JEE Units & Measurements Questions

First convert every dimension to centimetre (cm) so that the units are consistent:

Length: already $$10.5 \text{ cm}$$ (three significant figures).

Breadth: $$0.05 \text{ mm} = 0.05 \times 0.1 \text{ cm} = 0.005 \text{ cm}$$.
Only the digit 5 is significant here, so breadth has one significant figure.

Thickness: $$6.0 \text{ }\mu\text{m} = 6.0 \times 10^{-4} \text{ cm}$$ (two significant figures).

Volume of the strip is the product of the three dimensions:

$$V = 10.5 \times 0.005 \times 6.0 \times 10^{-4} \text{ cm}^3$$

Step-wise multiplication:
$$10.5 \times 0.005 = 0.0525$$
$$0.0525 \times 6.0 = 0.315$$
$$0.315 \times 10^{-4} = 3.15 \times 10^{-5} \text{ cm}^3$$

In multiplication, the result must carry the same number of significant figures as the factor with the fewest significant figures. Here the breadth (0.05 mm → 0.005 cm) has just 1 significant figure. Therefore the volume must be rounded to one significant figure:

$$3.15 \times 10^{-5} \;\text{cm}^3 \;\longrightarrow\; 3 \times 10^{-5} \;\text{cm}^3$$

Hence the volume, expressed with the correct number of significant figures, is:

Option D which is: $$3 \times 10^{-5}$$

Solution :

Given equation for real gas :

$$(P+\frac{a}{V^2})(V-b)=RT$$

Since quantities added inside bracket must have same dimensions,

$$\frac{a}{V^2}$$

has dimensions of pressure.

Therefore,

$$[a] = [P][V^2]$$

Pressure :

$$[P] = ML^{-1}T^{-2}$$

Volume :

$$[V] = L^3$$

Hence,

$$[a] = (ML^{-1}T^{-2})(L^6)$$

$$= ML^5T^{-2}$$

Also, since \(V-b\) is present,

$$[b] = [V]$$

$$= L^3$$

Now,

$$[ab^{-2}] = [a][b]^{-2}$$

$$= (ML^5T^{-2})(L^3)^{-2}$$

$$= ML^5T^{-2} \times L^{-6}$$

$$= ML^{-1}T^{-2}$$

This is the dimension of pressure.

Also,

Energy density

$$= \frac{\text{Energy}}{\text{Volume}}$$

Dimensions of energy :

$$[E] = ML^2T^{-2}$$

Therefore,

$$[\text{Energy density}] = \frac{ML^2T^{-2}}{L^3}$$

$$= ML^{-1}T^{-2}$$

Hence,

$$[ab^{-2}]$$

has dimensions of energy density.

Final Answer :

Energy density.

To find the dimension of $$\frac{ABC}{D}$$, observe that $$x(t) = A\sin t + B\cos^2 t + Ct^2 + D$$ represents a position, so each term on the right must have dimension of length $$[L]$$.

Since $$\sin t$$ and $$\cos^2 t$$ are dimensionless and $$[t^2] = [T^2]$$, we have $$[A\sin t] = [L]$$ implying $$[A] = [L]$$, $$[B\cos^2 t] = [L]$$ implying $$[B] = [L]$$, and $$[Ct^2] = [L]$$ so $$[C][T^2] = [L]$$ and thus $$[C] = [LT^{-2}]$$. The constant $$D$$ must satisfy $$[D] = [L].$$

Therefore,

$$\left[\frac{ABC}{D}\right] = \frac{[L][L][LT^{-2}]}{[L]} = \frac{L^3 T^{-2}}{L} = L^2 T^{-2}$$

The correct answer is Option (1): $$L^2T^{-2}$$.

First recall the symbols:
  • Planck’s constant  $$h$$
  • Angular momentum of an electron in Bohr orbit  $$L$$

Checking Statement (I): “The dimensions of Planck's constant and angular momentum are same.”

Write dimensions of Planck’s constant.

Planck’s constant is energy $$\times$$ time.
Energy has dimension $$ML^{2}T^{-2}$$, therefore

$$[h] = ML^{2}T^{-2}\times T = ML^{2}T^{-1}$$

Write dimensions of angular momentum.

For a particle of mass $$m$$ moving with speed $$v$$ in a circle of radius $$r$$,
$$L = mvr$$

Dimensions: $$[L] = M(LT^{-1})L = ML^{2}T^{-1}$$

Thus $$[h] = [L] = ML^{2}T^{-1}$$, so Statement (I) is correct.

Checking Statement (II): “In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant.”

Bohr’s postulate states
$$mvr = n\frac{h}{2\pi},\qquad n = 1,2,3,\dots$$ $$-(1)$$

Equation $$-(1)$$ shows that angular momentum is an integral multiple of $$\dfrac{h}{2\pi}$$ (i.e. $$\hbar$$), not of $$h$$ itself. Hence Statement (II) is incorrect.

Conclusion: Statement (I) is correct, Statement (II) is incorrect. This matches Option C.

Therefore, the most appropriate answer is Option C.

The expression is $$\sqrt{\dfrac{\mu_0}{\epsilon_0}}$$.
We must find its dimensional formula and compare it with the dimensions of the given physical quantities.

Step 1 : Dimensions of vacuum permeability $$\mu_0$$
In SI, $$\mu_0 = 4\pi \times 10^{-7}\;N\,A^{-2}$$, where $$N = kg \; m \; s^{-2}$$.
Hence, $$\mu_0$$ has dimensions
$$[\,\mu_0\,] = M^{1}\,L^{1}\,T^{-2}\,I^{-2}$$.

Step 2 : Dimensions of vacuum permittivity $$\epsilon_0$$
From Coulomb’s law $$F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q^{\,2}}{r^{\,2}}$$, we get
$$\epsilon_0 = \dfrac{q^{\,2}}{F\,r^{\,2}}$$.
Using $$q = I\,T$$ and $$F = M\,L\,T^{-2}$$:
$$[\,\epsilon_0\,] = \dfrac{(I\,T)^2}{M\,L\,T^{-2}\;L^{2}} = M^{-1}\,L^{-3}\,T^{4}\,I^{2}$$.

Step 3 : Dimensions of $$\dfrac{\mu_0}{\epsilon_0}$$
$$\left[\dfrac{\mu_0}{\epsilon_0}\right] = M^{1-(-1)}\,L^{1-(-3)}\,T^{-2-4}\,I^{-2-2}$$
$$= M^{2}\,L^{4}\,T^{-6}\,I^{-4}$$.

Step 4 : Taking the square root
$$\left[\sqrt{\dfrac{\mu_0}{\epsilon_0}}\right] = M^{1}\,L^{2}\,T^{-3}\,I^{-2}$$.

Step 5 : Dimensions of resistance
Resistance $$R = \dfrac{V}{I}$$.
Electrical power $$P = V\,I$$ has dimensions $$M\,L^{2}\,T^{-3}$$.
Therefore voltage $$V = \dfrac{P}{I}$$ has dimensions $$M\,L^{2}\,T^{-3}\,I^{-1}$$.
Thus,
$$[\,R\,] = \dfrac{[\,V\,]}{[\,I\,]} = M\,L^{2}\,T^{-3}\,I^{-2}$$.

Step 6 : Comparison
The dimension $$M\,L^{2}\,T^{-3}\,I^{-2}$$ obtained for $$\sqrt{\dfrac{\mu_0}{\epsilon_0}}$$ is exactly the same as that of resistance.

Hence, $$\sqrt{\dfrac{\mu_0}{\epsilon_0}}$$ has the dimensions of resistance.
Correct choice: Option D (Resistance).

Angular impulse:

Angular impulse = change in angular momentum

$$L=r\times p=r\cdot mv$$

$$[L]=[L]\cdot[MLT^{-1}]=[ML^2T^{-1}]$$

So, (A) matches with (III)

Latent heat:

Latent heat = heat per unit mass

$$L=\frac{Q}{m}$$

$$[Q]=[ML^2T^{-2}]$$

$$[L]=\frac{ML^2T^{-2}}{M}=[L^2T^{-2}]$$

So, (B) matches with (I)

Electrical resistivity:

$$R=\rho\frac{L}{A}\Rightarrow\rho=R\frac{A}{L}$$

Now,

$$R=\frac{V}{I}$$

$$[V]=[ML^2T^{-3}A^{-1}]$$

$$[R]=\frac{ML^2T^{-3}A^{-1}}{A}=[ML^2T^{-3}A^{-2}]$$

So,

$$[\rho]=[ML^2T^{-3}A^{-2}]\cdot\frac{L^2}{L}=[ML^3T^{-3}A^{-2}]$$

So, (C) matches with (IV)

Electromotive force (EMF):

EMF has same dimensions as potential difference

$$[EMF]=[ML^2T^{-3}A^{-1}]$$

So, (D) matches with (II)

Final matching:

(A)−(III), (B)−(I), (C)−(IV), (D)−(II)

Solution :

Dimensions of momentum :

$$[p] = MLT^{-1}$$

Given :

Charge,

$$[q] = AT$$

Current,

$$[I] = A$$

Permeability of vacuum :

From force per unit length between two current carrying wires,

$$\frac{\mu_0 I^2}{r}$$

has dimensions of force.

Therefore,

$$[\mu_0] = \frac{[F][r]}{[I]^2}$$

$$= \frac{(MLT^{-2})(L)}{A^2}$$

$$= MLT^{-2}A^{-2}$$

Now consider quantity :

$$\mu_0 qI$$

Its dimensions are :

$$[\mu_0 qI] = (MLT^{-2}A^{-2})(AT)(A)$$

$$= MLT^{-1}$$

which is the dimension of momentum.

Final Answer :

$$\mu_0 qI$$

Young’s modulus:

$$Y=\frac{\text{stress}}{\text{strain}}$$

Strain is dimensionless, so

$$[Y]=[\text{stress}]=\frac{\text{Force}}{\text{Area}}=\frac{MLT^{-2}}{L^2}=[ML^{-1}T^{-2}]$$

So, (A) matches with (II)

Torque:

$$\tau=r\times F$$

$$[\tau]=[L]\cdot[MLT^{-2}]=[ML^2T^{-2}]$$

So, (B) matches with (IV)

Coefficient of viscosity:

$$[\eta]=[ML^{-1}T^{-1}]$$

So, (C) matches with (I)

Gravitational constant:

From Newton’s law:

$$F=\frac{GMm}{r^2}\Rightarrow G=\frac{Fr^2}{Mm}$$

$$[G]=\frac{MLT^{-2}\cdot L^2}{M^2}=[M^{-1}L^3T^{-2}]$$

So, (D) matches with (III)

Final matching:

(A)−(II), (B)−(IV), (C)−(I), (D)−(III)

The four physical quantities are gravitational constant $$G$$, gravitational potential energy $$U$$, gravitational potential $$V$$ and acceleration due to gravity $$g$$. We first write the defining equations for each quantity and then extract their dimensional formulas.

Case A : Gravitational constant  $$G$$
Newton’s law gives the gravitational force between two masses $$m_1$$ and $$m_2$$ separated by distance $$r$$ as
$$F = \dfrac{G\,m_1\,m_2}{r^{2}}$$
Re-arranging,
$$G = \dfrac{F\,r^{2}}{m_1\,m_2}$$
Force has dimensions $$[MLT^{-2}]$$. Hence
$$[G] = \dfrac{[MLT^{-2}]\,L^{2}}{M\,M} = [M^{-1}L^{3}T^{-2}]$$

Case B : Gravitational potential energy  $$U$$
Potential energy near Earth’s surface is $$U = mgh$$.
Mass $$m$$ has $$[M]$$, $$g$$ has $$[LT^{-2}]$$ and height $$h$$ has $$[L]$$. Therefore
$$[U] = [M]\,[LT^{-2}]\,[L] = [ML^{2}T^{-2}]$$

Case C : Gravitational potential  $$V$$
Potential is energy per unit mass: $$V = \dfrac{U}{m}$$.
Divide the dimensions of $$U$$ obtained above by $$[M]$$:
$$[V] = \dfrac{[ML^{2}T^{-2}]}{[M]} = [L^{2}T^{-2}]$$

Case D : Acceleration due to gravity  $$g$$
Acceleration is rate of change of velocity: $$g = \dfrac{dv}{dt}$$.
Velocity $$v$$ has $$[LT^{-1}]$$, so acceleration has
$$[g] = \dfrac{[LT^{-1}]}{[T]} = [LT^{-2}]$$

Now compare with LIST-II:
I. $$[LT^{-2}]$$  II. $$[L^{2}T^{-2}]$$  III. $$[ML^{2}T^{-2}]$$  IV. $$[M^{-1}L^{3}T^{-2}]$$

Matching each quantity with its dimensional formula:
A. $$G$$ ⟶ IV  B. $$U$$ ⟶ III  C. $$V$$ ⟶ II  D. $$g$$ ⟶ I

Thus the correct set is A-IV, B-III, C-II, D-I, which corresponds to Option A.

The quantity is $$Q = X^{-2}\,Y^{-\,\frac{3}{2}}\,Z^{\frac{2}{5}}$$.

For a physical quantity of the form $$Q = X^{a}\,Y^{b}\,Z^{c}$$, the maximum fractional (relative) error is obtained by adding the absolute contributions of each factor:
$$\frac{\Delta Q}{Q}\Big|_{\max}=|a|\;\frac{\Delta X}{X}+|b|\;\frac{\Delta Y}{Y}+|c|\;\frac{\Delta Z}{Z}$$.

Here $$a=-2,\;b=-\frac{3}{2},\;c=\frac{2}{5}$$.
Their absolute values are $$|a|=2,\;|b|=\frac{3}{2}=1.5,\;|c|=\frac{2}{5}=0.4$$.

The given fractional errors are
$$\frac{\Delta X}{X}=0.1,\qquad \frac{\Delta Y}{Y}=0.2,\qquad \frac{\Delta Z}{Z}=0.5$$.

Substituting these values:
$$\frac{\Delta Q}{Q}\Big|_{\max}=2(0.1)+1.5(0.2)+0.4(0.5)$$

Calculate each term:
$$2(0.1)=0.2,\qquad 1.5(0.2)=0.3,\qquad 0.4(0.5)=0.2$$

Add them:
$$0.2+0.3+0.2 = 0.7$$

Therefore, the maximum fractional error in $$Q$$ is $$0.7$$.

Option C is correct.

Let us analyze each statement about the Moving Coil Galvanometer (MCG):

Statement A: The torsional constant has dimensions $$[ML^{2}T^{-2}]$$.

The restoring torque is $$\tau = k\theta$$, where $$k$$ is the torsional constant and $$\theta$$ is dimensionless (in radians).

So $$[k] = [\tau] = [ML^{2}T^{-2}]$$. Statement A is correct.

Statement B: Increasing the current sensitivity may not necessarily increase the voltage sensitivity.

Current sensitivity: $$I_s = \frac{NAB}{k}$$

Voltage sensitivity: $$V_s = \frac{I_s}{R} = \frac{NAB}{kR}$$

If we increase $$N$$ to increase current sensitivity, the resistance $$R$$ also increases (since $$R \propto N$$). So voltage sensitivity may not increase. Statement B is correct.

Statement C: If $$N$$ is doubled to $$2N$$, voltage sensitivity doubles.

When $$N$$ is doubled, $$R$$ also roughly doubles. So $$V_s = \frac{NAB}{kR}$$. Since both $$N$$ and $$R$$ double, $$V_s$$ remains approximately the same. Statement C is incorrect.

Statement D: MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel.

To convert a galvanometer into an ammeter, a small (low) resistance shunt is connected in parallel, not a large resistance. Statement D is incorrect.

Statement E: Current sensitivity depends inversely on number of turns.

Current sensitivity $$I_s = \frac{NAB}{k}$$, which is directly proportional to $$N$$, not inversely. Statement E is incorrect.

Therefore, only statements A and B are correct.

The correct answer is Option 4: A, B Only.

The dimensional formula of a physical quantity is obtained from the defining equation of that quantity in terms of the fundamental mechanical quantities $$M$$ (mass), $$L$$ (length) and $$T$$ (time).

Case (A) : Mass density

Mass density $$\rho$$ is mass per unit volume:
$$\rho = \frac{\text{mass}}{\text{volume}}$$.
Mass has dimension $$[M]$$ and volume has dimension $$[L^3]$$, so

$$[\rho] = \frac{[M]}{[L^3]} = [M L^{-3} T^{0}]$$.
Thus (A) corresponds to (IV).

Case (B) : Impulse

Impulse $$J$$ is defined as force multiplied by the time interval:
    $$J = F \, \Delta t$$.
Force has dimension $$[M L T^{-2}]$$ and time has dimension $$[T]$$, so

$$[J] = [M L T^{-2}] \,[T] = [M L T^{-1}]$$.
Thus (B) corresponds to (II).

Case (C) : Power

Power $$P$$ is work done per unit time:
    $$P = \frac{W}{t}$$.
Work (or energy) has dimension $$[M L^{2} T^{-2}]$$ and time has dimension $$[T]$$, hence

$$[P] = \frac{[M L^{2} T^{-2}]}{[T]} = [M L^{2} T^{-3}]$$.
Thus (C) corresponds to (I).

Case (D) : Moment of inertia

For a point mass $$m$$ at a distance $$r$$ from the axis, the moment of inertia $$I$$ is:
    $$I = m r^{2}$$.
Mass has dimension $$[M]$$ and distance squared has dimension $$[L^{2}]$$, so

$$[I] = [M] [L^{2}] = [M L^{2} T^{0}]$$.
Thus (D) corresponds to (III).

Collecting the matches:

(A) → (IV), (B) → (II), (C) → ( I ), (D) → (III).

This set of matches is given in Option C. Hence the correct answer is Option C.

Find the dimensions of $$B/\mu_0$$, where $$B$$ is the magnetic field and $$\mu_0$$ is the permeability of free space.

From the Lorentz force equation $$F = qvB$$, we can write $$B = \frac{F}{qv}$$.

$$[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{(AT)(LT^{-1})} = \frac{MLT^{-2}}{ALT^{-1} \cdot T} = \frac{M}{AT^2} = MT^{-2}A^{-1}$$

From the relation $$B = \mu_0 nI$$ (magnetic field inside a solenoid), where $$n$$ is the number of turns per unit length and $$I$$ is the current:

$$[\mu_0] = \frac{[B]}{[n][I]} = \frac{MT^{-2}A^{-1}}{L^{-1} \cdot A} = MLT^{-2}A^{-2}$$

$$\left[\frac{B}{\mu_0}\right] = \frac{MT^{-2}A^{-1}}{MLT^{-2}A^{-2}} = \frac{M}{M} \cdot \frac{T^{-2}}{T^{-2}} \cdot \frac{A^{-1}}{A^{-2}} \cdot \frac{1}{L} = L^{-1}A$$

Physical significance: $$B/\mu_0$$ has the same dimensions as magnetic field intensity $$H$$, which has dimensions of $$L^{-1}A$$ (equivalent to A/m in SI units).

The correct answer is Option (3): $$L^{-1}A$$.

The main scale of the travelling microscope has $$300$$ equal divisions spread over a length of $$15 \text{ cm}$$.

Length of one main scale division (MSD) is therefore
$$\text{MSD} = \frac{15 \text{ cm}}{300} = 0.05 \text{ cm}$$

The vernier scale has $$25$$ divisions that coincide with $$24$$ divisions of the main scale.
Thus, length of the entire vernier scale is
$$25 \text{ VSD} = 24 \text{ MSD}$$

Using $$\text{MSD} = 0.05 \text{ cm}$$,
$$25 \text{ VSD} = 24 \times 0.05 \text{ cm} = 1.2 \text{ cm}$$

Length of one vernier scale division (VSD) is then
$$\text{VSD} = \frac{1.2 \text{ cm}}{25} = 0.048 \text{ cm}$$

Formula for least count (LC) of a vernier instrument:
$$\text{LC} = \text{MSD} - \text{VSD}$$

Substituting the values,
$$\text{LC} = 0.05 \text{ cm} - 0.048 \text{ cm} = 0.002 \text{ cm}$$

Hence, the least count of the travelling microscope is $$0.002 \text{ cm}$$.

Option B is correct.

We need to find the dimensions of the product ABC, where $$v = At^2 + \frac{Bt}{C+t}$$.

Since $$C + t$$ must be dimensionally consistent, $$[C] = [t] = T$$.

From the first term $$At^2$$: $$[A][T^2] = [v] = LT^{-1}$$, so $$[A] = LT^{-3}$$.

From the second term $$\frac{Bt}{C+t}$$: for large $$t$$, this approaches $$B$$, so $$[B] = [v] = LT^{-1}$$.

(More rigorously: $$\frac{[B][T]}{[T]} = LT^{-1}$$, so $$[B] = LT^{-1}$$.)

Calculate $$[ABC]$$:

$$[ABC] = [A][B][C] = (LT^{-3})(LT^{-1})(T) = L^2T^{-3} = M^0L^2T^{-3}$$

The correct answer is Option (4): $$[M^0L^2T^{-3}]$$.

The dimensional formula for capacitance in farads (F) can be derived using the definition of capacitance. Capacitance $$C$$ is given by the formula:

$$ C = \frac{Q}{V} $$

where $$Q$$ is the charge and $$V$$ is the potential difference.

Charge $$Q$$ has the dimension $$[C]$$, so:

$$ [Q] = [C] $$

Potential difference $$V$$ is defined as work done per unit charge:

$$ V = \frac{W}{Q} $$

where $$W$$ is work done. Work done $$W$$ is force multiplied by displacement:

$$ W = F \cdot d $$

Force $$F$$ is mass times acceleration:

$$ F = m \cdot a $$

Acceleration $$a$$ has dimension $$[LT^{-2}]$$, so force $$F$$ has dimension:

$$ [F] = [M] \cdot [LT^{-2}] = [MLT^{-2}] $$

Displacement $$d$$ has dimension $$[L]$$, so work done $$W$$ has dimension:

$$ [W] = [F] \cdot [d] = [MLT^{-2}] \cdot [L] = [ML^2T^{-2}] $$

Therefore, potential difference $$V$$ has dimension:

$$ [V] = \frac{[W]}{[Q]} = \frac{[ML^2T^{-2}]}{[C]} = [ML^2T^{-2}C^{-1}] $$

Now, capacitance $$C$$ is:

$$ [C] = \frac{[Q]}{[V]} = \frac{[C]}{[ML^2T^{-2}C^{-1}]} $$

Simplifying the expression:

$$ [C] = [C] \times \frac{1}{[ML^2T^{-2}C^{-1}]} = [C] \times [M^{-1}L^{-2}T^{2}C] = [C^{1} \cdot C^{1}] \times [M^{-1}L^{-2}T^{2}] = [C^{2}M^{-1}L^{-2}T^{2}] $$

Thus, the dimensional formula for capacitance is $$[C^{2}M^{-1}L^{-2}T^{2}]$$.

Comparing with the options:

• A. $$[F]=[C^{2}M^{-1}L^{-2}T^{2}]$$
• B. $$[F]=[C^{2}M^{-2}L^{2}T^{2}]$$
• C. $$[F]=[CM^{-2}L^{-2}T^{-2}]$$
• D. $$[F]=[CM^{-1}L^{-2}T^{2}]$$

Option A matches the derived dimensional formula.

Therefore, the correct answer is A.

The dimensional analysis will be carried out for each physical quantity in LIST-I and then compared with the four alternatives given in LIST-II.

The average translational energy of one molecule in an ideal gas is $$E = \tfrac{3}{2}\,k_B\,T$$.
Hence $$k_B = \dfrac{E}{T}$$.
Energy has dimension $$ML^{2}T^{-2}$$ and temperature contributes the factor $$K$$ in the denominator.
Therefore $$[k_B] = \dfrac{ML^{2}T^{-2}}{K} = ML^{2}T^{-2}K^{-1}$$, which corresponds to LIST-II entry III.

Newton’s law of viscous flow states $$\tau = \eta \,\dfrac{dv}{dy}$$, where $$\tau$$ is shear stress.
Shear stress $$\tau = \dfrac{\text{Force}}{\text{Area}}$$ has dimension $$\dfrac{MLT^{-2}}{L^{2}} = ML^{-1}T^{-2}$$.
Velocity gradient $$\dfrac{dv}{dy}$$ has dimension $$T^{-1}$$.
Thus $$[\eta] = \dfrac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}$$, matching LIST-II entry IV.

The photon energy is $$E = h\nu$$, and frequency $$\nu$$ has dimension $$T^{-1}$$.
Hence $$h = \dfrac{E}{\nu}$$.
Taking $$[E] = ML^{2}T^{-2}$$ and dividing by $$T^{-1}$$ gives $$[h] = ML^{2}T^{-1}$$, i.e. LIST-II entry I.

Fourier’s law of heat conduction is $$\dfrac{Q}{t} = -K\,A\,\dfrac{\Delta T}{L}$$.
Heat conducted per unit time $$\dfrac{Q}{t}$$ is power with dimension $$ML^{2}T^{-3}$$.
Area $$A$$ contributes $$L^{2}$$ and temperature gradient $$\dfrac{\Delta T}{L}$$ contributes $$K\,L^{-1}$$ in the denominator.
Therefore

$$[K] = \dfrac{ML^{2}T^{-3}}{L^{2}\,K\,L^{-1}} = \dfrac{ML^{2}T^{-3}}{L^{1}K} = MLT^{-3}K^{-1}$$, corresponding to LIST-II entry II.

Collecting the results:
A → III, B → IV, C → I, D → II

This is exactly the pairing given in Option A.

Hence, the correct answer is Option A.

Write the three measured masses with their decimal places aligned:
$$\begin{array}{r} 435.42 \\[2pt] 226.3 \\[2pt] \;0.125\\ \hline \end{array}$$

Rule for addition/subtraction with significant figures:
The final result must be rounded off to the same number of decimal places as the quantity having the least number of decimal places.

Count decimal places in each measurement:
435.42 g  → 2 decimal places
226.3 g  → 1 decimal place (least among the three)
0.125 g  → 3 decimal places

Perform the addition without rounding first:
$$435.42 + 226.3 + 0.125 = 661.845$$ g

The least number of decimal places is 1, so retain only one digit after the decimal and round the result accordingly:
661.845 g  → 661.8 g (since the next digit 4 is less than 5)

Therefore, the correct sum expressed with proper significant figures is $$661.8$$ g.

Hence, the required option is Option C.

We need to find the pair of physical quantities that do NOT have the same dimensions.

Check each option:

(1) Pressure and Young's modulus: Both have dimensions $$[ML^{-1}T^{-2}]$$. Same. ✓

(2) Surface tension and Impulse:

Surface tension = Force/Length = $$[MLT^{-2}]/[L] = [MT^{-2}]$$
Impulse = Force × Time = $$[MLT^{-2}][T] = [MLT^{-1}]$$
These are different. ✗

(3) Torque and Energy: Both have dimensions $$[ML^2T^{-2}]$$. Same. ✓

(4) Angular momentum and Planck's constant: Both have dimensions $$[ML^2T^{-1}]$$. Same. ✓

The correct answer is Option (2): Surface tension and impulse.

We first write one standard defining relation for each physical quantity and then deduce its dimensional formula.

Case A: Coefficient of viscosity ($$\eta$$)
Shear stress $$\tau = \eta \dfrac{dv}{dy}\quad -(1)$$
  • Dimension of shear stress $$\tau$$ is the same as pressure: $$[M\,L^{-1}\,T^{-2}]$$.
  • Velocity gradient $$\dfrac{dv}{dy}$$ has dimension $$T^{-1}$$.
Therefore,
$$\eta = \dfrac{\tau}{dv/dy} \; \Rightarrow \; [M\,L^{-1}\,T^{-2}] \; / \; [T^{-1}] = [M\,L^{-1}\,T^{-1}]$$.
Hence $$\eta$$ corresponds to List-II (IV).

Case B: Intensity of wave ($$I$$)
Intensity $$I = \dfrac{\text{Power}}{\text{Area}} \quad -(2)$$
  • Power $$P = \dfrac{\text{Energy}}{\text{time}}$$. Energy has dimension $$[M\,L^{2}\,T^{-2}]$$, so
$$P = [M\,L^{2}\,T^{-3}]$$.
  • Area has dimension $$[L^{2}]$$.
Therefore,
$$I = \dfrac{[M\,L^{2}\,T^{-3}]}{[L^{2}]} = [M\,L^{0}\,T^{-3}]$$.
Hence $$I$$ corresponds to List-II (I).

Case C: Pressure gradient ($$\nabla P$$)
Pressure gradient means pressure per unit length:
$$\nabla P = \dfrac{P}{L} \quad -(3)$$
  • Pressure $$P$$ has dimension $$[M\,L^{-1}\,T^{-2}]$$.
  • Length $$L$$ has dimension $$[L]$$.
Therefore,
$$\nabla P = \dfrac{[M\,L^{-1}\,T^{-2}]}{[L]} = [M\,L^{-2}\,T^{-2}]$$.
Hence $$\nabla P$$ corresponds to List-II (II).

Case D: Compressibility ($$\beta$$)
Compressibility is the reciprocal of bulk modulus (pressure):
$$\beta = \dfrac{1}{P} \quad -(4)$$
Since the dimension of pressure is $$[M\,L^{-1}\,T^{-2}]$$, we get
$$\beta = [M^{-1}\,L\,T^{2}]$$.
Hence $$\beta$$ corresponds to List-II (III).

Final matching
(A) $$\eta$$  →  (IV) [ML$$^{-1}$$T$$^{-1}$$]
(B) $$I$$  →  (I) [ML$$^{0}$$T$$^{-3}$$]
(C) $$\nabla P$$  →  (II) [ML$$^{-2}$$T$$^{-2}$$]
(D) $$\beta$$  →  (III) [M$$^{-1}$$LT$$^{2}$$]

Thus the correct set is: (A)-(IV), (B)-(I), (C)-(II), (D)-(III), which is given in Option B.

Answer : Option B

Magnetic induction B:

SI unit is tesla, and in CGS it is gauss.

So, (A) matches with (III)

Magnetic intensity H:

$$H=\frac{B}{\mu}$$

Unit:

$$A/m$$

So, (B) matches with (IV)

Magnetic flux $$\Phi$$:

$$\Phi=B\cdot A$$

Unit:

$$\text{weber}$$

So, (C) matches with (II)

Magnetic moment:

$$\mu=I\cdot A$$

Unit:

$$\text{Ampere}\cdot\text{meter}^2$$

So, (D) matches with (I)

Final matching:

(A)−(III), (B)−(IV), (C)−(II), (D)−(I)

Given $$E(t) = \alpha^3 e^{-\beta t}$$ with $$\beta = 0.3$$ s⁻¹ and relative errors $$\Delta\alpha/\alpha = 1.2\%$$, $$\Delta t/t = 1.6\%$$, we want to find the maximum percentage error in $$E$$ at $$t = 5\,$$s}.

We begin by taking the natural logarithm of $$E$$ to prepare for error propagation:

$$\ln E = 3\ln\alpha - \beta t$$

By differentiating this relation, the relative error in $$E$$ becomes

$$\frac{\Delta E}{E} = 3\frac{\Delta\alpha}{\alpha} + \beta \Delta t$$

Since $$\Delta t/t = 1.6\%$$ at $$t = 5\,$$s}, we calculate

$$\Delta t = 0.016 \times 5 = 0.08\ \text{s}$$

Substituting $$\Delta\alpha/\alpha = 1.2\%$$ and the computed $$\Delta t$$ into the expression for $$\Delta E/E$$ and converting to a percentage gives

$$\frac{\Delta E}{E} \times 100 = 3 \times 1.2\% + \beta \times \Delta t \times 100$$

$$= 3.6\% + 0.3 \times 0.08 \times 100$$

$$= 3.6\% + 2.4\%$$

$$= 6\%$$

Therefore, the maximum percentage error in $$E$$ at $$t=5\,$$s} is 6\%, which corresponds to Option 1.

Density of the wire is given by $$\rho = \frac{m}{V}$$ where the volume of a cylindrical wire is $$V = \pi r^{2} \, l$$.

Therefore $$\rho = \frac{m}{\pi r^{2} l}$$.

For a function of several variables, the maximum fractional (relative) error is obtained by adding the absolute fractional errors of all factors:
$$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta (r^{2})}{r^{2}} + \frac{\Delta l}{l}$$.

Step 1: Fractional error in mass
Given $$m = 0.60 \text{ g} \pm 0.003 \text{ g}$$.
$$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005 = 0.5\%$$.

Step 2: Fractional error in radius
Given $$r = 0.50 \text{ cm} \pm 0.01 \text{ cm}$$.
$$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02 = 2\%$$.

Step 3: Fractional error in $$r^{2}$$
Since $$r^{2}$$ contains the radius twice, its fractional error is twice that of $$r$$:
$$\frac{\Delta (r^{2})}{r^{2}} = 2 \times \frac{\Delta r}{r} = 2 \times 2\% = 4\%$$.

Step 4: Fractional error in length
Given $$l = 10.00 \text{ cm} \pm 0.05 \text{ cm}$$.
$$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005 = 0.5\%$$.

Step 5: Total fractional (percentage) error in density
$$\frac{\Delta \rho}{\rho} = 0.5\% + 4\% + 0.5\% = 5\%$$.

Hence the maximum percentage error in the measurement of density is $$5\%$$.

Therefore, the correct option is Option B (5).

Permeability of free space:

Using relation:

$$B=\mu_0\frac{I}{L}$$

$$\mu_0=\frac{BL}{I}$$

Substitute dimensions:

$$[B]=[MT^{-2}A^{-1}]$$

$$[\mu_0]=\frac{(MT^{-2}A^{-1})\cdot L}{A}=[MLT^{-2}A^{-2}]$$

So, Permeability → (III)

Magnetic field:

From Lorentz force,

$$F=qvB$$

$$B=\frac{F}{qv}$$

Now substitute dimensions:

$$[q]=[AT]$$

$$\quad[v]=[LT^{-1}]$$

$$[B]=\frac{MLT^{-2}}{(AT)(LT^{-1})}=[MT^{-2}A^{-1}]$$

So, Magnetic field → (II)

Magnetic moment:

Magnetic moment is given by:

$$μ=IA$$

$$[\mu]=[A]\cdot[L^2]=[L^2A]$$

So, Magnetic moment → (IV)

Torsional constant (torsion constant) is defined from the relation:

$$C\theta\ =\tau$$

where $$τ=torque$$ and $$\theta$$ = angular displacement (dimensionless).

So,

$$C=\frac{\tau}{\theta}$$

Since θ\thetaθ is dimensionless:

$$[C]=[\tau]$$

Now, torque:

$$\tau=r\times F$$

$$[F]=[MLT^{-2}]$$

$$[\tau]=[L]\cdot[MLT^{-2}]=[ML^2T^{-2}]$$

So Torsional Constant →(I)

Final matching:

(A)−(III), (B)−(II), (C)−(IV), (D)−(I)

We need to compute $$y = \frac{32.3 \times 1125}{27.4}$$ and report it with the correct number of significant figures.

First, let us count the significant figures in each quantity:
$$32.3$$ has 3 significant figures.
$$1125$$ has 4 significant figures.
$$27.4$$ has 3 significant figures.

In multiplication and division, the final result must have the same number of significant figures as the quantity with the fewest significant figures. Here, the minimum is 3 significant figures.

Now let us compute the value step by step.

$$32.3 \times 1125 = 36337.5$$

$$y = \frac{36337.5}{27.4} = 1326.186...$$

We must round this to 3 significant figures. The first three significant digits of $$1326.186$$ are $$1$$, $$3$$, and $$2$$. The next digit is $$6$$, which is $$\ge 5$$, so we round up.

$$1326.186... \approx 1330$$

Hence, the correct answer is Option B.

We need to find the dimension of the ratio $$\frac{\text{Modulus of Elasticity}}{\text{Torque}}$$ and determine the value of $$c$$ given that $$b = -3$$.

To begin, we write the dimension of Modulus of Elasticity ($$E$$). Modulus of elasticity (Young's modulus) is defined as stress/strain. Since strain is dimensionless:

$$ [E] = [\text{Stress}] = \frac{[\text{Force}]}{[\text{Area}]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] $$

Next, we write the dimension of Torque ($$\tau$$). Torque is given by Force $$\times$$ distance (lever arm):

$$ [\tau] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}] $$

We then compute the dimension of $$E/\tau$$:

$$ \left[\frac{E}{\tau}\right] = \frac{[M L^{-1} T^{-2}]}{[M L^2 T^{-2}]} $$

Dividing each base dimension:

$$ M: M^1 / M^1 = M^0 $$

$$ L: L^{-1} / L^2 = L^{-1-2} = L^{-3} $$

$$ T: T^{-2} / T^{-2} = T^0 $$

Therefore,

$$ \left[\frac{E}{\tau}\right] = [M^0 L^{-3} T^0] $$

Comparing this result with $$[M^a L^b T^c]$$ shows that $$a = 0$$, $$b = -3$$, and $$c = 0$$. This confirms the given value $$b = -3$$ and yields the value of $$c = 0$$.

The answer is 0.

A physical quantity $$C$$ is given by $$C = \dfrac{pq^{2}}{r^{3}\sqrt{s}}$$.

Write every factor with its power explicitly:
$$C = p^{1}\,q^{2}\,r^{-3}\,s^{-1/2}$$.

For a quantity expressed as a product of powers,
$$C = \prod_{i} A_i^{n_i}$$,
the fractional (or percentage) error rule is
$$\frac{\Delta C}{C} = \sum_{i} |n_i|\,\frac{\Delta A_i}{A_i}$$, because errors always add algebraically in magnitude.

Apply the rule to each variable:

• For $$p$$: power $$= 1$$ ⇒ contribution $$= 1 \times (\text{error in }p)$$
• For $$q$$: power $$= 2$$ ⇒ contribution $$= 2 \times (\text{error in }q)$$
• For $$r$$: power $$= 3$$ ⇒ contribution $$= 3 \times (\text{error in }r)$$
• For $$s$$: power $$= \tfrac{1}{2}$$ ⇒ contribution $$= \tfrac{1}{2} \times (\text{error in }s)$$.

Insert the given percentage errors:
$$\frac{\Delta C}{C}\% = 1 \times 1\% \;+\; 2 \times 2\% \;+\; 3 \times 3\% \;+\; \frac{1}{2} \times 2\%$$.

Compute each term:
$$1 \times 1\% = 1\%$$
$$2 \times 2\% = 4\%$$
$$3 \times 3\% = 9\%$$
$$\frac{1}{2} \times 2\% = 1\%$$.

Add them:
$$1\% + 4\% + 9\% + 1\% = 15\%$$.

Therefore, the percentage error in measuring $$C$$ is $$15\%$$.

For $$Q = \frac{ab^4}{cd}$$, the percentage error is given by

$$ \frac{\Delta Q}{Q} \times 100 = \left(\frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}\right) \times 100 $$

Using the provided data, we calculate each relative error:

$$ \frac{\Delta a}{a} = \frac{3}{60} = 0.05 $$

$$ \frac{\Delta b}{b} = \frac{0.1}{20} = 0.005 $$

$$ \frac{\Delta c}{c} = \frac{0.2}{40} = 0.005 $$

$$ \frac{\Delta d}{d} = \frac{0.1}{50} = 0.002 $$

Substituting these results into the percentage error formula yields

$$ = (0.05 + 4 \times 0.005 + 0.005 + 0.002) \times 100 $$

$$ = (0.05 + 0.02 + 0.005 + 0.002) \times 100 $$

$$ = 0.077 \times 100 = 7.7\% $$

Since the percentage error can be expressed in the form $$\frac{x}{1000}$$, we set

$$ 7.7 = \frac{x}{1000} $$

$$ x = 7700 $$

The answer is 7700.

We need to find the percentage error in $$Q = \frac{a^4 b^3}{c^2}$$.

Recall the error propagation formula.

For a quantity $$Q = a^m b^n c^p$$, the maximum percentage error is:

$$\frac{\Delta Q}{Q} \times 100 = |m| \cdot \frac{\Delta a}{a} \times 100 + |n| \cdot \frac{\Delta b}{b} \times 100 + |p| \cdot \frac{\Delta c}{c} \times 100$$

Apply to the given expression.

$$Q = a^4 b^3 c^{-2}$$, so $$m = 4$$, $$n = 3$$, $$p = -2$$:

$$\% \text{error in } Q = 4 \times 3\% + 3 \times 4\% + 2 \times 5\%$$

$$= 12\% + 12\% + 10\% = 34\%$$

The correct answer is Option (3): 34%.

Use dimensional analysis.

Assume the relation:

$$T\propto G^aM^br^c$$

Write dimensions:

$$[T]=T$$

$$[G]=M^{-1}L^3T^{-2}$$

$$[M]=M$$

$$[r]=L$$

Substitute:

$$T=(M^{-1}L^3T^{-2})^a\cdot M^b\cdot L^c$$

$$=M^{-a+b}\cdot L^{3a+c}\cdot T^{-2a}$$

Now compare powers with LHS$$[T^1]$$:

For mass:

$$-a+b=0$$

For length:

$$3a+c=0$$

For time:

$$-2a=1\Rightarrow a=-\frac{1}{2}$$

Substitute aaa:

$$b=a=-\frac{1}{2}$$

$$3a+c=0\Rightarrow c=\frac{3}{2}$$

Final relation:

$$T\propto G^{-1/2}M^{-1/2}r^{3/2}$$

$$T\propto\sqrt{\frac{r^3}{GM}}$$

Correct expression:

$$T=k\sqrt{\frac{r^3}{GM}}$$

$$T^2 = \frac{k r^3}{GM}$$

We need to evaluate two statements about dimensions.

Statement I: Dimensions of specific heat is $$[L^2 T^{-2} K^{-1}]$$.

Specific heat = Energy / (mass × temperature).

Dimensions: $$\frac{[ML^2T^{-2}]}{[M][K]} = [L^2T^{-2}K^{-1}]$$

Statement I is correct. ✓

Statement II: Dimensions of gas constant is $$[ML^2T^{-1}K^{-1}]$$.

Gas constant R = Energy / (mol × temperature) (from PV = nRT).

Dimensions: $$\frac{[ML^2T^{-2}]}{[\text{mol}][K]} = [ML^2T^{-2}K^{-1}\text{mol}^{-1}]$$

The given dimension has $$T^{-1}$$ instead of $$T^{-2}$$. Statement II is incorrect. ✗

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

We need to check the dimensional correctness of both statements.

Statement (I): Planck's constant and angular momentum have the same dimensions.

Planck's constant $$h$$ has the relation $$E = h\nu$$, so:

$$[h] = \frac{[E]}{[\nu]} = \frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$$

Angular momentum $$L = mvr$$, so:

$$[L] = M \cdot LT^{-1} \cdot L = ML^2T^{-1}$$

Both have the same dimensions $$ML^2T^{-1}$$. Statement (I) is true.

Statement (II): Linear momentum and moment of force have the same dimensions.

Linear momentum $$p = mv$$:

$$[p] = MLT^{-1}$$

Moment of force (torque) $$\tau = r \times F$$:

$$[\tau] = L \cdot MLT^{-2} = ML^2T^{-2}$$

The dimensions are different: $$MLT^{-1} \neq ML^2T^{-2}$$. Statement (II) is false.

Therefore, Statement I is true but Statement II is false.

The correct answer is Option A.

We need to find the dimensions of $$\epsilon_0 E^2$$, where $$\epsilon_0$$ is the permittivity of free space and $$E$$ is the electric field.

The electrostatic energy density stored in an electric field is $$u = \frac{1}{2}\epsilon_0 E^2$$. Since $$\frac{1}{2}$$ is dimensionless, $$\epsilon_0 E^2$$ has the same dimensions as energy density.

Energy density is defined as energy per unit volume, so

$$[u] = \frac{[E]}{[V]} = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$$

Alternatively, we can verify this result by considering the dimensions of $$\epsilon_0$$ and $$E$$ separately. From Coulomb’s law, $$F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}$$, so

$$\epsilon_0 = \frac{q^2}{Fr^2} \implies [\epsilon_0] = \frac{A^2T^2}{MLT^{-2}\cdot L^2} = M^{-1}L^{-3}T^4A^2$$

The electric field has dimensions

$$[E] = \frac{[F]}{[q]} = \frac{MLT^{-2}}{AT} = MLT^{-3}A^{-1}$$

Therefore,

$$[\epsilon_0 E^2] = M^{-1}L^{-3}T^4A^2 \times M^2L^2T^{-6}A^{-2} = ML^{-1}T^{-2}$$

This confirms that $$[\epsilon_0 E^2] = ML^{-1}T^{-2}$$. The correct answer is Option (4): $$[ML^{-1}T^{-2}]$$.

Given $$m = kc^P G^{-1/2} h^{1/2}$$. Using dimensional analysis:

$$[m] = M$$, $$[c] = LT^{-1}$$, $$[G] = M^{-1}L^3T^{-2}$$, $$[h] = ML^2T^{-1}$$.

$$M = (LT^{-1})^P (M^{-1}L^3T^{-2})^{-1/2} (ML^2T^{-1})^{1/2}$$

$$= L^P T^{-P} \cdot M^{1/2}L^{-3/2}T^{1} \cdot M^{1/2}L^{1}T^{-1/2}$$

$$= M^{1} L^{P-3/2+1} T^{-P+1-1/2}$$

Matching M: $$1 = 1$$ ✓

Matching L: $$0 = P - 1/2 \Rightarrow P = 1/2$$

Matching T: $$0 = -P + 1/2 \Rightarrow P = 1/2$$ ✓

The answer is Option (1): $$\boxed{\frac{1}{2}}$$.

We need to find the percentage error in measuring the resistance of a wire, given percentage errors in length and diameter measurements.

The resistance of a wire is given by $$R = \frac{\rho L}{A}$$, where $$\rho$$ is the resistivity, $$L$$ is the length, and $$A$$ is the cross-sectional area. For a circular wire with diameter $$d$$, $$A = \frac{\pi d^2}{4}$$. Therefore, $$R = \frac{\rho L}{\frac{\pi d^2}{4}} = \frac{4\rho L}{\pi d^2}$$.

To determine the propagation of error for $$R = \frac{4\rho L}{\pi d^2}$$ (with $$\rho$$ treated as a constant), we take the natural logarithm: $$\ln R = \ln(4\rho/\pi) + \ln L - 2\ln d$$. Differentiating yields $$\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d}$$, where absolute values are taken to represent maximum error.

Given that $$\frac{\Delta L}{L}\times100 = 0.1\%$$ and $$\frac{\Delta d}{d}\times100 = 0.1\%$$, it follows that $$\frac{\Delta R}{R}\times100 = 0.1\% + 2\times0.1\% = 0.3\%$$.

The percentage error in measuring the resistance is 0.3%. Hence the correct answer is Option (2): 0.3%.

The lens formula relates the focal length f to the object distance u and the image distance v as $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$.

Differentiating this expression gives

$$ -\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2} .$$

Considering maximum errors in u and v leads to

$$ \frac{|\Delta f|}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} ,$$

which can be rearranged to

$$ \Delta f = f^2\left(\frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}\right) .$$

Therefore, the correct answer is Option 3: $$f^2\left[\frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}\right].$$

The order of magnitude of a number expressed in scientific notation as $$a \times 10^b$$ (where $$1 \leq a < 10$$) is defined based on the value of $$a$$. According to the convention used in this context:

If $$a \leq 5$$, then the order of magnitude is $$b$$.
If $$a > 5$$, then the order of magnitude is $$b+1$$.

Therefore, $$b$$ is the order of magnitude when $$a \leq 5$$.

Now, evaluating the options:

Option A states that $$b$$ is the order of magnitude for $$a \geq 5$$. This is incorrect because when $$a \geq 5$$, the order of magnitude is $$b$$ only if $$a \leq 5$$, but for $$a > 5$$, it is $$b+1$$.

Option B states that $$b$$ is the order of magnitude for $$a \leq 5$$. This is correct, as per the convention.

Option C states that $$a$$ is the order of magnitude for $$b \leq 5$$. This is incorrect because the order of magnitude is an exponent (a power of 10), not the coefficient $$a$$.

Option D states that $$b$$ is the order of magnitude for $$5 < a \leq 10$$. This is incorrect because when $$a > 5$$, the order of magnitude is $$b+1$$, not $$b$$.

Thus, the correct option is B.

1001:

All non-zero digits are significant, and zeros between non-zero digits are also significant.

So, 1001 has 4 significant figures.

(A) → (II)

010.1:

Leading zero is not significant. The zero between digits and decimal part is significant.

Digits: 1, 0, 1 → total 3 significant figures.

(B) → (I)

100.100:

Zeros between non-zero digits and trailing zeros after decimal are significant.

Digits: 1, 0, 0, 1, 0, 0 → total 6 significant figures.

(C) → (IV)

0.0010010:

Leading zeros are not significant. Zeros between digits and trailing zero after decimal are significant.

Digits: 1, 0, 0, 1, 0 → total 5 significant figures.

(D) → (III)

Final matching:

(A)−(II), (B)−(I), (C)−(IV), (D)−(III)

Coefficient of viscosity:

From Newton’s law of viscosity:

$$F=\eta A\frac{dv}{dx}$$

$$\eta=\frac{F}{A}\cdot\frac{dx}{dv}$$

$$[\eta]=\frac{MLT^{-2}}{L^2}\cdot\frac{L}{LT^{-1}}=[ML^{-1}T^{-1}]$$

So, (A) matches with (III)

Surface tension:

$$\text{Surface tension}=\frac{\text{Force}}{\text{length}}$$

$$\frac{MLT^{-2}}{L}=[MT^{-2}]$$

So, (B) matches with (IV)

Angular momentum:

$$L=r\times p=r\cdot mv$$

$$[L]=[L]\cdot[MLT^{-1}]=[ML^2T^{-1}]$$

So, (C) matches with (II)

Rotational kinetic energy:

$$E=\frac{1}{2}I\omega^2$$

$$[I]=ML^2,\quad[\omega]=T^{-1}$$

$$[E]=ML^2\cdot T^{-2}=[ML^2T^{-2}]$$

So, (D) matches with (I)

Final matching:

(A)−(III), (B)−(IV), (C)−(II), (D)−(I)

$$\lambda = \frac{h}{\sqrt{2mE}}$$, so $$h = \lambda\sqrt{2mE}$$.

Dimensions: $$[\lambda] = [L]$$, $$[m] = [M]$$, $$[E] = [ML^2T^{-2}]$$.

$$[h] = [L] \cdot [M \cdot ML^2T^{-2}]^{1/2} = [L] \cdot [M^2L^2T^{-2}]^{1/2} = [L][MLT^{-1}] = [ML^2T^{-1}]$$.

The correct answer is Option 1: $$[ML^2T^{-1}]$$.

Latent heat is defined as the heat energy per unit mass required for a phase change:

$$L = \frac{Q}{m}$$

The dimension of heat energy $$Q$$ is $$[ML^2T^{-2}]$$, and mass $$m$$ has dimension $$[M]$$.

Therefore, $$[L] = \frac{[ML^2T^{-2}]}{[M]} = [M^0L^2T^{-2}]$$.

The correct answer is Option 2: $$[M^0L^2T^{-2}]$$.

In the Van der Waals equation: $$\left(P + \frac{a}{V^2}\right)(V - b) = RT$$

$$\frac{a}{V^2}$$ has the same dimensions as $$P$$, so $$[a] = [PV^2]$$.

$$[b] = [V]$$.

$$\frac{a}{b^2} = \frac{PV^2}{V^2} = P$$

The dimensions of $$\frac{a}{b^2}$$ are the same as that of $$P$$.

The answer corresponds to Option (2).

ρ = Rl/(πr²). % error = ΔR/R + Δl/l + 2Δr/r = 10/100 + 0.2/15 + 2(0.05/0.35)

= 0.1 + 0.0133 + 0.2857 = 0.399 = 39.9%.

The correct answer is Option 2: 39.9%.

$$R = V/I$$. Percentage error in R:

$$\frac{\Delta R}{R} \times 100 = \frac{\Delta V}{V} \times 100 + \frac{\Delta I}{I} \times 100$$

$$= \frac{5}{200} \times 100 + \frac{0.2}{20} \times 100 = 2.5\% + 1\% = 3.5\%$$

The answer is $$3.5\%$$, which corresponds to Option (1).

A dimensionless quantity has zero net power of every base dimension $$\left(M,\,L,\,T,\,I\right)$$.
We therefore write $$e^{\alpha}\,\varepsilon_0^{\beta}\,h^{\gamma}\,c^{\delta}$$ and equate the exponents of each base dimension to zero.

Dimensions of the individual quantities
Electronic charge : $$[e]=I\,T$$
Permittivity of free space : $$[\varepsilon_0]=M^{-1}L^{-3}T^{4}I^{2}$$ (from Coulomb’s law)
Planck’s constant   : $$[h]=M\,L^{2}T^{-1}$$
Speed of light    : $$[c]=L\,T^{-1}$$

Total dimensional exponents after raising to respective powers
$$\begin{aligned} M:&\; -\beta+\gamma \\[2pt] L:&\; -3\beta+2\gamma+\delta \\[2pt] T:&\; \alpha+4\beta-\gamma-\delta \\[2pt] I:&\; \alpha+2\beta \end{aligned}$$

For the product to be dimensionless, each of the four sums must vanish:

$$-\beta+\gamma=0 \;\Longrightarrow\; \gamma=\beta$$

$$-3\beta+2\gamma+\delta=0 \;\Longrightarrow\; -3\beta+2\beta+\delta=0 \;\Longrightarrow\; \delta=\beta$$

$$\alpha+2\beta=0 \;\Longrightarrow\; \alpha=-2\beta$$

The time exponent yields the same relation $$\alpha+2\beta=0,$$ confirming consistency.

Let $$\beta=-n,$$ where $$n$$ is any non-zero integer (the question specifies an integer multiplier).
Then

$$\alpha=2n,\qquad \gamma=-n,\qquad \delta=-n$$

Thus $$(\alpha,\beta,\gamma,\delta)=(2n,\,-n,\,-n,\,-n),$$ which corresponds to
Option A: $$(2n,\,-n,\,-n,\,-n)$$

LC = 1/(20N) cm. MSD = 1mm = 1/10 cm. LC = MSD - VSD. VSD = MSD - LC = 1/10 - 1/(20N) = (2N-1)/(20N) cm. N×VSD = x×MSD. x = N×VSD/MSD = N×(2N-1)/(20N)÷(1/10) = (2N-1)/2.

Option (3).

$$F = ax^2 + bt^{1/2}$$. $$[F] = MLT^{-2}$$.

$$[a] = [F]/[x^2] = MLT^{-2}/L^2 = ML^{-1}T^{-2}$$.

$$[b] = [F]/[t^{1/2}] = MLT^{-2}/T^{1/2} = MLT^{-5/2}$$.

$$[b^2/a] = M^2L^2T^{-5}/(ML^{-1}T^{-2}) = ML^3T^{-3}$$.

The answer is Option (1): $$[ML^3T^{-3}]$$.

$$E=\frac{B-x^2}{At}$$. $$[B]=[x^2]=L^2$$. $$[A]=[B]/([E][t])=L^2/(ML^2T^{-2}\cdot T)=L^2/(ML^2T^{-1})=M^{-1}T$$.

$$[AB]=M^{-1}T\cdot L^2=L^2M^{-1}T^1$$.

The answer is Option (2): $$L^2M^{-1}T^1$$.

Torque:

Torque is given by force × distance.

$$\tau=r\times F$$

Dimensions of force:

$$[F]=[MLT^{-2}]$$

So,

$$[\tau]=[L]\cdot[MLT^{-2}]=[ML^2T^{-2}]$$

So, Torque → (IV)

Magnetic field:

From Lorentz force,

$$F=qvB$$

$$B=\frac{F}{qv}$$

Now substitute dimensions:

$$[q]=[AT]$$

$$\quad[v]=[LT^{-1}]$$

$$[B]=\frac{MLT^{-2}}{(AT)(LT^{-1})}=[MT^{-2}A^{-1}]$$

So, Magnetic field → (III)

Magnetic moment:

Magnetic moment is given by:

$$μ=IA$$

$$[\mu]=[A]\cdot[L^2]=[L^2A]$$

So, Magnetic moment → (II)

Permeability of free space:

Using relation:

$$B=\mu_0\frac{I}{L}$$

$$\mu_0=\frac{BL}{I}$$

Substitute dimensions:

$$[B]=[MT^{-2}A^{-1}]$$

$$[\mu_0]=\frac{(MT^{-2}A^{-1})\cdot L}{A}=[MLT^{-2}A^{-2}]$$

So, Permeability → (I)

Final matching:

(A)−(IV), (B)−(III), (C)−(II), (D)−(I)

We need to find the dimensional formula of angular impulse.

Angular impulse is the product of torque and time, giving Angular impulse = $$\tau \times t$$.

Torque ($$\tau$$) is force times the lever arm (perpendicular distance). The dimension of force is $$[M L T^{-2}]$$ and that of the lever arm is $$[L]$$, so $$[\tau] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$$.

Thus the dimension of angular impulse is $$[\text{Angular impulse}] = [\tau] \times [t] = [M L^2 T^{-2}] \times [T] = [M L^2 T^{-1}]$$.

The correct answer is Option D: $$[M L^2 T^{-1}]$$.

The equation is: $$y = 2a\sin\left(\frac{2\pi nt}{\lambda}\right)\cos\left(\frac{2\pi x}{\lambda}\right)$$

For the equation to be dimensionally consistent:

- $$\frac{2\pi x}{\lambda}$$ must be dimensionless, so $$[\lambda] = [L]$$ and $$[x] = [L]$$

- $$\frac{2\pi nt}{\lambda}$$ must be dimensionless, so $$[nt/\lambda]$$ is dimensionless

- This gives $$[n] = [\lambda/t] = [L/T] = [LT^{-1}]$$ (velocity dimension)

Now checking each option:

Option 1: Dimensions of $$n/\lambda = [LT^{-1}]/[L] = [T^{-1}]$$, NOT $$[T]$$. This is NOT correct. ✓ (This is the wrong statement)

Option 2: Dimensions of n is $$[LT^{-1}]$$. Correct.

Option 3: Dimensions of x is [L]. Correct.

Option 4: Dimensions of nt = $$[LT^{-1}][T] = [L]$$. Correct.

The correct answer is Option 1: The dimensions of $$n/\lambda$$ is [T] (this is NOT correct since it should be $$[T^{-1}]$$).

Given readings: $$4.62$$ s, $$4.632$$ s, $$4.6$$ s, $$4.64$$ s.

Calculate arithmetic mean.

$$ \text{Mean} = \frac{4.62 + 4.632 + 4.6 + 4.64}{4} = \frac{18.492}{4} = 4.623 \text{ s} $$

Apply significant figures rule.

When adding/averaging numbers, the result should have the same number of decimal places as the measurement with the fewest decimal places.

The readings have: 2, 3, 1, and 2 decimal places respectively. The least is 1 decimal place (from 4.6 s).

So the mean should be rounded to 1 decimal place: $$4.6$$ s.

The correct answer is Option (3): 4.6 s.

$$a/V^2$$ has units of pressure: $$[a]=PV^2=ML^{-1}T^{-2}\cdot L^6=ML^5T^{-2}$$. $$[b]=V=L^3$$.

$$[a/b]=ML^5T^{-2}/L^3=ML^2T^{-2}$$.

The answer is Option (3): $$ML^2T^{-2}$$.

We need to find the percentage error in Young's modulus $$Y = 49000\frac{M}{l}$$.

For $$Y = k\frac{M}{l}$$, taking the logarithm gives $$\ln Y = \ln k + \ln M - \ln l$$, and differentiating yields $$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l}$$.

From the graph paper, the smallest scale division along the load axis is 5 g, so $$\Delta M = 5$$ g, and along the extension axis it is 0.02 cm, giving $$\Delta l = 0.02$$ cm.

$$\frac{\Delta Y}{Y} = \frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01 = 0.02 = 2\%$$

The correct answer is Option (2): 2%.

Find the least count of a Vernier calliper where 10 MSD = 11 VSD and each MSD = 5 units.

MSD stands for Main Scale Division and VSD for Vernier Scale Division. Since 10 main-scale divisions span the same length as 11 vernier-scale divisions, we have $$ 10 \times 5 = 11 \times \text{(1 VSD)} $$. It follows that $$ 1 \text{ VSD} = \frac{50}{11} \text{ units} $$. The least count (LC) of a Vernier instrument is the difference between one main-scale division and one vernier-scale division, so $$ LC = 1 \text{ MSD} - 1 \text{ VSD} = 5 - \frac{50}{11} = \frac{55 - 50}{11} = \frac{5}{11} $$. Therefore, the correct answer is Option D: $$\frac{5}{11}$$.

We are given a vernier callipers where 20 divisions on the vernier scale coincide with 19 divisions on the main scale, and the least count is 0.1 mm. We need to find the value of one main scale division (MSD).

The given condition is $$20 \text{ VSD} = 19 \text{ MSD}$$ which implies $$1 \text{ VSD} = \frac{19}{20} \text{ MSD}$$.

The least count (LC) is defined as the difference between one main scale division and one vernier scale division: $$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD}$$.

Substituting the expression for VSD from the previous relation gives $$\text{LC} = 1 \text{ MSD} - \frac{19}{20} \text{ MSD} = \text{MSD} \left(1 - \frac{19}{20}\right) = \frac{1}{20} \text{ MSD}$$.

Since the least count is given as $$\text{LC} = 0.1$$ mm, one can write $$0.1 = \frac{1}{20} \times \text{MSD}$$ which yields $$\text{MSD} = 0.1 \times 20 = 2 \text{ mm}$$.

Therefore, one main scale division is equal to 2 mm. The correct answer is Option (2): 2.

We need to evaluate the Assertion and Reason about Vernier callipers.

Assertion (A): In a Vernier calliper, if positive zero error exists, the reading taken will be more than the actual reading.

This is true. Positive zero error means the zero of the vernier scale is to the right of the zero of the main scale when the jaws are closed. This adds an extra amount to every reading, making measured values larger than actual values. The correction requires subtracting the zero error.

Reason (R): The zero error in a Vernier calliper might have happened due to manufacturing defect or due to rough handling.

This is true. Manufacturing imperfections or damage from rough use can cause the jaws to not align properly at zero.

However, R merely explains the cause of zero error in general, not why positive zero error specifically leads to higher readings. R does not explain the mechanism described in A.

The correct answer is Option 2: Both (A) and (R) are correct but (R) is not the correct explanation of (A).

A) Radius of curvature of concave surface
A spherometer is specifically designed to measure radius of curvature (both concave and convex).
So this can be measured.

B) Specific rotation of liquids
This is an optical property measured using a polarimeter, not a spherometer.
Spherometer only measures mechanical dimensions (height, curvature).
So this cannot be measured.

C) Thickness of thin plates
Yes, a spherometer can measure very small thickness by comparing readings.
So this can be measured.

D) Radius of curvature of convex surface
Again, one of the main uses of a spherometer.
So this can be measured.

Final conclusion:
Only option B is not measurable using a spherometer.

Answer: Specific rotation of liquids

50 Vernier divisions = 49 main scale divisions.

1 VSD = $$\frac{49}{50}$$ MSD.

Least count = 1 MSD - 1 VSD = $$1 - \frac{49}{50} = \frac{1}{50}$$ MSD = $$\frac{0.5}{50}$$ mm = $$0.01$$ mm.

The answer is Option (4): $$\boxed{0.01 \text{ mm}}$$.

We need to find how many main scale divisions are in 1 cm for a vernier caliper.

First, the least count LC is defined as one main scale division (MSD) minus one vernier scale division (VSD), and since 50 VSD equal 49 MSD, we have 1 VSD = 49/50 MSD. Moreover, the zero error equals the number of coinciding VSD multiplied by the least count.

Expressing the least count in terms of MSD gives $$\text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}$$

Since the zero of the vernier has shifted left (indicating a positive zero error) and the fourth division on the vernier coincides with a main scale division, the zero error is four times the least count: $$\text{Zero error} = 4 \times \text{LC} = 4 \times \frac{1}{50}\,\text{MSD} = \frac{4}{50}\,\text{MSD}$$

We are told this zero error measures 0.04 mm, so $$\frac{4}{50}\,\text{MSD} = 0.04\,\text{mm} \implies \text{MSD} = 0.04 \times \frac{50}{4} = 0.04 \times 12.5 = 0.5\,\text{mm}$$

Hence, the number of main scale divisions in 1 cm (10 mm) is $$\frac{10\,\text{mm}}{0.5\,\text{mm}} = 20$$

The correct answer is Option (3): 20.

One main scale division (MSD) = $$m$$ units.

The $$n^{th}$$ division of the main scale coincides with the $$(n+1)^{th}$$ division of the vernier scale.

This means: $$n$$ MSD = $$(n+1)$$ VSD (vernier scale divisions).

Therefore: 1 VSD = $$\frac{n}{n+1}$$ MSD = $$\frac{nm}{n+1}$$ units.

Least count = 1 MSD - 1 VSD = $$m - \frac{nm}{n+1} = m\left(1 - \frac{n}{n+1}\right) = m \cdot \frac{1}{n+1} = \frac{m}{n+1}$$.

The correct answer is Option 3: $$\frac{m}{n+1}$$.

We need to find the density of a sphere measured with a vernier caliper.

9 main scale divisions = 10 vernier scale divisions. Smallest main scale division = 1 mm.

$$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 - \frac{9}{10} = 0.1 \text{ mm} = 0.01 \text{ cm}$$

Main scale reading = 2 cm. Vernier coincidence = 2nd division.

$$d = \text{MSR} + \text{VSR} \times \text{LC} = 2 + 2 \times 0.01 = 2.02 \text{ cm}$$

Radius: $$r = 1.01$$ cm.

$$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.01)^3$$

$$(1.01)^3 \approx 1.0303$$

$$V = \frac{4}{3} \times 3.1416 \times 1.0303 \approx 4.32 \text{ cm}^3$$

$$\rho = \frac{m}{V} = \frac{8.635}{4.32} \approx 2.0 \text{ g/cm}^3$$

The correct answer is Option (1): 2.0 g/cm$$^3$$.

We need to find the diameter of a wire measured using a screw gauge that has a positive zero error. Since the least count of a screw gauge is given by $$\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$$, we have determined the smallest measurable length increment.

The problem states that with no measuring quantity between the jaws, the zero of the circular scale lies 5 divisions below the reference line. From this we infer a positive zero error, which is calculated as $$\text{Zero error} = +5 \times \text{LC} = +5 \times 0.01 = +0.05 \text{ mm}$$. Since a positive zero error indicates that the instrument reads more than the actual value, we must subtract it from the measured reading.

Next, the observed reading consists of a main scale reading and a circular scale reading. For the main scale reading, 4 linear scale divisions are visible, and since the pitch is 1 mm, each linear division equals 1 mm. Hence $$\text{MSR} = 4 \times 1 = 4 \text{ mm}$$. The circular scale reading corresponds to the 60th division coinciding with the reference line, giving $$\text{CSR} = 60 \times \text{LC} = 60 \times 0.01 = 0.60 \text{ mm}$$. Altogether, the observed reading is $$\text{Observed reading} = \text{MSR} + \text{CSR} = 4 + 0.60 = 4.60 \text{ mm}$$.

Finally, applying the zero error correction yields $$\text{Corrected reading} = \text{Observed reading} - \text{Zero error} = 4.60 - 0.05 = 4.55 \text{ mm}$$. The correct answer is Option (3): 4.55 mm.

We need to find the diameter of a wire measured using a screw gauge. Recall that the screw gauge reading is given by: $$ \text{Reading} = \text{MSR} + \text{CSR} \times \text{Least Count} $$ where MSR = Main Scale Reading, CSR = Circular Scale Reading, and Least Count = Pitch / Number of divisions on circular scale.

Since the pitch is 1 mm and the circular scale has 100 divisions, the least count is: $$ LC = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} $$. Substituting in the formula gives the diameter: $$ d = 1 + 42 \times 0.01 = 1 + 0.42 = 1.42 \text{ mm} $$. Expressing 1.42 mm in fractional form: $$ d = 1.42 = \frac{142}{100} = \frac{71}{50} \text{ mm} $$. Therefore x = 71. The correct answer is Option (3): 71.

Let angle between $$\vec{A}$$ and $$\vec{B}$$ be $$\theta$$. Resultant $$\vec{R} = \vec{A} + \vec{B}$$ is perpendicular to $$\vec{A}$$.

$$\vec{R} \cdot \vec{A} = 0$$: $$A^2 + AB\cos\theta = 0 \implies \cos\theta = -A/B$$.

$$|\vec{R}| = B/2$$: $$A^2 + B^2 + 2AB\cos\theta = B^2/4$$.

$$A^2 + B^2 - 2A^2 = B^2/4 \implies B^2 - A^2 = B^2/4 \implies A^2 = 3B^2/4$$.

$$\cos\theta = -\frac{A}{B} = -\frac{\sqrt{3}}{2} \implies \theta = 150°$$.

The answer is $$\boxed{150}$$°.

The volume of a right circular cone in terms of the measured base diameter $$d$$ and height $$h$$ is

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi\left(\frac{d}{2}\right)^{2}h=\frac{\pi}{12}\,d^{2}h$$

If a quantity $$Q$$ depends on the variables $$x,\,y,\,z$$ as $$Q=x^{m}y^{n}z^{p}$$, the maximum fractional (relative) error in $$Q$$ is

$$\frac{\Delta Q}{Q}=m\frac{\Delta x}{x}+n\frac{\Delta y}{y}+p\frac{\Delta z}{z}$$

Comparing $$V=\dfrac{\pi}{12}d^{2}h$$ with the general form, we have $$m=2$$ for $$d$$ and $$n=1$$ for $$h$$ (the constant $$\pi/12$$ does not affect errors).

Therefore, the maximum fractional error in volume is

$$\frac{\Delta V}{V}=2\frac{\Delta d}{d}+1\frac{\Delta h}{h}$$ $$-(1)$$

The measurements are:

Diameter $$d = 20.0\text{ cm}$$     Height $$h = 20.0\text{ cm}$$

The least count of the scale is $$2\text{ mm}=0.2\text{ cm}$$. In error calculations we take the full least count as the maximum absolute error, so

$$\Delta d = 0.2\text{ cm}, \qquad \Delta h = 0.2\text{ cm}$$

Compute the fractional errors:

$$\frac{\Delta d}{d}=\frac{0.2}{20.0}=0.01$$   (that is, $$1\%$$)

$$\frac{\Delta h}{h}=\frac{0.2}{20.0}=0.01$$   (that is, $$1\%$$)

Substitute into $$(1)$$:

$$\frac{\Delta V}{V}=2(0.01)+0.01=0.03$$

Converting to percentage:

Maximum percentage error $$=0.03\times100\% = 3\%$$

Hence, the maximum percentage error in determining the volume of the cone is 3 %.

In Young’s double slit experiment, intensity at a point is given by:

$$I = I_{\max}\cos^2\left(\frac{\phi}{2}\right)$$

When $$I = \frac{I_{\max}}{4}$$, we have

$$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4} \implies \cos\left(\frac{\phi}{2}\right) = \frac{1}{2} \implies \frac{\phi}{2} = \frac{\pi}{3} \implies \phi = \frac{2\pi}{3}$$

The phase difference is related to the path difference by $$\phi = \frac{2\pi}{\lambda}\Delta$$, where $$\Delta = \frac{yd}{D}$$.

Thus,

$$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \cdot \frac{yd}{D}$$

Rearranging to solve for $$y$$ gives

$$y = \frac{\lambda D}{3d} = \frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}} = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} = 200 \times 10^{-6} \text{ m} = 200 \text{ }\mu\text{m}$$

Therefore, the answer is $$\boxed{200}$$ μm.

This question tests knowledge of the SI definition of the candela, the base unit of luminous intensity.

The Official SI Definition of the Candela:

The candela is defined as the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency $$540 \times 10^{12}$$ hertz and that has a radiant intensity in that direction of $$\frac{1}{683}$$ watt per steradian.

Why these specific values?

- The frequency $$540 \times 10^{12}$$ Hz corresponds to green light with a wavelength of approximately 555 nm. This is chosen because the human eye is most sensitive to light at this wavelength (the peak of the photopic luminosity function).

- The factor $$\frac{1}{683}$$ W/sr is chosen to maintain consistency with the historical definition of the candela, ensuring backward compatibility with previous measurement standards.

Matching with the question format:

The question states the frequency is $$A \times 10^{12}$$ Hz, so $$A = 540$$.

The question states the radiant intensity is $$\frac{1}{B}$$ W/sr, so $$B = 683$$.

The correct answer is Option 1: 540 and 683.

We need to determine how Young's modulus changes with temperature.

What is Young’s modulus? Young’s modulus ($$Y$$) is a measure of the stiffness of a material defined as the ratio of stress to strain within the elastic limit: $$ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} $$. It quantifies how much a material resists deformation under an applied force.

Effect of temperature on interatomic bonds: Young’s modulus depends on the strength of interatomic bonds, and as temperature increases, atoms vibrate with greater amplitude about their equilibrium positions. This increased vibration causes thermal expansion, increasing the average interatomic distance. Consequently, the interatomic forces become weaker because the atoms are, on average, further apart and lie on a flatter part of the potential energy curve. Since the restoring force per unit displacement decreases, the material becomes less stiff.

Therefore, Young’s modulus decreases with increasing temperature.

The correct answer is Option B: decreases.

For a planet revolving around the sun in a circular orbit of radius aaa:

The gravitational force provides the centripetal force.

$$\frac{GMm}{a^2}=\frac{mv^2}{a}$$

From this, we get the orbital velocity:

$$v^2=\frac{GM}{a}$$

Now kinetic energy of the planet is:

$$K=\frac{1}{2}mv^2=\frac{1}{2}m\cdot\frac{GM}{a}=\frac{GMm}{2a}$$

So, (A) matches with (II)

Gravitational potential energy of the sun-planet system is:

$$U=-\frac{GMm}{a}$$

So, (B) matches with (I)

Total mechanical energy is:

$$E=K+U=\frac{GMm}{2a}-\frac{GMm}{a}=-\frac{GMm}{2a}​$$​

So, (C) matches with (IV)

Escape energy per unit mass at the surface of a planet is:

$$\ \frac{\ 1}{2}​v_e^2​=\ \frac{\ GM}{r}$$

$$v_e=\sqrt{\frac{2GM}{r}}$$

Energy per unit mass:

$$v_e^2=\frac{GM}{r}$$

So, (D) matches with (III)

Thus, the correct matching is:

(A)−(II), (B)−(I), (C)−(IV), (D)−(III)

The density of a cylindrical wire is: $$\rho = \frac{m}{\pi r^2 L}$$

The maximum percentage error in density is:

$$ \frac{\Delta\rho}{\rho} \times 100 = \frac{\Delta m}{m} \times 100 + 2\frac{\Delta r}{r} \times 100 + \frac{\Delta L}{L} \times 100 $$

Substituting the given values:

$$ = \frac{0.01}{0.4} \times 100 + 2 \times \frac{0.03}{6} \times 100 + \frac{0.04}{8} \times 100 $$

$$ = 2.5\% + 1.0\% + 0.5\% = 4\% $$

The maximum error in density is 4%.

We need to find the SI units of $$A$$ and $$B$$ in the electric field expression $$\vec{E} = \frac{A}{x^2}\hat{i} + \frac{B}{y^3}\hat{j}$$.

Finding the unit of A:

The term $$\frac{A}{x^2}$$ represents the x-component of electric field.

$$[A] = [E] \times [x^2] = \text{N C}^{-1} \times \text{m}^2 = \text{N m}^2 \text{ C}^{-1}$$

Finding the unit of B:

The term $$\frac{B}{y^3}$$ represents the y-component of electric field.

$$[B] = [E] \times [y^3] = \text{N C}^{-1} \times \text{m}^3 = \text{N m}^3 \text{ C}^{-1}$$

Therefore, $$A$$ has SI unit N m$$^2$$ C$$^{-1}$$ and $$B$$ has SI unit N m$$^3$$ C$$^{-1}$$.

The correct answer is Option 2.

Statement I: Astronomical unit (Au), Parsec (Pc) and Light year (ly) are all units used for measuring astronomical distances. This is correct.

Statement II: Au < Parsec (Pc) < ly

The actual values are:

$$1 \text{ AU} \approx 1.496 \times 10^{11}$$ m

$$1 \text{ ly} \approx 9.461 \times 10^{15}$$ m

$$1 \text{ Pc} \approx 3.086 \times 10^{16}$$ m

So the correct order is: $$\text{Au} < \text{ly} < \text{Pc}$$

Statement II says Au < Pc < ly, which places Parsec before Light year. This is incorrect since 1 Pc is actually larger than 1 ly.

Therefore, Statement I is correct but Statement II is incorrect, which is Option B.

Resistance $$R$$, inductive reactance $$X_L$$, and capacitive reactance $$X_C$$ all have the same dimensions of electrical resistance (Ohm, $$\Omega$$).

For a quantity to be dimensionless, the dimensions in the numerator and denominator must cancel.

Checking option 2: $$\dfrac{R}{\sqrt{X_L X_C}}$$

Dimensions of numerator: $$[\Omega]$$

Dimensions of denominator: $$\sqrt{[\Omega][\Omega]} = [\Omega]$$

$$ \dfrac{R}{\sqrt{X_L X_C}} = \dfrac{[\Omega]}{[\Omega]} = \text{dimensionless} $$

This is the quality factor related quantity, and it is indeed dimensionless.

We need to express mass $$M$$ in terms of $$c$$, $$G$$, and $$h$$.

Dimensions: $$[c] = LT^{-1}$$, $$[G] = M^{-1}L^3T^{-2}$$, $$[h] = ML^2T^{-1}$$

Let $$[M] = [h]^a [c]^b [G]^d$$. Comparing dimensions:

Mass: $$a - d = 1$$ ... (1)

Length: $$2a + b + 3d = 0$$ ... (2)

Time: $$-a - b - 2d = 0$$ ... (3)

From (1): $$a = 1 + d$$

From (3): $$b = -a - 2d = -(1+d) - 2d = -1 - 3d$$

Substituting in (2): $$2(1+d) + (-1-3d) + 3d = 0 \Rightarrow 1 + 2d = 0 \Rightarrow d = -\frac{1}{2}$$

Therefore: $$a = \frac{1}{2}$$, $$b = \frac{1}{2}$$, $$d = -\frac{1}{2}$$

We need to match each physical quantity with its correct SI unit.

A. Surface Tension

Surface tension = Force per unit length = $$\frac{\text{N}}{\text{m}} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{m}} = \text{kg s}^{-2}$$

This matches IV: kg s$$^{-2}$$.

B. Pressure

Pressure = Force per unit area = $$\frac{\text{N}}{\text{m}^2} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{m}^2} = \text{kg m}^{-1}\text{s}^{-2}$$

This matches III: kg m$$^{-1}$$ s$$^{-2}$$.

C. Viscosity (Dynamic)

Viscosity = Pressure $$\times$$ Time = Pa $$\cdot$$ s = kg m$$^{-1}$$ s$$^{-2}$$ $$\times$$ s = kg m$$^{-1}$$ s$$^{-1}$$

This matches I: kg m$$^{-1}$$ s$$^{-1}$$.

D. Impulse

Impulse = Force $$\times$$ Time = N $$\cdot$$ s = kg $$\cdot$$ m/s$$^2$$ $$\times$$ s = kg m s$$^{-1}$$

This matches II: kg m s$$^{-1}$$.

Summary: A-IV, B-III, C-I, D-II

The correct answer is Option B.

We need to find the dimensions of each quantity in List I and match with List II.

A. Torque:

Torque = Force × Distance = $$MLT^{-2} \times L = ML^2T^{-2}$$

This matches with II.

B. Stress:

Stress = Force / Area = $$\frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$$

This matches with IV.

C. Pressure gradient:

Pressure gradient = Pressure / Distance = $$\frac{ML^{-1}T^{-2}}{L} = ML^{-2}T^{-2}$$

This matches with I.

D. Coefficient of viscosity:

From Newton's law of viscosity: $$F = \eta A \frac{dv}{dx}$$

$$\eta = \frac{F}{A} \times \frac{dx}{dv} = \frac{MLT^{-2}}{L^2} \times \frac{L}{LT^{-1}} = ML^{-1}T^{-1}$$

This matches with III.

Therefore: A-II, B-IV, C-I, D-III

The correct answer is Option 3.

Matching physical quantities with their dimensional formulae:

A. Young's Modulus (Y): Stress/Strain = (Force/Area) = [MLT⁻²]/[L²] = [ML⁻¹T⁻²] → III

B. Coefficient of Viscosity (η): [ML⁻¹T⁻¹] → I

C. Planck's Constant (h): E = hν → h = E/ν = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹] → II

D. Work Function (φ): Energy = [ML²T⁻²] → IV

Matching: A-III, B-I, C-II, D-IV

The correct answer is Option 2.

We need to find the dimensional formula for each physical quantity and match with the given list.

A. Pressure gradient:

Pressure gradient = $$\frac{\text{Pressure}}{\text{Distance}}$$

Dimension of Pressure = $$[M^1L^{-1}T^{-2}]$$

Dimension of Pressure gradient = $$\frac{[M^1L^{-1}T^{-2}]}{[L]} = [M^1L^{-2}T^{-2}]$$

This matches with III.

B. Energy density:

Energy density = $$\frac{\text{Energy}}{\text{Volume}}$$

Dimension of Energy = $$[M^1L^2T^{-2}]$$

Dimension of Energy density = $$\frac{[M^1L^2T^{-2}]}{[L^3]} = [M^1L^{-1}T^{-2}]$$

This matches with II.

C. Electric Field:

Electric Field = $$\frac{\text{Force}}{\text{Charge}}$$

Dimension of Force = $$[M^1L^1T^{-2}]$$

Dimension of Charge = $$[A^1T^1]$$

Dimension of Electric Field = $$\frac{[M^1L^1T^{-2}]}{[A^1T^1]} = [M^1L^1T^{-3}A^{-1}]$$

This matches with IV.

D. Latent heat:

Latent heat = $$\frac{\text{Heat Energy}}{\text{Mass}}$$

Dimension of Latent heat = $$\frac{[M^1L^2T^{-2}]}{[M^1]} = [M^0L^2T^{-2}]$$

This matches with I.

Therefore: A-III, B-II, C-IV, D-I

The correct answer is Option 3.

Planck’s constant h 

$$E=h\nu$$

$$[h]=\frac{[E]}{[\nu]}=\frac{ML^2T^{-2}}{T^{-1}}=[ML^2T^{-1}]$$

So, (A) matches with (III)

Stopping potential Vs:

Potential difference has dimensions:

$$[V]=\frac{\text{work}}{\text{charge}}=\frac{ML^2T^{-2}}{AT}=[ML^2T^{-3}A^{-1}]$$

So, (B) matches with (IV)

Work function $$\phi$$:

Work function is energy:

$$[\phi]=[ML^2T^{-2}]$$

So, (C) matches with (I)

Momentum p:

$$p=mv$$

$$[p]=M\cdot LT^{-1}=[MLT^{-1}]$$

So, (D) matches with (II)

Final matching:

(A)−(III), (B)−(IV), (C)−(I), (D)−(II)

We are given the equation of state $$\left(P + \frac{a}{V^2}\right)(V - b) = RT$$.

Since $$b$$ is subtracted from $$V$$, it must have the same dimensions as volume, so $$[b] = [V] = L^3$$.

Now, since $$\frac{a}{V^2}$$ is added to $$P$$, it must have the same dimensions as pressure. This gives us

$$[a] = [P][V^2] = ML^{-1}T^{-2} \times L^6 = ML^5T^{-2}$$

We can now find the dimensional formula of $$\frac{b^2}{a}$$:

$$\left[\frac{b^2}{a}\right] = \frac{L^6}{ML^5T^{-2}} = M^{-1}L^1T^{2}$$

This is the dimensional formula of the reciprocal of pressure (since compressibility is defined as $$\frac{1}{\text{Bulk modulus}}$$ and has dimensions $$M^{-1}LT^2$$).

Hence, the correct answer is Compressibility.

We need to find the dimensions of $$A$$ and $$B$$ in the equation $$(x - At)^2 + \left(y - \frac{t}{B}\right)^2 = a^2$$, where the dimension of $$t$$ is $$[T^{-1}]$$.

Principle of Dimensional Homogeneity: Each term in an equation must have the same dimensions.

Finding dimension of A:

The term $$At$$ must have the same dimension as $$x$$, which is $$[L]$$.

$$[At] = [L]$$

$$[A][T^{-1}] = [L]$$

$$[A] = [LT]$$

Finding dimension of B:

The term $$\frac{t}{B}$$ must have the same dimension as $$y$$, which is $$[L]$$.

$$\left[\frac{t}{B}\right] = [L]$$

$$\frac{[T^{-1}]}{[B]} = [L]$$

$$[B] = \frac{[T^{-1}]}{[L]} = [L^{-1}T^{-1}]$$

Therefore, $$A = [LT]$$ and $$B = [L^{-1}T^{-1}]$$.

The correct answer is Option 2.

We have two resistances $$R_1 = (10 \pm 0.5)\;\Omega$$ and $$R_2 = (15 \pm 0.5)\;\Omega$$ connected in parallel, and we need to find the percentage error in the equivalent resistance.

For resistances in parallel, the equivalent resistance is:

$$R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\;\Omega$$

Now, to find the error propagation, we start from $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$. Differentiating both sides and taking absolute values (since errors always add), the maximum absolute error in $$R_{eq}$$ is:

$$\Delta R_{eq} = R_{eq}^2 \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)$$

This follows because differentiating $$R_{eq}^{-1}$$ with respect to $$R_1$$ gives $$-R_1^{-2}$$, and similarly for $$R_2$$. Substituting the values with $$R_{eq}^2 = 36$$:

$$\Delta R_{eq} = 36 \left(\frac{0.5}{100} + \frac{0.5}{225}\right) = 36 \left(0.005 + 0.002222\right) = 36 \times 0.007222 = 0.26\;\Omega$$

So the percentage error is:

$$\%\;\text{error} = \frac{\Delta R_{eq}}{R_{eq}} \times 100 = \frac{0.26}{6} \times 100 = 4.33\%$$

Hence, the correct answer is Option 4.

Given: $$\vec{B} - \vec{A} = 2\hat{j}$$

$$\vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$$

$$\vec{B} = \vec{A} + 2\hat{j} = 2\hat{i} + 3\hat{j} + 2\hat{k} + 2\hat{j} = 2\hat{i} + 5\hat{j} + 2\hat{k}$$

$$|\vec{B}| = \sqrt{2^2 + 5^2 + 2^2} = \sqrt{4 + 25 + 4} = \sqrt{33}$$

The dimensional formulae of the physical quantities are:

• Young’s modulus (stress): $$Y \rightarrow M^{1}L^{-1}T^{-2}$$
• Speed of light: $$c \rightarrow L^{1}T^{-1}$$
• Planck’s constant: $$h \rightarrow M^{1}L^{2}T^{-1}$$
• Gravitational constant: $$G \rightarrow M^{-1}L^{3}T^{-2}$$

Assume $$Y = c^{\alpha}h^{\beta}G^{\gamma}$$. Equate the dimensions on both sides.

Step 1: Write combined dimensions on the right-hand side.
$$ \begin{aligned} c^{\alpha}&:&& L^{\alpha}T^{-\alpha}\\ h^{\beta}&:&& M^{\beta}L^{2\beta}T^{-\beta}\\ G^{\gamma}&:&& M^{-\gamma}L^{3\gamma}T^{-2\gamma} \end{aligned} $$ Multiplying, $$ M^{\,\beta-\gamma}\;L^{\,\alpha+2\beta+3\gamma}\;T^{-\alpha-\beta-2\gamma} $$

Step 2: Equate powers of $$M,\,L,\,T$$ with those of $$Y$$.

$$ \begin{aligned} \text{M:}&\quad \beta-\gamma &=&\; 1\; \quad -(1)\\ \text{L:}&\quad \alpha+2\beta+3\gamma &=&\; -1\quad -(2)\\ \text{T:}&\quad -\alpha-\beta-2\gamma &=&\; -2\quad -(3) \end{aligned} $$

Step 3: Solve the simultaneous equations.

From $$(1):\; \beta = 1+\gamma$$.

Substitute in $$(3):\; -\alpha-(1+\gamma)-2\gamma = -2$$
$$\Rightarrow -\alpha-1-3\gamma = -2$$
$$\Rightarrow \alpha + 3\gamma = 1 \quad -(4)$$

Substitute $$\beta = 1+\gamma$$ in $$(2):$$
$$\alpha + 2(1+\gamma)+3\gamma = -1$$
$$\Rightarrow \alpha + 2 + 5\gamma = -1$$
$$\Rightarrow \alpha + 5\gamma = -3 \quad -(5)$$

Subtract $$(4)$$ from $$(5):$$
$$(\alpha + 5\gamma) - (\alpha + 3\gamma) = -3 - 1$$
$$2\gamma = -4 \;\Rightarrow\; \gamma = -2$$

Insert $$\gamma = -2$$ into $$(4):$$
$$\alpha + 3(-2) = 1 \;\Rightarrow\; \alpha = 7$$

Insert $$\gamma = -2$$ into $$\beta = 1+\gamma$$:
$$\beta = 1 - 2 = -1$$

Step 4: State the exponents.
$$\alpha = 7,\; \beta = -1,\; \gamma = -2$$

Hence, the correct choice is:
Option A which is: $$\alpha = 7,\; \beta = -1,\; \gamma = -2$$

We need to express density in terms of force (F), velocity (V), and time (T).

Dimensions: $$[F] = MLT^{-2}$$, $$[V] = LT^{-1}$$, $$[T] = T$$

From these:

$$[L] = [V][T] = LT^{-1} \cdot T = L$$

$$[M] = \frac{[F]}{[L][T]^{-2}} = \frac{[F]}{[V][T] \cdot [T]^{-2}} = \frac{[F]}{[V][T]^{-1}} = [F][V]^{-1}[T]$$

Density $$= [M][L]^{-3}$$:

$$[\rho] = [F][V]^{-1}[T] \cdot ([V][T])^{-3} = [F][V]^{-1}[T] \cdot [V]^{-3}[T]^{-3}$$

$$= [F][V]^{-4}[T]^{-2}$$

The dimensional formula of density is $$FV^{-4}T^{-2}$$.

A. Spring constant (k): $$[F] = [k][x]$$ → $$[k] = [MT^{-2}]$$ → (II)

B. Angular speed (ω): $$[\omega] = [T^{-1}]$$ → (I)

C. Angular momentum (L): $$[L] = [ML^2T^{-1}]$$ → (IV)

D. Moment of Inertia (I): $$[I] = [ML^2]$$ → (III)

Answer: A-II, B-I, C-IV, D-III

Using dimensional analysis, the frequency $$\nu = r^a \rho^b s^c$$.

Dimensions of each quantity:

$$[\nu] = T^{-1}$$

$$[r] = L$$

$$[\rho] = ML^{-3}$$

$$[s] = MT^{-2}$$ (surface tension = force per unit length)

Writing the dimensional equation:

$$T^{-1} = L^a \cdot (ML^{-3})^b \cdot (MT^{-2})^c$$

$$T^{-1} = M^{b+c} \cdot L^{a-3b} \cdot T^{-2c}$$

Comparing powers of each dimension:

For $$M$$: $$0 = b + c$$ ... (i)

For $$L$$: $$0 = a - 3b$$ ... (ii)

For $$T$$: $$-1 = -2c$$ ... (iii)

From (iii): $$c = \frac{1}{2}$$

From (i): $$b = -c = -\frac{1}{2}$$

From (ii): $$a = 3b = 3 \times \left(-\frac{1}{2}\right) = -\frac{3}{2}$$

Therefore, $$(a, b, c) = \left(-\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}\right)$$

The correct answer is Option 1.

Using dimensional analysis: $$v = \lambda^a g^b \rho^c$$

Dimensions: $$[v] = [LT^{-1}]$$, $$[\lambda] = [L]$$, $$[g] = [LT^{-2}]$$, $$[\rho] = [ML^{-3}]$$

$$[LT^{-1}] = [L]^a [LT^{-2}]^b [ML^{-3}]^c = M^c L^{a+b-3c} T^{-2b}$$

Equating powers:

M: $$c = 0$$

T: $$-1 = -2b$$, so $$b = \frac{1}{2}$$

L: $$1 = a + b - 3c = a + \frac{1}{2}$$, so $$a = \frac{1}{2}$$

$$a = \frac{1}{2}, b = \frac{1}{2}, c = 0$$

This matches option 4.

Solution :

Refractive index of glass slab is :

$$\mu = \frac{\text{Real thickness}}{\text{Apparent thickness}}$$

Given :

$$t_r = 5.25\text{ mm}$$

$$t_a = 5.00\text{ mm}$$

Therefore,

$$\mu = \frac{5.25}{5.00}$$

$$= 1.05$$

Main scale :

$$20\text{ divisions} = 1\text{ cm}$$

Hence,

$$1\text{ MSD} = \frac{1}{20}\text{ cm}$$

$$= 0.05\text{ cm}$$

$$= 0.5\text{ mm}$$

Given,

$$50\text{ VSD} = 49\text{ MSD}$$

Least count :

$$LC = 1\text{ MSD} - 1\text{ VSD}$$

$$= \frac{1}{50}\text{ MSD}$$

$$= \frac{0.5}{50}\text{ mm}$$

$$= 0.01\text{ mm}$$

Fractional error in refractive index :

$$\frac{\Delta \mu}{\mu} = \frac{\Delta t_r}{t_r} + \frac{\Delta t_a}{t_a}$$

Substituting values :

$$\frac{\Delta \mu}{1.05} = \frac{0.01}{5.25} + \frac{0.01}{5.00}$$

$$\approx 0.0039$$

Therefore,

$$\Delta \mu = 1.05 \times 0.0039$$

$$\approx 0.0041$$

$$= 4.1 \times 10^{-3}$$

Given uncertainty is of the form :

$$\frac{x}{10} \times 10^{-3}$$

Comparing,

$$\frac{x}{10} = 4.1$$

$$x = 41$$

Final Answer :

$$41$$

Zero error: +0.06 cm (positive, since zero of vernier is to the right of main scale zero, with 6th division coinciding).

Observed reading: MSR = 3.2 cm, VSR = 4 × 0.01 = 0.04 cm. Total observed = 3.24 cm.

Corrected reading = 3.24 - 0.06 = 3.18 cm.

The correct answer is Option 2: 3.18 cm.

The focal length obtained from the measured object-distance $$u$$ and image-distance $$v$$ is found with the thin-lens formula

$$\frac{1}{f}= \frac{1}{u}+\frac{1}{v}\qquad\Longrightarrow\qquad f=\frac{uv}{u+v}\,.$$

The measured values and their maximum absolute errors are
    $$u = 10\;\text{cm},\; \Delta u = \pm 0.1\;\text{cm}$$
    $$v = 20\;\text{cm},\; \Delta v = \pm 0.2\;\text{cm}$$

To find the error in $$f$$, use logarithmic differentiation (propagation of errors).

Taking natural logarithm of $$f = \dfrac{uv}{u+v}$$:

$$\ln f = \ln u + \ln v - \ln(u+v)$$

Differentiating:

$$\frac{\Delta f}{f}= \frac{\Delta u}{u}+\frac{\Delta v}{v}-\frac{\Delta u+\Delta v}{u+v} \quad -(1)$$

Because $$\Delta u$$ and $$\Delta v$$ may be positive or negative, we are interested in the maximum possible fractional error. Equation $$(1)$$ can be rewritten as

$$\frac{\Delta f}{f}= \Delta u\!\left(\frac{1}{u}-\frac{1}{u+v}\right)+\Delta v\!\left(\frac{1}{v}-\frac{1}{u+v}\right).$$

Calculate the two coefficients once for the given $$u$$ and $$v$$:

$$\frac{1}{u}-\frac{1}{u+v}= \frac{1}{10}-\frac{1}{30}= \frac{2}{30}= \frac{1}{15}=0.0667,$$
$$\frac{1}{v}-\frac{1}{u+v}= \frac{1}{20}-\frac{1}{30}= \frac{1}{60}=0.0167.$$

Hence

$$\frac{\Delta f}{f}= 0.0667\,\Delta u + 0.0167\,\Delta v \quad -(2)$$

The numerical coefficients are positive, so the maximum fractional error occurs when both $$\Delta u$$ and $$\Delta v$$ take their maximum positive values (+0.1 cm and +0.2 cm):

$$\left|\frac{\Delta f}{f}\right|_{\text{max}} = 0.0667(0.1) + 0.0167(0.2) = 0.00667 + 0.00333 = 0.0100.$$

Thus the maximum percentage error in the focal length is

$$n = 0.0100 \times 100 = 1\%.$$

Therefore, the error in the determination of the focal length of the lens is 1 %.

We know that the speed of light in vacuum is:

$$ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} $$

Therefore:

$$ \frac{1}{\mu_0 \varepsilon_0} = c^2 $$

The dimension of $$c$$ (speed) is $$[LT^{-1}]$$.

Therefore, the dimension of $$\frac{1}{\mu_0\varepsilon_0}$$ is:

$$ [c^2] = [L^2T^{-2}] $$

The correct answer is $$L^2 T^{-2}$$.

A screw gauge has 100 circular scale divisions and a pitch of 0.5 mm, meaning the main scale moves 0.5 mm per complete rotation. The least count is $$\frac{0.5}{100} = 0.005$$ mm.

Because the zero of the circular scale lies 6 divisions below the reference line, there is a positive zero error of $$+6 \times 0.005 = +0.03$$ mm.

When measuring the wire, the main scale reading is 4 divisions $$\times$$ 0.5 mm = 2.00 mm, and the circular scale reading is 46 $$\times$$ 0.005 = 0.23 mm. Thus the observed reading is 2.00 + 0.23 = 2.23 mm, and the corrected reading is 2.23 - 0.03 = 2.20 mm, which can also be written as $$= 220 \times 10^{-2}$$ mm.

Hence, the diameter of the wire is $$\boxed{220} \times 10^{-2}$$ mm.

We need to find the maximum percentage error in the measurement of density of the silver wire.

The density is given by:

$$\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}$$

The percentage error in density is:

$$\frac{\Delta \rho}{\rho} \times 100 = \frac{\Delta m}{m} \times 100 + 2 \times \frac{\Delta r}{r} \times 100 + \frac{\Delta l}{l} \times 100$$

Substituting the given values:

$$\frac{\Delta m}{m} \times 100 = \frac{0.006}{0.6} \times 100 = 1\%$$

$$2 \times \frac{\Delta r}{r} \times 100 = 2 \times \frac{0.005}{0.5} \times 100 = 2\%$$

$$\frac{\Delta l}{l} \times 100 = \frac{0.04}{4} \times 100 = 1\%$$

Therefore, the maximum percentage error in density:

$$\frac{\Delta \rho}{\rho} \times 100 = 1\% + 2\% + 1\% = 4\%$$

The correct answer is Option C.

We need to find the net accuracy of torque measured by a torque meter calibrated with 5% accuracy in mass, length, and time.

The dimensional formula of torque is derived from torque = force $$\times$$ distance = (mass $$\times$$ acceleration) $$\times$$ distance, which yields $$\tau = M L^2 T^{-2}$$.

For a quantity $$\tau = M^a L^b T^c$$, the maximum percentage error is given by $$\frac{\Delta \tau}{\tau} \times 100 = a\left(\frac{\Delta M}{M}\right) \times 100 + b\left(\frac{\Delta L}{L}\right) \times 100 + |c|\left(\frac{\Delta T}{T}\right) \times 100$$.

Here $$a = 1$$, $$b = 2$$, $$|c| = 2$$, and each accuracy is $$5\%$$, so $$\frac{\Delta \tau}{\tau} \times 100 = 1 \times 5\% + 2 \times 5\% + 2 \times 5\% = 5\% + 10\% + 10\% = 25\%$$.

The correct answer is Option B: $$25\%$$.

We need to find the dimensional formula of the constant $$\beta$$ from the given expression for energy density.

The given equation is:

$$u = \alpha \beta \sin\left(\frac{\alpha x}{kt}\right)$$

1. Analyze the Argument of the Sine Function

Trigonometric functions like sine take dimensionless arguments. Therefore, the term inside the parenthesis must be dimensionless ($$[M^0 L^0 T^0]$$):

$$\left[\frac{\alpha x}{kt}\right] = [1]$$

Let's find the dimensions of the variables involved:

• Displacement ($$x$$) = $$[L]$$
• The product of the Boltzmann constant and temperature ($$kt$$) represents thermal energy, which has the dimensions of energy/work:

  $$[kt] = [M L^2 T^{-2}]$$

Equating the dimensions to find the dimensional formula for $$\alpha$$:

$$[\alpha] \cdot \frac{[L]}{[M L^2 T^{-2}]} = [1]$$

$$[\alpha] = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$$

2. Analyze the Dimensions of Energy Density ($u$)

Energy density is defined as energy per unit volume:

$$[u] = \frac{[\text{Energy}]}{[\text{Volume}]} = \frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]$$

3. Determine the Dimensional Formula of $$\beta$$

Since the sine function itself is completely dimensionless, the dimensions of the left side of the equation must equal the dimensions of the coefficients on the right side:

$$[u] = [\alpha] \cdot [\beta]$$

Substitute the dimensions of $$u$$ and $$\alpha$$ to isolate $$[\beta]$$:

$$[M L^{-1} T^{-2}] = [M L T^{-2}] \cdot [\beta]$$

$$[\beta] = \frac{[M L^{-1} T^{-2}]}{[M L T^{-2}]}$$

Simplifying the terms by canceling mass ($$M$$) and time ($$T$$):

$$[\beta] = [L^{-1 - 1}] = [L^{-2}]$$

In standard base form, this is written as:

$$[\beta] = [M^0 L^{-2} T^0]$$

Correction Note

(Option D) contains a sign typo regarding the exponent of length ($$L$$). Following proper dimensional analysis rules, the true mathematical output is $$[M^0 L^{-2} T^0]$$.

We need to check both the Assertion and the Reason.

Checking Assertion A: Dimension of P × t

Pressure $$P$$ has dimensions: $$[P] = [ML^{-1}T^{-2}]$$

Time $$t$$ has dimensions: $$[t] = [T]$$

So, $$[P \times t] = [ML^{-1}T^{-2}] \times [T] = [ML^{-1}T^{-1}]$$

Dimension of coefficient of viscosity ($$\eta$$):

From the formula: $$\eta = \frac{F}{A \times \text{velocity gradient}}$$

Velocity gradient = $$\frac{dv}{dx}$$ has dimensions $$[T^{-1}]$$

So, $$[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]} = \frac{[MLT^{-2}]}{[L^2T^{-1}]} = [ML^{-1}T^{-1}]$$

Since $$[P \times t] = [ML^{-1}T^{-1}] = [\eta]$$, the Assertion is TRUE.

Checking Reason R:

The Reason states: Coefficient of viscosity = $$\frac{\text{Force}}{\text{Velocity gradient}}$$

The correct formula is: $$F = \eta A \frac{dv}{dx}$$, which gives $$\eta = \frac{F}{A \times \frac{dv}{dx}}$$

The Reason is missing the area $$A$$ in the denominator. So the Reason is FALSE.

Since Assertion A is true but Reason R is false, the correct answer is Option C.

We are given that the efficiency of a Carnot engine is expressed as $$\eta = \frac{\alpha\beta}{\sin\theta} \log_e\frac{\beta x}{kT}$$. Since efficiency $$\eta$$ is dimensionless and $$\sin\theta$$ is also dimensionless, the product $$\alpha\beta$$ must be dimensionless. Also, the argument of the logarithm $$\frac{\beta x}{kT}$$ must be dimensionless.

Since $$\alpha\beta$$ is dimensionless, $$\alpha$$ and $$\beta$$ must have reciprocal dimensions: $$[\alpha] = [\beta]^{-1}$$.

Now, from the condition that $$\frac{\beta x}{kT}$$ is dimensionless, we get $$[\beta] = \frac{[kT]}{[x]}$$. The Boltzmann constant $$k$$ has dimensions of energy per temperature, i.e., $$[k] = M L^2 T^{-2} K^{-1}$$. So $$[kT] = M L^2 T^{-2}$$ (energy). Since $$x$$ has dimensions of length $$[L]$$, we get $$[\beta] = \frac{M L^2 T^{-2}}{L} = M L T^{-2}$$, which is the dimension of force.

So the dimensions of $$\beta$$ are the same as that of force. This makes Option A correct.

Now, since $$[\alpha] = [\beta]^{-1} = M^{-1} L^{-1} T^{2}$$, let us check $$[\alpha^{-1} x]$$. We have $$[\alpha^{-1}] = M L T^{-2}$$ and $$[x] = L$$, so $$[\alpha^{-1} x] = M L^2 T^{-2}$$, which is the dimension of energy. This makes Option B correct.

For Option C, since $$\eta$$ is dimensionless and $$\sin\theta$$ is dimensionless, $$\eta^{-1}\sin\theta$$ is dimensionless. We already established that $$\alpha\beta$$ is dimensionless. So $$[\eta^{-1}\sin\theta] = [\alpha\beta]$$, both being dimensionless. This makes Option C correct.

For Option D, we found $$[\alpha] = M^{-1} L^{-1} T^{2}$$ and $$[\beta] = M L T^{-2}$$. These are clearly not the same dimensions (in fact they are reciprocals of each other). So Option D is incorrect.

Hence, the correct answer is Option D.

We need to check whether the expression $$T = K\sqrt{\dfrac{\rho r^3}{S^{3/2}}}$$ is dimensionally correct, where $$\rho$$ is density, $$r$$ is radius, $$S$$ is surface tension, and $$K$$ is dimensionless.

The dimensions of each quantity are: density $$[\rho] = ML^{-3}$$, radius $$[r] = L$$, and surface tension $$[S] = MT^{-2}$$ (force per unit length).

Now, the expression inside the square root has dimensions:

$$\dfrac{[\rho][r]^3}{[S]^{3/2}} = \dfrac{ML^{-3} \cdot L^3}{(MT^{-2})^{3/2}} = \dfrac{M}{M^{3/2}T^{-3}} = \dfrac{T^3}{M^{1/2}}$$

Taking the square root: $$\sqrt{\dfrac{T^3}{M^{1/2}}} = \dfrac{T^{3/2}}{M^{1/4}}$$

The dimension of the R.H.S. is $$\dfrac{T^{3/2}}{M^{1/4}}$$, which is NOT equal to $$[T] = T$$.

So, the Assertion (A) is false — the given expression is NOT dimensionally correct.

The Reason (R) states that using dimensional analysis, the R.H.S. has different dimensions than the time period. Since we showed that the R.H.S. has dimensions $$T^{3/2}M^{-1/4}$$ instead of $$T$$, the Reason is true.

Hence, the correct answer is Option D: (A) is false but (R) is true.

We need to identify which pair of physical quantities have the same dimensions.

For option A, velocity gradient and decay constant are considered. Velocity gradient is defined as the rate of change of velocity with respect to distance: $$\text{Velocity gradient} = \frac{dv}{dx}$$ Its dimensions are: $$\frac{[LT^{-1}]}{[L]} = [T^{-1}]$$ Decay constant $$\lambda$$ appears in the radioactive decay law $$N = N_0 e^{-\lambda t}$$ and since $$\lambda t$$ must be dimensionless, $$[\lambda] = [T^{-1}]$$ Both have dimensions $$[T^{-1}]$$ and therefore match.

For option B, angular frequency $$\omega$$ has dimensions $$[T^{-1}]$$, while angular momentum $$L = I\omega$$ has dimensions $$[ML^2T^{-1}]$$. These are different.

For option C, wave number $$k = 1/\lambda$$ has dimensions $$[L^{-1}]$$, while Avogadro number $$N_A$$ has dimensions $$[\text{mol}^{-1}]$$. These are different.

For option D, Wien's displacement constant $$b$$ has dimensions $$[LK]$$ (from $$\lambda_{max} T = b$$), while Stefan's constant $$\sigma$$ has dimensions $$[MT^{-3}K^{-4}]$$ (from $$P = \sigma A T^4$$). These are different.

Therefore, the correct answer is Option A: Velocity gradient and decay constant both have dimensions $$[T^{-1}]$$.

We need to identify the pair of physical quantities that have different dimensions.

Option A: Wave number and Rydberg's constant

Wave number $$k = \frac{1}{\lambda}$$, so its dimension is $$[L^{-1}]$$.

Rydberg's constant $$R$$ also has dimension $$[L^{-1}]$$ (it appears in $$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$).

Both have the same dimensions.

Option B: Stress and Coefficient of elasticity

Stress $$= \frac{F}{A}$$, dimension $$= [ML^{-1}T^{-2}]$$.

Coefficient of elasticity (Young's modulus) $$= \frac{\text{Stress}}{\text{Strain}}$$. Since strain is dimensionless, its dimension is also $$[ML^{-1}T^{-2}]$$.

Both have the same dimensions.

Option C: Coercivity and Magnetisation

Both coercivity and magnetisation have the dimension of magnetic field intensity $$H$$, which is $$[AL^{-1}]$$.

Both have the same dimensions.

Option D: Specific heat capacity and Latent heat

Specific heat capacity $$c = \frac{Q}{m \Delta T}$$, dimension $$= \frac{[ML^2T^{-2}]}{[M][K]} = [L^2T^{-2}K^{-1}]$$.

Latent heat $$L = \frac{Q}{m}$$, dimension $$= \frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}]$$.

These have different dimensions — specific heat capacity has an extra factor of $$K^{-1}$$.

Therefore, the correct answer is Option D.

We are given $$Z = \frac{A^2 B^3}{C^4}$$ and need to find the relative error in $$Z$$.

Apply the error propagation rule: For a quantity expressed as a product or quotient of powers, the relative error is found by taking the sum of the absolute values of (power multiplied by relative error) for each variable.

If $$Z = \frac{A^a \cdot B^b}{C^c}$$, then the relative error in $$Z$$ is:

$$\frac{\Delta Z}{Z} = a \cdot \frac{\Delta A}{A} + b \cdot \frac{\Delta B}{B} + c \cdot \frac{\Delta C}{C}$$

Substitute the powers from the given expression: Here $$a = 2$$, $$b = 3$$, and $$c = 4$$. Substituting these values:

$$\frac{\Delta Z}{Z} = 2 \cdot \frac{\Delta A}{A} + 3 \cdot \frac{\Delta B}{B} + 4 \cdot \frac{\Delta C}{C}$$

Note that in error propagation, all terms are added with positive signs because errors always accumulate in the worst case, regardless of whether the variable appears in the numerator or denominator.

The correct answer is Option A.

Torque:

$$\tau=r\times F$$

Unit:

$$\text{N}\cdot\text{m}$$

So, (A) matches with (III)

Stress:

$$\text{Stress}=\frac{\text{Force}}{\text{Area}}$$

$$=\frac{\text{N}}{\text{m}^2}=\text{N m}^{-2}$$

So, (B) matches with (IV)

Latent heat:

Latent heat is heat per unit mass:

$$L=\frac{Q}{m}$$

$$=\frac{\text{J}}{\text{kg}}$$

So, (C) matches with (II)

Power:

$$P=\frac{\text{Work}}{\text{Time}}=\frac{\text{J}}{\text{s}}=\text{W}$$

Also,

$$\text{W}=\text{N m s}^{-1}$$

So, (D) matches with (I)

Final matching:

(A)−(III), (B)−(IV), (C)−(II), (D)−(I)

We need to find the dimensions of mutual inductance.

The emf induced in a coil is related to mutual inductance by:

$$\varepsilon = M \frac{dI}{dt}$$

So: $$M = \frac{\varepsilon \cdot dt}{dI}$$

The dimension of emf ($$\varepsilon$$) is the same as voltage:

$$[\varepsilon] = [V] = ML^2T^{-3}A^{-1}$$

The dimension of $$\frac{dt}{dI}$$ is:

$$\left[\frac{dt}{dI}\right] = \frac{T}{A} = TA^{-1}$$

Therefore, the dimension of mutual inductance is:

$$[M] = ML^2T^{-3}A^{-1} \times TA^{-1} = ML^2T^{-2}A^{-2}$$

Hence, the correct answer is Option B.

We need to find the dimensions of $$\frac{B^2}{\mu_0}$$.

From the Lorentz force equation $$F = qvB$$, we get:

$$[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{AT \cdot LT^{-1}} = MT^{-2}A^{-1}$$

Meanwhile, using $$B = \mu_0 \frac{I}{2\pi r}$$ leads to

$$[\mu_0] = \frac{[B][L]}{[I]} = \frac{MT^{-2}A^{-1} \cdot L}{A} = MLT^{-2}A^{-2}$$

Combining these results gives the dimension of $$\frac{B^2}{\mu_0}$$ as

$$\left[\frac{B^2}{\mu_0}\right] = \frac{[MT^{-2}A^{-1}]^2}{MLT^{-2}A^{-2}} = \frac{M^2T^{-4}A^{-2}}{MLT^{-2}A^{-2}} = ML^{-1}T^{-2}$$

This matches the dimension of pressure or energy density, consistent with the interpretation of $$\frac{B^2}{2\mu_0}$$ as magnetic energy density. Therefore, the correct answer is Option C: $$[ML^{-1}T^{-2}]$$.

We are given that the distance of the Sun from Earth is $$R = 1.5 \times 10^{11}$$ m and its angular diameter is $$\theta = 2000''$$ (arc-seconds).

Convert angular diameter to radians.

We know that $$1'' = \frac{1}{206265}$$ rad.

$$\theta = 2000'' = \frac{2000}{206265} \approx 9.7 \times 10^{-3}$$ rad

Calculate the diameter of the Sun.

For small angles, the diameter $$D$$ is related to distance $$R$$ and angular diameter $$\theta$$ by:

$$D = R \times \theta$$

$$D = 1.5 \times 10^{11} \times 9.7 \times 10^{-3}$$

$$D = 1.45 \times 10^{9}$$ m

Therefore, the diameter of the Sun is $$1.45 \times 10^{9}$$ m.

The correct answer is Option C.

The heat dissipated in a resistor is given by:

$$H = I^2 R t$$

Taking the logarithm of both sides and differentiating yields the formula for the percentage error in H:

$$\frac{\Delta H}{H} \times 100 = 2\left(\frac{\Delta I}{I} \times 100\right) + \frac{\Delta R}{R} \times 100 + \frac{\Delta t}{t} \times 100$$

Since $$\frac{\Delta R}{R} \times 100 = 1\%$$, $$\frac{\Delta I}{I} \times 100 = 2\%$$, and $$\frac{\Delta t}{t} \times 100 = 3\%$$, substituting these values gives:

$$\frac{\Delta H}{H} \times 100 = 2(2\%) + 1\% + 3\% = 4\% + 1\% + 3\% = 8\%$$

Therefore, the maximum percentage error in the dissipated heat is $$8\%$$, which corresponds to Option D.

We are given the relations between velocity and acceleration in two systems of units: $$v_2 = \frac{n}{m^2} v_1$$ and $$a_2 = \frac{a_1}{mn}$$.

Velocity has dimensions $$[LT^{-1}]$$ and acceleration has dimensions $$[LT^{-2}]$$, so $$\frac{L_2}{T_2} = \frac{n}{m^2} \cdot \frac{L_1}{T_1}$$ ... (i) and $$\frac{L_2}{T_2^2} = \frac{1}{mn} \cdot \frac{L_1}{T_1^2}$$ ... (ii).

Dividing equation (i) by (ii) yields $$\frac{L_2/T_2}{L_2/T_2^2} = \frac{\frac{n}{m^2} \cdot \frac{L_1}{T_1}}{\frac{1}{mn} \cdot \frac{L_1}{T_1^2}}$$ and thus $$T_2 = \frac{n}{m^2} \times mn \times T_1 = \frac{n^2}{m} T_1$$.

Substituting back into equation (i) gives $$L_2 = \frac{n}{m^2} \cdot \frac{L_1 \cdot T_2}{T_1} = \frac{n}{m^2} \cdot L_1 \cdot \frac{n^2/m}{1} = \frac{n^3}{m^3} L_1$$.

Therefore, $$L_2 = \frac{n^3}{m^3} L_1$$ and $$T_2 = \frac{n^2}{m} T_1$$.

The correct answer is Option A.

We have a travelling microscope whose main scale has 20 divisions per cm. This means each main scale division (MSD) is $$\dfrac{1}{20}$$ cm $$= 0.05$$ cm.

We are told that 25 Vernier scale divisions (VSD) coincide with 24 main scale divisions. Therefore, $$25 \text{ VSD} = 24 \text{ MSD}$$, which gives $$1 \text{ VSD} = \dfrac{24}{25} \text{ MSD}$$.

The least count (LC) of a Vernier instrument is the difference between one main scale division and one Vernier scale division: $$\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ MSD} - \dfrac{24}{25}\text{ MSD} = \dfrac{1}{25}\text{ MSD}$$.

Now substituting the value of one MSD: $$\text{LC} = \dfrac{1}{25} \times 0.05 \text{ cm} = \dfrac{0.05}{25} = 0.002 \text{ cm}$$.

Hence, the correct answer is Option C.

We need to express the coefficient of viscosity in terms of momentum (P), area (A), and time (T) as fundamental quantities.

The coefficient of viscosity $$\eta$$ is defined from Newton's law of viscosity: $$F = \eta A \dfrac{dv}{dx}$$, where $$F$$ is the viscous force, $$A$$ is the area, $$dv$$ is the velocity change, and $$dx$$ is the distance perpendicular to the flow. Therefore $$\eta = \dfrac{F}{A} \cdot \dfrac{dx}{dv}$$, and in SI dimensions $$[\eta] = \dfrac{[MLT^{-2}]}{[L^2]} \cdot \dfrac{[L]}{[LT^{-1}]} = [ML^{-1}T^{-1}].$$

The dimensions of momentum, area, and time are respectively $$[P] = [MLT^{-1}]$$, $$[A] = [L^2]$$, and $$[T] = [T].$$

Assuming a relation of the form $$[\eta] = [P]^a [A]^b [T]^c$$, substitution gives $$[ML^{-1}T^{-1}] = [MLT^{-1}]^a \cdot [L^2]^b \cdot [T]^c = [M^a \cdot L^{a+2b} \cdot T^{-a+c}].$$

Equating exponents for each fundamental dimension yields for mass $$a = 1$$, for length $$a + 2b = -1$$ which with $$a = 1$$ gives $$b = -1$$, and for time $$-a + c = -1$$ which with $$a = 1$$ gives $$c = 0$$.

Therefore $$[\eta] = P^1 A^{-1} T^0 = PA^{-1}T^0$$. Verification shows $$PA^{-1}T^0 = [MLT^{-1}][L^{-2}][1] = [ML^{-1}T^{-1}]$$, matching the dimensions of viscosity. Hence, the correct answer is Option A.

We need to find the dimensional formula of a physical quantity whose SI unit is Pascal-second (Pa·s).

Write the dimensional formula of Pascal.

Pascal is the unit of pressure:

$$[\text{Pa}] = \frac{[\text{Force}]}{[\text{Area}]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$$

Multiply by the dimension of time.

$$[\text{Pa} \cdot \text{s}] = ML^{-1}T^{-2} \times T = ML^{-1}T^{-1}$$

This is the dimensional formula of dynamic viscosity (coefficient of viscosity).

The correct answer is Option C.

We are given a screw gauge with pitch = 0.5 mm and 50 divisions on the circular scale.

Step 1: Find the Least Count (LC)

$$\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions}} = \dfrac{0.5}{50} = 0.01 \text{ mm}$$

Step 2: Find the diameter of the wire

$$d = \text{MSR} + \text{CSR} \times \text{LC} = 1.5 + 7 \times 0.01 = 1.57 \text{ mm}$$

Step 3: Calculate the curved surface area

The curved surface area of a cylinder is:

$$A = \pi d L$$

where $$d = 1.57 \text{ mm} = 0.157 \text{ cm}$$ and $$L = 6.8 \text{ cm}$$.

$$A = \pi \times 0.157 \times 6.8 = \pi \times 1.0676 = 3.3543 \text{ cm}^2$$

Step 4: Apply significant figures

The diameter (1.57 mm) has 3 significant figures, and the length (6.8 cm) has 2 significant figures. The result should be rounded to 2 significant figures:

$$A \approx 3.4 \text{ cm}^2$$

Therefore, the correct answer is Option B.

Given that 10 divisions of the Vernier scale equal 9 divisions of the main scale, and 1 main scale division (MSD) = 1 mm.

Since each Vernier scale division (VSD) is $$\tfrac{9}{10}$$ of an MSD, it follows that $$1 \text{ VSD} = \tfrac{9}{10} \text{ MSD} = 0.9 \text{ mm}.$$ Therefore the least count (LC) is the difference between one MSD and one VSD: $$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 - 0.9 = 0.1 \text{ mm}.$$

The zero of the Vernier scale is shifted to the left of the main scale zero, indicating a negative zero error. Because the 4th Vernier division coincides with the main scale, one finds $$\text{Zero error} = -(4 \times \text{LC}) = -(4 \times 0.1) = -0.4 \text{ mm}.$$

The main scale reading (MSR) is 30 mm (since the Vernier zero lies between 30 and 31), and the 6th Vernier division coincides, giving the Vernier scale reading (VSR). Hence $$\text{Observed reading} = \text{MSR} + \text{VSR} \times \text{LC} = 30 + 6 \times 0.1 = 30.6 \text{ mm}.$$

Subtracting the zero error from the observed reading yields the corrected measurement: $$\text{Corrected reading} = \text{Observed reading} - \text{Zero error} = 30.6 - (-0.4) = 30.6 + 0.4 = 31.0 \text{ mm} = 3.10 \text{ cm}.$$ Thus, the diameter of the spherical body is $$3.10 \text{ cm}.$$

The correct answer is Option C.

We need to determine the correct diameter of the wire using the observations recorded from the screw gauge experiment.

The total measurement from a screw gauge is calculated using the formula:

$$\text{Total Reading} = \text{Main Scale Reading (MSR)} + \text{Circular Scale Reading (CSR)} - \text{Zero Error}$$

Step 1: Calculate the Least Count (LC) of the Screw Gauge

The least count is the smallest distance the screw can measure accurately, which is defined as the pitch divided by the total number of divisions on the circular scale.

• Pitch: The distance moved on the main scale in one complete rotation = $$0.5\text{ mm}$$.
• Circular scale divisions: $$50$$

$$LC = \frac{\text{Pitch}}{\text{Total Circular Scale Divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}$$

Step 2: Calculate the Circular Scale Reading (CSR)

The circular scale reading is obtained by multiplying the coinciding division line on the pitch axis by the least count.

• Coinciding division: $$45^{\text{th}}\text{ division}$$

$$\text{CSR} = 45 \times LC = 45 \times 0.01\text{ mm} = 0.45\text{ mm}$$

Step 3: Account for the Zero Error

The instrument is given to have a negative error:

• Zero Error: $$-0.03\text{ mm}$$

When computing the true reading, a negative zero error must be subtracted, which mathematically amounts to adding its absolute magnitude to the observed reading.

Step 4: Calculate the Total Diameter of the Wire

Given the Main Scale Reading ($\text{MSR}$) is $$2.5\text{ mm}$$, we substitute all the values into our main formula:

$$\text{Diameter} = \text{MSR} + \text{CSR} - \text{Zero Error}$$

$$\text{Diameter} = 2.5\text{ mm} + 0.45\text{ mm} - (-0.03\text{ mm})$$

$$\text{Diameter} = 2.5\text{ mm} + 0.45\text{ mm} + 0.03\text{ mm} = 2.98\text{ mm}$$

Therefore, the correct diameter of the wire is 2.98 mm, which corresponds to Option C.

We consider the quantity $$z = a^2 x^3 y^{1/2}$$, where $$a$$ is a constant. The percentage errors in $$x$$ and $$y$$ are 4\% and 12\%, respectively.

Since $$z = a^2 x^3 y^{1/2}$$, the percentage error in $$z$$ follows the propagation rule for products and powers. Specifically, $$\frac{\Delta z}{z} \times 100 = 3\left(\frac{\Delta x}{x} \times 100\right) + \frac{1}{2}\left(\frac{\Delta y}{y} \times 100\right),$$ and because $$a$$ is constant it contributes no error.

Substituting the given errors into this formula gives $$\% \text{ error in } z = 3 \times 4\% + \frac{1}{2} \times 12\% = 12\% + 6\% = 18\%,$$ so the percentage error in $$z$$ is 18\%.

Young's modulus is given by:

$$Y = \dfrac{FL}{\pi r^2 \Delta L} = \dfrac{4FL}{\pi d^2 \Delta L}$$

We start by substituting the values $$F = mg = 1 \times 9.8 = 9.8 \text{ N}$$, $$L = 1 \text{ m}$$, $$d = 0.4 \text{ mm} = 4 \times 10^{-4} \text{ m}$$, and $$\Delta L = 0.4 \text{ mm} = 4 \times 10^{-4} \text{ m}$$ into the expression for Y:

$$Y = \dfrac{4 \times 9.8 \times 1}{\pi \times (4 \times 10^{-4})^2 \times 4 \times 10^{-4}}$$

This gives

$$Y = \dfrac{39.2}{\pi \times 16 \times 10^{-8} \times 4 \times 10^{-4}} = \dfrac{39.2}{\pi \times 6.4 \times 10^{-11}}$$

Therefore,

$$Y = \dfrac{39.2}{2.0106 \times 10^{-10}} = 1.95 \times 10^{11} \text{ N m}^{-2}$$

Next, the fractional error in Y is expressed as

$$\dfrac{\Delta Y}{Y} = \dfrac{\Delta F}{F} + \dfrac{\Delta L}{L} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta(\Delta L)}{\Delta L}$$

Since F and L have no uncertainty mentioned, this reduces to

$$\dfrac{\Delta Y}{Y} = 2 \times \dfrac{0.01}{0.4} + \dfrac{0.02}{0.4} = 0.05 + 0.05 = 0.10$$

Hence, the absolute error in Y becomes

$$\Delta Y = 0.10 \times 1.95 \times 10^{11} = 1.95 \times 10^{10} \approx 2 \times 10^{10} \text{ N m}^{-2}$$

Therefore, the value of $$x$$ is 2.

We are given a simple pendulum experiment where the time period is $$T = 0.5 \text{ s}$$ (measured from 100 oscillations using a watch with 1 s resolution), and the length is $$L = 10 \text{ cm}$$ known to 1 mm accuracy. We need to find the percentage accuracy in the determination of $$g$$.

The formula relating $$g$$ to the pendulum parameters is $$g = 4\pi^2 \frac{L}{T^2}$$. Taking the relative error, we get $$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$$.

Now, the error in length is $$\Delta L = 1 \text{ mm} = 0.1 \text{ cm}$$, so $$\frac{\Delta L}{L} = \frac{0.1}{10} = 0.01$$.