frac{3}{1}$$

We are given that the ratio is $$\frac{3}{x}$$, so:

$$\frac{3}{x} = 3 \implies x = 1$$

Hence, the answer is $$1$$.

Viscous Force $$F = \eta A \frac{v}{l}$$

$$v = \frac{F \times l}{\eta \times A}$$

$$v = \frac{0.1 \times (0.25 \times 10^{-3})}{(5.0 \times 10^{-3}) \times 0.20}$$

$$v = \frac{0.025}{1.0} = 0.025\text{ m s}^{-1}$$

$$0.025\text{ m s}^{-1} = 25 \times 10^{-3}\text{ m s}^{-1}$$

The value is 25.

A spherical ball has radius $$r = 1$$ mm $$= 10^{-3}$$ m and density $$\rho_b = 10.5$$ g/cc $$= 10500$$ kg/m³. The glycerine has coefficient of viscosity $$\eta = 9.8$$ poise $$= 0.98$$ Pa·s and density $$\rho_g = 1.5$$ g/cc $$= 1500$$ kg/m³. Taking $$g = 9.8$$ m/s² and $$\pi = \frac{22}{7}$$.

Since at terminal velocity the net force is zero, the weight of the ball equals the sum of the buoyant force and the viscous force, so the viscous force is given by

$$\text{Viscous force} = \text{Weight} - \text{Buoyant force}.$$

Substituting the expressions for weight and buoyancy into this relation yields

$$F_{\text{viscous}} = \frac{4}{3}\pi r^3 (\rho_b - \rho_g) g$$

and hence

$$F_{\text{viscous}} = \frac{4}{3} \times \frac{22}{7} \times (10^{-3})^3 \times (10500 - 1500) \times 9.8 = \frac{4}{3} \times \frac{22}{7} \times 10^{-9} \times 9000 \times 9.8.$$

Now evaluating each factor:

$$\frac{4}{3} \times \frac{22}{7} = \frac{88}{21},\qquad 9000 \times 9.8 = 88200,$$

so that

$$F = \frac{88}{21} \times 88200 \times 10^{-9} = \frac{88 \times 88200}{21} \times 10^{-9} = \frac{7761600}{21} \times 10^{-9} = 369600 \times 10^{-9} = 3696 \times 10^{-7}\,\text{N}.$$

Therefore, comparing with the given form $$3696 \times 10^{-x}\,\text{N}$$ shows that $$x = \mathbf{7}.$$

We need to find the coefficient of viscosity using Stokes' law for a rising air bubble.

At terminal velocity, the net force is zero. For a bubble rising in a liquid (neglecting the density of air), the buoyant force equals the viscous drag:

$$6\pi \eta r v = \frac{4}{3}\pi r^3 \rho g$$

Solve for $$\eta$$: $$\eta = \frac{2r^2 \rho g}{9v}$$

Substitute the values: $$r = \frac{6}{2} = 3$$ mm $$= 3 \times 10^{-3}$$ m

$$\rho = 1750$$ kg m$$^{-3}$$

$$g = 10$$ m s$$^{-2}$$

$$v = 0.35$$ cm s$$^{-1}$$ $$= 3.5 \times 10^{-3}$$ m s$$^{-1}$$

$$\eta = \frac{2 \times (3 \times 10^{-3})^2 \times 1750 \times 10}{9 \times 3.5 \times 10^{-3}}$$

$$= \frac{2 \times 9 \times 10^{-6} \times 17500}{9 \times 3.5 \times 10^{-3}}$$

$$= \frac{2 \times 9 \times 17500 \times 10^{-6}}{31.5 \times 10^{-3}}$$

$$= \frac{315000 \times 10^{-6}}{31.5 \times 10^{-3}} = \frac{0.315}{0.0315} = 10 \text{ Pa s}$$

The correct answer is 10 Pa s.

We need to find the velocity of the liquid at the outlet of a tube when a piston pushes it through. We begin by noting that $$A_1 = 2.0 \text{ cm}^2$$ is the inlet area, $$A_2 = 10 \text{ mm}^2$$ is the outlet area, and $$v_1 = 4 \text{ cm/s}$$ is the piston speed.

Next, we convert $$A_2$$ to the same units as the inlet by writing $$A_2 = 10 \text{ mm}^2 = 10 \times (0.1 \text{ cm})^2 = 10 \times 0.01 \text{ cm}^2 = 0.10 \text{ cm}^2$$.

Then, applying the equation of continuity for an incompressible fluid yields $$A_1 v_1 = A_2 v_2$$.
This equality follows from conservation of mass, since the fluid entering per unit time must equal the fluid leaving per unit time.

Finally, solving for $$v_2$$ gives $$v_2 = \frac{A_1 v_1}{A_2} = \frac{2.0 \times 4}{0.10} = \frac{8.0}{0.10} = 80 \text{ cm/s}$$. Therefore, the speed of the liquid at the outlet is 80 cm/s.

We need to find the value of $$x$$ where $$\frac{dh}{dt} = x \times 10^{-3}$$ m/s.

The cross-sectional area of the tank is $$A = 750$$ cm$$^2 = 750 \times 10^{-4}$$ m$$^2$$, the cross-sectional area of the tap is $$a = 500$$ mm$$^2 = 500 \times 10^{-6}$$ m$$^2$$, and the speed of water at the tap is $$v = 30$$ cm/s $$= 0.30$$ m/s.

By the equation of continuity, the volume flow rate out of the tap equals the rate of decrease of water volume in the tank: $$A \cdot \left|\frac{dh}{dt}\right| = a \cdot v$$.

Solving for $$\left|\frac{dh}{dt}\right|$$ gives $$\left|\frac{dh}{dt}\right| = \frac{a \cdot v}{A} = \frac{500 \times 10^{-6} \times 0.30}{750 \times 10^{-4}} = \frac{150 \times 10^{-6}}{750 \times 10^{-4}} = \frac{150}{750} \times 10^{-2} = 0.2 \times 10^{-2} = 2 \times 10^{-3} \text{ m/s}$$.

Since $$\frac{dh}{dt} = x \times 10^{-3}$$ m/s, it follows that $$x = \boxed{2}$$.

Initially the air above the liquid is at atmospheric pressure $$P_0 = 10^{5}\,\text{N m}^{-2}$$ and has volume $$V_0$$. When the air isothermally compressed to $$\dfrac{100}{101}V_0$$, the ideal-gas relation $$P V = \text{constant}$$ gives

$$P_1 V_1 = P_0 V_0 \quad\Rightarrow\quad P_1 = P_0\dfrac{V_0}{V_1} = P_0\dfrac{V_0}{\dfrac{100}{101}V_0} = P_0 \dfrac{101}{100}$$

Hence the air pressure inside the container becomes
$$P_1 = \dfrac{101}{100}\,P_0 = \dfrac{101}{100}\,(10^{5}) = 1.01\times10^{5}\,\text{Pa}$$

The top of the capillary tube is open to the atmosphere, so the air pressure above the meniscus is still $$P_0$$. Let $$h$$ be the height of the liquid column that rises in the capillary, and $$r$$ the inner radius of the tube.

Across a concave meniscus (contact angle $$\theta = 0^\circ$$), the Laplace pressure difference is $$\Delta P = \dfrac{2T\cos\theta}{r} = \dfrac{2T}{r}$$.

Pressure just below the meniscus inside the liquid: $$P_\text{liq} = P_0 - \dfrac{2T}{r}$$    (pressure is lower in the liquid for a concave surface)

Pressure at the base of the capillary (same level as the reservoir surface): $$P_\text{base} = P_\text{liq} + \rho g h$$

This pressure must equal the air pressure on the free surface inside the container, $$P_1$$. Therefore,

$$P_1 = \left(P_0 - \dfrac{2T}{r}\right) + \rho g h$$

Solving for $$h$$:

$$h = \dfrac{P_1 - P_0 + \dfrac{2T}{r}}{\rho g}$$

Substituting the given data $$T = 0.075\,\text{N m}^{-1}, \; r = 0.1\,\text{mm} = 1\times10^{-4}\,\text{m},$$ $$\rho = 10^{3}\,\text{kg m}^{-3}, \; g = 10\,\text{m s}^{-2},$$ and $$P_1 - P_0 = 1.01\times10^{5} - 1.00\times10^{5} = 1.0\times10^{3}\,\text{Pa},$$ we obtain

$$\dfrac{2T}{r} = \dfrac{2\,(0.075)}{1\times10^{-4}} = 1.5\times10^{3}\,\text{Pa}$$

$$h = \dfrac{1.0\times10^{3} + 1.5\times10^{3}}{(10^{3})(10)} = \dfrac{2.5\times10^{3}}{1.0\times10^{4}} = 0.25\,\text{m}$$

Converting to centimetres, $$h = 0.25\,\text{m} = 25\,\text{cm}$$

Hence, the height of the liquid column in the capillary is 25 cm.

Eight equal drops of water falling with steady speed (terminal velocity) of 10 cm/s coalesce into one drop. We need to find the new terminal velocity.

First, use conservation of volume to find the new radius.

When 8 equal spherical drops (each of radius $$r$$) merge into one larger drop (radius $$R$$), the total volume is conserved:

$$ \frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 $$

$$ R^3 = 8r^3 \implies R = 2r $$

Next, recall how terminal velocity depends on radius.

For a sphere falling through a viscous fluid, the terminal velocity is given by Stokes' law:

$$ v_t = \frac{2r^2(\rho - \sigma)g}{9\eta} $$

where $$\rho$$ is the density of the sphere, $$\sigma$$ is the density of the fluid, $$\eta$$ is the viscosity, and $$g$$ is gravitational acceleration. The key dependence is:

$$ v_t \propto r^2 $$

Now, calculate the new terminal velocity.

$$ \frac{v_{\text{new}}}{v_{\text{old}}} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4 $$

$$ v_{\text{new}} = 4 \times 10 = 40\;\text{cm/s} $$

The new terminal velocity is 40 cm/s.

The correct answer is Option 2: 40 cm s$$^{-1}$$.

We have a vehicle of mass $$m = 5000$$ kg and the area of cross section of the load-carrying cylinder is $$A = 250$$ cm$$^2 = 250 \times 10^{-4}$$ m$$^2$$, with $$g = 10$$ m/s$$^2$$.

In a hydraulic lift, the pressure is the same throughout the fluid (by Pascal's law). So the maximum pressure the smaller piston would have to bear equals the pressure in the larger cylinder:

$$P = \frac{F}{A} = \frac{mg}{A} = \frac{5000 \times 10}{250 \times 10^{-4}} = \frac{50000}{0.025} = 2000000 \text{ Pa} = 2 \times 10^6 \text{ Pa}$$

Hence, the correct answer is Option 4.

Assertion A: When you squeeze one end of a tube to get toothpaste out from the other end, Pascal's principle is observed.

This is correct. The pressure applied at one end is transmitted through the toothpaste (which behaves approximately as a fluid) to the other end, pushing it out.

Reason R: A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container.

This is correct. This is the statement of Pascal's law/principle.

R correctly explains A: the toothpaste comes out because the pressure applied by squeezing is transmitted undiminished throughout the enclosed paste (acting as a fluid), which is exactly Pascal's principle.

The correct answer is Both A and R are correct and R is the correct explanation of A.

Statement I: Pressure in a reservoir of water is same at all points at the same level of water.

This is true. This follows from the hydrostatic pressure principle: $$P = P_0 + \rho gh$$, which is the same for all points at the same depth (same level).

Statement II: The pressure applied to enclosed water is transmitted in all directions equally.

This is true. This is Pascal's law - pressure applied to an enclosed fluid is transmitted undiminished in all directions.

Both statements are true.

The correct answer is Option 4.

We have a ball of mass $$M$$ and density $$\rho$$ dropped in a viscous liquid of density $$\rho_0$$. At terminal velocity, the net force on the ball is zero.

The forces acting on the ball are: the weight $$W = Mg$$ acting downward, the buoyant force $$F_b = V\rho_0 g$$ acting upward (where $$V = \frac{M}{\rho}$$ is the volume of the ball), and the viscous force $$F$$ acting upward.

At terminal velocity, these forces balance, so

$$Mg = F_b + F$$

$$F = Mg - F_b = Mg - \frac{M}{\rho}\rho_0 g$$

Factoring out $$Mg$$ (since the buoyant force reduces the effective weight by the density ratio):

$$F = Mg\left(1 - \frac{\rho_0}{\rho}\right)$$

Hence, the correct answer is $$F = Mg\left(1 - \dfrac{\rho_0}{\rho}\right)$$.

We need to find the volume of an air bubble when it reaches the surface from 40 m depth.

Apply Boyle's Law (constant temperature). Since the temperature is the same at the bottom and surface:

$$P_1 V_1 = P_2 V_2$$

Calculate pressure at the bottom: $$P_1 = P_{\text{atm}} + \rho g h = 1 \times 10^5 + 1000 \times 10 \times 40 = 1 \times 10^5 + 4 \times 10^5 = 5 \times 10^5 \text{ Pa}$$

Pressure at the surface: $$P_2 = P_{\text{atm}} = 1 \times 10^5 \text{ Pa}$$

Calculate the volume at the surface: $$V_2 = V_1 \times \frac{P_1}{P_2} = 1 \times \frac{5 \times 10^5}{1 \times 10^5} = 5 \text{ cm}^3$$

The correct answer is Option D: 5 cm$$^3$$.

Inside the furnace the air is heated at the constant (atmospheric) pressure $$P_a$$ from $$T_a = 300\;{\rm K}$$ to $$T = 360\;{\rm K}$$.
For an ideal gas, $$\rho \,T = \rho_a \,T_a$$ at the same pressure, therefore the density of the hot air is

$$\rho = \rho_a\;\dfrac{T_a}{T}=1.2\;\dfrac{300}{360}=1.0\;{\rm kg\,m^{-3}}$$

With the chimney open, the upward motion of the hot air is produced by the difference between the weights of the two vertical columns of air (cold outside, hot inside) of the same height $$(H+h)=1+9=10\;{\rm m}$$. The driving pressure head is

$$\Delta P_{\text{drive}}=(\rho_a-\rho)\,g\,(H+h) =(1.2-1.0)\times10\times10 =20\;{\rm N\,m^{-2}}$$

Neglecting viscous losses, Bernoulli’s theorem for the stream‐line from the furnace inlet (very large area, negligible speed) to the chimney exit gives

$$\frac{1}{2}\,\rho\,v^{2}= \Delta P_{\text{drive}} \quad\Longrightarrow\quad v=\sqrt{\dfrac{2\,\Delta P_{\text{drive}}}{\rho}} =\sqrt{\dfrac{2\times20}{1.0}} =6\;{\rm m\,s^{-1}}$$

This velocity corresponds to a dynamic pressure (stagnation over static) of

$$\Delta P_{\text{dyn}}=\dfrac{1}{2}\,\rho\,v^{2}=20\;{\rm N\,m^{-2}}$$

Now a cap is fixed at the top of the chimney. The rising air jet strikes the lower face of the cap and comes to rest there; its dynamic pressure therefore appears as an additional static pressure on the underside of the cap. Besides this, the column of hot air inside the furnace of height $$H=1\;{\rm m}$$ still exerts its usual hydrostatic pressure difference

$$\Delta P_{\text{hyd}}=\rho\,g\,H =1.0\times10\times1 =10\;{\rm N\,m^{-2}}$$

Hence the net pressure difference between the top (outer) and the bottom (inner) faces of the cap after it is closed is

$$\Delta P =\Delta P_{\text{dyn}}+\Delta P_{\text{hyd}} =20+10 =30\;{\rm N\,m^{-2}}$$

Therefore, the required pressure difference is
30 N m−2.

A small spherical ball falls freely through height h, then enters water and continues at the same constant velocity (terminal velocity). We need to find h. The radius of the ball is $$r = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} = 10^{-4} \text{ m}$$, its density is $$\rho = 10^4 \text{ kg/m}^3$$, the density of water is $$\sigma = 10^3 \text{ kg/m}^3$$, the viscosity of water is $$\eta = 1.0 \times 10^{-5} \text{ N·s/m}^2$$, and the acceleration due to gravity is $$g = 10 \text{ m/s}^2$$.

At terminal velocity, the net downward force (weight minus buoyancy) equals the viscous drag force: $$\frac{4}{3}\pi r^3 (\rho - \sigma) g = 6\pi \eta r v_t$$. Solving for $$v_t$$ gives $$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$$.

Substituting the values into this expression yields $$v_t = \frac{2 \times (10^{-4})^2 \times (10^4 - 10^3) \times 10}{9 \times 1.0 \times 10^{-5}}$$. Calculating step by step: $$r^2 = (10^{-4})^2 = 10^{-8} \text{ m}^2$$, $$\rho - \sigma = 10^4 - 10^3 = 9000 \text{ kg/m}^3$$, hence the numerator is $$\text{Numerator} = 2 \times 10^{-8} \times 9000 \times 10 = 2 \times 10^{-8} \times 9 \times 10^4 = 18 \times 10^{-4}$$ and the denominator is $$\text{Denominator} = 9 \times 10^{-5}$$, giving $$v_t = \frac{18 \times 10^{-4}}{9 \times 10^{-5}} = \frac{18}{9} \times \frac{10^{-4}}{10^{-5}} = 2 \times 10 = 20 \text{ m/s}$$.

As the ball falls freely from rest through height h, using $$v^2 = u^2 + 2gh$$ with $$u = 0$$, we have $$v_t^2 = 2gh$$ and therefore $$h = \frac{v_t^2}{2g} = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20 \text{ m}$$. The answer is 20 m.

A large tank has a cross-section area of $$A = 0.5$$ m$$^2$$, a narrow opening near the bottom with area $$a = 1$$ cm$$^2 = 1 \times 10^{-4}$$ m$$^2$$, a load of 25 kg on the water at the top, and a water height of $$h = 40$$ cm = 0.4 m above the bottom.

At the surface of the water, in addition to atmospheric pressure, there is extra pressure due to the applied load: $$P_{load} = \frac{mg}{A} = \frac{25 \times 10}{0.5} = 500 \text{ Pa}$$

Let the top surface be point 1 and the opening be point 2. Both are exposed to atmospheric pressure $$P_0$$ (at the opening) and $$P_0 + P_{load}$$ (at the surface due to the load). Applying Bernoulli’s equation between these points and neglecting the speed of water at the top surface (since $$A \gg a$$) gives

$$P_0 + P_{load} + \rho g h = P_0 + \frac{1}{2}\rho v^2$$

Rearranging this equation yields

$$P_{load} + \rho g h = \frac{1}{2}\rho v^2$$

Substituting the values, the pressure due to the water column is

$$\rho g h = 1000 \times 10 \times 0.4 = 4000 \text{ Pa}$$

Therefore,

$$500 + 4000 = \frac{1}{2} \times 1000 \times v^2$$

which simplifies to

$$4500 = 500 \times v^2$$

and hence

$$v^2 = 9$$

so that

$$v = 3 \text{ m/s} = 300 \text{ cm/s}$$

The velocity of water coming out of the opening is 300 cm s$$^{-1}$$.

The density of the liquid is $$\rho = 750$$ kg/m^3, the cross-sectional area of the wider section is $$A_1 = 1.2 \times 10^{-2}$$ m^2, that of the narrow section is $$A_2 = \frac{A_1}{2} = 0.6 \times 10^{-2}$$ m^2, and the pressure difference is $$P_1 - P_2 = 4500$$ Pa.

Using the equation of continuity, $$A_1 v_1 = A_2 v_2$$, we have $$v_2 = \frac{A_1}{A_2} v_1 = 2v_1$$.

Applying Bernoulli's equation between the wide and narrow sections, $$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$, we get $$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$. Substituting $$v_2 = 2v_1$$ yields $$4500 = \frac{1}{2} \times 750 \times (4v_1^2 - v_1^2)$$, which becomes $$4500 = \frac{1}{2} \times 750 \times 3v_1^2 = 1125 \, v_1^2$$. Therefore, $$v_1^2 = \frac{4500}{1125} = 4$$ and $$v_1 = 2 \text{ m/s}$$.

The volume flow rate is $$Q = A_1 v_1 = 1.2 \times 10^{-2} \times 2 = 2.4 \times 10^{-2} \text{ m}^3/\text{s} = 24 \times 10^{-3} \text{ m}^3/\text{s}$$.

Hence, the answer is 24.

We have a small ball of mass $$m = 0.3$$ g $$= 3 \times 10^{-4}$$ kg and density $$\rho_b = 8$$ g/cc $$= 8000$$ kg/m$$^3$$, dropped in glycerine of density $$\rho_g = 1.3$$ g/cc $$= 1300$$ kg/m$$^3$$. When the ball reaches terminal velocity, its acceleration is zero, so the net force on the ball is zero.

The forces acting on the ball at terminal velocity are: (1) the weight $$W = mg$$ acting downward, (2) the buoyant force $$F_b$$ acting upward, and (3) the viscous force $$F_v$$ acting upward. At terminal velocity, these balance: $$mg = F_b + F_v$$

The volume of the ball is $$V = \frac{m}{\rho_b}$$. The buoyant force is $$F_b = \rho_g V g = \rho_g \cdot \frac{m}{\rho_b} \cdot g$$.

Now, the viscous force is: $$F_v = mg - F_b = mg - \frac{\rho_g}{\rho_b} mg = mg\left(1 - \frac{\rho_g}{\rho_b}\right)$$

Substituting the values: $$F_v = 3 \times 10^{-4} \times 10 \times \left(1 - \frac{1.3}{8}\right) = 3 \times 10^{-3} \times \left(1 - 0.1625\right) = 3 \times 10^{-3} \times 0.8375$$

$$F_v = 2.5125 \times 10^{-3}$$ N $$= 25.125 \times 10^{-4}$$ N.

Rounding to the nearest integer, $$x = 25$$.

Hence, the correct answer is 25.

We use Stokes' law for terminal velocity of a sphere rising through a viscous medium. For an air bubble rising in a liquid, the buoyant force drives the motion and viscous drag opposes it.

Given data:

Diameter of bubble = 2 mm, so radius $$r = 1$$ mm = $$0.1$$ cm

Density of solution $$\rho = 1750$$ kg/m$$^3$$ = 1.75 g/cm$$^3$$

Terminal velocity $$v = 0.35$$ cm/s

Mass of the bubble is negligible, so density of bubble $$\sigma \approx 0$$

Using Stokes' law for terminal velocity:

Since $$\sigma \approx 0$$:

Solving for viscosity $$\eta$$:

Substituting the values in CGS units ($$g = 980$$ cm/s$$^2$$):

Rounding to the nearest integer:

Therefore, the coefficient of viscosity is 11 Poise.

Velocity of the upper layer is given as $$v = 36$$ km/h, the shearing stress is $$\tau = 10^{-3}$$ N/m$$^2$$, and the coefficient of viscosity is $$\eta = 10^{-2}$$ Pa·s. Converting this velocity to metres per second yields $$v = 36 \times \frac{5}{18} = 10 \text{ m/s}$$.

According to Newton’s law of viscosity, the shearing stress is related to the velocity gradient by $$\tau = \eta \frac{dv}{dx}$$. The velocity gradient $$\frac{dv}{dx}$$ represents the change in velocity over the depth $$d$$ of the river. Assuming the bottom layer is stationary and the top layer moves at velocity $$v$$, we have $$\frac{dv}{dx} = \frac{v}{d}$$.

Substituting into the law of viscosity gives $$\tau = \eta \frac{v}{d}$$, which can be rearranged to solve for the depth:

$$d = \frac{\eta v}{\tau} = \frac{10^{-2} \times 10}{10^{-3}} = \frac{10^{-1}}{10^{-3}} = 10^{2} = 100 \text{ m}$$.

Thus, the depth of the river is 100 m.

For a thin soap bubble in air, the excess pressure due to surface tension is $$\dfrac{4S}{r}$$ (two surfaces of the film). Hence at any instant

$$P = P_a + \dfrac{4S}{r}$$

where $$P_a$$ is the atmospheric pressure, $$P$$ the gas pressure inside the bubble and $$r$$ the instantaneous radius.

Case 1: Surface is a perfect heat insulator (adiabatic process)

The enclosed gas obeys the adiabatic relation

$$PV^{\gamma}= \text{constant}$$

with $$\gamma = \dfrac{5}{3}$$ and $$V=\dfrac{4\pi r^{3}}{3} \;\;(\therefore V\propto r^{3})$$.

Therefore

$$\bigl(P_a+\dfrac{4S}{r}\bigr)\;r^{3\gamma}= \text{constant}$$

$$\Rightarrow\; \bigl(P_{a1}+\dfrac{4S}{r_1}\bigr)\,r_1^{5} =\bigl(P_{a2}+\dfrac{4S}{r_2}\bigr)\,r_2^{5}$$

$$\Rightarrow\; \left(\dfrac{r_1}{r_2}\right)^{5}= \dfrac{P_{a2}+ \dfrac{4S}{r_2}}{P_{a1}+ \dfrac{4S}{r_1}}$$

• Option A uses $$2S/r$$ instead of $$4S/r$$, so Option A is wrong.

Next, relate temperature using the ideal-gas law $$PV = nRT$$.

Combining $$PV = nRT$$ with $$PV^{\gamma}= \text{constant}$$ gives the standard result for an adiabatic change

$$TV^{\gamma-1}= \text{constant}$$

Since $$\gamma-1=\dfrac{2}{3}$$, we get $$T r^{2}= \text{constant}$$, i.e.

$$\dfrac{T_2}{T_1}= \left(\dfrac{r_1}{r_2}\right)^{2}$$

Raising both sides to the power $$\dfrac{5}{2}$$ and substituting the radius relation derived earlier:

$$\left(\dfrac{T_2}{T_1}\right)^{5/2}= \left(\dfrac{r_1}{r_2}\right)^{5}= \dfrac{P_{a2}+ \dfrac{4S}{r_2}}{P_{a1}+ \dfrac{4S}{r_1}}$$

This is exactly the statement in Option D, so Option D is correct.

The total energy of the system is

$$U_{\text{total}} = U_{\text{gas}} + E_{\text{surface}} = nC_{V}T + 4\pi r^{2}S$$

Both $$T$$ and $$r$$ change during the adiabatic compression/expansion, hence $$U_{\text{total}}$$ is not constant. Therefore Option B is wrong.

Case 2: Surface is a perfect heat conductor (isothermal process)

If heat flows freely through the film while the surrounding temperature remains nearly unchanged, the gas undergoes an isothermal change so that $$PV=\text{constant}.$$ Using $$V\propto r^{3}$$:

$$\bigl(P_a+\dfrac{4S}{r}\bigr)\,r^{3}= \text{constant}$$

$$\Rightarrow\; \left(\dfrac{r_1}{r_2}\right)^{3}= \dfrac{P_{a2}+ \dfrac{4S}{r_2}}{P_{a1}+ \dfrac{4S}{r_1}}$$

This matches Option C exactly, so Option C is correct.

Hence the correct statements are:
Option C and Option D.

The mass flow rate is $$\alpha = 0.8 \text{ kg s}^{-1}$$. At the lower end: $$A_1 = 0.1 \text{ m}^2 , \; \rho_1 = 0.2 \text{ kg m}^{-3}$$.

1. Velocity at the lower end
Continuity gives $$\alpha = \rho_1 A_1 v_1 \; \Rightarrow \; v_1 = \frac{\alpha}{\rho_1 A_1} = \frac{0.8}{0.2 \times 0.1}=40 \text{ m s}^{-1}$$.

2. Gas constant and specific heat
Ideal-gas relation at the lower end: $$R_s = \frac{P_1}{\rho_1 T_1} = \frac{600}{0.2 \times 300}=10 \text{ J kg}^{-1}\text{K}^{-1}$$.
With $$\gamma = 2$$, $$C_p = \frac{\gamma R_s}{\gamma-1}= \frac{2 \times 10}{1}=20 \text{ J kg}^{-1}\text{K}^{-1}$$.

3. Density at the upper end from adiabatic relation
For an adiabatic (reversible) flow of an ideal gas $$P\rho^{-\gamma}= \text{constant} \quad\Rightarrow\quad \frac{P_1}{\rho_1^{\gamma}}=\frac{P_2}{\rho_2^{\gamma}} \; -(1)$$

The ideal-gas law also gives $$P_2 = \rho_2 R_s T_2$$.
Substitute into (1): $$\frac{600}{(0.2)^2}= \frac{\rho_2 R_s T_2}{\rho_2^{2}} \; \Longrightarrow\; 600\left(\frac{\rho_2}{0.2}\right)^2 = \rho_2 R_s T_2$$
$$\Rightarrow 600\frac{\rho_2^2}{0.04}= \rho_2(10)(150)$$ $$\Rightarrow 15000\,\rho_2 = 1500\,\rho_2 \; \Longrightarrow\; \rho_2 = 0.1 \text{ kg m}^{-3}$$.

4. Velocity at the upper end from continuity
$$v_2 = \frac{\alpha}{\rho_2 A_2}= \frac{0.8}{0.1 \times 0.4}=20 \text{ m s}^{-1}$$.

5. Pressure at the upper end
$$P_2 = \rho_2 R_s T_2 = 0.1 \times 10 \times 150 = 150 \text{ Pa}$$ (not 300 Pa).

6. Height of the chimney
Steady-flow energy equation for adiabatic flow (per unit mass): $$C_p T + \frac{v^2}{2} + g z = \text{constant}$$
Between lower (1) and upper (2) ends: $$C_p T_1 + \frac{v_1^2}{2}= C_p T_2 + \frac{v_2^2}{2}+ g h$$
$$20(300)+\frac{40^2}{2}=20(150)+\frac{20^2}{2}+10h$$ $$6000+800 = 3000+200+10h$$ $$10h = 3600 \;\Rightarrow\; h = 360 \text{ m}$$ (not 590 m).

7. Summary of the options
A. $$P_2 = 150\text{ Pa}\neq 300\text{ Pa}$$ ⇒ wrong.
B. $$v_1 = 40\text{ m s}^{-1},\; v_2 = 20\text{ m s}^{-1}$$ ⇒ correct.
C. Height obtained is 360 m, not 590 m ⇒ wrong.
D. $$\rho_2 = 0.1\text{ kg m}^{-3}\neq 0.05\text{ kg m}^{-3}$$ ⇒ wrong.

Hence, only Option B is correct.

Option B which is: The velocity of the gas at the lower end of the chimney is 40 ms$$^{-1}$$ and at the upper end is 20 ms$$^{-1}$$.

We are given a pressure-pump with a horizontal tube of cross-sectional area $$A = 10 \text{ cm}^2 = 10 \times 10^{-4} \text{ m}^2$$ and outflow speed $$v = 20 \text{ m/s}$$. A vertical wall in front of the tube stops the water completely.

The force exerted on the wall can be found using the rate of change of momentum. The mass flow rate of water is $$\frac{dm}{dt} = \rho A v$$, where $$\rho = 1000 \text{ kg/m}^3$$.

Since the water is brought to rest (from speed $$v$$ to 0), the force on the wall is $$F = \frac{dm}{dt} \times v = \rho A v^2$$.

Substituting the values: $$F = 1000 \times 10 \times 10^{-4} \times (20)^2 = 1000 \times 10^{-3} \times 400 = 400 \text{ N}$$.

Hence, the correct answer is Option D.

The Reynolds number is a dimensionless quantity used to predict the flow pattern in a fluid.

The Reynolds number is defined as the ratio of inertial forces to viscous forces:

$$R_e = \frac{\text{Inertial forces}}{\text{Viscous forces}} = \frac{\rho v d}{\eta}$$

where:

• $$\rho$$ = density of fluid
• $$v$$ = velocity of flow
• $$d$$ = diameter of pipe
• $$\eta$$ = coefficient of viscosity

$$[\rho] = ML^{-3}, \quad [v] = LT^{-1}, \quad [d] = L, \quad [\eta] = ML^{-1}T^{-1}$$

$$[R_e] = \frac{ML^{-3} \times LT^{-1} \times L}{ML^{-1}T^{-1}} = \frac{ML^{-1}T^{-1}}{ML^{-1}T^{-1}} = M^0L^0T^0$$

This confirms $$R_e$$ is dimensionless.

Hence, the correct answer is Option C.

We need to determine how the terminal velocity $$v_t$$ of a spherical raindrop depends on its radius $$r$$.

When a spherical raindrop falls through air at terminal velocity, three forces act on it: the weight (downward) $$W = \frac{4}{3}\pi r^3 \rho g$$, where $$\rho$$ is the density of water; the buoyant force (upward) $$F_b = \frac{4}{3}\pi r^3 \sigma g$$, where $$\sigma$$ is the density of air; and the viscous drag (upward) given by Stokes’ law as $$F_d = 6\pi \eta r v_t$$, where $$\eta$$ is the viscosity of air.

At terminal velocity, the net force is zero, so $$W = F_b + F_d$$ which implies $$\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v_t$$.

Rearranging to solve for the terminal velocity yields $$6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho - \sigma) g$$ and hence $$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$$.

Since $$\rho$$, $$\sigma$$, $$g$$, and $$\eta$$ are constants for a given system, it follows that $$v_t \propto r^2$$. The correct answer is Option B: $$r^2$$.

Two cylindrical vessels of equal cross-section $$A = 16 \text{ cm}^2$$ contain water at heights 100 cm and 150 cm. We need the work done by gravity when they are interconnected.

Since both vessels have equal cross-sectional area, the final height is the average:

$$h_f = \frac{100 + 150}{2} = 125 \text{ cm}$$

Work done by gravity = loss in potential energy of the system.

The centre of mass of water in vessel 1 was at $$h_1/2 = 50$$ cm and goes to $$125/2 = 62.5$$ cm.

The centre of mass of water in vessel 2 was at $$h_2/2 = 75$$ cm and goes to $$62.5$$ cm.

Mass of water in vessel 1: $$m_1 = \rho A h_1 = 1000 \times 16 \times 10^{-4} \times 1 = 1.6 \text{ kg}$$

Mass of water in vessel 2: $$m_2 = \rho A h_2 = 1000 \times 16 \times 10^{-4} \times 1.5 = 2.4 \text{ kg}$$

Change in PE of vessel 1 (rises from 50 cm to 62.5 cm): $$\Delta PE_1 = m_1 g \Delta h_1 = 1.6 \times 10 \times 0.125 = 2 \text{ J}$$

Change in PE of vessel 2 (falls from 75 cm to 62.5 cm): $$\Delta PE_2 = -m_2 g \Delta h_2 = -2.4 \times 10 \times 0.125 = -3 \text{ J}$$

Net change in PE = $$2 + (-3) = -1 \text{ J}$$

Work done by gravity = $$-\Delta PE = -(-1) = 1 \text{ J}$$

The correct answer is Option B: $$1 \text{ J}$$.

Given: radius $$r = 1 \, \mu$$m $$= 10^{-6}$$ m, coefficient of viscosity $$\eta = 1.8 \times 10^{-5}$$ N s m$$^{-2}$$, density of water $$\rho = 10^6$$ g m$$^{-3}$$ $$= 10^3$$ kg m$$^{-3}$$, $$g = 10$$ m s$$^{-2}$$. Buoyant force is negligible.

Write the formula for terminal velocity (neglecting buoyancy).

When buoyancy is negligible, at terminal velocity the drag force equals the weight:

$$6\pi\eta r v_t = \frac{4}{3}\pi r^3 \rho g$$

$$v_t = \frac{2r^2 \rho g}{9\eta}$$

Substitute the values.

$$v_t = \frac{2 \times (10^{-6})^2 \times 10^3 \times 10}{9 \times 1.8 \times 10^{-5}}$$

$$v_t = \frac{2 \times 10^{-12} \times 10^4}{16.2 \times 10^{-5}}$$

$$v_t = \frac{2 \times 10^{-8}}{16.2 \times 10^{-5}}$$

$$v_t = \frac{2}{16.2} \times 10^{-3} = 0.1234 \times 10^{-3}$$

$$v_t = 123.4 \times 10^{-6} \text{ m s}^{-1}$$

The correct answer is Option B.

When the ball reaches terminal (constant) velocity, the net force on it is zero. The three forces acting on the ball are:

1. Weight (downward): $$W = mg$$

2. Buoyant force (upward): $$F_b = \frac{m}{\rho_1} \times d_2 \times g = mg\frac{d_2}{d_1}$$

Here, the volume of ball $$= \frac{m}{d_1}$$, so buoyant force $$= \frac{m}{d_1} \times d_2 \times g$$.

3. Viscous force (upward): $$F_v$$

At terminal velocity, the net force is zero:

$$mg = F_b + F_v$$

$$mg = mg\frac{d_2}{d_1} + F_v$$

$$F_v = mg - mg\frac{d_2}{d_1}$$

$$F_v = mg\left(1 - \frac{d_2}{d_1}\right)$$

The correct answer is Option B.

We have a tank in which the water surface is $$12 \text{ m}$$ above the ground. A small orifice (hole) is pierced in the vertical wall at a depth $$h$$ below this surface. Our task is to find that value of $$h$$ for which the water jet, after leaving the hole, lands on the ground at the greatest possible horizontal distance (range) from the wall.

First we determine the speed with which water emerges from the hole. For a small orifice in the side of a large tank, Torricelli’s theorem states:

$$v = \sqrt{2 g h}$$

where $$v$$ is the exit speed of the water, $$g$$ is the acceleration due to gravity, and $$h$$ is the vertical depth of the orifice below the free surface. We explicitly see that the deeper the hole, the faster the jet comes out.

Next we identify the vertical drop available for the jet after it leaves the hole. The hole is situated $$h$$ metres below the surface, so its height above the ground is

$$\text{height of hole above ground} = 12 \text{ m} - h.$$

Once the water leaves the hole horizontally with speed $$v$$, it behaves like a projectile that has:

- horizontal component of velocity $$v_x = v = \sqrt{2 g h},$$
- vertical component of velocity $$v_y = 0$$ (because it emerges horizontally),
- vertical drop $$y = 12 - h.$$

The time $$t$$ taken for the jet to fall this vertical distance is obtained from the standard equation of uniformly accelerated motion (starting from rest in the vertical direction):

$$y = \frac{1}{2} g t^2.$$

Substituting $$y = 12 - h$$ gives

$$12 - h = \frac{1}{2} g t^2.$$

Solving for $$t$$:

$$t = \sqrt{\frac{2(12 - h)}{g}}.$$

The horizontal range $$R$$ is the horizontal speed multiplied by this time:

$$R = v_x \, t = \bigl(\sqrt{2 g h}\bigr)\;\bigl(\sqrt{\tfrac{2(12 - h)}{g}}\bigr).$$

Simplifying inside the radicals and cancelling $$g$$:

$$R = 2 \sqrt{h(12 - h)}.$$

Thus the range depends only on $$h$$ through the function

$$f(h) = h(12 - h).$$

To maximise $$R$$, it suffices to maximise $$f(h).$$ We expand:

$$f(h) = 12h - h^2.$$

This is a downward-opening quadratic in $$h$$. The maximum of a quadratic $$ax^2 + bx + c$$ with $$a < 0$$ occurs at

$$h = -\frac{b}{2a}.$$

In our expression $$f(h) = -h^2 + 12h$$ we have $$a = -1$$ and $$b = 12$$, so

$$h_{\text{max}} = -\frac{12}{2(-1)} = 6 \text{ m}.$$

We also check that $$0 \le h \le 12,$$ and clearly $$h = 6 \text{ m}$$ lies within this permissible interval. Therefore the water jet attains its greatest horizontal range when the orifice is placed $$6 \text{ m}$$ below the surface of the water.

Hence, the correct answer is Option C (numerical value $$6 \text{ m}$$).

We have a hydraulic press where a mass $$m$$ on the smaller piston can lift 100 kg on the larger piston. By Pascal's law, the pressure is transmitted equally, so $$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$.

Let the original diameters of the smaller and larger pistons be $$d$$ and $$D$$ respectively. Then $$\frac{mg}{A_1} = \frac{100g}{A_2}$$, which gives $$\frac{A_2}{A_1} = \frac{100}{m}$$.

Now, the diameter of the larger piston is increased by 4 times to $$4D$$, and the diameter of the smaller piston is decreased by 4 times to $$\frac{d}{4}$$.

The new area of the larger piston is $$A_2' = \frac{\pi(4D)^2}{4} = 16 \times \frac{\pi D^2}{4} = 16A_2$$.

The new area of the smaller piston is $$A_1' = \frac{\pi(d/4)^2}{4} = \frac{1}{16} \times \frac{\pi d^2}{4} = \frac{A_1}{16}$$.

Using Pascal's law with the same mass $$m$$ on the smaller piston: $$\frac{mg}{A_1'} = \frac{Mg}{A_2'}$$, where $$M$$ is the new mass that can be lifted.

$$M = m \times \frac{A_2'}{A_1'} = m \times \frac{16A_2}{A_1/16} = m \times 256 \times \frac{A_2}{A_1}$$

We know $$m \times \frac{A_2}{A_1} = 100$$, so $$M = 256 \times 100 = 25600$$ kg.

So, the answer is $$25600$$.

The nature of flow is determined by the Reynolds number $$Re = \frac{\rho v D}{\eta}$$, where $$\rho$$ is the density, $$v$$ is the flow velocity, $$D$$ is the diameter, and $$\eta$$ is the viscosity.

The radius of the tap is $$r = 0.5$$ cm $$= 5 \times 10^{-3}$$ m, so the diameter $$D = 0.01$$ m and the cross-sectional area is $$A = \pi r^2 = \pi \times 25 \times 10^{-6}$$ m$$^2$$.

For the first flow rate $$Q_1 = 0.18$$ L/min $$= \frac{0.18 \times 10^{-3}}{60} = 3 \times 10^{-6}$$ m$$^3$$/s, the velocity is $$v_1 = \frac{Q_1}{A} = \frac{3 \times 10^{-6}}{\pi \times 25 \times 10^{-6}} = \frac{3}{25\pi} \approx 0.0382$$ m/s. The Reynolds number is $$Re_1 = \frac{10^3 \times 0.0382 \times 0.01}{10^{-3}} \approx 382$$.

For the second flow rate $$Q_2 = 0.48$$ L/min $$= \frac{0.48 \times 10^{-3}}{60} = 8 \times 10^{-6}$$ m$$^3$$/s, the velocity is $$v_2 = \frac{Q_2}{A} = \frac{8}{25\pi} \approx 0.1019$$ m/s. The Reynolds number is $$Re_2 = \frac{10^3 \times 0.1019 \times 0.01}{10^{-3}} \approx 1019$$.

Since $$Re_1 \approx 382$$ is well below the critical value (around 1000 for flow from a tap), the initial flow is steady (laminar). However, $$Re_2 \approx 1019$$ exceeds this threshold, indicating a transition to unsteady flow.

Therefore, the flow changes from steady flow to unsteady flow.

In Millikan’s oil-drop experiment each drop very quickly reaches its terminal (steady) velocity. At this stage the forces on the uncharged drop balance exactly, so the downward gravitational force is matched by the upward viscous (Stokes) force. Neglecting the tiny buoyant force of the air, we may therefore write

$$\text{Viscous force } F_v \;=\; \text{Weight of the drop } mg.$$

We first determine the mass $$m$$ of the spherical oil drop. The formula for the volume of a sphere is

$$V \;=\;\frac{4}{3}\,\pi r^{3},$$

where $$r$$ is the radius. Substituting $$r = 2.0 \times 10^{-5}\,\text{m}$$, we get

$$V \;=\;\frac{4}{3}\,\pi\,(2.0 \times 10^{-5})^{3}.$$ Now, $$\left(2.0 \times 10^{-5}\right)^{3} = 2^{3} \times 10^{-15} = 8 \times 10^{-15}.$$ So

$$V \;=\;\frac{4}{3}\,\pi \,\left(8 \times 10^{-15}\right) = \frac{32}{3}\,\pi \times 10^{-15}$$ $$\qquad = 10.666\ldots\,\pi \times 10^{-15}$$ $$\qquad \approx 33.51 \times 10^{-15}\;\text{m}^{3}$$ $$\qquad = 3.351 \times 10^{-14}\;\text{m}^{3}.$$

The mass is obtained from $$m = \rho V,$$ where the density of oil is $$\rho = 1.2 \times 10^{3}\,\text{kg m}^{-3}.$$ Hence

$$m \;=\; \left(1.2 \times 10^{3}\right)\left(3.351 \times 10^{-14}\right) = 4.021 \times 10^{-11}\;\text{kg}.$$

The weight of the drop is $$mg,$$ with $$g = 9.8\;\text{m s}^{-2}:$$

$$mg \;=\; (4.021 \times 10^{-11})(9.8) = 3.94 \times 10^{-10}\;\text{N}.$$

Because the drop is falling at terminal velocity, this weight is opposed by an equal viscous drag, so the magnitude of the viscous force is

$$F_v \;=\; 3.9 \times 10^{-10}\;\text{N}.$$

Hence, the correct answer is Option B.

For a very small spherical body falling slowly through a viscous fluid, Stokes established that the viscous drag acting opposite to the motion is

$$F_{\text{drag}} = 6\pi \,\eta\, r\, v$$

where $$\eta$$ is the coefficient of viscosity of the fluid, $$r$$ is the radius of the sphere and $$v$$ is its speed relative to the fluid.

At terminal speed the net force on the sphere becomes zero. The downward forces are its weight and (if considered) the upward buoyant force. Because the problem tells us to neglect buoyancy, the force balance at terminal speed $$v_t$$ is simply

$$\text{weight} = \text{viscous drag}$$

$$\rho_w \, V \, g = 6\pi \,\eta\, r\, v_t$$

where $$V$$ is the volume $$\left(\dfrac{4}{3}\pi r^3\right)$$ of the raindrop and $$\rho_w$$ is the density of water.

Substituting the volume of the sphere we have

$$\rho_w \left(\dfrac{4}{3}\pi r^3\right) g = 6\pi \eta r v_t$$

Dividing both sides by $$\pi r$$ and then simplifying, we get

$$\dfrac{4}{3}\,\rho_w \, r^2 \, g = 6 \eta v_t$$

Next, dividing both sides by $$6\eta$$ gives

$$v_t = \dfrac{4}{18}\,\dfrac{\rho_w\, r^2 g}{\eta}$$

Recognising that $$\dfrac{4}{18} = \dfrac{2}{9}$$, the standard Stokes-law expression for terminal speed when buoyancy is neglected becomes

$$v_t = \dfrac{2 r^{2} g \rho_w}{9 \eta}$$

More generally, if buoyancy is included it is $$v_t = \dfrac{2 r^{2} g (\rho_w - \rho_a)}{9 \eta}$$, and because $$\rho_a$$ is so much smaller than $$\rho_w$$ the same result is obtained numerically whether or not we subtract it. For completeness we shall keep the subtraction:

$$v_t = \dfrac{2 r^{2} g (\rho_w - \rho_a)}{9 \eta}$$

Now we substitute the numerical values. The radius is given as

$$r = R = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m} = 2.0 \times 10^{-4}\ \text{m}$$

First we calculate $$r^2$$ :

$$r^2 = \left(2.0 \times 10^{-4}\right)^2 = 4.0 \times 10^{-8}\ \text{m}^2$$

The difference in densities is

$$(\rho_w - \rho_a) = 1000\ \text{kg m}^{-3} - 1.2\ \text{kg m}^{-3} = 998.8\ \text{kg m}^{-3}$$

The factor $$2 r^{2} g (\rho_w - \rho_a)$$ therefore equals

$$2 \times (4.0 \times 10^{-8}) \times 10 \times 998.8$$

We multiply step by step:

$$2 \times 4.0 \times 10^{-8} = 8.0 \times 10^{-8}$$

Multiplying by $$10$$ gives

$$8.0 \times 10^{-7}$$

Finally, multiplying by $$998.8$$ gives

$$8.0 \times 998.8 \times 10^{-7} = 7990.4 \times 10^{-7} = 7.9904 \times 10^{-4}$$

Now we find $$9\eta$$ :

$$\eta = 1.8 \times 10^{-5}\ \text{N s m}^{-2}$$

$$9\eta = 9 \times 1.8 \times 10^{-5} = 16.2 \times 10^{-5} = 1.62 \times 10^{-4}$$

Putting numerator and denominator together, the terminal speed is

$$v_t = \dfrac{7.9904 \times 10^{-4}}{1.62 \times 10^{-4}}$$

Dividing the powers of ten first we have $$10^{-4} / 10^{-4} = 1$$, so we simply divide the coefficients:

$$v_t = \dfrac{7.9904}{1.62}$$

Carrying out the division,

$$v_t \approx 4.93\ \text{m s}^{-1}$$

The tabulated options list 4.94 m s$$^{-1}$$, which is the same to two significant figures.

Hence, the correct answer is Option C.

We begin by noting that the metal (or glass) sphere is first allowed to fall through air. We are told to neglect the viscosity of air, so the only significant force acting on the ball in air is its weight. Consequently the ball performs free-fall motion.

For a freely falling body that starts from rest and drops through a vertical distance $$h$$, the kinematic relation

$$v^2 \;=\; u^2 \;+\; 2\,g\,h$$

applies (where $$u=0$$ because the ball is released from rest). Hence the speed of the sphere just before it touches the water surface is

$$v_{\text{air}} \;=\; \sqrt{2\,g\,h}.$$

Once the sphere enters the water, its motion is governed by viscous drag. For slow motion of a small sphere in a viscous fluid the drag force is given by Stokes’ law, and the sphere quickly attains a constant (terminal) speed. Stokes’ law states explicitly that the terminal velocity $$v_t$$ of a sphere of radius $$r$$ descending in a fluid of viscosity $$\eta$$ is

$$v_t \;=\; \dfrac{2}{9}\,\dfrac{r^{2}\,g\,(\rho-\rho_w)}{\eta},$$

where

• $$\rho$$ is the density of the sphere,

• $$\rho_w$$ is the density of water,

• $$g$$ is the acceleration due to gravity.

The problem states that the terminal velocity attained inside water is numerically equal to the velocity that the sphere already possessed on just entering the water. Symbolically,

$$v_t \;=\; v_{\text{air}}.$$

Substituting the two expressions derived above, we have

$$\dfrac{2}{9}\,\dfrac{r^{2}\,g\,(\rho-\rho_w)}{\eta} \;=\; \sqrt{2\,g\,h}.$$

Our aim is to solve this equality for $$h$$ in terms of $$r$$ and note the resulting proportionality. First, square both sides to remove the square root:

$$\left(\dfrac{2}{9}\,\dfrac{r^{2}\,g\,(\rho-\rho_w)}{\eta}\right)^{2} \;=\; 2\,g\,h.$$

Now isolate $$h$$ by dividing both sides by $$2\,g$$:

$$h \;=\; \dfrac{1}{2\,g}\,\left(\dfrac{2}{9}\,\dfrac{r^{2}\,g\,(\rho-\rho_w)}{\eta}\right)^{2}.$$

We next expand the square in the numerator step by step:

$$h \;=\; \dfrac{1}{2\,g}\;\Bigg[\;\dfrac{4}{81}\;r^{4}\;g^{2}\;(\rho-\rho_w)^{2}\;\dfrac{1}{\eta^{2}}\;\Bigg].$$

Combining the constants outside, we obtain

$$h \;=\; \dfrac{4}{81}\,\dfrac{g^{2}}{2\,g}\;\dfrac{(\rho-\rho_w)^{2}}{\eta^{2}}\;r^{4}.$$

Simplifying $$g^{2}/(2\,g)=g/2$$ and $$4/(81)\times 1/2 = 2/81$$, we find

$$h \;=\; \dfrac{2}{81}\;\dfrac{g\,(\rho-\rho_w)^{2}}{\eta^{2}}\;r^{4}.$$

Every term enclosed in the large bracket is independent of the radius $$r$$, whereas $$r^{4}$$ appears as an explicit factor. Therefore

$$h \;\propto\; r^{4}.$$

Hence, the correct answer is Option A.

We consider the steady, streamline flow of an incompressible fluid through a horizontal pipe. For such a situation we can apply Bernoulli’s theorem, which states that for any two points 1 and 2 along the same streamline

$$P_1 \;+\; \frac{1}{2}\,\rho\,v_1^{\,2} \;+\; \rho g h_1 \;=\; P_2 \;+\; \frac{1}{2}\,\rho\,v_2^{\,2} \;+\; \rho g h_2.$$

Because the pipe is horizontal the heights are equal, so $$h_1 = h_2$$ and the gravitational potential energy terms cancel. Hence, in the horizontal case the relation reduces to

$$P_1 \;+\; \frac{1}{2}\,\rho\,v_1^{\,2} \;=\; P_2 \;+\; \frac{1}{2}\,\rho\,v_2^{\,2}.$$

At the first point the pressure is given as $$P$$ and the speed as $$v$$. At the second point the pressure is given as $$\dfrac{P}{2}$$ and the speed is $$V$$. Substituting these values into the simplified Bernoulli equation we obtain

$$P \;+\; \frac{1}{2}\,\rho\,v^{2} \;=\; \frac{P}{2} \;+\; \frac{1}{2}\,\rho\,V^{2}.$$

We next bring all pressure terms to one side and all kinetic-energy terms to the other side. First subtract $$\dfrac{P}{2}$$ from both sides:

$$P \;-\; \frac{P}{2} \;+\; \frac{1}{2}\,\rho\,v^{2} \;=\; \frac{1}{2}\,\rho\,V^{2}.$$

Simplifying the pressure difference $$P - \dfrac{P}{2}$$ gives $$\dfrac{P}{2}$$, so we have

$$\frac{P}{2} \;+\; \frac{1}{2}\,\rho\,v^{2} \;=\; \frac{1}{2}\,\rho\,V^{2}.$$

Now eliminate the common factor $$\frac{1}{2}$$ on both sides by multiplying the entire equation by 2:

$$P \;+\; \rho\,v^{2} \;=\; \rho\,V^{2}.$$

We want to isolate $$V^{2}$$, so divide every term by $$\rho$$:

$$\frac{P}{\rho} \;+\; v^{2} \;=\; V^{2}.$$

Finally, take the positive square root (speed is positive) to find $$V$$:

$$V \;=\; \sqrt{\frac{P}{\rho} \;+\; v^{2}}.$$

This expression matches Option D, which is written as $$\sqrt{\frac{P}{\rho} + v^{2}}.$$

Hence, the correct answer is Option D.

We have a thin‐walled hollow sphere whose outer radius is $$R$$ and inner radius is $$r$$. Its material has specific gravity $$\dfrac{27}{8}$$ with respect to water. Specific gravity is defined as

$$\text{specific gravity}=\dfrac{\text{density of substance}}{\text{density of water}}.$$

So, if we denote the density of water by $$\rho$$, the density of the shell material is

$$\rho_{\!s}=\dfrac{27}{8}\,\rho.$$

Because the sphere floats just submerged, the whole external volume $$\left(\dfrac{4}{3}\pi R^{3}\right)$$ is under water. According to Archimedes’ principle, the buoyant force equals the weight of the water displaced, i.e.

$$F_{\text{buoyant}}=\rho\,g\left(\dfrac{4}{3}\pi R^{3}\right).$$

The weight of the body that must be supported is simply the weight of the shell material (the air inside has negligible mass). Its volume is the difference between the outer and inner volumes, so

$$F_{\text{weight}}=\rho_{\!s}\,g\left[\dfrac{4}{3}\pi\left(R^{3}-r^{3}\right)\right]=\dfrac{27}{8}\rho\,g\left[\dfrac{4}{3}\pi\left(R^{3}-r^{3}\right)\right].$$

For floating equilibrium we set buoyant force equal to weight:

$$\rho\,g\left(\dfrac{4}{3}\pi R^{3}\right)=\dfrac{27}{8}\rho\,g\left[\dfrac{4}{3}\pi\left(R^{3}-r^{3}\right)\right].$$

Now we cancel the common factors $$\rho,\,g,$$ and $$\dfrac{4}{3}\pi$$ from both sides:

$$R^{3}=\dfrac{27}{8}\left(R^{3}-r^{3}\right).$$

Multiplying through by $$8$$ gives

$$8R^{3}=27\left(R^{3}-r^{3}\right).$$

Expanding the right-hand side,

$$8R^{3}=27R^{3}-27r^{3}.$$

Rearranging to isolate $$r^{3},$$

$$27r^{3}=27R^{3}-8R^{3}=19R^{3},$$

hence

$$r^{3}=\dfrac{19}{27}R^{3}.$$

Taking the cube root of both sides we obtain

$$\dfrac{r}{R}=\left(\dfrac{19}{27}\right)^{1/3}\approx0.889.$$

The numerical value $$0.889$$ is practically equal to $$\dfrac{8}{9}=0.888\dots$$ Therefore,

$$r\;\approx\;\dfrac{8}{9}R.$$

Among the alternatives, this matches Option A.

Hence, the correct answer is Option A.

We begin by noting the basic principle of flotation: a body floats when the buoyant force exerted by the liquid equals the weight of the body. Mathematically, for a floating object, we have the equality $$\text{Buoyant force}= \text{Weight of the body}.$$

The buoyant force is given by Archimedes’ principle, which states that $$\text{Buoyant force}= \rho g V_{\text{sub}},$$ where $$\rho$$ is the density of the liquid, $$g$$ is the acceleration due to gravity, and $$V_{\text{sub}}$$ is the volume of the liquid displaced, i.e. the submerged volume of the object.

Let us denote:

$$\rho_0=$$ density of water at $$0^\circ\text{C},$$

$$\rho_4=$$ density of water at $$4^\circ\text{C},$$

$$A=$$ cross‑sectional area of the cylinder (constant),

$$L=1\ \text{m}=100\ \text{cm},$$ the length of the cylinder (constant, since the metal has negligible expansion).

The cylinder stands vertically and floats. Therefore the submerged length at each temperature can be found by subtracting the height that projects above the water level from the total length.

At $$0^\circ\text{C}:$$ the height above water is $$20\ \text{cm}=0.20\ \text{m}.$$ Hence the submerged length is

$$L_{\text{sub,0}}=L-0.20\ \text{m}=1.00\ \text{m}-0.20\ \text{m}=0.80\ \text{m}.$$

The submerged volume at $$0^\circ\text{C}$$ is therefore

$$V_{\text{sub,0}}=A\,L_{\text{sub,0}}=A(0.80\ \text{m}).$$

Applying Archimedes’ principle, the buoyant force equals the weight of the cylinder, so

$$\rho_0 g V_{\text{sub,0}}=mg,$$

which simplifies to

$$\rho_0 g \,(0.80A)=mg.$$

From this we can express the mass of the cylinder as

$$m=\rho_0(0.80A).$$

At $$4^\circ\text{C}:$$ the height above water becomes $$21\ \text{cm}=0.21\ \text{m}.$$ The submerged length is now

$$L_{\text{sub,4}}=L-0.21\ \text{m}=1.00\ \text{m}-0.21\ \text{m}=0.79\ \text{m}.$$

The submerged volume at $$4^\circ\text{C}$$ is correspondingly

$$V_{\text{sub,4}}=A\,L_{\text{sub,4}}=A(0.79\ \text{m}).$$

Again using Archimedes’ principle at $$4^\circ\text{C},$$ we have

$$\rho_4 g V_{\text{sub,4}}=mg.$$

Substituting the expressions we have just obtained gives

$$\rho_4 g\,(0.79A)=mg.$$

But we already expressed $$mg$$ from the $$0^\circ\text{C}$$ condition as $$\rho_0 g(0.80A).$$ Therefore we can write

$$\rho_4 g\,(0.79A)=\rho_0 g\,(0.80A).$$

We may now cancel the common factors $$g$$ and $$A,$$ yielding the simple ratio

$$\rho_4\,\bigl(0.79\bigr)=\rho_0\,\bigl(0.80\bigr).$$

Hence,

$$\frac{\rho_4}{\rho_0}=\frac{0.80}{0.79}.$$

Evaluating this fraction gives

$$\frac{\rho_4}{\rho_0}=\frac{80}{79}\approx1.012658\ldots\approx1.01.$$

This value of approximately $$1.01$$ corresponds to a relative increase of about $$1\%$$ in the density of water from $$0^\circ\text{C}$$ to $$4^\circ\text{C}.$$

Hence, the correct answer is Option C.

We have an air bubble in water. The radius of the bubble is given as $$r = 1\ \text{cm}$$, the upward (net) acceleration is $$a = 9.8\ \text{cm s}^{-2}$$, the density of water is $$\rho_w = 1\ \text{g cm}^{-3}$$ and the acceleration due to gravity in the CGS system is $$g = 980\ \text{cm s}^{-2}$$. Water drag is said to be negligible.

First we calculate the volume of the spherical bubble. The formula for the volume of a sphere is stated as

$$V \;=\; \frac{4}{3}\,\pi\,r^{3}.$$

Substituting $$r = 1\ \text{cm}$$, we get

$$V \;=\; \frac{4}{3}\,\pi\,(1\ \text{cm})^{3} \;=\; \frac{4\pi}{3}\ \text{cm}^{3}.$$

Evaluating the numerical value with $$\pi \approx 3.1416$$,

$$V \;=\; \frac{4 \times 3.1416}{3} \;=\; 4.18879\ \text{cm}^{3}.$$

Next we write the forces acting on the bubble. The upward buoyant force is equal to the weight of the displaced water. Hence,

$$F_{\text{buoyant}} \;=\; \rho_w\,g\,V.$$

The downward weight of the bubble itself is

$$W \;=\; m\,g,$$

where $$m$$ is the (unknown) mass of the bubble. Because drag is negligible, the net upward force on the bubble must provide the upward acceleration $$a$$. Therefore, using Newton’s second law,

$$F_{\text{net}} \;=\; F_{\text{buoyant}} \;-\; W \;=\; m\,a.$$

Substituting the expressions for the buoyant force and the weight, we have

$$\rho_w\,g\,V \;-\; m\,g \;=\; m\,a.$$

Now we collect the terms containing $$m$$ on one side:

$$\rho_w\,g\,V \;=\; m\,g \;+\; m\,a \;=\; m\,(g + a).$$

Solving for $$m$$ gives

$$m \;=\; \frac{\rho_w\,g\,V}{g + a}.$$

We now substitute the numerical values:

$$m \;=\; \frac{(1\ \text{g cm}^{-3})\,(980\ \text{cm s}^{-2})\,(4.18879\ \text{cm}^{3})}{980\ \text{cm s}^{-2} + 9.8\ \text{cm s}^{-2}}.$$

First compute the numerator:

$$980 \times 4.18879 \;=\; 4105.0162\ \text{g cm s}^{-2}.$$

Next compute the denominator:

$$980 + 9.8 \;=\; 989.8\ \text{cm s}^{-2}.$$

Now divide:

$$m \;=\; \frac{4105.0162}{989.8}\ \text{g}.$$

Carrying out the division,

$$m \;\approx\; 4.147\ \text{g}.$$

Rounding to three significant figures,

$$m \;\approx\; 4.15\ \text{g}.$$

Hence, the correct answer is Option C.

First, we recall the continuity equation for an ideal (incompressible) fluid flowing steadily through a pipe: $$A\,v=\text{constant},$$ where $$A$$ is the cross-sectional area of the pipe at any section and $$v$$ is the fluid speed at that section.

For a circular pipe the area is given by the well-known geometric formula $$A=\dfrac{\pi d^{2}}{4},$$ because the radius is $$d/2$$ and the area of a circle is $$\pi r^{2}.$$

Let us denote:

• $$d_{\text{max}} = 6.4\ \text{cm}$$ (the largest diameter),
• $$d_{\text{min}} = 4.8\ \text{cm}$$ (the smallest diameter),
• $$v_{\text{min}}$$ = fluid speed in the wider section (corresponding to $$d_{\text{max}}$$),
• $$v_{\text{max}}$$ = fluid speed in the narrower section (corresponding to $$d_{\text{min}}$$).

Using continuity, we write

$$A_{\text{max}}\,v_{\text{min}} = A_{\text{min}}\,v_{\text{max}}.$$

Substituting the area formula $$A=\dfrac{\pi d^{2}}{4}$$ into each term gives

$$\left(\dfrac{\pi d_{\text{max}}^{2}}{4}\right)v_{\text{min}} = \left(\dfrac{\pi d_{\text{min}}^{2}}{4}\right)v_{\text{max}}.$$

The common factors $$\pi/4$$ cancel out, leaving

$$d_{\text{max}}^{2}\,v_{\text{min}} = d_{\text{min}}^{2}\,v_{\text{max}}.$$

We need the ratio $$\dfrac{v_{\text{min}}}{v_{\text{max}}}.$$ Rearranging the previous relation, we obtain

$$\dfrac{v_{\text{min}}}{v_{\text{max}}} = \dfrac{d_{\text{min}}^{2}}{d_{\text{max}}^{2}}.$$

Now we substitute the given diameters:

$$\dfrac{v_{\text{min}}}{v_{\text{max}}} = \left(\dfrac{d_{\text{min}}}{d_{\text{max}}}\right)^{2} = \left(\dfrac{4.8\ \text{cm}}{6.4\ \text{cm}}\right)^{2}.$$

First simplify the fraction inside the brackets:

$$\dfrac{4.8}{6.4} = \dfrac{48}{64} = \dfrac{3}{4}.$$

Squaring this result gives

$$\left(\dfrac{3}{4}\right)^{2} = \dfrac{9}{16}.$$

Thus, the required ratio of the minimum velocity to the maximum velocity is

$$\dfrac{v_{\text{min}}}{v_{\text{max}}} = \dfrac{9}{16}.$$

Hence, the correct answer is Option A.

We are given a solid sphere of radius $$R$$ whose mass density varies with the radial distance $$r$$ from the centre according to the relation

$$\rho(r)=\rho_0\left(1-\dfrac{r^2}{R^2}\right),\qquad 0<r\le R.$$

To decide whether the sphere will float in a liquid we compare its average (bulk) density with the density of the liquid. Archimedes’ principle tells us that the sphere will just float when

$$\text{density of liquid}=\text{average density of sphere}.$$

Hence we first find the total mass $$M$$ of the sphere and then its average density $$\bar\rho=\dfrac{M}{V}$$, where $$V$$ is the volume of the sphere.

The infinitesimal mass of a thin spherical shell of radius $$r$$ and thickness $$dr$$ is obtained from

$$dM=\rho(r)\,dV,$$

and for a spherical shell the elemental volume is

$$dV=4\pi r^{2}\,dr.$$

Therefore, using the given density function,

$$dM=4\pi r^{2}\,\rho_0\!\left(1-\dfrac{r^{2}}{R^{2}}\right)\!dr.$$

To obtain the total mass we integrate from the centre $$r=0$$ to the surface $$r=R$$:

$$M=\int_{0}^{R}4\pi\rho_0\bigl(r^{2}-\dfrac{r^{4}}{R^{2}}\bigr)\,dr.$$

We now evaluate the two integrals separately, remembering the standard results

$$\int r^{2}\,dr=\dfrac{r^{3}}{3}\quad\text{and}\quad \int r^{4}\,dr=\dfrac{r^{5}}{5}.$$

So we have

$$\begin{aligned} M &=4\pi\rho_0\left[\int_{0}^{R}r^{2}\,dr-\dfrac{1}{R^{2}}\int_{0}^{R}r^{4}\,dr\right] \\ &=4\pi\rho_0\left[\left.\dfrac{r^{3}}{3}\right|_{0}^{R}-\dfrac{1}{R^{2}}\left.\dfrac{r^{5}}{5}\right|_{0}^{R}\right] \\ &=4\pi\rho_0\left[\dfrac{R^{3}}{3}-\dfrac{1}{R^{2}}\cdot\dfrac{R^{5}}{5}\right] \\ &=4\pi\rho_0\left[\dfrac{R^{3}}{3}-\dfrac{R^{3}}{5}\right] \\ &=4\pi\rho_0R^{3}\left(\dfrac{1}{3}-\dfrac{1}{5}\right). \end{aligned}$$

Inside the brackets the numerical subtraction gives

$$\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5-3}{15}=\dfrac{2}{15}.$$

Substituting this result,

$$M=4\pi\rho_0R^{3}\cdot\dfrac{2}{15}=\dfrac{8\pi\rho_0R^{3}}{15}.$$

The geometric volume of a sphere is well-known:

$$V=\dfrac{4}{3}\pi R^{3}.$$

Now we form the average density $$\bar\rho$$:

$$\bar\rho=\dfrac{M}{V} =\dfrac{\dfrac{8\pi\rho_0R^{3}}{15}}{\dfrac{4\pi R^{3}}{3}} =\dfrac{8}{15}\cdot\dfrac{3}{4}\,\rho_0 =\dfrac{24}{60}\,\rho_0 =\dfrac{2\rho_0}{5}.$$

Thus the least (minimum) density a liquid must have for the sphere to be able to float is

$$\rho_{\text{liq,min}}=\dfrac{2\rho_0}{5}.$$

Looking at the options provided, this matches Option C.

Hence, the correct answer is Option C.

Let us consider an element of the liquid situated at a distance $$r$$ from the axis of the cylindrical vessel which is rotating with a uniform angular speed $$\omega$$ rad s$$^{-1}$$. Because of the rotation, the element experiences a centrifugal acceleration $$\omega^{2}r$$ directed horizontally outward, while the usual gravitational acceleration $$g$$ acts vertically downward.

On the free surface of the liquid the pressure is the same everywhere (equal to the atmospheric pressure). For any two points on that free surface the total change in hydrostatic pressure must therefore be zero. We can write this balance in differential form. If the surface rises by a vertical amount $$dz$$ when we move radially outward by an infinitesimal distance $$dr$$, the pressure change due to gravity is $$\rho g\,dz$$ downward, and the pressure change due to the centrifugal field is $$\rho\,\omega^{2}r\,dr$$ outward. Equating the two (and cancelling the common factor $$\rho$$) we obtain

$$g\,dz \;=\; \omega^{2}r\,dr.$$

We now integrate this differential relation from the axis ($$r=0$$) to any general point on the surface at radius $$r$$. The left-hand side integrates over the corresponding vertical rise $$z$$ (measured from the level at the axis), while the right-hand side integrates over the radial distance:

$$\displaystyle\int_{0}^{z}\! g\,dz \;=\; \int_{0}^{r}\! \omega^{2}r\,dr.$$

Carrying out the integrations term by term, we get

$$g\,z \;=\; \omega^{2}\,\frac{r^{2}}{2}.$$

Solving for the height $$z$$ of the liquid surface above the centre level at a distance $$r$$ gives the well-known parabolic profile

$$z \;=\; \frac{\omega^{2}r^{2}}{2g}.$$

Now we are interested in the difference in height between the surface at the extreme side of the vessel ($$r = R$$) and the surface at the axis ($$r = 0$$). At the centre, $$r = 0$$ so $$z = 0$$. At the wall, $$r = R = 5\;{\rm cm}$$. Substituting this value into the expression for $$z$$, we obtain

$$h \;=\; z(R) - z(0) \;=\; \frac{\omega^{2}R^{2}}{2g} \;=\; \frac{\omega^{2}(5\;{\rm cm})^{2}}{2g}.$$

Evaluating the square of the radius,

$$(5\;{\rm cm})^{2} = 25\;{\rm cm}^{2},$$

and substituting back, we have

$$h \;=\; \frac{25\,\omega^{2}}{2g}\;{\rm cm}.$$

This expression matches Option C in the given list.

Hence, the correct answer is Option C.

Let us denote the density of the liquid by $$d$$ and the acceleration due to gravity by $$g$$. Each cylindrical vessel has the same base-area $$S$$, but the initial heights of the liquid columns are different: one is at height $$x_1$$ while the other is at height $$x_2$$.

For a uniform vertical column of liquid of height $$h$$, the centre of mass lies exactly at its mid-height $$\dfrac{h}{2}$$. Therefore, the gravitational potential energy $$U$$ of such a column is obtained from the general expression $$U = mgh_{\text{cm}}$$. Substituting the mass $$m = \rho V = d(S\,h)$$ and $$h_{\text{cm}}=\dfrac{h}{2}$$, we have

$$ U = (d\,S\,h)\,g\;\frac{h}{2} = \frac12\,d\,g\,S\,h^{2}. $$

We first compute the total gravitational potential energy of the system before the two vessels are connected.

For the vessel with height $$x_1$$ the energy is

$$U_1 = \frac12\,d\,g\,S\,x_1^{2},$$

and for the vessel with height $$x_2$$ the energy is

$$U_2 = \frac12\,d\,g\,S\,x_2^{2}.$$

Hence the initial total energy is

$$ U_{\text{initial}} = U_1 + U_2 = \frac12\,d\,g\,S\bigl(x_1^{2}+x_2^{2}\bigr). $$

Now we join the vessels by a tube of negligible volume fixed very near their bottoms. Because the liquid can now flow freely, it redistributes itself until both columns attain a common final height, which we shall call $$h_f$$.

The total volume of liquid is conserved. Initially the volume is

$$ V_{\text{total}} = Sx_1 + Sx_2 = S\,(x_1 + x_2). $$

After equilibrium the liquid occupies two cylinders, each of base area $$S$$, so the combined base area is $$2S$$. Therefore the common final height is obtained from

$$ 2S\,h_{f}=S\,(x_1 + x_2) \;\;\Longrightarrow\;\; h_{f} = \frac{x_1 + x_2}{2}. $$

For each vessel at this stage the liquid column has height $$h_f$$, so using the same energy formula we write

$$ U_{\text{one\,column\,(final)}} = \frac12\,d\,g\,S\,h_f^{2}. $$

Because there are two identical columns, the total final energy is

$$ U_{\text{final}} = 2\left(\frac12\,d\,g\,S\,h_f^{2}\right) = d\,g\,S\,h_f^{2}. $$

Next, we find the change in gravitational potential energy of the system:

$$ \Delta U = U_{\text{final}} - U_{\text{initial}} = d\,g\,S\,h_f^{2} - \frac12\,d\,g\,S\,(x_1^{2}+x_2^{2}). $$

We now substitute $$h_f = \dfrac{x_1 + x_2}{2}$$. First evaluate $$h_f^{2}$$:

$$ h_f^{2} = \left(\frac{x_1 + x_2}{2}\right)^{2} = \frac{(x_1 + x_2)^{2}}{4}. $$

Putting this into the expression for $$\Delta U$$ gives

$$ \Delta U = d\,g\,S\left[\frac{(x_1 + x_2)^{2}}{4} - \frac{1}{2}(x_1^{2}+x_2^{2})\right]. $$

To simplify the bracket, we write everything with the same denominator:

$$ \frac{(x_1 + x_2)^{2}}{4} - \frac{1}{2}(x_1^{2}+x_2^{2}) = \frac{(x_1 + x_2)^{2}}{4} - \frac{2(x_1^{2}+x_2^{2})}{4}. $$

Expanding $$ (x_1 + x_2)^{2} = x_1^{2} + 2x_1x_2 + x_2^{2} $$, we have

$$ \frac{x_1^{2} + 2x_1x_2 + x_2^{2} - 2x_1^{2} - 2x_2^{2}}{4} = \frac{-\,x_1^{2} + 2x_1x_2 - x_2^{2}}{4}. $$

Recognising the numerator as the negative of a perfect square,

$$ -\,x_1^{2} + 2x_1x_2 - x_2^{2} = -\,(x_1^{2} - 2x_1x_2 + x_2^{2}) = -\,(x_1 - x_2)^{2}. $$

Hence the entire bracket simplifies to

$$ \frac{-\,(x_1 - x_2)^{2}}{4} = -\,\frac{(x_1 - x_2)^{2}}{4}. $$

Substituting this back, the change in energy becomes

$$ \Delta U = d\,g\,S \left[-\,\frac{(x_1 - x_2)^{2}}{4}\right] = -\,\frac{1}{4}\,d\,g\,S\,(x_1 - x_2)^{2}. $$

The negative sign shows that energy has decreased. The magnitude of this decrease is

$$ \bigl|\Delta U\bigr| = \frac{1}{4}\,d\,g\,S\,(x_2 - x_1)^{2}, $$

which exactly matches Option D.

Hence, the correct answer is Option D.

We are asked to estimate the Reynolds number for water flowing through a pipe. The Reynolds number for flow in a circular pipe is defined by the formula

$$\displaystyle \text{Re} \;=\;\frac{\rho\,v\,D}{\eta},$$

where $$\rho$$ is the density of the fluid, $$v$$ is the average speed of flow, $$D$$ is the internal diameter of the pipe and $$\eta$$ is the dynamic (shear) viscosity.

First we turn every given quantity into S.I. units. The volumetric flow rate is given as 100 litres per minute. Because $$1\;\text{litre}=10^{-3}\;\text{m}^3$$ and $$1\;\text{min}=60\;\text{s}$$, we have

$$Q \;=\;100\times10^{-3}\,\text{m}^3\ \text{per}\ 60\ \text{s} \;=\;\frac{0.100}{60}\,\text{m}^3\!\big/\!\text{s} \;=\;1.6667\times10^{-3}\,\text{m}^3\!\big/\!\text{s}.$$

The radius of the pipe is 5 cm, that is $$r=5\;\text{cm}=0.05\;\text{m}.$$ Hence the diameter is

$$D \;=\;2r \;=\;2\times0.05\;\text{m}=0.10\;\text{m}.$$

Next we find the cross-sectional area of the pipe. Using $$A=\pi r^{2},$$

$$A \;=\;\pi\,(0.05\;\text{m})^{2} \;=\;\pi\,(2.5\times10^{-3}\,\text{m}^2) \;=\;7.8539\times10^{-3}\,\text{m}^2.$$

The average flow speed $$v$$ is the volume flow rate divided by the cross-sectional area, so

$$v \;=\;\frac{Q}{A} \;=\;\frac{1.6667\times10^{-3}}{7.8539\times10^{-3}} \;\text{m}\!/\!\text{s} \;=\;0.212\;\text{m/s}\;(\text{approximately}).$$

Now we substitute every value into the Reynolds number formula. For ordinary water the density is practically $$\rho=1000\;\text{kg/m}^{3}$$ and the viscosity is $$\eta=1\;\text{mPa·s}=1\times10^{-3}\;\text{Pa·s}$$ (remember $$1\;\text{Pa·s}=1\;\text{kg·m}^{-1}\text{s}^{-1}$$). Therefore

$$\text{Re} \;=\;\frac{\rho\,v\,D}{\eta} \;=\;\frac{(1000\;\text{kg/m}^3)(0.212\;\text{m/s})(0.10\;\text{m})} {1\times10^{-3}\;\text{Pa·s}} \;=\;\frac{21.2\;\text{kg·m}^{-1}\text{s}^{-1}}{10^{-3}\;\text{kg·m}^{-1}\text{s}^{-1}} \;=\;2.12\times10^{4}.$$

The number $$2.12\times10^{4}$$ is clearly of the order of $$10^{4}.$$ Hence, the correct answer is Option B.

First, recall Archimedes’ Principle: “When a body floats in a fluid, the weight of the body equals the weight of the fluid displaced.” We translate this into the mathematical form

$$\text{Weight of block} \;=\; (\text{density of water}) \times (\text{volume submerged}) \times g.$$

We have a cube whose edge length is 0.5 m, so its total geometric volume is

$$V_{\text{cube}} \;=\; (0.5\ \text{m})^{3} \;=\; 0.5 \times 0.5 \times 0.5 \;=\; 0.125\ \text{m}^{3}.$$

At the moment the cube is floating with only 30 % of its volume under water. Therefore the submerged volume is

$$V_{\text{sub, now}} \;=\; 0.30 \times V_{\text{cube}} \;=\; 0.30 \times 0.125 \;=\; 0.0375\ \text{m}^{3}.$$

Using Archimedes’ Principle the weight of the cube itself equals the weight of 0.0375 m³ of water. Taking the density of water to be $$\rho_{\text{w}} = 10^{3}\ \text{kg m}^{-3},$$ we write

$$\text{Mass of cube}\; (m_{\text{cube}}) =\; \rho_{\text{w}} \, V_{\text{sub, now}} =\; 10^{3} \times 0.0375 =\; 37.5\ \text{kg}.$$

Now imagine loading the cube until it is just about to be completely immersed, but not more. In that limiting situation the entire volume 0.125 m³ is displaced. The corresponding maximum supporting (buoyant) mass is

$$m_{\text{max\,disp}} =\; \rho_{\text{w}} \, V_{\text{cube}} =\; 10^{3} \times 0.125 =\; 125\ \text{kg}.$$

This 125 kg must balance the combined mass of the cube plus the additional load. Writing this balance condition, we get

$$m_{\text{cube}} \;+\; m_{\text{load, max}} =\; m_{\text{max\,disp}}.$$

Substituting the numerical values,

$$37.5 \;+\; m_{\text{load, max}} =\; 125.$$

Solving for the unknown maximum load,

$$m_{\text{load, max}} =\; 125 \;-\; 37.5 =\; 87.5\ \text{kg}.$$

Hence, the correct answer is Option A.

We begin at the mouth of the tap, where the water has an initial speed $$v_1 = 1.0\ \mathrm{m\,s^{-1}}$$ and a cross-sectional area $$A_1 = 10^{-4}\ \mathrm{m^2}$$. The water then falls vertically downward through a distance $$h = 0.15\ \mathrm{m}$$ under gravity. Because the motion is steadily downward in air, the pressure everywhere in the thin stream is the same atmospheric pressure; hence the pressure term in Bernoulli’s equation will cancel out between any two points along the streamline.

First, we apply Bernoulli’s theorem, which states

$$P + \frac{1}{2}\rho v^2 + \rho g z = \text{constant along a streamline},$$

where $$P$$ is pressure, $$\rho$$ is the density of the fluid, $$v$$ is speed and $$z$$ is the vertical height (measured upward from an arbitrary reference level).

Taking point 1 at the tap opening and point 2 at the position $$0.15\ \mathrm{m}$$ below the tap, we note that $$P_1 = P_2$$ (both equal to atmospheric pressure). Therefore, equating the remaining terms, we have

$$\frac{1}{2}\rho v_1^2 + \rho g z_1 = \frac{1}{2}\rho v_2^2 + \rho g z_2.$$ Because $$z_2 = z_1 - 0.15$$ (point 2 is lower), we substitute $$z_2 = z_1 - 0.15$$:

$$\frac{1}{2}\rho v_1^2 + \rho g z_1 = \frac{1}{2}\rho v_2^2 + \rho g (z_1 - 0.15).$$

Now we cancel the common $$\rho g z_1$$ term from both sides:

$$\frac{1}{2}\rho v_1^2 = \frac{1}{2}\rho v_2^2 - \rho g (0.15).$$

Dividing every term by $$\rho$$ and multiplying by 2 to clear the fractions, we get

$$v_1^2 = v_2^2 - 2g(0.15).$$

Rearranging to make $$v_2^2$$ the subject,

$$v_2^2 = v_1^2 + 2g(0.15).$$

Now we substitute the given numerical values $$v_1 = 1.0\ \mathrm{m\,s^{-1}}$$ and $$g = 10\ \mathrm{m\,s^{-2}}$$:

$$v_2^2 = (1.0)^2 + 2 \times 10 \times 0.15 = 1 + 3 = 4.$$

Hence,

$$v_2 = \sqrt{4} = 2\ \mathrm{m\,s^{-1}}.$$

Next, we invoke the equation of continuity for incompressible, steady flow, which states

$$A_1 v_1 = A_2 v_2,$$

where $$A_2$$ is the (unknown) cross-sectional area of the water stream at the lower point. Solving for $$A_2$$ gives

$$A_2 = \frac{A_1 v_1}{v_2}.$$

Substituting the known quantities $$A_1 = 10^{-4}\ \mathrm{m^2},\; v_1 = 1.0\ \mathrm{m\,s^{-1}},\; v_2 = 2\ \mathrm{m\,s^{-1}},$$ we obtain

$$A_2 = \frac{10^{-4} \times 1.0}{2} = 0.5 \times 10^{-4}\ \mathrm{m^2}.$$

Expressing the numerical factor in standard scientific notation,

$$A_2 = 5 \times 10^{-5}\ \mathrm{m^2}.$$

Hence, the correct answer is Option D.

To relate pressure and depth in a fluid we begin with the well-known hydrostatic formula

$$P \;=\; P_0 + \rho\,g\,h$$

where $$P$$ is the absolute pressure at depth $$h$$ below the free surface, $$P_0$$ is the pressure at the surface (atmospheric pressure), $$\rho$$ is the density of the fluid and $$g$$ is the acceleration due to gravity.

For two different depths $$d_1$$ and $$d_2$$ the corresponding pressures are

$$P_1 = P_0 + \rho g d_1$$

$$P_2 = P_0 + \rho g d_2$$

We are interested in the difference of the two depths, so we subtract the first equation from the second. The atmospheric term $$P_0$$ cancels out automatically:

$$P_2 - P_1 = \rho g (d_2 - d_1)$$

Rearranging for the depth difference gives

$$d_2 - d_1 = \frac{P_2 - P_1}{\rho g}$$

Now we substitute the numerical values given in the question. The submarine feels

$$P_1 = 5.05 \times 10^6 \text{ Pa}$$

$$P_2 = 8.08 \times 10^6 \text{ Pa}$$

and the data for water and gravity are

$$\rho = 10^3 \text{ kg m}^{-3}, \qquad g = 10 \text{ m s}^{-2}$$

First compute the pressure difference:

$$P_2 - P_1 = 8.08 \times 10^6 - 5.05 \times 10^6 = 3.03 \times 10^6 \text{ Pa}$$

Next form the product $$\rho g$$ appearing in the denominator:

$$\rho g = 10^3 \times 10 = 10^4 \text{ N m}^{-3}$$

Now divide to obtain the required depth difference:

$$d_2 - d_1 = \frac{3.03 \times 10^6}{10^4} = 3.03 \times 10^2 \text{ m} = 303 \text{ m}$$

This numerical result is very close to $$300$$ m.

Hence, the correct answer is Option C.

We begin by recalling what happens to the free surface of a liquid when its container spins steadily about a vertical axis. The surface becomes a paraboloid whose height, measured upward from the lowest point, is given by the equation

$$z \;=\; \frac{\omega^{2} r^{2}}{2g} + C,$$

where $$\omega$$ is the angular speed of the vessel, $$r$$ is the radial distance from the axis, $$g$$ is the acceleration due to gravity and $$C$$ is a constant representing the level at the centre. Hence, the rise of the liquid at any radius is

$$z(r) - z(0) \;=\; \frac{\omega^{2} r^{2}}{2g}.$$

We are interested in the maximum rise, i.e. the difference in height between the wall (radius $$R = 5 \text{ cm}$$) and the centre ($$r = 0$$). Putting $$r = R$$ gives

$$\Delta h \;=\; \frac{\omega^{2} R^{2}}{2g}.$$

Now we calculate each quantity one by one.

The vessel completes $$n = 2$$ full rotations every second, so the angular speed is

$$\omega \;=\; 2\pi n \;=\; 2\pi \times 2 \;=\; 4\pi \; \text{rad s}^{-1}.$$

Next, convert the radius to metres so that all SI units are consistent:

$$R = 5 \text{ cm} = 0.05 \text{ m}.$$

Square both $$\omega$$ and $$R$$ so we can place them into the formula:

$$\omega^{2} = (4\pi)^{2} = 16\pi^{2},$$

$$R^{2} = (0.05)^{2} = 0.0025.$$

Multiply these two squared quantities:

$$\omega^{2} R^{2} = 16\pi^{2} \times 0.0025.$$

Using $$\pi^{2} \approx 9.8696$$ we have

$$16\pi^{2} = 16 \times 9.8696 \approx 157.9136,$$

so

$$\omega^{2} R^{2} \approx 157.9136 \times 0.0025 \approx 0.3948.$$

We take $$g = 9.8 \text{ m s}^{-2}$$. Substituting everything into the formula for $$\Delta h$$ we obtain

$$\Delta h \;=\; \frac{0.3948}{2 \times 9.8}.$$

The denominator is $$2g = 2 \times 9.8 = 19.6$$, so

$$\Delta h = \frac{0.3948}{19.6} \approx 0.02016 \text{ m}.$$

Convert this height back into centimetres:

$$0.02016 \text{ m} = 0.02016 \times 100 \text{ cm} = 2.016 \text{ cm}.$$

Thus the difference in height between the liquid level at the wall and at the centre is essentially $$2.0 \text{ cm}$$ (to the precision of the given data).

Hence, the correct answer is Option B.

We have a horizontal jet emerging from a hose of radius $$a$$. The cross-sectional area of the jet is

$$A=\pi a^{2}\;.$$

Density of the liquid is $$\rho$$ and its speed is $$v$$, so the total mass that leaves the pipe per second (the mass flow rate) is, by definition,

$$\dot m=\rho A v=\rho\pi a^{2}v\;.$$

The mesh divides this stream into three parts:

• 50 % of the mass passes straight through without any change in velocity.
• 25 % of the mass is completely stopped, that is, its final velocity becomes $$0$$.
• 25 % of the mass reverses direction and returns with the same speed $$v$$ (so its final velocity is $$-v$$, opposite to the incident direction).

The force exerted on the mesh equals the rate of change of linear momentum of the fluid. In symbols,

$$F=\sum \dot m_i\,(v_{f,i}-v_{i,i})$$

where the summation is taken over the three portions of the jet. We now evaluate each contribution separately.

1. 50 % passes through unchanged

The mass flow in this part is $$\dot m_1=0.50\,\dot m$$, the initial velocity is $$v$$, and the final velocity is also $$v$$. Therefore

$$v_{f,1}-v_{i,1}=v-v=0\quad\Longrightarrow\quad \dot m_1\,(v_{f,1}-v_{i,1})=0\;.$$

This part gives no force.

2. 25 % is stopped

Here the mass flow rate is $$\dot m_2=0.25\,\dot m$$. Initial velocity is $$v$$ and final velocity is $$0$$, hence

$$v_{f,2}-v_{i,2}=0-v=-v\;.$$

The momentum change per second for this portion is

$$\dot m_2\,(v_{f,2}-v_{i,2})=0.25\,\dot m\,(-v)=-0.25\,\rho A v^{2}\;.$$

The negative sign shows that the fluid loses forward momentum; the mesh receives an equal and opposite impulse, so the force magnitude from this part is

$$F_2=0.25\,\rho A v^{2}\;.$$

3. 25 % rebounds with speed $$v$$

This portion has $$\dot m_3=0.25\,\dot m$$, initial velocity $$v$$, and final velocity $$-v$$. Hence

$$v_{f,3}-v_{i,3}=-v-v=-2v\;,$$

and the momentum change per second is

$$\dot m_3\,(v_{f,3}-v_{i,3})=0.25\,\dot m\,(-2v)=-0.50\,\rho A v^{2}\;.$$

Again the negative sign merely signals direction; the force on the mesh has magnitude

$$F_3=0.50\,\rho A v^{2}\;.$$

Total force on the mesh

Adding the magnitudes (the contributions act in the same direction against the mesh), we get

$$F=F_2+F_3=\left(0.25+0.50\right)\rho A v^{2}=0.75\,\rho A v^{2}=\frac{3}{4}\,\rho A v^{2}\;.$$

Pressure on the mesh

Pressure is force divided by the area over which the force acts. The jet area is $$A=\pi a^{2}$$, so

$$P=\frac{F}{A}=\frac{0.75\,\rho A v^{2}}{A}=0.75\,\rho v^{2} =\frac{3}{4}\,\rho v^{2}\;.$$

Hence, the correct answer is Option B.

For a small solid sphere moving slowly through a viscous fluid, we first recall Stokes’-law expression for the terminal (constant) velocity. The formula is

$$v = \frac{2}{9}\,\frac{r^{2}\,(\rho_{s}-\rho_{f})\,g}{\eta},$$

where $$r$$ is the radius of the sphere, $$\rho_{s}$$ its density, $$\rho_{f}$$ the density of the fluid, $$g$$ the acceleration due to gravity, and $$\eta$$ the coefficient of viscosity of the fluid.

When the original sphere of radius $$R$$ falls, its terminal velocity is therefore

$$v_{1}= \frac{2}{9}\,\frac{R^{2}\,(\rho_{s}-\rho_{f})\,g}{\eta}.$$

Now the same sphere is broken into 27 identical smaller spheres. Because mass (and therefore volume) is conserved, we equate the total volume before and after breaking.

The volume of the original sphere is

$$V_{\text{initial}} = \frac{4}{3}\pi R^{3}.$$

Each new sphere, of unknown radius $$r$$, has volume

$$V_{\text{one}} = \frac{4}{3}\pi r^{3}.$$

Since there are 27 such spheres, the total volume afterwards is

$$V_{\text{final}} = 27\left(\frac{4}{3}\pi r^{3}\right).$$

Setting the two volumes equal, we have

$$\frac{4}{3}\pi R^{3} = 27\left(\frac{4}{3}\pi r^{3}\right).$$

First we cancel the common factor $$\frac{4}{3}\pi$$ on both sides to obtain

$$R^{3}=27\,r^{3}.$$

Now taking the cube root of both sides, we get

$$R = 3\,r \quad\Longrightarrow\quad r = \frac{R}{3}.$$

Thus the radius of each small sphere is one-third that of the original sphere.

Applying Stokes’ law again, the terminal velocity of one of the small spheres is

$$v_{2}= \frac{2}{9}\,\frac{r^{2}\,(\rho_{s}-\rho_{f})\,g}{\eta}.$$

Substituting $$r = \dfrac{R}{3}$$ gives

$$v_{2}= \frac{2}{9}\,\frac{\left(\dfrac{R}{3}\right)^{2}\,(\rho_{s}-\rho_{f})\,g}{\eta}.$$

Evaluating the square, we have

$$\left(\dfrac{R}{3}\right)^{2}= \dfrac{R^{2}}{9},$$

so

$$v_{2}= \frac{2}{9}\,\frac{\dfrac{R^{2}}{9}\,(\rho_{s}-\rho_{f})\,g}{\eta} = \left(\frac{2}{9}\,\frac{R^{2}\,(\rho_{s}-\rho_{f})\,g}{\eta}\right)\frac{1}{9} = v_{1}\,\frac{1}{9}.$$

Finally, we form the requested ratio:

$$\frac{v_{1}}{v_{2}} = \frac{v_{1}}{v_{1}/9} = 9.$$

Hence, the correct answer is Option D.

We have an open tank, so the water surface and the jet emerging from the circular opening are both exposed to atmospheric pressure. For an ideal (non-viscous, incompressible) fluid this allows the use of Torricelli’s theorem, which states first that the speed with which the liquid emerges is the same as that which a body would acquire in freely falling through the vertical distance $$h$$ between the liquid surface and the centre of the orifice. Mathematically,

$$v \;=\; \sqrt{2 g h},$$

where $$g$$ is the acceleration due to gravity and $$h$$ is the required depth.

If the radius of the opening is $$r$$, its cross-sectional area $$A$$ is

$$A \;=\; \pi r^{2}.$$

The volume flow rate, or discharge $$Q$$, is the volume that leaves the tank per unit time. For steady flow we simply multiply the area by the exit velocity:

$$Q \;=\; A \, v.$$

Substituting $$v = \sqrt{2 g h}$$ into the discharge relation gives

$$Q \;=\; A \, \sqrt{2 g h}.$$

Now we solve algebraically for $$h$$. First, isolate the square root:

$$\sqrt{2 g h} \;=\; \dfrac{Q}{A}.$$

Next square both sides:

$$2 g h \;=\; \left(\dfrac{Q}{A}\right)^{2}.$$

Finally, divide by $$2g$$ to obtain the required depth:

$$h \;=\; \dfrac{1}{2 g}\,\left(\dfrac{Q}{A}\right)^{2}.$$

We now substitute the numerical values step by step.

Radius of the opening:

$$r = 2 \text{ cm} = 0.02 \text{ m}.$$

Area of the opening:

$$A = \pi r^{2} = \pi (0.02)^{2} = \pi \times 0.0004 = 0.001256637 \text{ m}^{2}.$$

Discharge in cubic metres per second (convert from “per minute”):

$$Q = 0.74 \text{ m}^{3}\text{/min} = \dfrac{0.74}{60} \text{ m}^{3}\text{/s} = 0.012333\text{ m}^{3}\text{/s}.$$

Compute the ratio $$Q/A$$:

$$\dfrac{Q}{A} = \dfrac{0.012333}{0.001256637} = 9.813 \text{ m/s}.$$

Insert this value into the depth formula (take $$g = 9.8 \text{ m/s}^{2}$$):

$$h = \dfrac{1}{2 \times 9.8}\,(9.813)^{2} \;=\; \dfrac{1}{19.6} \times 96.280 \;=\; 4.913 \text{ m}.$$

The value obtained is very close to $$4.8 \text{ m}$$ when rounded to the nearest tenth of a metre, matching one of the given choices.

Hence, the correct answer is Option B.

Let the density of water be $$\rho_{w}$$, the density of the wooden block be $$\rho_{b}$$ and the density of oil be $$\rho_{o}$$. We denote the total volume of the block by $$V$$ and take $$g$$ as the acceleration due to gravity.

First, consider the initial situation when only water is present. The block floats with $$\dfrac{4}{5}$$ of its volume submerged. For a floating body, the condition of equilibrium is given by the principle of flotation:

$$\text{(buoyant force)} = \text{(weight of block)}.$$

The buoyant force is the weight of the displaced water, which equals $$\rho_{w}g \left(\dfrac{4}{5}V\right)$$, while the weight of the block is $$\rho_{b}gV$$. Setting these equal, we have

$$\rho_{w}g \left(\dfrac{4}{5}V\right)=\rho_{b}gV.$$

The factor $$gV$$ cancels from both sides, giving

$$\rho_{b}=\dfrac{4}{5}\rho_{w}.$$

Now oil is gently poured on top of the water until the block just touches the oil-air interface. At this stage, exactly half the block’s volume is under water and the remaining half is in oil. Thus the volumes in each liquid are $$\dfrac{V}{2}$$.

The total buoyant force is now the sum of the forces due to the displaced water and the displaced oil, so we write

$$\rho_{w}g\left(\dfrac{V}{2}\right)+\rho_{o}g\left(\dfrac{V}{2}\right)=\rho_{b}gV.$$

Again, we can cancel the common factor $$g$$ and also factor out $$\dfrac{V}{2}$$ from the left-hand side:

$$\dfrac{V}{2}\left(\rho_{w}+\rho_{o}\right)=\rho_{b}V.$$

Dividing both sides by $$\dfrac{V}{2}$$ yields

$$\rho_{w}+\rho_{o}=2\rho_{b}.$$

Substituting the value $$\rho_{b}=\dfrac{4}{5}\rho_{w}$$ obtained earlier, we get

$$\rho_{w}+\rho_{o}=2\left(\dfrac{4}{5}\rho_{w}\right)=\dfrac{8}{5}\rho_{w}.$$

Solving for $$\rho_{o}$$, we find

$$\rho_{o}=\dfrac{8}{5}\rho_{w}-\rho_{w}=\left(\dfrac{8}{5}-1\right)\rho_{w}=\dfrac{3}{5}\rho_{w}.$$

The ratio $$\dfrac{\rho_{o}}{\rho_{w}}$$ is therefore

$$\dfrac{\rho_{o}}{\rho_{w}}=\dfrac{3}{5}=0.6.$$

Hence, the correct answer is Option D.

We are told that water enters the tank with a volume flow rate $$Q_{\text{in}} = 10^{-4}\, \text{m}^3\text{s}^{-1}$$. At the same time, water leaves through a small hole in the bottom.

The cross-sectional area of this hole is given as $$1\; \text{cm}^2$$. Converting to SI units, $$1\; \text{cm}^2 = 1 \times 10^{-4}\; \text{m}^2$$.

For the speed of efflux of a liquid through a small orifice at depth $$h$$ below the free surface, we use Torricelli’s theorem, which states:

$$v = \sqrt{2gh},$$

where $$g$$ is the acceleration due to gravity (take $$g = 9.8\; \text{m s}^{-2}$$) and $$h$$ is the height of the liquid column above the hole.

The volume flow rate of water leaking out is therefore

$$Q_{\text{out}} = A\,v = (1 \times 10^{-4}) \sqrt{2gh}\; \text{m}^3\text{s}^{-1}.$$

Because the water level is steady, the inflow and outflow rates are equal:

$$Q_{\text{in}} = Q_{\text{out}}.$$

Substituting the expressions for the two flow rates, we have

$$10^{-4} = (1 \times 10^{-4}) \sqrt{2gh}.$$

Dividing both sides by $$1 \times 10^{-4}$$ gives

$$1 = \sqrt{2gh}.$$

Now we square both sides to remove the square root:

$$1^2 = ( \sqrt{2gh} )^2 \;\;\Longrightarrow\;\; 1 = 2gh.$$

Solving for $$h$$:

$$h = \frac{1}{2g}.$$

Substituting $$g = 9.8\; \text{m s}^{-2}$$, we obtain

$$h = \frac{1}{2 \times 9.8} = \frac{1}{19.6} \approx 0.05102\; \text{m}.$$

Converting meters to centimeters:

$$0.05102\; \text{m} = 0.05102 \times 100\; \text{cm} \approx 5.1\; \text{cm}.$$

Hence, the correct answer is Option A.

The bubble is a pocket of gas. During its slow ascent we can safely assume that the temperature of this gas does not change, so the gas obeys Boyle’s law, which states first: $$P_1V_1 = P_2V_2,$$ where $$P_1,\,V_1$$ are the pressure and volume at the lake bottom and $$P_2,\,V_2$$ are the corresponding quantities at the surface.

We denote by $$h$$ the depth of the lake. Let $$\rho$$ be the density of water and $$g$$ the acceleration due to gravity. At the bottom, the total pressure acting on the bubble is the sum of the atmospheric pressure and the hydrostatic pressure of the water column above it. Hence we have

$$P_1 = P_{\text{atm}} + \rho g h.$$

At the surface the bubble is exposed only to the atmospheric pressure, so

$$P_2 = P_{\text{atm}}.$$

The problem tells us that the atmospheric pressure is equivalent to the pressure of a 10 m column of water, that is

$$P_{\text{atm}} = \rho g \times 10\;\text{m} = 10\rho g.$$

Now we examine the volumes. The volume of a sphere varies as the cube of its radius. If the initial radius is $$r$$ and the final radius is $$\dfrac{5r}{4},$$ then

$$\frac{V_2}{V_1} = \left(\frac{5r/4}{\,r}\right)^3 = \left(\frac{5}{4}\right)^3 = \frac{125}{64}.$$

With Boyle’s law $$P_1V_1 = P_2V_2,$$ we substitute the pressures and the volume ratio:

$$\bigl(P_{\text{atm}} + \rho g h\bigr)\,V_1 \;=\; P_{\text{atm}}\,V_2 \;=\; P_{\text{atm}}\left(\frac{125}{64}V_1\right).$$

The factor $$V_1$$ appears on both sides, so it cancels out, leaving

$$P_{\text{atm}} + \rho g h = P_{\text{atm}}\left(\frac{125}{64}\right).$$

We now isolate $$\rho g h$$:

$$\rho g h = P_{\text{atm}}\left(\frac{125}{64} - 1\right) = P_{\text{atm}}\left(\frac{125 - 64}{64}\right) = P_{\text{atm}}\left(\frac{61}{64}\right).$$

Because $$P_{\text{atm}} = 10\rho g,$$ we substitute this value:

$$\rho g h = 10\rho g \times \frac{61}{64}.$$

Both $$\rho$$ and $$g$$ cancel out, leaving a simple numerical result:

$$h = 10 \times \frac{61}{64} = \frac{610}{64}\,\text{m} \approx 9.53\,\text{m}.$$

The depth of the lake is therefore very close to 9.5 m, which matches the numerical choice in the options.

Hence, the correct answer is Option 4.

We deal with viscous flow through narrow tubes, so we start with the Poiseuille relation. The formula for the pressure difference $$P$$ required to push a steady (stream-line) volume flow rate $$Q$$ through a cylindrical tube of length $$l$$ and radius $$r$$ is stated as

$$P \;=\;\dfrac{8\eta\,l\,Q}{\pi\,r^{4}}$$

where $$\eta$$ is the viscosity of the liquid. We notice that $$\eta$$, $$\pi$$ and the common volume flow rate $$Q$$ are the same for both tubes because the tubes are connected in series; whatever liquid flows through tube 1 per unit time must immediately flow through tube 2 per unit time.

Writing the Poiseuille expression separately for the two tubes, we have

For tube 1: $$P_{1} \;=\;\dfrac{8\eta\,l_{1}\,Q}{\pi\,r_{1}^{4}}$$

For tube 2: $$P_{2} \;=\;\dfrac{8\eta\,l_{2}\,Q}{\pi\,r_{2}^{4}}$$

Now we are given two experimental facts:

$$P_{2}=4P_{1} \quad\text{and}\quad l_{2}=\dfrac{l_{1}}{4}$$

To connect these facts with the radii, we divide the second Poiseuille equation by the first. Doing this cancels the common factors $$8\eta Q/\pi$$ and yields

$$\dfrac{P_{2}}{P_{1}} =\dfrac{\;l_{2}/r_{2}^{4}\;}{\;l_{1}/r_{1}^{4}\;} =\dfrac{l_{2}\,r_{1}^{4}}{l_{1}\,r_{2}^{4}}$$

Substituting the known ratio $$P_{2}/P_{1}=4$$ and the length relation $$l_{2}=l_{1}/4$$, we obtain

$$4=\dfrac{\bigl(l_{1}/4\bigr)\,r_{1}^{4}}{l_{1}\,r_{2}^{4}}$$

The lengths $$l_{1}$$ cancel out, leaving

$$4=\dfrac{r_{1}^{4}}{4\,r_{2}^{4}}$$

Multiplying both sides by $$4\,r_{2}^{4}$$ gives

$$16\,r_{2}^{4}=r_{1}^{4}$$

Now we take the fourth root of both sides to extract the radius ratio:

$$r_{2}=\dfrac{r_{1}}{\,\sqrt[4]{16}\,}=\dfrac{r_{1}}{2}$$

Thus the radius of the second tube is half the radius of the first tube. Among the options provided, this corresponds to Option D.

Hence, the correct answer is Option D.

At the hole (height $$h$$): Velocity = $$v_1$$, Radius = $$r$$, Area = $$\pi r^2$$.

At the ground (height 0): Velocity = $$v_2$$, Radius = $$x$$, Area = $$\pi x^2$$.

$$v_1 = \sqrt{2gH}$$

Using $$v^2 = u^2 + 2as$$: $$v_2^2 = v_1^2 + 2gh$$

$$v_2^2 = 2gH + 2gh = 2g(H + h)$$

$$v_2 = \sqrt{2g(H + h)}$$

Using equation of continuity, $$A_1 v_1 = A_2 v_2$$, 

$$\pi r^2 v_1 = \pi x^2 v_2$$

$$x^2 = r^2 \left( \frac{v_1}{v_2} \right)$$

$$x^2 = r^2 \frac{\sqrt{2gH}}{\sqrt{2g(H + h)}}$$

$$x^2 = r^2 \sqrt{\frac{H}{H + h}} = r^2 \left( \frac{H}{H + h} \right)^{1/2}$$

$$x = r \left( \frac{H}{H + h} \right)^{1/4}$$

The block is cylindrical, so its volume is equal to the base area multiplied by the height. We first convert all the given quantities to SI units so that we can use them directly in the standard formulas.

The base area is given as $$30\ \text{cm}^2$$. Since $$1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$$, we have $$A = 30 \times 10^{-4}\ \text{m}^2 = 3.0 \times 10^{-3}\ \text{m}^2.$$

The height is given as $$54\ \text{cm}$$. Using $$1\ \text{cm} = 10^{-2}\ \text{m}$$, we write $$h = 54 \times 10^{-2}\ \text{m} = 0.54\ \text{m}.$$

Now the volume of the cylinder is $$V = A \times h = (3.0 \times 10^{-3})\ \text{m}^2 \times 0.54\ \text{m} = 1.62 \times 10^{-3}\ \text{m}^3.$$

The density of the wooden block is $$\rho_b = 650\ \text{kg m}^{-3}$$, so its mass is $$m = \rho_b V = 650\ \text{kg m}^{-3} \times 1.62 \times 10^{-3}\ \text{m}^3 = 1.053\ \text{kg}.$$

The liquid in which the block floats has density $$\rho_l = 900\ \text{kg m}^{-3}$$. When the block is pushed down by a small vertical displacement $$x$$ and released, the extra volume of liquid displaced is $$A x$$, and therefore the additional buoyant force is

$$F_b = \rho_l g (A x).$$

This force acts upward and tends to restore the block to equilibrium. Using Newton’s second law, the equation of motion for the block becomes

$$m \frac{d^2x}{dt^2} + \rho_l g A x = 0.$$

This is the standard form of a simple harmonic motion equation, $$\frac{d^2x}{dt^2} + \omega^2 x = 0,$$ where the angular frequency $$\omega$$ is given by

$$\omega^2 = \frac{\rho_l g A}{m}.$$

Hence $$\omega = \sqrt{\frac{\rho_l g A}{m}}.$$

The time period of oscillation is related to the angular frequency by the formula $$T = \frac{2\pi}{\omega}.$$ Substituting $$\omega$$ from above we get

$$T = 2\pi \sqrt{\frac{m}{\rho_l g A}}.$$

Now we substitute the numerical values: $$m = 1.053\ \text{kg},\quad \rho_l = 900\ \text{kg m}^{-3},\quad g = 9.8\ \text{m s}^{-2},\quad A = 3.0 \times 10^{-3}\ \text{m}^2.$$

First calculate the denominator inside the square root: $$\rho_l g A = 900 \times 9.8 \times 3.0 \times 10^{-3} = 8820 \times 3.0 \times 10^{-3} = 26.46.$$

Next compute the ratio inside the square root: $$\frac{m}{\rho_l g A} = \frac{1.053}{26.46} \approx 0.0398.$$

Taking the square root gives $$\sqrt{0.0398} \approx 0.1995.$$

Therefore the time period is $$T = 2\pi \times 0.1995 \approx 6.283 \times 0.1995 \approx 1.25\ \text{s}.$$

We are asked to find the length $$L$$ of a simple pendulum that has the same time period. For a simple pendulum, the time period is given by the well-known formula

$$T = 2\pi \sqrt{\frac{L}{g}}.$$

Equating the two expressions for the period, we write $$1.25 = 2\pi \sqrt{\frac{L}{g}}.$$

First isolate the square root term: $$\sqrt{\frac{L}{g}} = \frac{1.25}{2\pi} = \frac{1.25}{6.283} \approx 0.199.$$

Now square both sides: $$\frac{L}{g} = (0.199)^2 \approx 0.0396.$$

Finally multiply by $$g = 9.8\ \text{m s}^{-2}$$ to obtain $$L$$: $$L = 0.0396 \times 9.8 \approx 0.388\ \text{m}.$$

Converting metres to centimetres, $$L = 0.388\ \text{m} \times 100\ \text{cm m}^{-1} \approx 38.8\ \text{cm}.$$

The value is very close to $$39\ \text{cm}$$, which matches option C.

Hence, the correct answer is Option C.

The Reynolds number is a dimensionless quantity used to predict flow patterns in fluid dynamics. It is given by the formula:

$$ \text{Re} = \frac{\rho v d}{\eta} $$

where:

• $$\rho$$ is the density of the fluid,
• $$v$$ is the flow velocity,
• $$d$$ is the diameter of the pipe,
• $$\eta$$ is the dynamic viscosity.

Given:

• Time to fill the bucket, $$t = 5$$ minutes = $$5 \times 60 = 300$$ seconds,
• Volume of bucket, $$V = 15$$ litres = $$15 \times 10^{-3}$$ m$$^3$$ (since 1 litre = $$10^{-3}$$ m$$^3$$),
• Diameter of tap, $$d = \frac{2}{\sqrt{\pi}}$$ cm,
• Density of water, $$\rho = 10^3$$ kg/m$$^3$$,
• Viscosity of water, $$\eta = 10^{-3}$$ Pa.s.

First, convert the diameter to meters because SI units are required. Since 1 cm = 0.01 m,

$$ d = \frac{2}{\sqrt{\pi}} \times 10^{-2} \text{ m} = \frac{0.02}{\sqrt{\pi}} \text{ m}. $$

Next, calculate the cross-sectional area $$A$$ of the tap. Since the tap is circular,

$$ A = \pi \left( \frac{d}{2} \right)^2. $$

Substitute $$d$$:

$$ \frac{d}{2} = \frac{0.02}{2\sqrt{\pi}} = \frac{0.01}{\sqrt{\pi}} \text{ m}, $$

so,

$$ A = \pi \left( \frac{0.01}{\sqrt{\pi}} \right)^2 = \pi \times \frac{(0.01)^2}{\pi} = \pi \times \frac{0.0001}{\pi} = 0.0001 \text{ m}^2 = 10^{-4} \text{ m}^2. $$

Now, find the volume flow rate $$Q$$, which is volume divided by time:

$$ Q = \frac{V}{t} = \frac{15 \times 10^{-3}}{300} = \frac{0.015}{300} = 0.00005 \text{ m}^3/\text{s} = 5 \times 10^{-5} \text{ m}^3/\text{s}. $$

The flow velocity $$v$$ is given by $$Q = A \times v$$, so

$$ v = \frac{Q}{A} = \frac{5 \times 10^{-5}}{10^{-4}} = \frac{5 \times 10^{-5}}{1 \times 10^{-4}} = 5 \times 10^{-1} = 0.5 \text{ m/s}. $$

Now, substitute all values into the Reynolds number formula:

$$ \text{Re} = \frac{\rho v d}{\eta} = \frac{(10^3) \times (0.5) \times \left( \frac{0.02}{\sqrt{\pi}} \right)}{10^{-3}}. $$

First, compute the numerator:

$$ \rho v d = 1000 \times 0.5 \times \frac{0.02}{\sqrt{\pi}} = 500 \times \frac{0.02}{\sqrt{\pi}} = \frac{10}{\sqrt{\pi}}. $$

Then, divide by $$\eta$$:

$$ \text{Re} = \frac{\frac{10}{\sqrt{\pi}}}{10^{-3}} = \frac{10}{\sqrt{\pi}} \times \frac{1}{10^{-3}} = \frac{10}{\sqrt{\pi}} \times 10^3 = \frac{10000}{\sqrt{\pi}}. $$

Now, compute the numerical value. Using $$\pi \approx 3.1416$$, $$\sqrt{\pi} \approx \sqrt{3.1416} \approx 1.77245$$, so

$$ \text{Re} = \frac{10000}{1.77245} \approx 5641.89. $$

Comparing with the options:

• A. 5500
• B. 550
• C. 1100
• D. 11000

The value 5641.89 is closest to 5500, with a difference of approximately 141.89, while the other options are farther away (550 is too small, 1100 is smaller, and 11000 is much larger). Therefore, the Reynolds number is close to 5500.

Hence, the correct answer is Option A.

We have a straight glass tube whose bore (cross-sectional area) is uniform. Initially both ends are open and the tube is dipped into a mercury bath in such a way that $$8\ \text{cm}$$ of its length projects above the outside mercury level. Hence the column inside the tube that lies above the mercury in the tube is an $$8\ \text{cm}$$ long column of air which is at the atmospheric pressure $$P_{\text{atm}} = 76\ \text{cm of Hg}$$, because the upper end is still open to the atmosphere.

Now the upper (projecting) end is closed and instantly sealed. At that instant the trapped air has

$$P_1 = 76\ \text{cm Hg}, \qquad V_1 \propto 8\ \text{cm}.$$

Next the whole tube is lifted vertically upward by an additional $$46\ \text{cm}$$ while its lower end remains immersed in the same mercury bath. Consequently, the top of the tube now stands

$$8 + 46 = 54\ \text{cm}$$

above the outside mercury surface.

Let the new length of the trapped air column be $$L\ \text{cm}$$ (this is what we have to find). Then the mercury inside the tube must have risen above the outside level by

$$H = 54 - L\ \text{cm}.$$

To relate the new air pressure to this rise, we use the hydrostatic pressure relation:

Moving upward through a column of mercury of height $$H$$ reduces the pressure by $$H\ \text{cm of Hg}$$. Hence the pressure just beneath the mercury-air interface inside the tube is

$$P_{\text{interface}} = P_{\text{atm}} - H = 76 - (54 - L) = 76 - 54 + L = 22 + L\ \text{cm Hg}.$$

This pressure is the same as the pressure $$P_2$$ of the trapped air above the mercury, because the air is in direct contact with the mercury surface. Therefore,

$$P_2 = 22 + L\ \text{cm Hg}.$$

The process occurs slowly enough for us to treat the trapped air as undergoing an isothermal change (temperature is effectively constant). Hence we can apply Boyle’s law, which states

$$P_1 V_1 = P_2 V_2.$$

Because the cross-sectional area of the tube is constant, the volume is proportional to the length of the air column, so we have

$$P_1 \times (\text{length }8) = P_2 \times (\text{length }L).$$

Substituting the known values,

$$76 \times 8 = (22 + L)\, L.$$

That gives the quadratic equation

$$L^2 + 22L - 608 = 0.$$

Solving it by the quadratic formula,

$$L = \frac{-22 \pm \sqrt{22^2 + 4 \times 608}}{2} = \frac{-22 \pm \sqrt{484 + 2432}}{2} = \frac{-22 \pm \sqrt{2916}}{2} = \frac{-22 \pm 54}{2}.$$

The negative root is not physically meaningful (it would give a negative length). Taking the positive root,

$$L = \frac{32}{2} = 16\ \text{cm}.$$

Hence the length of the trapped air column after the tube has been raised is $$16\ \text{cm}$$.

Hence, the correct answer is Option A.

The pressure difference between the two points is determined by the height difference $$\Delta h = 5\text{ cm}$$ in the manometer:

$$\Delta P = \rho g \Delta h = 1\text{ g/cc} \times 10^3\text{ cm/s}^2 \times 5\text{ cm} = 5 \times 10^3\text{ dyne/cm}^2$$

Using the equation of continuity $$A_1v_1 = A_2v_2$$: $$6v_A = 10v_B \implies v_A = \frac{5}{3}v_B$$

Substituting this into Bernoulli's equation, $$\frac{1}{2}\rho(v_A^2 - v_B^2) = P_B - P_A$$:

$$\frac{1}{2} \times 1 \left[ \left(\frac{5}{3}v_B\right)^2 - v_B^2 \right] = 5 \times 10^3$$

$$\frac{1}{2} \left( \frac{16}{9}v_B^2 \right) = 5 \times 10^3 \implies v_B = \frac{3}{4} \times 10^2\text{ cm/s}$$

$$\text{Volume flow rate} = v_B A_B = \left(\frac{3}{4} \times 10^2\right) \times (10 \times 10^{-2}) = 7.5\text{ cc/s}$$

To solve this problem, we need to find the minimum force required to stop the flow of water using your palm. The water flows at a speed of 1.5 m/s through a tube with a cross-sectional area of $$10^{-2}$$ m², and it stops immediately upon hitting the palm. The density of water is $$10^3$$ kg/m³.

The force exerted by the palm must counteract the momentum of the incoming water per unit time. Force is defined as the rate of change of momentum. When the water hits the palm and stops, its velocity changes from 1.5 m/s to 0 m/s. The change in velocity ($$\Delta v$$) is therefore $$1.5 - 0 = 1.5$$ m/s.

First, we calculate the volume flow rate, which is the volume of water passing through the tube per second. This is given by the product of the cross-sectional area ($$A$$) and the flow velocity ($$v$$):

Volume flow rate = $$A \times v = 10^{-2} \, \text{m}^2 \times 1.5 \, \text{m/s} = 0.015 \, \text{m}^3/\text{s}$$.

Next, we find the mass flow rate, which is the mass of water flowing per second. This is obtained by multiplying the volume flow rate by the density of water ($$\rho$$):

Mass flow rate = $$\rho \times \text{volume flow rate} = 10^3 \, \text{kg/m}^3 \times 0.015 \, \text{m}^3/\text{s} = 15 \, \text{kg/s}$$.

The rate of change of momentum is equal to the mass flow rate multiplied by the change in velocity ($$\Delta v$$), since the water comes to a complete stop:

Rate of change of momentum = mass flow rate $$\times \Delta v = 15 \, \text{kg/s} \times 1.5 \, \text{m/s} = 22.5 \, \text{kg} \cdot \text{m/s}^2$$.

Since $$1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2$$, the force required is 22.5 N. This is the minimum force needed to stop the water flow, assuming an immediate stop with no other losses.

Comparing with the options: A. 33.7 N, B. 45 N, C. 15 N, D. 22.5 N. Hence, the correct answer is Option D.

The average mass of each rain drop is $$ m = 3.0 \times 10^{-5} $$ kg, and the average terminal velocity is $$ v = 9 $$ m/s. The energy transferred by each rain drop when it hits the surface is due to its kinetic energy. The kinetic energy (KE) of one rain drop is given by:

$$ KE = \frac{1}{2} m v^2 $$

Substituting the values:

$$ KE = \frac{1}{2} \times (3.0 \times 10^{-5}) \times (9)^2 $$

First, calculate $$ 9^2 $$:

$$ 9^2 = 81 $$

Then multiply by the mass:

$$ m \times 81 = 3.0 \times 10^{-5} \times 81 = 243 \times 10^{-5} = 2.43 \times 10^{-3} $$

Now multiply by $$ \frac{1}{2} $$:

$$ KE = \frac{1}{2} \times 2.43 \times 10^{-3} = 1.215 \times 10^{-3} \text{ J} $$

So, each rain drop transfers $$ 1.215 \times 10^{-3} $$ J of energy.

The place receives 100 cm of rain in a year. Convert this to meters:

$$ 100 \text{ cm} = 1 \text{ m} $$

This is the height of the water column over an area of 1 square meter. The volume of rain falling on 1 m² area is:

$$ \text{Volume} = \text{area} \times \text{height} = 1 \times 1 = 1 \text{ m}^3 $$

The density of water is 1000 kg/m³, so the total mass of rain falling on 1 m² area is:

$$ \text{Mass} = \text{volume} \times \text{density} = 1 \times 1000 = 1000 \text{ kg} $$

The mass of one rain drop is $$ m = 3.0 \times 10^{-5} $$ kg. The number of rain drops falling on 1 m² area is:

$$ \text{Number of drops} = \frac{\text{Total mass}}{\text{Mass per drop}} = \frac{1000}{3.0 \times 10^{-5}} $$

Calculate:

$$ \frac{1000}{3.0 \times 10^{-5}} = \frac{1000}{0.00003} = \frac{1000 \times 10^5}{3} = \frac{10^8}{3} = 3.333 \times 10^7 $$

The total energy transferred to 1 m² area is the number of drops multiplied by the energy per drop:

$$ E_{\text{total}} = \text{Number of drops} \times KE = \left( \frac{1000}{3.0 \times 10^{-5}} \right) \times (1.215 \times 10^{-3}) $$

Alternatively, we can use the formula for total kinetic energy since all drops have the same terminal velocity:

$$ E_{\text{total}} = \frac{1}{2} \times \text{Total mass} \times v^2 = \frac{1}{2} \times M \times v^2 $$

Substituting $$ M = 1000 $$ kg and $$ v = 9 $$ m/s:

$$ E_{\text{total}} = \frac{1}{2} \times 1000 \times (9)^2 = \frac{1}{2} \times 1000 \times 81 $$

First, $$ \frac{1}{2} \times 1000 = 500 $$:

$$ 500 \times 81 = 40500 \text{ J} $$

Which is $$ 4.05 \times 10^4 $$ J.

Hence, the energy transferred by rain to each square metre of the surface is $$ 4.05 \times 10^4 $$ J.

Comparing with the options:

A. $$ 3.5 \times 10^5 $$ J

B. $$ 4.05 \times 10^4 $$ J

C. $$ 3.0 \times 10^5 $$ J

D. $$ 9.0 \times 10^4 $$ J

So, the answer is Option B.

The velocity of water near the surface is given as 18 km h$$^{-1}$$, and the river depth is 5 m. To find the shearing stress between horizontal layers of water, we use Newton's law of viscosity, which states that the shearing stress (τ) is equal to the coefficient of viscosity (η) multiplied by the velocity gradient (dv/dy). The formula is τ = η × (dv/dy).

First, we need to convert the velocity from km h$$^{-1}$$ to m s$$^{-1}$$ because SI units are required. We know that 1 km = 1000 m and 1 hour = 3600 seconds. So, 18 km h$$^{-1}$$ = 18 × (1000 m) / (3600 s) = 18 × (1000 ÷ 3600) m s$$^{-1}$$. Simplifying 1000 ÷ 3600 gives 5/18, so 18 × (5/18) = 5 m s$$^{-1}$$. Therefore, the velocity at the surface is 5 m s$$^{-1}$$.

At the bottom of the river (depth = 5 m), the velocity of water is zero due to the no-slip condition. The velocity gradient dv/dy is the change in velocity per unit depth. Assuming a linear velocity profile from the bottom to the surface, dv/dy = (velocity at surface - velocity at bottom) ÷ depth = (5 m s$$^{-1}$$ - 0 m s$$^{-1}$$) ÷ 5 m = 5 ÷ 5 = 1 s$$^{-1}$$.

The coefficient of viscosity η is given as 10$$^{-2}$$ poise. Poise is a CGS unit, so we convert it to SI units (Pa·s or N s m$$^{-2}$$). We know that 1 poise = 0.1 Pa·s. Therefore, η = 10$$^{-2}$$ poise = 10$$^{-2}$$ × 0.1 Pa·s = 10$$^{-3}$$ Pa·s = 10$$^{-3}$$ N s m$$^{-2}$$.

Now, substituting into the formula for shearing stress: τ = η × (dv/dy) = (10$$^{-3}$$ N s m$$^{-2}$$) × (1 s$$^{-1}$$) = 10$$^{-3}$$ N m$$^{-2}$$.

Comparing with the options, 10$$^{-3}$$ N m$$^{-2}$$ corresponds to option B. Hence, the correct answer is Option B.

$$h_1 = R[1 - \sin\alpha]; \quad h_2 = R[1 - \cos\alpha]$$

$$h_2^1 = R\sin\alpha + R\cos\alpha$$

$$h_1 d_1 g = h_2 d_1 g + h_2^1 d_2 g$$

$$d_1 R[1 - \sin\alpha] = d_1 R[1 - \cos\alpha] + d_2 R[\sin\alpha + \cos\alpha]$$

$$d_1 [\cos\alpha - \sin\alpha] = d_2 [\cos\alpha + \sin\alpha]$$

$$\frac{d_1}{d_2} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$

We are given a cylindrical vessel with cross-sectional area $$A$$ filled with water to a height $$h$$. There is a hole at the bottom with radius $$a$$, so the area of the hole is $$\pi a^2$$. We need to find the time taken for the vessel to empty completely.

According to Torricelli's theorem, the speed of efflux $$v$$ of water through the hole is given by $$v = \sqrt{2gh}$$, where $$g$$ is the acceleration due to gravity and $$h$$ is the height of water above the hole at any instant.

The volume flow rate of water through the hole is the product of the area of the hole and the speed of efflux. Since the volume is decreasing, we write:

$$\frac{dV}{dt} = - (\pi a^2) \sqrt{2gh}$$

Here, $$V$$ is the volume of water in the vessel. For a cylinder, $$V = A h$$. Differentiating both sides with respect to time $$t$$:

$$\frac{dV}{dt} = A \frac{dh}{dt}$$

Setting the two expressions for $$\frac{dV}{dt}$$ equal:

$$A \frac{dh}{dt} = - \pi a^2 \sqrt{2gh}$$

We can rearrange this equation to separate the variables $$h$$ and $$t$$. First, express $$\sqrt{2gh}$$ as $$\sqrt{2g} \sqrt{h}$$:

$$A \frac{dh}{dt} = - \pi a^2 \sqrt{2g} \sqrt{h}$$

Bring terms involving $$h$$ to one side and $$t$$ to the other:

$$A dh = - \pi a^2 \sqrt{2g} \sqrt{h} dt$$

Divide both sides by $$\sqrt{h}$$ and rearrange:

$$dt = - \frac{A}{\pi a^2 \sqrt{2g}} \frac{dh}{\sqrt{h}}$$

Simplify the right side:

$$dt = - \frac{A}{\pi a^2 \sqrt{2g}} h^{-1/2} dh$$

To find the total time $$T$$ for the vessel to empty, integrate both sides. At $$t = 0$$, $$h = h$$ (initial height). At $$t = T$$, $$h = 0$$ (vessel empty). So:

$$\int_{0}^{T} dt = - \frac{A}{\pi a^2 \sqrt{2g}} \int_{h}^{0} h^{-1/2} dh$$

The left side integrates to $$T$$. For the right side, reversing the limits of integration removes the negative sign:

$$T = \frac{A}{\pi a^2 \sqrt{2g}} \int_{0}^{h} h^{-1/2} dh$$

Now, integrate $$h^{-1/2}$$:

$$\int h^{-1/2} dh = \frac{h^{1/2}}{1/2} = 2 h^{1/2}$$

Evaluate the definite integral:

$$\int_{0}^{h} h^{-1/2} dh = \left[ 2 \sqrt{h} \right]_{0}^{h} = 2\sqrt{h} - 2\sqrt{0} = 2\sqrt{h}$$

Substitute back:

$$T = \frac{A}{\pi a^2 \sqrt{2g}} \times 2 \sqrt{h}$$

Simplify:

$$T = \frac{2A \sqrt{h}}{\pi a^2 \sqrt{2g}}$$

Express $$\sqrt{h}$$ and $$\sqrt{2g}$$ in terms of $$\sqrt{\frac{h}{g}}$$:

$$T = \frac{2A}{\pi a^2} \cdot \frac{\sqrt{h}}{\sqrt{2g}} = \frac{2A}{\pi a^2} \sqrt{\frac{h}{2g}}$$

Now, $$\sqrt{\frac{h}{2g}} = \frac{1}{\sqrt{2}} \sqrt{\frac{h}{g}}$$, so:

$$T = \frac{2A}{\pi a^2} \cdot \frac{1}{\sqrt{2}} \sqrt{\frac{h}{g}} = \frac{2A}{\sqrt{2} \pi a^2} \sqrt{\frac{h}{g}}$$

Simplify $$\frac{2}{\sqrt{2}}$$:

$$\frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$$

Thus:

$$T = \frac{\sqrt{2} A}{\pi a^2} \sqrt{\frac{h}{g}}$$

Comparing with the options, this matches option B.

Hence, the correct answer is Option B.

Pressure from Kerosene ($$P_k$$): $$\rho_k g h_k = (0.8 \times 10^3) \times 10 \times 2 = 16,000\text{ Pa}$$

Pressure from Water ($$P_w$$): $$\rho_w g h_w = 10^3 \times 10 \times 3 = 30,000\text{ Pa}$$

Total Gauge Pressure: $$P_{total} = 16,000 + 30,000 = 46,000\text{ Pa}$$

$$P_{total} = \frac{1}{2} \rho_w v^2$$

$$46,000 = \frac{1}{2} \times 1000 \times v^2$$

$$v = \sqrt{92} \approx 9.59\text{ m/s}$$

At the equilibrium position the cylinder remains at rest, so the net force acting on it must be zero. We therefore equate the upward forces to the downward forces.

The downward force is simply the weight of the cylinder, given by the well-known expression $$W = Mg,$$ where $$M$$ is the mass of the cylinder and $$g$$ is the acceleration due to gravity.

The upward forces consist of two parts.

First, the restoring force of the massless spring. According to Hooke’s law, the magnitude of the spring force is $$F_{\text{spring}} = kx_0,$$ where $$k$$ is the spring constant and $$x_0$$ is the extension of the spring. This force acts upward because the cylinder pulls the spring downward.

Second, the buoyant force exerted by the liquid. Archimedes’ principle states that the buoyant force equals the weight of the displaced liquid: $$F_{\text{buoyant}} = (\text{mass of displaced liquid})\,g = (\rho V_{\text{sub}})g,$$ where $$\rho$$ (here given as $$\sigma$$) is the density of the liquid and $$V_{\text{sub}}$$ is the submerged volume.

The problem tells us that at equilibrium the cylinder is half submerged. The full volume of the cylinder is $$V_{\text{cyl}} = AL,$$ with $$A$$ as cross-sectional area and $$L$$ as length. Hence the submerged volume is $$V_{\text{sub}} = \frac{1}{2}AL.$$ Substituting this into the buoyant-force formula gives $$F_{\text{buoyant}} = \sigma \left(\frac{1}{2}AL\right) g \;=\; \frac{\sigma AL g}{2}.$$

Now we impose equilibrium. Taking upward forces as positive, we write $$F_{\text{spring}} + F_{\text{buoyant}} - W = 0.$$ Substituting the expressions already obtained, we have $$kx_0 + \frac{\sigma AL g}{2} - Mg = 0.$$

We isolate the term involving $$x_0$$:

Factoring out $$g$$ from the right-hand side,

Finally, dividing both sides by $$k$$ gives the required extension,

Pulling out a common factor $$\displaystyle\frac{Mg}{k}$$ we write

This expression matches Option A in the list provided.

Hence, the correct answer is Option A.

In the experiment, the steel ball falls downward through the liquid at a constant speed of 10 cm/s. Since the speed is constant, the acceleration is zero, meaning the net force acting on the ball is zero. Let's denote the mass of the ball as $$ m $$ and the acceleration due to gravity as $$ g $$. The weight of the ball acting downward is $$ mg $$. There is an upward buoyant force, $$ F_b $$, due to the liquid. Additionally, when the ball moves, a viscous force opposes the motion. For downward motion, the viscous force acts upward.

At constant downward speed of 10 cm/s, the net force is zero. The effective weight (the net gravitational force) is $$ mg - F_b $$, which acts downward. This is balanced by the upward viscous force, $$ F_v $$. Therefore:

$$ mg - F_b = F_v $$

Let $$ W_{\text{eff}} = mg - F_b $$, so:

$$ W_{\text{eff}} = F_v $$

Given that the downward speed is 10 cm/s, we have $$ F_v = k \times 10 $$, where $$ k $$ is the constant of proportionality for viscous force (assuming Stokes' law, where viscous force is proportional to speed). Thus:

$$ W_{\text{eff}} = k \times 10 $$

Now, the ball is pulled upward with a force equal to twice its effective weight, so the applied force upward is $$ 2W_{\text{eff}} $$. When moving upward, the forces acting on the ball are:

• Upward forces: Applied force $$ 2W_{\text{eff}} $$ and buoyant force $$ F_b $$
• Downward forces: Weight $$ mg $$ and viscous force $$ F_v' $$ (which now acts downward because it opposes the upward motion)

The net force upward is:

$$ F_{\text{net, upward}} = (2W_{\text{eff}} + F_b) - (mg + F_v') $$

Substitute $$ W_{\text{eff}} = mg - F_b $$:

$$ F_{\text{net, upward}} = 2(mg - F_b) + F_b - mg - F_v' $$

Simplify step by step:

$$ F_{\text{net, upward}} = 2mg - 2F_b + F_b - mg - F_v' $$

$$ F_{\text{net, upward}} = (2mg - mg) + (-2F_b + F_b) - F_v' $$

$$ F_{\text{net, upward}} = mg - F_b - F_v' $$

$$ F_{\text{net, upward}} = W_{\text{eff}} - F_v' $$

When the ball moves upward at a constant speed, the net force becomes zero (no acceleration). Therefore:

$$ W_{\text{eff}} - F_v' = 0 $$

$$ F_v' = W_{\text{eff}} $$

The viscous force $$ F_v' $$ is proportional to the upward speed $$ v' $$, so $$ F_v' = k v' $$. From earlier, $$ W_{\text{eff}} = k \times 10 $$. Substituting:

$$ k v' = k \times 10 $$

Divide both sides by $$ k $$ (assuming $$ k \neq 0 $$):

$$ v' = 10 \text{ cm/s} $$

Thus, the ball moves upward at a constant speed of 10 cm/s.

Hence, the correct answer is Option C.

To solve this problem, we need to find the pressure difference between the upper and lower sides of the aeroplane wings. We are given that the density of air, ρ, is 1.2 kg m⁻³. The speed of air above the wing is 150 m s⁻¹, and below the wing is 100 m s⁻¹.

We use Bernoulli's theorem, which states that for an ideal fluid in steady flow, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline. Since the wings are horizontal, the height difference is negligible, so we can ignore the potential energy term. Therefore, Bernoulli's equation simplifies to:

$$ P + \frac{1}{2} \rho v^2 = \text{constant} $$

Applying this equation to points above and below the wing:

$$ P_{\text{above}} + \frac{1}{2} \rho v_{\text{above}}^2 = P_{\text{below}} + \frac{1}{2} \rho v_{\text{below}}^2 $$

We rearrange this equation to find the pressure difference, $$ P_{\text{below}} - P_{\text{above}} $$:

$$ P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \rho v_{\text{above}}^2 - \frac{1}{2} \rho v_{\text{below}}^2 $$

Factor out $$\frac{1}{2} \rho$$:

$$ P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \rho \left( v_{\text{above}}^2 - v_{\text{below}}^2 \right) $$

Now substitute the given values: ρ = 1.2 kg m⁻³, $$ v_{\text{above}} = 150 $$ m s⁻¹, and $$ v_{\text{below}} = 100 $$ m s⁻¹.

First, calculate $$ v_{\text{above}}^2 $$:

$$ v_{\text{above}}^2 = (150)^2 = 22500 $$

Next, calculate $$ v_{\text{below}}^2 $$:

$$ v_{\text{below}}^2 = (100)^2 = 10000 $$

Now find the difference:

$$ v_{\text{above}}^2 - v_{\text{below}}^2 = 22500 - 10000 = 12500 $$

Substitute into the equation:

$$ P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \times 1.2 \times 12500 $$

First, multiply 1.2 and 12500:

$$ 1.2 \times 12500 = 1.2 \times (125 \times 100) = (1.2 \times 125) \times 100 $$

$$ 1.2 \times 125 = 1.2 \times (100 + 25) = (1.2 \times 100) + (1.2 \times 25) = 120 + 30 = 150 $$

Then, $$ 150 \times 100 = 15000 $$, so:

$$ 1.2 \times 12500 = 15000 $$

Now multiply by $$\frac{1}{2}$$:

$$ \frac{1}{2} \times 15000 = 7500 $$

Therefore, the pressure difference is 7500 N m⁻².

Comparing with the options:

A. 60 Nm⁻²

B. 180 Nm⁻²

C. 7500 Nm⁻²

D. 12500 Nm⁻²

Hence, the correct answer is Option C.