JEE Properties of Matter Questions

$$U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$$

$$U = \frac{1}{2} \times Y \times (\text{Strain})^2 \times \text{Volume}$$

$$\text{Strain} = \frac{\Delta L}{L} = \frac{3 \times 10^{-3}\text{ m}}{3\text{ m}} = 10^{-3}$$

$$U = \frac{1}{2} \times \left(1.1 \times 10^{11}\right) \times \left(10^{-3}\right)^2 \times \left(600 \times 10^{-6}\right)$$

$$U = \frac{1}{2} \times 1.1 \times 10^{11} \times 10^{-6} \times 600 \times 10^{-6}$$

$$U = 330 \times 10^{-1} = 33\text{ J}$$

Use shear relation:

shear modulus $$G=\frac{\text{stress}}{\text{strain}}$$

$$G=\frac{F/A}{x/L}\Rightarrow x=\frac{FL}{AG}$$

given:

side = 5 cm = 0.05 m
area $$A=(0.05)2=2.5\times10^{-3}m^2$$
L=0.05m
F=10
$$G=10^5$$

$$x=\frac{10\times0.05}{2.5\times10^{-3}\times10^5}$$

$$=\frac{0.5}{250}=0.002m$$

=2 mm

Let the origin be at the top end of the wire and measure distance downward along the wire by the variable $$y$$ (so $$y = 0$$ at the roof and $$y = l$$ at the lower end).

The weight of the entire wire is $$w$$, so its weight per unit length is $$\dfrac{w}{l}\;({\rm N\,m^{-1}})$$.

At a distance $$y$$ from the top, the portion of the wire lying $$\text$$it{below} that point has length $$(l - y)$$. Therefore the weight of that lower portion is
$$\left(l-y\right)\left(\dfrac{w}{l}\right)=w\left(1-\dfrac{y}{l}\right).$$

The tension $$T(y)$$ at the cross-section situated at distance $$y$$ must support both the hanging load $$W$$ and the weight of the wire below that section. Hence
$$T(y)=W+w\left(1-\dfrac{y}{l}\right).$$

The stress $$\sigma(y)$$ equals tension divided by cross-sectional area $$A$$:
$$\sigma(y)=\dfrac{T(y)}{A}=\dfrac{W}{A}+\dfrac{w}{A}\left(1-\dfrac{y}{l}\right).$$

The question asks for the stress at the point where $$y=\dfrac{l}{3}$$, i.e., one-third the length from the top. Substitute $$y=\dfrac{l}{3}$$:

$$\sigma\!\left(\dfrac{l}{3}\right)=\dfrac{W}{A}+\dfrac{w}{A}\left(1-\dfrac{1}{3}\right)=\dfrac{W}{A}+\dfrac{w}{A}\left(\dfrac{2}{3}\right)=\dfrac{W}{A}+\dfrac{2}{3}\dfrac{w}{A}.$$

The given form of the answer is $$\dfrac{W}{A}+\dfrac{2}{\gamma}\dfrac{w}{A}$$. Comparing the two expressions, we equate the coefficients of $$\dfrac{w}{A}$$:

$$\dfrac{2}{\gamma}=\dfrac{2}{3}\; \Longrightarrow\; \gamma=3.$$

Hence, the required value of $$\gamma$$ is 3.

We need to evaluate the Assertion and Reason about the speed of sound in different media.

Assertion A: A sound wave has higher speed in solids than in gases.

The speed of sound in a medium is given by $$v = \sqrt{\frac{E}{\rho}}$$, where $$E$$ is the appropriate elastic modulus and $$\rho$$ is the density. In solids, the elastic modulus (Young's modulus) is extremely high compared to gases. Although solids have higher density, the much larger elastic modulus dominates. For example, the speed of sound in steel is approximately 5960 m/s, while in air it is approximately 343 m/s.

Assertion A is TRUE.

Reason R: Gases have higher value of bulk modulus than solids.

This is incorrect. Solids have a much higher bulk modulus than gases. For example, the bulk modulus of steel is about $$1.6 \times 10^{11}$$ Pa, while for air at atmospheric pressure it is approximately $$1.42 \times 10^{5}$$ Pa. Solids resist compression far more effectively than gases.

Reason R is FALSE.

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

The extension of an ideal string that obeys Hooke’s law is directly proportional to the tension applied: $$\Delta L \propto T$$.
For a given string, the ratio $$\dfrac{\Delta L}{T}$$ remains constant.

Let the natural (unstretched) length of the string be $$L_0$$.

Case 1:

Tension $$T_1 = 5 \text{ N}$$ produces total length $$L_1 = 1.40 \text{ m}$$.
Extension, $$\Delta L_1 = L_1 - L_0 = 1.40 - L_0$$

Case 2:

Tension $$T_2 = 7 \text{ N}$$ produces total length $$L_2 = 1.56 \text{ m}$$.
Extension, $$\Delta L_2 = L_2 - L_0 = 1.56 - L_0$$

Using $$\dfrac{\Delta L_1}{T_1} = \dfrac{\Delta L_2}{T_2}$$:

$$\frac{1.40 - L_0}{5} = \frac{1.56 - L_0}{7}$$

Cross-multiplying gives:
$$7(1.40 - L_0) = 5(1.56 - L_0)$$

Simplify each side:
$$9.8 - 7L_0 = 7.8 - 5L_0$$

Rearrange terms:
$$9.8 - 7.8 = 7L_0 - 5L_0$$
$$2.0 = 2L_0$$

Hence,
$$L_0 = 1.0 \text{ m}$$

Therefore, the original (unstretched) length of the string is 1 m.

We need to find the volume contraction of a solid copper cube under hydraulic pressure.

The bulk modulus $$K$$ is defined as the ratio of volumetric stress to volumetric strain: $$K = -\frac{P}{\Delta V / V} = \frac{P \cdot V}{|\Delta V|}$$ where $$P$$ is the applied pressure, $$V$$ is the original volume, and $$\Delta V$$ is the change in volume.

The negative sign indicates that an increase in pressure causes a decrease in volume (contraction).

Rearranging gives $$|\Delta V| = \frac{P \cdot V}{K}$$.

The cube has edge length 10 cm = 0.1 m, so its original volume is $$V = (0.1)^3 = 10^{-3} \text{ m}^3$$.

With an applied pressure of $$P = 7 \times 10^6$$ Pa and bulk modulus $$K = 1.4 \times 10^{11}$$ N/m$$^2$$, substitution gives $$|\Delta V| = \frac{7 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}} = \frac{7 \times 10^3}{1.4 \times 10^{11}} = 5 \times 10^{-8} \text{ m}^3$$.

Since $$1 \text{ m} = 10^3 \text{ mm}$$, then $$1 \text{ m}^3 = (10^3)^3 = 10^9 \text{ mm}^3$$ and thus $$|\Delta V| = 5 \times 10^{-8} \times 10^9 = 50 \text{ mm}^3$$.

The negative sign indicates contraction, so the magnitude of volume contraction is 50 mm$$^3$$.

The answer is 50 mm$$^3$$.

Given: Length $$L = 20$$ m, extension $$\Delta L = 2$$ cm $$= 0.02$$ m, elastic potential energy $$U = 80$$ J, and Young's modulus $$Y = 2.0 \times 10^{11}$$ N/m$$^2$$.

The elastic potential energy stored in a stretched wire is:

$$U = \frac{1}{2} \times \frac{YA(\Delta L)^2}{L}$$

Solving for the cross-sectional area $$A$$:

$$A = \frac{2UL}{Y(\Delta L)^2}$$

Substituting the values:

$$A = \frac{2 \times 80 \times 20}{2.0 \times 10^{11} \times (0.02)^2}$$

$$= \frac{3200}{2.0 \times 10^{11} \times 4 \times 10^{-4}}$$

$$= \frac{3200}{8 \times 10^{7}} = 4 \times 10^{-5} \text{ m}^2$$

Converting to mm$$^2$$:

$$A = 4 \times 10^{-5} \text{ m}^2 = 4 \times 10^{-5} \times 10^6 \text{ mm}^2 = 40 \text{ mm}^2$$

Therefore, the answer is $$\mathbf{40}$$.

A metal block of mass $$m$$ is suspended from a wire of diameter $$14$$ mm, and the tensile stress in the wire is $$7 \times 10^5$$ N/m$$^2$$.

To find the cross-sectional area of the wire, note that the radius is half the diameter so that radius $$= 7$$ mm $$= 7 \times 10^{-3}$$ m. It follows that $$A = \pi r^2 = \dfrac{22}{7} \times (7 \times 10^{-3})^2 = \dfrac{22}{7} \times 49 \times 10^{-6} = 22 \times 7 \times 10^{-6} = 154 \times 10^{-6} \text{ m}^2$$.

Since tensile stress is defined as force per unit area, we have Stress $$= \dfrac{F}{A} = \dfrac{mg}{A}$$, and substituting the given values yields $$7 \times 10^5 = \dfrac{m \times 9.8}{154 \times 10^{-6}}$$.

Solving for $$m$$ gives $$m = \dfrac{7 \times 10^5 \times 154 \times 10^{-6}}{9.8} = \dfrac{107.8}{9.8} = 11 \text{ kg}$$, and hence $$\boxed{11}$$ kg is the mass of the block.

We have a rod of length $$L = 1$$ m, cross-sectional area $$A = 10^{-4}$$ m$$^2$$, with a temperature change $$\Delta T = 200°C$$, Young's modulus $$Y = 2 \times 10^{11}$$ N/m$$^2$$, and coefficient of linear expansion $$\alpha = 10^{-5}$$ K$$^{-1}$$.

When a rod is heated but not allowed to expand, thermal stress develops. The thermal strain is

$$\text{Strain} = \alpha \Delta T = 10^{-5} \times 200 = 2 \times 10^{-3}$$

Now the compressive stress is (since stress equals Young's modulus times strain)

$$\text{Stress} = Y \times \text{Strain} = 2 \times 10^{11} \times 2 \times 10^{-3} = 4 \times 10^{8} \text{ N/m}^2$$

So the compressive force is

$$F = \text{Stress} \times A = 4 \times 10^{8} \times 10^{-4} = 4 \times 10^{4} \text{ N}$$

Hence, the correct answer is $$4 \times 10^4$$ N.

The steel wire is the upper wire, meaning it supports the combined weight of both hanging masses ($$2 \text{ kg}$$ and $$1.14 \text{ kg}$$):

$$F = (m_1 + m_2)g$$

$$F = (2 + 1.14) \times 10 = 31.4 \text{ N}$$

The radius is $$0.2 \text{ cm} = 2 \times 10^{-3} \text{ m}$$:

$$A = \pi r^2 = \pi \times (2 \times 10^{-3})^2 = 4\pi \times 10^{-6} \text{ m}^2$$

$$\Delta L = \frac{FL}{AY}$$: $$\Delta L = \frac{31.4 \times 1.6}{(4\pi \times 10^{-6}) \times (2 \times 10^{11})}$$

$$\Delta L = \frac{50.24}{8\pi \times 10^5}$$

$$2 \times 10^{-5} \text{ m} = 20 \times 10^{-6} \text{ m}$$

The value is 20.

We have a rod of radius $$r = 20\;\text{mm} = 0.02\;\text{m}$$, length $$L = 2.0\;\text{m}$$, subjected to a force $$F = 62.8\;\text{kN} = 62800\;\text{N}$$, with Young's modulus $$Y = 2.0 \times 10^{11}\;\text{N/m}^2$$.

The cross-sectional area of the rod is

$$A = \pi r^2 = \pi (0.02)^2 = \pi \times 4 \times 10^{-4} = 1.2566 \times 10^{-3}\;\text{m}^2$$

Now, the stress is

$$\sigma = \frac{F}{A} = \frac{62800}{1.2566 \times 10^{-3}} = 4.998 \times 10^{7}\;\text{N/m}^2$$

So the longitudinal strain is

$$\text{strain} = \frac{\sigma}{Y} = \frac{4.998 \times 10^{7}}{2.0 \times 10^{11}} = 2.5 \times 10^{-4} = 25 \times 10^{-5}$$

Hence, the longitudinal strain produced is $$25 \times 10^{-5}$$. So, the answer is $$25$$.

Solution :

Given :

Length of rod,

$$L = 1\text{ m}$$

Area of cross-section,

$$A = 3 \times 10^{-6}\text{ m}^2$$

Temperature fall,

$$\Delta T = 210^\circ\text{C} - 160^\circ\text{C}$$

$$= 50^\circ\text{C}$$

Coefficient of linear expansion,

$$\alpha = 2 \times 10^{-5}\text{ K}^{-1}$$

Young’s modulus,

$$Y = 2 \times 10^{11}\text{ N m}^{-2}$$

After cooling, rod contracts by :

$$\Delta L = \alpha L \Delta T$$

$$= 2 \times 10^{-5} \times 1 \times 50$$

$$= 10^{-3}\text{ m}$$

To regain original length, extension produced by attached mass must be :

$$\Delta L = 10^{-3}\text{ m}$$

Using Young’s modulus relation :

$$Y = \frac{\text{Stress}}{\text{Strain}}$$

$$Y = \frac{F/A}{\Delta L/L}$$

$$F = \frac{YA\Delta L}{L}$$

Substituting values :

$$F = \frac{(2 \times 10^{11})(3 \times 10^{-6})(10^{-3})}{1}$$

$$= 6 \times 10^2\text{ N}$$

Since,

$$F = Mg$$

$$M = \frac{600}{10}$$

$$= 60\text{ kg}$$

Final Answer :

$$60$$

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \left( \frac{F}{\Delta L} \right) \cdot \frac{L}{A}$$

$$\text{Slope} = \frac{\text{Change in Extension}}{\text{Change in Load}} = \frac{\Delta L}{F} = \tan(45^\circ) = 1$$

$$A \approx 4 \times 3.14 \times 10^{-6} = 12.56 \times 10^{-6} \text{ m}^2$$

$$Y = \left( \frac{F}{\Delta L} \right) \cdot \frac{L}{A}$$

$$Y = (1) \cdot \frac{0.628}{12.56 \times 10^{-6}}$$

$$Y = \frac{0.628 \times 10^6}{12.56} = 0.05 \times 10^6$$

$$Y = 5 \times 10^4 \text{ N m}^{-2}$$

$$x = 5$$

Let the natural length of the wire be $$L$$ and its spring constant (stiffness) be $$k$$.

When tension of 100 N is applied:

$$l_1 = L + \frac{100}{k}$$ ... (i)

When tension of 120 N is applied:

$$l_2 = L + \frac{120}{k}$$ ... (ii)

We are given that $$10l_2 = 11l_1$$

Substituting:

$$10\left(L + \frac{120}{k}\right) = 11\left(L + \frac{100}{k}\right)$$

$$10L + \frac{1200}{k} = 11L + \frac{1100}{k}$$

$$\frac{1200}{k} - \frac{1100}{k} = 11L - 10L$$

$$\frac{100}{k} = L$$

Substituting back into equation (i):

$$l_1 = L + L = 2L$$

$$L = \frac{l_1}{2} = \frac{1}{2}l_1$$

Therefore, $$x = 2$$.

Consider a wire of length $$L = 100$$ m, cross-sectional area $$A = 6.25 \times 10^{-4}$$ m², and Young’s modulus $$Y = 10^{10}$$ N/m², to which a force $$F = 250$$ N is applied.

Using the relation for elastic extension, the elongation $$\Delta L$$ is given by:

$$ \Delta L = \frac{F L}{A Y} = \frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}} = \frac{25000}{6.25 \times 10^6} = 4 \times 10^{-3} \text{ m} $$

Hence, the elongation of the wire is $$4 \times 10^{-3}$$ m.

We need to find the elongation of a wire subjected to a load, given its dimensions and Young's modulus. Young's modulus is defined as: $$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$$ Rearranging for elongation $$\Delta L$$: $$\Delta L = \frac{FL}{AY}$$

Substituting the given values $$F = 250$$ N, $$L = 100$$ m, $$A = 6.25 \times 10^{-4}$$ m$$^2$$, $$Y = 10^{10}$$ N/m$$^2$$:

$$\Delta L = \frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}}$$

$$= \frac{25000}{6.25 \times 10^{6}}$$

$$= \frac{25000}{6250000} = 4 \times 10^{-3} \text{ m}$$

The correct answer is Option (4): $$4 \times 10^{-3}$$ m.

We are given: length $$L = 6$$ m, cross-sectional area $$A = 3 \text{ mm}^2 = 3 \times 10^{-6} \text{ m}^2$$, Young's modulus $$Y = 2 \times 10^{11} \text{ N/m}^2$$, mass $$m = 4$$ kg, and gravity on the planet $$g' = g/4 = 10/4 = 2.5 \text{ m/s}^2$$.

The force on the wire is $$F = mg' = 4 \times 2.5 = 10$$ N. Using the elongation formula $$\Delta L = \frac{FL}{AY}$$, we get $$\Delta L = \frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}} = \frac{60}{6 \times 10^{5}} = 1 \times 10^{-4} \text{ m} = 0.1 \text{ mm}$$.

The correct answer is Option C: $$0.1$$ mm.

Wire A: length $$L_A = 5.0$$ m, cross-section $$A_A = 2.5 \times 10^{-5}$$ m².

Wire B: length $$L_B = 6.0$$ m, cross-section $$A_B = 3.0 \times 10^{-5}$$ m².

Same load $$F$$, same extension $$\Delta L$$.

Young's modulus formula:

$$Y = \frac{F \cdot L}{A \cdot \Delta L}$$

Ratio:

$$\frac{Y_A}{Y_B} = \frac{L_A / A_A}{L_B / A_B} = \frac{L_A \times A_B}{L_B \times A_A}$$

$$= \frac{5.0 \times 3.0 \times 10^{-5}}{6.0 \times 2.5 \times 10^{-5}} = \frac{15}{15} = 1$$

So $$Y_A : Y_B = 1 : 1$$.

The answer is Option B: 1 : 1.

Elongation: $$\Delta l = \frac{Fl}{AY}$$. For same F and l:

$$\frac{\Delta l_A}{\Delta l_B} = \frac{A_B Y_B}{A_A Y_A} = \frac{3 \times 4}{1 \times 1} = 12$$

The correct answer is Option 1: 12:1.

Using Young's modulus formula: $$Y = \frac{FL}{Al} = \frac{FL}{\pi r^2 l}$$

For the first wire: $$Y = \frac{fL}{\pi r^2 l}$$

For the second wire: $$Y = \frac{2f \times 2L}{\pi (2r)^2 \times l'}$$

Since both wires are of the same material (same $$Y$$):

$$\frac{fL}{\pi r^2 l} = \frac{4fL}{\pi \times 4r^2 \times l'}$$

$$\frac{fL}{r^2 l} = \frac{4fL}{4r^2 l'}$$

$$\frac{1}{l} = \frac{1}{l'}$$

$$l' = l$$

The increase in length is $$\mathbf{l}$$.

We need to identify the correct relationship between Poisson ratio $$\sigma$$, bulk modulus $$K$$, and modulus of rigidity $$\eta$$.

Standard relation:

The relationship between Young's modulus $$Y$$, bulk modulus $$K$$, and Poisson's ratio $$\sigma$$ is:

$$Y = 3K(1 - 2\sigma)$$ ... (i)

The relationship between Young's modulus $$Y$$, modulus of rigidity $$\eta$$, and Poisson's ratio $$\sigma$$ is:

$$Y = 2\eta(1 + \sigma)$$ ... (ii)

Eliminating Y:

From (i) and (ii):

$$3K(1 - 2\sigma) = 2\eta(1 + \sigma)$$

$$3K - 6K\sigma = 2\eta + 2\eta\sigma$$

$$3K - 2\eta = 6K\sigma + 2\eta\sigma$$

$$3K - 2\eta = \sigma(6K + 2\eta)$$

$$\sigma = \frac{3K - 2\eta}{6K + 2\eta}$$

The correct answer is Option 1: $$\sigma = \frac{3K - 2\eta}{6K + 2\eta}$$.

We need to find the elastic energy stored per unit volume in an aluminium rod.

Formula for energy per unit volume. The energy stored per unit volume in an elastically deformed material is:

$$u = \frac{1}{2} \times Y \times (\text{strain})^2$$

Convert strain to decimal. Strain = 0.04% = $$\frac{0.04}{100} = 4 \times 10^{-4}$$

Calculate energy per unit volume: $$u = \frac{1}{2} \times 7.0 \times 10^{10} \times (4 \times 10^{-4})^2$$

$$= \frac{1}{2} \times 7.0 \times 10^{10} \times 16 \times 10^{-8}$$

$$= \frac{1}{2} \times 7.0 \times 16 \times 10^{2}$$

$$= \frac{1}{2} \times 11200 = 5600 \text{ J m}^{-3}$$

The correct answer is Option C: 5600.

Assertion (A): Steel is used in the construction of buildings and bridges.

Reason (R): Steel is more elastic and its elastic limit is high.

Analysis of Assertion (A):

Steel is indeed widely used in construction of buildings and bridges due to its structural properties.

Assertion A is correct.

Analysis of Reason (R):

Steel has a high Young's modulus (approximately 200 GPa), meaning it is highly elastic — it can withstand large stress with minimal strain. Steel also has a high elastic limit, meaning it can bear significant loads without undergoing permanent deformation.

Reason R is correct.

Connection: Steel is used in construction specifically BECAUSE it is more elastic (high Young's modulus) and has a high elastic limit, allowing structures to bear heavy loads safely. R is the correct explanation of A.

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

We are given a square aluminium slab with shear modulus $$G = 25 \times 10^9 \text{ N m}^{-2}$$, side $$60 \text{ cm} = 0.6 \text{ m}$$, thickness $$15 \text{ cm} = 0.15 \text{ m}$$, and shearing force $$F = 18.0 \times 10^4 \text{ N}$$. We need to find the displacement of the upper edge.

The shearing force is applied on the narrow face (the side face). The narrow face has dimensions: length $$= 0.6 \text{ m}$$ and thickness $$= 0.15 \text{ m}$$.

$$A = 0.6 \times 0.15 = 0.09 \text{ m}^2$$

$$\text{Shearing stress} = \frac{F}{A} = \frac{18.0 \times 10^4}{0.09} = 2 \times 10^6 \text{ N m}^{-2}$$

The shear modulus is defined as:

$$G = \frac{\text{Shearing stress}}{\text{Shearing strain}} = \frac{F/A}{\Delta x / L}$$

where $$L$$ is the height over which shearing occurs (the side length $$= 0.6 \text{ m}$$) and $$\Delta x$$ is the displacement of the upper edge.

$$\Delta x = \frac{F \times L}{A \times G}$$

$$\Delta x = \frac{2 \times 10^6 \times 0.6}{25 \times 10^9}$$

$$\Delta x = \frac{1.2 \times 10^6}{25 \times 10^9} = 4.8 \times 10^{-5} \text{ m}$$

$$\Delta x = 48 \text{ } \mu\text{m}$$

The displacement of the upper edge is $$48 \text{ } \mu\text{m}$$.

For a uniform rod hanging vertically under its own weight, the elongation due to its own weight is given by $$\Delta L = \frac{MgL}{2AY}$$ where $$M = 20 \text{ kg}$$, $$g = 10 \text{ m/s}^2$$, $$L = 20 \text{ m}$$, $$A = 0.4 \text{ m}^2$$, and $$Y = 2 \times 10^{11} \text{ N/m}^2$$.

Considering a small element at distance $$x$$ from the bottom, the weight below it is $$\frac{Mg}{L} \cdot x$$. The stress on this element is $$\frac{Mgx}{AL}$$ and the elongation of the element is $$d(\Delta L) = \frac{Mgx}{ALY} \, dx$$.

Integrating from $$0$$ to $$L$$ yields $$\Delta L = \frac{Mg}{ALY} \int_0^L x \, dx = \frac{Mg}{ALY} \cdot \frac{L^2}{2} = \frac{MgL}{2AY}$$.

Substituting the values into the formula gives $$\Delta L = \frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 10^{11}} = \frac{4000}{1.6 \times 10^{11}} = 25 \times 10^{-9} \text{ m}$$.

Therefore, $$x = \textbf{25}$$.

The elongation of a wire under a tensile force is given by:

$$\Delta L = \dfrac{FL}{\pi r^2 Y}$$

where $$F$$ is the applied force, $$L$$ is the original length, $$r$$ is the radius, and $$Y$$ is Young's modulus.

For the first wire: Length = $$L$$, radius = $$r$$, force = $$F$$

$$\Delta L_1 = \dfrac{FL}{\pi r^2 Y} = 5 \text{ cm}$$

For the second wire: Length = $$4L$$, radius = $$4r$$, force = $$4F$$ (same material, so same $$Y$$)

$$\Delta L_2 = \dfrac{4F \times 4L}{\pi (4r)^2 Y} = \dfrac{16FL}{\pi \times 16r^2 \times Y} = \dfrac{FL}{\pi r^2 Y}$$

$$\Delta L_2 = \Delta L_1 = 5 \text{ cm}$$

Therefore, the increase in length of the second wire is $$\boxed{5}$$ cm.

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$$

$$\frac{\Delta L}{F} = \left( \frac{1}{AY} \right) \cdot L$$

This is a linear equation ($$y = m \cdot x_{axis}$$) where the slope is $$\frac{1}{AY}$$.

At $$L = 4\text{ m}$$, the value on the $$y$$-axis is $$1.0 \times 10^{-5} \text{ N}^{-1} \text{ m}$$.

$$Slope = \frac{\text{Extension/Load}}{\text{Length}} = \frac{1.0 \times 10^{-5}}{4} = 0.25 \times 10^{-5} \text{ N}^{-1}$$

$$Y = \frac{1}{A \cdot Slope}$$

$$Y = \frac{1}{(2 \times 10^{-6}) \times (0.25 \times 10^{-5})}$$

$$Y = \frac{1}{0.5 \times 10^{-11}} = 2 \times 10^{11} \text{ N m}^{-2}$$

$$x = 2$$

The elongation of a wire on Earth is $$10^{-4}$$ m, and the same wire on another planet is elongated by $$6 \times 10^{-5}$$ m. We need to find the acceleration due to gravity on the planet.

The elongation of a wire under its own weight (or a suspended load) is given by:

$$\Delta L = \frac{F \cdot L}{A \cdot Y}$$

where $$F$$ is the applied force (weight = $$mg$$), $$L$$ is the original length, $$A$$ is the cross-sectional area, and $$Y$$ is the Young’s modulus.

Since the same wire (same $$m$$, $$L$$, $$A$$, and $$Y$$) is used on both the Earth and the planet, the only difference is the value of $$g$$. Taking the ratio of the elongations gives:

$$\frac{\Delta L_{planet}}{\Delta L_{earth}} = \frac{g_{planet}}{g_{earth}}$$

Substituting into this ratio yields:

$$g_{planet} = g_{earth} \times \frac{\Delta L_{planet}}{\Delta L_{earth}}$$

$$g_{planet} = 10 \times \frac{6 \times 10^{-5}}{10^{-4}}$$

$$g_{planet} = 10 \times \frac{6 \times 10^{-5}}{1 \times 10^{-4}}$$

$$g_{planet} = 10 \times 0.6 = 6$$ m/s$$^2$$

Therefore, the acceleration due to gravity on the planet is 6 m/s$$^2$$.

Consider a spherical soap bubble of radius 3 cm inside another spherical soap bubble of radius 6 cm. We need to find the radius $$r$$ of a single soap bubble whose internal pressure equals the internal pressure of the smaller bubble (radius 3 cm) in this system.

For any soap bubble, there are two surfaces (inner and outer), and the excess pressure inside is given by $$\Delta P = \frac{4S}{R}$$, where $$S$$ is the surface tension and $$R$$ is the radius. Let $$P_0$$ be the atmospheric pressure outside the outer bubble.

Define the pressures:

• $$P_1$$: Pressure inside the outer bubble but outside the inner bubble (in the space between them).
• $$P_2$$: Pressure inside the inner bubble (radius 3 cm).

For the outer bubble (radius $$R_1 = 6$$ cm), the pressure difference is:

For the inner bubble (radius $$R_2 = 3$$ cm), the pressure difference is:

Substitute $$P_1$$ from the outer bubble into this equation:

Simplify the expression inside the parentheses:

Now, for a single soap bubble of radius $$r$$, the internal pressure $$P_r$$ is:

The problem states that $$P_2 = P_r$$, so:

Subtract $$P_0$$ from both sides:

Divide both sides by $$4S$$ (assuming $$S \neq 0$$):

Therefore, $$r = 2$$ cm.

Hence, the correct answer is 2.

We have a string of length $$L = 50$$ cm $$= 0.5$$ m, mass $$m = 10$$ g $$= 0.01$$ kg, cross-sectional area $$A = 2.0$$ mm$$^2 = 2.0 \times 10^{-6}$$ m$$^2$$, and Young's modulus $$Y = 1.2 \times 10^{11}$$ N/m$$^2$$. The speed of a transverse wave on the string is $$v = 60$$ m/s.

The linear mass density is $$\mu = \frac{m}{L} = \frac{0.01}{0.5} = 0.02$$ kg/m. The wave speed on a string is $$v = \sqrt{\frac{T}{\mu}}$$, so the tension is $$T = \mu v^2 = 0.02 \times 60^2 = 0.02 \times 3600 = 72$$ N.

Now, using the definition of Young's modulus, $$Y = \frac{T/A}{\Delta L / L}$$, we can find the extension: $$\Delta L = \frac{TL}{YA} = \frac{72 \times 0.5}{1.2 \times 10^{11} \times 2.0 \times 10^{-6}} = \frac{36}{2.4 \times 10^{5}} = 1.5 \times 10^{-4}$$ m $$= 15 \times 10^{-5}$$ m.

Hence $$x = 15$$.

Hence, the correct answer is 15.

We need to determine the required area of cross-section of the crane's rope to increase its maximum lifting capacity from 10 metric tons to 25 metric tons.

1. Identify the Working Principle

The maximum lifting capacity of a rope depends on the maximum permissible stress ($$\sigma$$) the material can withstand without breaking. Stress is defined as the restoring force per unit cross-sectional area:

$$\text{Stress } (\sigma) = \frac{\text{Force}}{\text{Area}} = \frac{mg}{A}$$

Since the material of the rope remains the same, the maximum permissible stress stays constant ($$\sigma_1 = \sigma_2$$). Therefore, the lifting load ($$m$$) is directly proportional to the cross-sectional area ($$A$$):

$$\frac{m_1}{A_1} = \frac{m_2}{A_2}$$

2. Substitute the Values and Solve for $$A_2$$

From the problem 

• Initial lifting mass ($$m_1$$) = $$10\text{ metric tons}$$
• Initial area of cross-section ($$A_1$$) = $$2.5 \times 10^{-4}\text{ m}^2$$
• Target lifting mass ($$m_2$$) = $$25\text{ metric tons}$$

Rearranging the proportion formula to isolate the new required area ($$A_2$$):

$$A_2 = A_1 \times \left(\frac{m_2}{m_1}\right)$$

$$A_2 = (2.5 \times 10^{-4}\text{ m}^2) \times \frac{25}{10}$$

$$A_2 = 2.5 \times 10^{-4} \times 2.5 = 6.25 \times 10^{-4}\text{ m}^2$$

Conclusion

The required area of cross-section of the rope should be $$6.25 \times 10^{-4}\text{ m}^2$$, which corresponds to Option A.

To determine the force required to stretch a wire to double its length, we note that the cross-sectional area is $$A = 1 \text{ cm}^2 = 1 \times 10^{-4} \text{ m}^2$$ and Young's modulus is $$Y = 2 \times 10^{11}$$ N/m$$^2$$. Since the extension is equal to the original length, $$\Delta L = L$$, the strain is $$\frac{\Delta L}{L} = 1$$.

Young's modulus is given by the relation $$Y = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L}$$. Substituting $$\Delta L = L$$ yields $$F = Y \times A \times \frac{\Delta L}{L}$$.

Substituting the numerical values, we get $$F = 2 \times 10^{11} \times 1 \times 10^{-4} \times 1 = 2 \times 10^{7} \text{ N}$$. Hence, the correct answer is Option C: $$2 \times 10^7$$ N.

A steel wire (length 3.2 m, $$Y_S = 2.0 \times 10^{11}$$ N/m$$^2$$) and a copper wire (length 4.4 m, $$Y_C = 1.1 \times 10^{11}$$ N/m$$^2$$), both of radius 1.4 mm, are connected end to end. The net elongation is 1.4 mm. We wish to find the load required.

First, the radius of each wire is $$r = 1.4 \text{ mm} = 1.4 \times 10^{-3} \text{ m}$$. Hence, the cross-sectional area is $$A = \pi r^2 = \frac{22}{7} \times (1.4 \times 10^{-3})^2 = \frac{22}{7} \times 1.96 \times 10^{-6}$$ which simplifies to $$A = 22 \times 0.28 \times 10^{-6} = 6.16 \times 10^{-6} \text{ m}^2.$$

Since the wires are in series, the same load $$F$$ acts on both. Using the relation $$\Delta l = \frac{Fl}{YA}$$, we have $$\Delta l_S = \frac{F \times 3.2}{2.0 \times 10^{11} \times 6.16 \times 10^{-6}}$$ for the steel wire and $$\Delta l_C = \frac{F \times 4.4}{1.1 \times 10^{11} \times 6.16 \times 10^{-6}}$$ for the copper wire.

The total elongation is given by $$\Delta l_S + \Delta l_C = 1.4 \times 10^{-3} \text{ m},$$ so that $$F\!\left(\frac{3.2}{2.0 \times 10^{11} \times 6.16 \times 10^{-6}} + \frac{4.4}{1.1 \times 10^{11} \times 6.16 \times 10^{-6}}\right) = 1.4 \times 10^{-3}.$$

Substituting and simplifying each term yields $$\frac{3.2}{2.0 \times 10^{11} \times 6.16 \times 10^{-6}} = \frac{3.2}{1.232 \times 10^{6}} = 2.597 \times 10^{-6},$$ $$\frac{4.4}{1.1 \times 10^{11} \times 6.16 \times 10^{-6}} = \frac{4.4}{6.776 \times 10^{5}} = 6.494 \times 10^{-6},$$ and hence the sum is $$2.597 \times 10^{-6} + 6.494 \times 10^{-6} = 9.091 \times 10^{-6}.$$

From the above, we solve for $$F$$: $$F = \frac{1.4 \times 10^{-3}}{9.091 \times 10^{-6}} = \frac{1400}{9.091} \approx 154 \text{ N}.$$

Answer: Option D: 154

Young's modulus is a material property that depends only on the nature of the material, not on the dimensions of the specimen. It is defined as $$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$$, where $$F$$ is the applied force, $$A$$ is the cross-sectional area, $$\Delta L$$ is the change in length, and $$L$$ is the original length.

When the length is doubled and the radius is halved, these are changes to the geometry of the wire, not to its material composition. Since Young's modulus is an intrinsic property of the material, it does not change when the dimensions of the wire are altered.

Hence, the correct answer is Option A.

Let the cross-sectional area of the wire be $$A$$, its Young’s modulus be $$Y$$, and the acceleration due to gravity be $$g$$.

When a mass of $$1\text{ kg}$$ is suspended, by Hooke’s law the extension is $$\Delta L_1 = \dfrac{(1\cdot g)\,L}{A\,Y}$$.
Hence the new length is $$L_1 = L + \frac{g\,L}{A\,Y}\quad-(1)$$

When a mass of $$2\text{ kg}$$ is suspended, the extension is $$\Delta L_2 = \dfrac{(2\cdot g)\,L}{A\,Y}$$.
Hence the new length is $$L_2 = L + \frac{2\,g\,L}{A\,Y}\quad-(2)$$

Subtracting $$(1)$$ from $$(2)$$ gives $$L_2 - L_1 = \frac{2\,g\,L}{A\,Y} - \frac{g\,L}{A\,Y} = \frac{g\,L}{A\,Y} \quad-(3)$$

Substitute equation $$(3)$$ into equation $$(1)$$:
$$L_1 = L + \frac{g\,L}{A\,Y} = L + (L_2 - L_1)$$

Rearrange to obtain $$L_1 = L + L_2 - L_1 \quad\Longrightarrow\quad 2\,L_1 = L + L_2$$ $$\therefore\quad L = 2\,L_1 - L_2$$

This matches Option C.

The elongation of a wire under a linear force $$F$$ is given by $$\Delta L = \frac{FL}{AY}$$, where $$L$$ is the length of the wire, $$A$$ is the cross-sectional area, and $$Y$$ is the Young's modulus of the material.

For the original wire with length $$L$$, diameter $$d$$, and area $$A = \frac{\pi d^2}{4}$$, the elongation is $$\Delta L_1 = \frac{FL}{AY} = 0.04$$ m.

When both the length and diameter are doubled, the new length is $$2L$$ and the new diameter is $$2d$$, giving a new cross-sectional area of $$A' = \frac{\pi (2d)^2}{4} = 4A$$.

The new elongation under the same force $$F$$ is $$\Delta L_2 = \frac{F(2L)}{(4A)Y} = \frac{2FL}{4AY} = \frac{1}{2} \cdot \frac{FL}{AY} = \frac{\Delta L_1}{2}$$.

Substituting the given value: $$\Delta L_2 = \frac{0.04}{2} = 0.02$$ m $$= 2$$ cm.

Therefore, the elongation is $$2$$ cm.

We have to find that depth in sea water where the pressure becomes high enough to compress the rubber ball so that its volume falls by 0.5 %. The datum values supplied are:

Bulk modulus of rubber $$B = 9.8 \times 10^{8}\ \text{N m}^{-2}$$,   density of sea-water $$\rho = 10^{3}\ \text{kg m}^{-3}$$,   and the acceleration due to gravity $$g = 9.8\ \text{m s}^{-2}$$.

The fractional change in volume is given as $$0.5\%$$, i.e.

$$\frac{\Delta V}{V} = -0.5\% = -\frac{0.5}{100} = -0.005.$$

For bulk deformation the defining relation is first stated:

$$B = -\frac{\Delta P}{\dfrac{\Delta V}{V}}.$$

Here $$\Delta P$$ is the increase in external pressure which produces the fractional volume change $$\dfrac{\Delta V}{V}$$. In our situation the sign merely tells the direction of change; we can work with magnitudes:

$$B = \frac{\Delta P}{\left|\dfrac{\Delta V}{V}\right|}.$$

The pressure increase at a depth $$h$$ in a fluid of density $$\rho$$ is furnished by the hydrostatic relation

$$\Delta P = \rho\,g\,h.$$

Substituting this expression for $$\Delta P$$ into the bulk-modulus formula gives

$$B \;=\; \frac{\rho\,g\,h}{\left|\dfrac{\Delta V}{V}\right|}.$$

Solving for the depth $$h$$ yields

$$h \;=\; \frac{B\,\left|\dfrac{\Delta V}{V}\right|}{\rho\,g}.$$

Next we substitute the numerical quantities step by step:

$$h = \frac{\displaystyle 9.8 \times 10^{8}\ \text{N m}^{-2} \;\times\; 0.005}{\displaystyle 10^{3}\ \text{kg m}^{-3} \;\times\; 9.8\ \text{m s}^{-2}}.$$

First multiply the numerator:

$$9.8 \times 10^{8} \times 0.005 \;=\; 9.8 \times 5 \times 10^{8} \times 10^{-3} = 49 \times 10^{5} = 4.9 \times 10^{6}.$$

Now compute the denominator:

$$10^{3} \times 9.8 \;=\; 9.8 \times 10^{3}.$$

Dividing the two magnitudes, we have

$$h = \frac{4.9 \times 10^{6}}{9.8 \times 10^{3}} = \left(\frac{4.9}{9.8}\right) \times 10^{3} = 0.5 \times 10^{3} = 5.0 \times 10^{2} \ \text{m}.$$

Therefore

$$h = 500 \ \text{m}.$$

So, the answer is $$500\ \text{m}$$.

We begin with the idea that, when a rod is heated but its length is not allowed to change, the thermal tendency to expand is opposed by an internal restoring force. The result is a thermal stress. For a uniform rod that is perfectly constrained, the magnitude of this stress is given by the well-known relation

$$\sigma = Y \, \alpha \, \Delta T,$$

where $$\sigma$$ is the stress, $$Y$$ is Young’s modulus, $$\alpha$$ is the coefficient of linear expansion, and $$\Delta T$$ is the change in temperature.

We have the numerical data: $$Y = 2.0 \times 10^{11}\, \text{N m}^{-2}, \quad \alpha = 10^{-5}\, ^\circ\text{C}^{-1}, \quad \Delta T = 400^\circ\text{C}.$$ Substituting these values,

$$\sigma = \left(2.0 \times 10^{11}\right)\!\left(10^{-5}\right)\!\left(400\right).$$

First multiply $$10^{-5}$$ and $$400$$:

$$10^{-5} \times 400 = 4 \times 10^{-3}.$$

Next multiply this result by $$2.0 \times 10^{11}$$:

$$\sigma = 2.0 \times 10^{11} \times 4 \times 10^{-3} = 8 \times 10^{8}\, \text{N m}^{-2}.$$

So the thermal stress inside the rod is $$\sigma = 8 \times 10^{8}\, \text{N m}^{-2}.$$

The tension (force) in the rod is obtained from the definition $$\sigma = F/A,$$ where $$A$$ is the cross-sectional area. Rearranging, $$F = \sigma A.$$ The given area is $$10 \text{ cm}^2.$$ Converting to square metres,

$$10 \text{ cm}^2 = 10 \times 10^{-4} \text{ m}^2 = 10^{-3} \text{ m}^2.$$

Now substitute $$\sigma = 8 \times 10^{8}\, \text{N m}^{-2}$$ and $$A = 10^{-3} \text{ m}^2$$ into $$F = \sigma A$$:

$$F = \left(8 \times 10^{8}\right) \left(10^{-3}\right) = 8 \times 10^{5}\, \text{N}.$$

The question states that this force can be written as $$x \times 10^{5}\, \text{N}.$$ Comparing, we see $$x = 8.$$

Hence, the correct answer is Option 8.

The ratio of hydraulic stress to hydraulic strain is the bulk modulus $$B$$. The hydraulic stress (pressure) at a depth of 2 km beneath the water surface is $$P = \rho g h = 1000 \times 9.8 \times 2000 = 1.96 \times 10^7$$ N m$$^{-2}$$.

The hydraulic strain is the fractional compression $$\frac{\Delta V}{V} = 1.36\% = 0.0136$$.

Therefore the bulk modulus is $$B = \frac{P}{\Delta V / V} = \frac{1.96 \times 10^7}{0.0136} = 1.44 \times 10^9$$ N m$$^{-2}$$.

We have two wires, each of original length $$L$$ and the same radius $$r$$, therefore each has the same cross-sectional area $$A = \pi r^2$$. They are joined end to end, so the total length of the combination is $$2L$$. The same tensile force $$F$$ acts through the whole system because the wires are in series.

For any wire, Young’s modulus is defined by the formula

$$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta l / l}.$$

Applying this separately to the two wires, we write the extensions $$\Delta l_1$$ and $$\Delta l_2$$ produced in the two materials:

First wire:

$$Y_1=\frac{F/A}{\Delta l_1/L}\qquad\Rightarrow\qquad \Delta l_1=\frac{F\,L}{A\,Y_1}.$$

Second wire:

$$Y_2=\frac{F/A}{\Delta l_2/L}\qquad\Rightarrow\qquad \Delta l_2=\frac{F\,L}{A\,Y_2}.$$

The total extension of the composite wire is simply the sum of the individual extensions because the wires are connected in series, so

$$\Delta l=\Delta l_1+\Delta l_2 =\frac{F\,L}{A\,Y_1}+\frac{F\,L}{A\,Y_2} =\frac{F\,L}{A}\left(\frac{1}{Y_1}+\frac{1}{Y_2}\right).$$

Now, let the combination behave like a single uniform wire of total length $$2L$$, radius $$r$$, and some effective Young’s modulus $$Y$$. Using the same definition of Young’s modulus for this equivalent single wire, we have

$$Y=\frac{F/A}{\Delta l/(2L)}=\frac{F}{A}\cdot\frac{2L}{\Delta l} =\frac{2F\,L}{A\,\Delta l}.$$

Substituting the value of $$\Delta l$$ obtained above, we get

$$Y=\frac{2F\,L}{A\left[\dfrac{F\,L}{A}\left(\dfrac{1}{Y_1}+\dfrac{1}{Y_2}\right)\right]} =\frac{2}{\dfrac{1}{Y_1}+\dfrac{1}{Y_2}}.$$

Combining the fractions in the denominator,

$$\frac{1}{Y_1}+\frac{1}{Y_2}=\frac{Y_1+Y_2}{Y_1Y_2}.$$

So,

$$Y=\frac{2}{\dfrac{Y_1+Y_2}{Y_1Y_2}} =\frac{2Y_1Y_2}{Y_1+Y_2}.$$

Thus the effective Young’s modulus of the composite wire is

$$Y=\frac{2Y_1Y_2}{Y_1+Y_2}.$$

Hence, the correct answer is Option B.

We need to determine the radius of curvature of the common interface formed when two soap bubbles of different radii coalesce under isothermal conditions.

1. Understand Excess Pressure in a Soap Bubble

A soap bubble has two free surfaces (inner and outer) in contact with air. The excess pressure ($$\Delta P$$) inside a soap bubble of radius $$r$$ with surface tension $$T$$ is given by the formula:

$$\Delta P = \frac{4T}{r}$$

Let the two individual soap bubbles have radii $$a$$ and $$b$$, with given condition $$b > a$$.

• Excess pressure inside the smaller bubble ($$P_a$$): $$P_a = \frac{4T}{a}$$
• Excess pressure inside the larger bubble ($$P_b$$): $$P_b = \frac{4T}{b}$$

Since $$b > a$$, the internal pressure of the smaller bubble is greater than that of the larger bubble ($$P_a > P_b$$).

2. Analyze the Common Interface Surface

When the two bubbles coalesce and form a common internal boundary, the pressure difference across this shared wall determines its curvature.

Because the smaller bubble has a higher internal pressure, the common surface bulges into the larger bubble, becoming concave towards the smaller bubble.

The net excess pressure ($$\Delta P_{\text{common}}$$) acting across this common surface of radius $$R$$ is:

$$\Delta P_{\text{common}} = P_a - P_b$$

3. Deriving the Radius of Curvature ($$R$$)

Substitute the structural excess pressure formulas into the balance equation:

$$\frac{4T}{R} = \frac{4T}{a} - \frac{4T}{b}$$

Dividing the entire equation by $$4T$$ simplifies it to:

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

Find a common denominator on the right-hand side:

$$\frac{1}{R} = \frac{b - a}{ab}$$

Taking the reciprocal to isolate $$R$$ gives:

$$R = \frac{ab}{b - a}$$

Conclusion

The radius of curvature of the common surface is $$\frac{ab}{b - a}$$, which corresponds to Option A.

The total mass of the structure is given as $$m = 50 \times 10^{3}\ \text{kg}$$. The weight is the gravitational force $$mg$$ acting downward. Because the load is uniformly distributed over four identical columns, each column supports exactly one‐fourth of this weight.

First we calculate the total weight:

$$W_{\text{total}} = mg = \left( 50 \times 10^{3}\ \text{kg} \right)\left( 9.8\ \text{m s}^{-2} \right) = 4.9 \times 10^{5}\ \text{N}.$$

The load per column is therefore

$$F = \frac{W_{\text{total}}}{4} = \frac{4.9 \times 10^{5}\ \text{N}}{4} = 1.225 \times 10^{5}\ \text{N}.$$

Each column is a hollow cylinder with outer radius $$R = 100\ \text{cm} = 1.0\ \text{m}$$ and inner radius $$r = 50\ \text{cm} = 0.50\ \text{m}.$$ The cross-sectional area of a hollow cylinder is

$$A = \pi \left( R^{2} - r^{2} \right).$$

Substituting the radii,

$$A = \pi \left[ (1.0\ \text{m})^{2} - (0.50\ \text{m})^{2} \right] = \pi \left( 1.00 - 0.25 \right)\ \text{m}^{2} = 0.75\pi\ \text{m}^{2}.$$

Numerically,

$$A = 0.75 \times 3.1416\ \text{m}^{2} \approx 2.356\ \text{m}^{2}.$$

Now we compute the compressive stress on each column. By definition,

$$\sigma = \frac{\text{Force}}{\text{Area}} = \frac{F}{A}.$$

Hence,

$$\sigma = \frac{1.225 \times 10^{5}\ \text{N}}{2.356\ \text{m}^{2}} \approx 5.20 \times 10^{4}\ \text{Pa}.$$

The relation between Young’s modulus $$Y$$, stress $$\sigma$$ and longitudinal strain $$\varepsilon$$ is

$$Y = \frac{\sigma}{\varepsilon} \quad\Longrightarrow\quad \varepsilon = \frac{\sigma}{Y}.$$

With $$Y = 2.0 \times 10^{11}\ \text{Pa},$$ we have

$$\varepsilon = \frac{5.20 \times 10^{4}\ \text{Pa}} {2.0 \times 10^{11}\ \text{Pa}} = 2.60 \times 10^{-7}.$$

Thus the compression strain in each steel column is

$$\boxed{2.60 \times 10^{-7}}.$$

Hence, the correct answer is Option B.

The standard relation between Young's modulus $$Y$$, bulk modulus $$K$$, and modulus of rigidity $$\eta$$ is $$Y = \frac{9K\eta}{3K + \eta}$$.

We rearrange this to express $$K$$ in terms of $$Y$$ and $$\eta$$. Cross-multiplying, we get $$Y(3K + \eta) = 9K\eta$$.

Expanding, $$3YK + Y\eta = 9K\eta$$.

Collecting terms with $$K$$, we get $$9K\eta - 3YK = Y\eta$$, so $$K(9\eta - 3Y) = Y\eta$$.

Therefore, $$K = \frac{Y\eta}{9\eta - 3Y}$$ N m$$^{-2}$$.

Hence, the correct answer is Option C.

We need to determine the minimum radius of the metal wire supporting two suspended masses in an Atwood machine setup without exceeding its breaking stress.

1. Identify the Given Parameters

From the problem , the system consists of two masses connected across a smooth pulley by a metal wire:

• Mass of the first block ($$m_1$$) = $$3\text{ kg}$$
• Mass of the second block ($$m_2$$) = $$5\text{ kg}$$
• Breaking stress of the metal ($$\sigma_{\text{breaking}}$$) = $$\frac{24}{\pi} \times 10^2\text{ N m}^{-2}$$
• Acceleration due to gravity ($$g$$) = $$10\text{ m s}^{-2}$$

2. Calculate the Tension ($$T$$) in the Wire

For a standard ideal pulley system containing two unequal masses, the dynamic tension ($$T$$) developed in the connecting string while the blocks are in motion is given by the formula:

$$T = \frac{2 m_1 m_2}{m_1 + m_2} g$$

Substituting the given mass values and acceleration due to gravity:

$$T = \frac{2 \times 3 \times 5}{3 + 5} \times 10 = \frac{30}{8} \times 10 = \frac{300}{8} = 37.5\text{ N}$$

3. Relate Stress to the Minimum Radius ($$r$$)

The internal mechanical stress experienced by a wire of a circular cross-section is defined as the restoring force per unit area. To prevent the wire from snapping, the operational stress must not exceed the breaking limit:

$$\sigma_{\text{breaking}} = \frac{\text{Tension}}{\text{Cross-sectional Area}} = \frac{T}{\pi r^2}$$

Rearranging the expression to solve for the radius ($$r$$):

$$r^2 = \frac{T}{\pi \cdot \sigma_{\text{breaking}}}$$

Substituting the computed values of tension ($$T = 37.5\text{ N}$$) and breaking stress into the equation:

$$r^2 = \frac{37.5}{\pi \times \left(\frac{24}{\pi} \times 10^2\right)} = \frac{37.5}{24 \times 10^2} = \frac{1.5625}{10^2} = 0.015625\text{ m}^2$$

Taking the square root on both sides to find the radius in meters:

$$r = \sqrt{0.015625} = 0.125\text{ m}$$

4. Convert the Unit to Centimeters

To match the final options provided in the problem layout, we convert the minimum required radius from meters to centimeters ($$1\text{ m} = 100\text{ cm}$$):

$$r = 0.125 \times 100\text{ cm} = 12.5\text{ cm}$$

Conclusion

The minimum radius of the wire required to support the moving masses without breaking is 12.5 cm.

Let the original (natural) length of the wire be $$l_0$$ and the cross-sectional area be $$A$$. When a tension $$T$$ is applied, the extension is $$\Delta l = \frac{Tl_0}{YA}$$, where $$Y$$ is Young's modulus.

When tension $$T_1$$ is applied, the length becomes $$l_1 = l_0 + \frac{T_1 l_0}{YA}$$, which gives $$l_1 - l_0 = \frac{T_1 l_0}{YA}$$ ... (i).

When tension $$T_2$$ is applied, the length becomes $$l_2 = l_0 + \frac{T_2 l_0}{YA}$$, which gives $$l_2 - l_0 = \frac{T_2 l_0}{YA}$$ ... (ii).

Dividing equation (i) by equation (ii): $$\frac{l_1 - l_0}{l_2 - l_0} = \frac{T_1}{T_2}$$.

Cross-multiplying: $$T_2(l_1 - l_0) = T_1(l_2 - l_0)$$, which gives $$T_2 l_1 - T_2 l_0 = T_1 l_2 - T_1 l_0$$. Rearranging: $$l_0(T_1 - T_2) = T_1 l_2 - T_2 l_1$$, so $$l_0 = \frac{T_1 l_2 - T_2 l_1}{T_1 - T_2} = \frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}$$.

Let $$\ell_0$$ be the natural length and $$Y$$ be Young's modulus, $$A$$ the cross-sectional area. By Hooke's law, extension $$= \frac{T \ell_0}{YA}$$, so the stretched length is $$\ell = \ell_0 + \frac{T \ell_0}{YA} = \ell_0\left(1 + \frac{T}{YA}\right)$$.

For the two cases: $$\ell_1 = \ell_0\left(1 + \frac{T_1}{YA}\right)$$ and $$\ell_2 = \ell_0\left(1 + \frac{T_2}{YA}\right)$$.

Let $$k = \frac{\ell_0}{YA}$$, so $$\ell_1 = \ell_0 + kT_1$$ and $$\ell_2 = \ell_0 + kT_2$$. From these two equations: $$\ell_2 - \ell_1 = k(T_2 - T_1)$$, giving $$k = \frac{\ell_2 - \ell_1}{T_2 - T_1}$$.

Substituting back: $$\ell_1 = \ell_0 + \frac{(\ell_2 - \ell_1)T_1}{T_2 - T_1}$$, so $$\ell_0 = \ell_1 - \frac{(\ell_2 - \ell_1)T_1}{T_2 - T_1} = \frac{\ell_1(T_2 - T_1) - T_1(\ell_2 - \ell_1)}{T_2 - T_1} = \frac{\ell_1 T_2 - \ell_2 T_1}{T_2 - T_1}$$.

Therefore the natural length of the wire is $$\ell_0 = \dfrac{\ell_1 T_2 - \ell_2 T_1}{T_2 - T_1}$$.

Let us denote the atmospheric (external) pressure by $$P_{0}$$. For a thin soap bubble the excess pressure inside the bubble is given by the well-known formula

$$\Delta P \;=\; P_{\text{inside}} - P_{0} \;=\; \dfrac{4T}{r}$$

Here $$T$$ is the surface tension of the liquid film and $$r$$ is the radius of the spherical bubble. This expression contains the factor $$4T$$ (and not $$2T$$) because a soap bubble has two liquid surfaces in contact with air, one on the inside and one on the outside.

We are told that the absolute internal pressures of the two bubbles are

$$P_{1}=1.01\;\text{atm},\qquad P_{2}=1.02\;\text{atm}$$

and, naturally, both bubbles are in the same atmosphere so

$$P_{0}=1.00\;\text{atm}$$

Using the formula for each bubble, we first calculate their respective excess pressures:

For the first bubble $$\Delta P_{1}=P_{1}-P_{0}=1.01-1.00=0.01\;\text{atm}$$

For the second bubble $$\Delta P_{2}=P_{2}-P_{0}=1.02-1.00=0.02\;\text{atm}$$

Now we substitute these excess pressures into the relation $$\Delta P=\dfrac{4T}{r}$$ to connect each pressure to the corresponding radius.

For the first bubble $$\dfrac{4T}{r_{1}}=0.01\;\text{atm}\qquad\Longrightarrow\qquad r_{1}=\dfrac{4T}{0.01\;\text{atm}}$$

For the second bubble $$\dfrac{4T}{r_{2}}=0.02\;\text{atm}\qquad\Longrightarrow\qquad r_{2}=\dfrac{4T}{0.02\;\text{atm}}$$

To compare the radii, we take the ratio of the two expressions:

$$\dfrac{r_{2}}{r_{1}} =\dfrac{\dfrac{4T}{0.02}}{\dfrac{4T}{0.01}} =\dfrac{0.01}{0.02} =\dfrac{1}{2}$$

So we find $$r_{2}=\dfrac{r_{1}}{2}$$ meaning the second bubble (with the higher internal pressure) has half the radius of the first bubble.

Each bubble is spherical, so its volume is given by the standard geometric formula

$$V=\dfrac{4}{3}\pi r^{3}$$

Therefore, the ratio of their volumes becomes

$$\dfrac{V_{1}}{V_{2}} =\dfrac{\dfrac{4}{3}\pi r_{1}^{3}}{\dfrac{4}{3}\pi r_{2}^{3}} =\left(\dfrac{r_{1}}{r_{2}}\right)^{3} =\left(\dfrac{r_{1}}{\,r_{1}/2}\right)^{3} =(2)^{3}=8$$

Thus

$$V_{1}:V_{2}=8:1$$

The option that matches this ratio is Option C.

Hence, the correct answer is Option C.

We have two wires, one of brass and one of steel, each of length $$L = 1 \text{ m}$$ and cross-sectional area $$A = 1 \text{ mm}^2 = 1 \times 10^{-6}\,\text{m}^2$$. The wires are joined in series, so when a tensile force is applied the same force $$F$$ acts on both wires. Because the area is the same for both, the stress $$\sigma$$ in each wire is also the same, where

$$\sigma = \dfrac{F}{A}.$$

The elongation of a wire obeys Hooke’s law for small strains:

$$\Delta L = \dfrac{\sigma L}{Y},$$

where $$Y$$ is Young’s modulus. First we write down the given values:

$$Y_{\text{steel}} = 120 \times 10^{9}\,\text{N m}^{-2},\qquad Y_{\text{brass}} = 60 \times 10^{9}\,\text{N m}^{-2}.$$

For the brass wire the extension is

$$\Delta L_{\text{brass}} = \dfrac{\sigma L}{Y_{\text{brass}}},$$

and for the steel wire

$$\Delta L_{\text{steel}} = \dfrac{\sigma L}{Y_{\text{steel}}}.$$

The two wires are in series, so the total elongation is the sum of the individual elongations:

$$\Delta L_{\text{total}} = \Delta L_{\text{brass}} + \Delta L_{\text{steel}} = \dfrac{\sigma L}{Y_{\text{brass}}} + \dfrac{\sigma L}{Y_{\text{steel}}} = \sigma L\left(\dfrac{1}{Y_{\text{brass}}} + \dfrac{1}{Y_{\text{steel}}}\right).$$

The experiment specifies a required total elongation of

$$\Delta L_{\text{total}} = 0.2 \text{ mm} = 0.2 \times 10^{-3}\,\text{m}.$$

Substituting $$L = 1 \text{ m}$$ and solving for $$\sigma$$ gives

$$\sigma = \dfrac{\Delta L_{\text{total}}} {L\left(\dfrac{1}{Y_{\text{brass}}} + \dfrac{1}{Y_{\text{steel}}}\right)}.$$

Now we evaluate the bracketed term:

$$\dfrac{1}{Y_{\text{brass}}} + \dfrac{1}{Y_{\text{steel}}} = \dfrac{1}{60 \times 10^{9}} + \dfrac{1}{120 \times 10^{9}} = \dfrac{2}{120 \times 10^{9}} + \dfrac{1}{120 \times 10^{9}} = \dfrac{3}{120 \times 10^{9}} = \dfrac{1}{40 \times 10^{9}} = 2.5 \times 10^{-11}\,\text{m}^2\text{N}^{-1}.$$

Putting everything together, we obtain

$$\sigma = \dfrac{0.2 \times 10^{-3}} {1 \times 2.5 \times 10^{-11}} = \dfrac{0.2}{2.5} \times 10^{8} = 0.08 \times 10^{8} = 8.0 \times 10^{6}\,\text{N m}^{-2}.$$

Hence, the correct answer is Option A.

For a uniform wire that is stretched within the elastic limit, the extension produced by a load is given by the standard elastic‐string formula

$$\Delta L=\frac{F\,L}{A\,Y},$$

where $$F$$ is the applied force, $$L$$ is the original length, $$A$$ is the cross-sectional area and $$Y$$ is Young’s modulus of the material.

Because the two wires are subjected to the same load and are said to “stretch by the same length”, we have for wires A and B

$$\frac{F\,L_A}{A_A\,Y_A}=\frac{F\,L_B}{A_B\,Y_B}.$$

The force $$F$$ and the factor $$\pi$$ hidden inside each area will cancel, so we can write

$$\frac{L_A}{r_A^{\,2}\,Y_A}=\frac{L_B}{r_B^{\,2}\,Y_B},$$

where $$r_A=R$$ (unknown) and $$r_B=2\;\text{mm}$$ (given).

The ratio of Young’s moduli is given as $$Y_A:Y_B=7:4$$, so we may set $$Y_B=\dfrac{4}{7}\,Y_A$$.

Substituting this value together with the given lengths $$L_A=2\;\text{m}$$ and $$L_B=1.5\;\text{m}$$ gives

$$\frac{2}{R^{2}\,Y_A}=\frac{1.5}{(2\;\text{mm})^{2}\,\left(\dfrac{4}{7}Y_A\right)}.$$

The factor $$Y_A$$ now cancels out completely, leading to

$$\frac{2}{R^{2}}=\frac{1.5\times7}{(2\;\text{mm})^{2}\times4}.$$

Taking the reciprocal of both sides so that $$R^{2}$$ stands alone, we obtain

$$R^{2}=\frac{2\times4}{1.5\times7}\,(2\;\text{mm})^{2}.$$

Simplifying each numerical factor step by step:

$$\frac{2}{1.5}=\frac{4}{3}, \quad \text{so} \quad R^{2}=\frac{4}{3}\times\frac{4}{7}\times(2\;\text{mm})^{2}.$$

Next, multiply the first two fractions:

$$\frac{4}{3}\times\frac{4}{7}=\frac{16}{21}.$$

Because $$(2\;\text{mm})^{2}=4\;\text{mm}^{2},$$ we have

$$R^{2}=\frac{16}{21}\times4\;\text{mm}^{2}=\frac{64}{21}\;\text{mm}^{2}.$$

Evaluating the fraction numerically,

$$\frac{64}{21}\approx3.0476\;\text{mm}^{2}.$$

Taking the square root finally yields

$$R=\sqrt{3.0476}\;\text{mm}\approx1.7457\;\text{mm}.$$

The value rounds to about $$1.7\;\text{mm}$$, which matches the fourth option in the list.

Hence, the correct answer is Option D.

In Searle’s apparatus the extension in the steel wire is obtained from Young’s modulus formula

$$\Delta L=\frac{F\,L}{A\,Y}$$

where $$\Delta L$$ is the increase in length, $$F$$ the tensile force in the wire, $$L$$ the original length, $$A$$ the cross-sectional area and $$Y$$ Young’s modulus of steel. For a given wire $$L,\;A$$ and $$Y$$ are fixed, so the extension is directly proportional to the applied force:

$$\Delta L\;\propto\;F$$

Initially the wire supports the load in air only. The force is simply the weight

$$F_1 = M\,g$$

and the corresponding extension is given as

$$\Delta L_1 = 4.0\;\text{mm}$$

Now the load is completely immersed in a liquid. The liquid exerts an upthrust (buoyant force) that reduces the effective weight. We determine this buoyant force step by step.

The relative density (specific gravity) of the load is 8, so its density is

$$\rho_{\text{load}} = 8\,\rho_{\text{water}}$$

The relative density of the liquid is 2, so the liquid’s density is

$$\rho_{\text{liq}} = 2\,\rho_{\text{water}}$$

The volume of the load is

$$V = \frac{M}{\rho_{\text{load}}} = \frac{M}{8\,\rho_{\text{water}}}$$

The buoyant force equals the weight of the displaced liquid:

$$F_b = \rho_{\text{liq}}\,V\,g = (2\,\rho_{\text{water}})\left(\frac{M}{8\,\rho_{\text{water}}}\right)g = \frac{2}{8}\,M\,g = \frac{1}{4}\,M\,g$$

Hence the apparent (effective) weight of the load while immersed is

$$F_2 = F_1 - F_b = M\,g - \frac{1}{4}\,M\,g = \frac{3}{4}\,M\,g$$

Because extension is proportional to force, the new extension $$\Delta L_2$$ satisfies

$$\frac{\Delta L_2}{\Delta L_1} = \frac{F_2}{F_1} = \frac{\frac{3}{4}M\,g}{M\,g} = \frac{3}{4}$$

Substituting $$\Delta L_1 = 4.0\;\text{mm}$$, we obtain

$$\Delta L_2 = \frac{3}{4}\times 4.0\;\text{mm} = 3.0\;\text{mm}$$

Hence, the correct answer is Option D.

Strain energy stored in the stretched rubber cord is

$$U=\frac{1}{2}\times \text{stress} \times \text{strain} \times \text{volume}$$

Using,

$$U=\frac{1}{2}\left(\frac{\Delta \ell}{\ell}\right)^2 Y A \ell$$

This stored energy converts into kinetic energy of the stone.

Hence,

$$\frac{1}{2}\left(\frac{\Delta \ell}{\ell}\right)^2 Y A\ell = \frac{1}{2}mv^2$$

Given,

$$\ell=42\ \text{cm}=42\times10^{-2}\ \text{m}$$$$\Delta \ell=20\ \text{cm}=20\times10^{-2}\ \text{m}$$

Radius of cord,

$$r=3\ \text{mm}=3\times10^{-3}\ \text{m}$$

Area of cross-section,

$$A=\pi r^2=\pi(3\times10^{-3})^2$$

Mass of stone,

$$m=0.02\ \text{kg}=2\times10^{-2}\ \text{kg}$$

Velocity,

$$v=20\ \text{m s}^{-1}$$

Substituting,

$$\frac{1}{2}\left(\frac{20}{42}\right)^2 Y\times \pi(3\times10^{-3})^2 \times 42\times10^{-2} = \frac{1}{2}\times2\times10^{-2}\times(20)^2$$

Solving,

$$Y \approx 3\times10^6\ \text{N m}^{-2}$$

Hence,

$$\boxed{Y \approx 3\times10^6\ \text{N m}^{-2}}$$

We have a brass wire whose un-stretched length is given as $$l = 0.2\text{ m}$$ and whose radius is $$r = 1\text{ mm}=1\times 10^{-3}\text{ m}$$. The cross-sectional area is therefore

$$A = \pi r^{2}= \pi\,(1\times 10^{-3})^{2}= \pi\times 10^{-6}\;{\rm m^{2}}.$$

The wire is first at $$40^{\circ}{\rm C}$$, a mass $$M$$ is hung on it, and then the system is cooled to $$20^{\circ}{\rm C}$$. The information that “it regains its original length” tells us that the increase in length caused by the load is exactly cancelled by the decrease in length caused by cooling. Hence the magnitude of the elastic extension equals the magnitude of the thermal contraction:

$$\Delta l_{\text{load}}=\Delta l_{\text{thermal}}.$$

For a wire, the elastic (Young’s-law) extension produced by a force $$F$$ is

$$\Delta l_{\text{load}}=\frac{F\,l}{A\,Y},$$

where $$Y$$ is Young’s modulus. Here $$F=Mg$$, so

$$\boxed{\;\Delta l_{\text{load}}=\dfrac{Mg\,l}{A\,Y}\;}.$$

The thermal contraction when the temperature falls by $$\Delta T$$ is obtained from the linear expansion formula

$$\Delta l_{\text{thermal}}=\alpha\,l\,\Delta T,$$

where $$\alpha$$ is the coefficient of linear expansion. Because the wire is cooled, $$\Delta T = -20^{\circ}\text{C}$$, and the magnitude is

$$|\Delta l_{\text{thermal}}|=\alpha\,l\,(20).$$

Equating the two magnitudes we get

$$\frac{Mg\,l}{A\,Y} \;=\; \alpha\,l\,(20).$$

The original length $$l$$ cancels from both sides, giving a direct expression for the unknown mass:

$$\boxed{\;M = \frac{\alpha\,(20)\,A\,Y}{g}\;}.$$

Now we substitute the numerical data supplied in the question:

Coefficient of linear expansion: $$\alpha = 1.0\times 10^{-5}\;{\rm /^\circ C}$$
Temperature drop: $$20^{\circ}{\rm C}$$
Area: $$A = \pi\times 10^{-6}\;{\rm m^{2}}$$
Young’s modulus: $$Y = 1.0\times 10^{11}\;{\rm N/m^{2}}$$
Acceleration due to gravity: $$g = 10\;{\rm m/s^{2}}$$.

Putting these numbers in:

$$M = \frac{(1.0\times 10^{-5})(20)\,(\pi\times 10^{-6})\,(1.0\times 10^{11})}{10}.$$

Simplifying step by step, we first multiply the powers of ten:

$$1.0\times 10^{-5}\times 20 = 2.0\times 10^{-4},$$

and

$$(\pi\times 10^{-6})\times (1.0\times 10^{11}) = \pi\times 10^{5}.$$

Hence the numerator becomes

$$ (2.0\times 10^{-4})\times (\pi\times 10^{5}) = 2\pi\times 10^{1}=20\pi. $$

Finally we divide by $$g=10$$:

$$ M = \frac{20\pi}{10}=2\pi\;{\rm kg}\approx 6.3\;{\rm kg}. $$

The mass required is therefore about $$6\;{\rm kg}$$. Looking at the four choices provided, the value that lies closest to this calculation is $$9\;{\rm kg}$$.

Hence, the correct answer is Option D.

We have a vertical wire that is supporting a stationary load, so the force producing the stress is simply the weight of the load.

First, recall the definition of tensile stress.
$$\text{Tensile stress} \; (\sigma) \;=\; \dfrac{\text{Force}}{\text{Cross-sectional area}}$$

The force is the weight $$W$$ of the 4 kg mass. By Newton’s second law, weight is mass times acceleration due to gravity:

$$W \;=\; m\,g$$ where $$m = 4\ \text{kg}$$ and $$g = 3.1\pi\ \text{m s}^{-2}$$.

Substituting the numbers,

$$W \;=\; 4 \times (3.1\pi) \;=\; 12.4\pi\ \text{N}.$$

Next we need the cross-sectional area $$A$$ of the wire. The radius is given as 2.0 mm, so

$$r \;=\; 2.0\ \text{mm} \;=\; 2.0 \times 10^{-3}\ \text{m}.$$

For a circle, the area formula is

$$A \;=\; \pi r^{2}.$$

Substituting the value of $$r$$,

$$A \;=\; \pi \left(2.0 \times 10^{-3}\right)^{2} \;=\; \pi \left(4.0 \times 10^{-6}\right) \;=\; 4.0\pi \times 10^{-6}\ \text{m}^{2}.$$

Now we divide the force by the area to obtain the stress:

$$\sigma \;=\; \dfrac{W}{A} \;=\; \dfrac{12.4\pi}{4.0\pi \times 10^{-6}}.$$

The factor $$\pi$$ appears in both numerator and denominator, so it cancels out:

$$\sigma \;=\; \dfrac{12.4}{4.0} \times 10^{6} \;=\; 3.1 \times 10^{6}\ \text{N m}^{-2}.$$

Hence, the correct answer is Option D.

Let the linear coefficient of thermal expansion of the material be denoted by $$\alpha$$ and its coefficient of volume expansion by $$\beta$$. For any isotropic solid we have the standard relation $$\beta = 3\alpha$$ because uniform heating produces the same fractional change along all three mutually perpendicular directions.

When the temperature of the rod is raised by $$T$$, the natural (unconstrained) increase in its length would be given by the well-known thermal expansion formula

$$\Delta L_{\text{thermal}} = \alpha L T.$$

Simultaneously, the rod is subjected to a uniform compressive longitudinal force $$F$$. The mechanical (elastic) shortening produced by this force is obtained from the definition of Young’s modulus. First, the longitudinal stress is

$$\text{Stress} = \frac{F}{A},$$

where $$A$$ is the cross-sectional area. For a cylinder of radius $$r$$ we have $$A = \pi r^{2}.$$

Young’s modulus $$Y$$ is defined by the relation

$$Y = \frac{\text{Stress}}{\text{Strain}} \;\;\; \Longrightarrow \;\;\; \text{Strain} = \frac{\text{Stress}}{Y}.$$

The longitudinal strain is the fractional change in length, so

$$\text{Strain} = \frac{\Delta L_{\text{elastic}}}{L} = \frac{F}{A Y}.$$

Since the force is compressive, this change in length is a decrease, hence

$$\Delta L_{\text{elastic}} = -\,\frac{F L}{A Y} = -\,\frac{F L}{\pi r^{2} Y}.$$

The problem states that the rod’s overall length does not change at all, which means the algebraic sum of the thermal increase and the elastic decrease is zero:

$$\Delta L_{\text{thermal}} + \Delta L_{\text{elastic}} = 0.$$

Substituting the two expressions we have just obtained,

$$\alpha L T \;+\;\left(-\,\frac{F L}{\pi r^{2} Y}\right) = 0.$$

The common factor $$L$$ can be cancelled on both sides, giving

$$\alpha T - \frac{F}{\pi r^{2} Y} = 0.$$

Solving for $$\alpha$$ we obtain

$$\alpha = \frac{F}{\pi r^{2} Y T}.$$

Finally, using the earlier relation $$\beta = 3\alpha$$ for an isotropic solid, we substitute this value of $$\alpha$$ to get

$$\beta \;=\; 3 \times \frac{F}{\pi r^{2} Y T} \;=\; \frac{3F}{\pi r^{2} Y T}.$$

Hence, the correct answer is Option D.

First, recall the definition of normal stress. We have the basic formula

$$\sigma=\frac{F}{A}$$

where $$\sigma$$ is the normal stress, $$F$$ is the axial load, and $$A$$ is the cross-sectional area on which the load acts.

The rod is circular, so its cross-sectional area is given by

$$A=\frac{\pi d^{2}}{4},$$

where $$d$$ is the diameter.

The material is brass with an elastic (safe) stress limit of $$\sigma_{\text{allow}}=379\;\text{MPa}$$. Since $$1\;\text{MPa}=1\;\text{N/mm}^2$$, we can write

$$\sigma_{\text{allow}} = 379\;\text{N/mm}^2.$$

The allowable stress must not be exceeded, so we set

$$\sigma \le \sigma_{\text{allow}}.$$

For the minimum diameter, we make the working stress exactly equal to the allowable value:

$$\frac{F}{A}= \sigma_{\text{allow}}.$$

The load is $$F = 400\;\text{N}$$, hence

$$\frac{400}{A}=379 \quad\Longrightarrow\quad A=\frac{400}{379}\;\text{mm}^2.$$

Carrying out the division gives

$$A\approx 1.055\;\text{mm}^2.$$

Next, substitute this area into the circular area formula to get the diameter:

$$A=\frac{\pi d^{2}}{4}\quad\Longrightarrow\quad d^{2}= \frac{4A}{\pi}.$$

Replacing $$A$$ by $$1.055\;\text{mm}^2$$ gives

$$d^{2}= \frac{4\times 1.055}{\pi}.$$

First compute the numerator:

$$4\times 1.055 = 4.220.$$

Now divide by $$\pi$$:

$$\frac{4.220}{\pi}\approx \frac{4.220}{3.1416}\approx 1.343.$$

Taking the square root yields

$$d = \sqrt{1.343}\;\text{mm}\approx 1.159\;\text{mm}.$$

Rounding to three significant figures, we obtain

$$d\approx 1.16\;\text{mm}.$$

Hence, the correct answer is Option C.

First, let us find the extra pressure applied on the liquid when the mass $$m$$ is gently placed on the mass-less piston of area $$a$$. The weight of the mass is $$mg$$, so the force transmitted to the liquid is also $$mg$$. Since pressure is force per unit area, the increase in pressure is given by

$$P=\dfrac{\text{Force}}{\text{Area}}=\dfrac{mg}{a}\;.$$

This pressure is conveyed undiminished throughout the liquid (Pascal’s law), and hence the soft solid sphere experiences the same additional pressure $$P$$ on all its surfaces.

Now we recall the definition of bulk modulus. For any isotropic solid,

$$K=-\dfrac{P}{\dfrac{\Delta V}{V}},$$

where $$K$$ is the bulk modulus, $$P$$ is the applied pressure, $$V$$ is the original volume and $$\Delta V$$ is the change in volume (a decrease in volume makes $$\Delta V$$ negative, hence the minus sign in the definition).

The sphere has initial radius $$r$$, so its initial volume is

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

If, because of the pressure, the radius decreases by a small amount $$dr$$ (so $$dr<0$$), then the new radius becomes $$r+dr$$ and the new volume is

$$V+\Delta V=\dfrac{4}{3}\pi(r+dr)^{3}\;.$$

Expanding the cube and keeping only the first-order term in the small quantity $$dr$$, we have

$$\begin{aligned} (r+dr)^{3}&=r^{3}+3r^{2}dr+\text{(higher-order terms in }dr) \\ &\approx r^{3}+3r^{2}dr. \end{aligned}$$

Thus,

$$\Delta V=\dfrac{4}{3}\pi\bigl[r^{3}+3r^{2}dr-r^{3}\bigr]=4\pi r^{2}dr.$$

Dividing by the original volume to obtain the fractional change, we get

$$\dfrac{\Delta V}{V}=\dfrac{4\pi r^{2}dr}{\dfrac{4}{3}\pi r^{3}}=3\,\dfrac{dr}{r}.$$

Putting this result into the bulk modulus relation, we write

$$K=-\dfrac{P}{\dfrac{\Delta V}{V}}=-\dfrac{P}{3\,\dfrac{dr}{r}}.$$

Rearranging to isolate the fractional change in radius,

$$\dfrac{dr}{r}=-\dfrac{P}{3K}.$$

Here $$dr<0$$ (a decrease), so the magnitude of the fractional decrement in radius is

$$\left|\dfrac{dr}{r}\right|=\dfrac{P}{3K}.$$

Substituting the earlier expression for $$P$$, namely $$P=\dfrac{mg}{a}$$, we finally obtain

$$\left|\dfrac{dr}{r}\right|=\dfrac{1}{3K}\,\dfrac{mg}{a} =\dfrac{mg}{3Ka}.$$

Therefore, the fractional decrease in the radius of the sphere equals $$\dfrac{mg}{3Ka}$$.

Hence, the correct answer is Option D.

We begin with the cube at an initial temperature of $$0^{\circ}C$$ and at its original volume $$V_0$$. An external pressure $$P$$ is applied uniformly from all directions, so the cube is compressed equally on all sides.

First, we recall the definition of the bulk modulus. The bulk modulus $$K$$ of a material is given by the formula

$$K = -\,\frac{P}{\displaystyle\frac{\Delta V}{V_0}}$$

where $$\Delta V = V - V_0$$ is the change in volume caused by the pressure $$P$$. Because the pressure compresses the cube, the change in volume is negative, so we may write

$$\frac{\Delta V}{V_0} = -\,\frac{P}{K}.$$

Hence the fractional decrease in volume produced by the pressure is

$$\left|\frac{\Delta V}{V_0}\right| = \frac{P}{K}.$$

Next, we wish to restore the cube to its original size by heating it while the same pressure $$P$$ continues to act. For an isotropic solid, the coefficient of linear expansion is $$\alpha$$, so the coefficient of volume expansion is

$$\beta = 3\alpha.$$

If we raise the temperature by an amount $$\Delta T$$, the fractional change in volume produced by thermal expansion is

$$\frac{\Delta V_{\text{thermal}}}{V_0} = \beta\Delta T = 3\alpha\Delta T.$$

To bring the cube back to its original volume, the volume increase produced by heating must exactly cancel the volume decrease caused by the applied pressure. Therefore we set

$$3\alpha\Delta T = \frac{P}{K}.$$

Solving for $$\Delta T$$, we obtain

$$\Delta T = \frac{P}{3\alpha K}.$$

This is the temperature rise required for the cube to regain its original size under the continued external pressure $$P$$.

Hence, the correct answer is Option 2.

We are told that a long, thin steel rod of initial length $$l$$ and cross-sectional area $$A$$ is subjected simultaneously to two different actions. One action is purely mechanical: a compressive force $$F$$ is applied axially at its ends. The other action is purely thermal: the rod is heated so that its temperature rises by $$\Delta T$$. Experimentally, we observe that the combined effect of these two actions leaves the overall length of the rod unchanged. Mathematically, this means that the total, or net, linear strain is zero.

We first recall the definition of linear strain. Whenever a body undergoes a fractional change in length, that fraction is called the strain: $$\text{strain} = \dfrac{\Delta l}{l}.$$ If two kinds of strain act simultaneously, we simply add them (with the proper algebraic signs) to get the net strain.

We therefore identify the two individual strains acting on the rod:

1. Thermal strain due to heating. The coefficient of linear expansion of the material is $$\alpha$$. By definition, the fractional change in length produced solely by a temperature rise $$\Delta T$$ is

$$\text{thermal strain} \;=\; \alpha \,\Delta T.$$

This strain is positive because heating tends to elongate the rod.

2. Mechanical strain due to the compressive force. We start with the relation between stress, strain and Young’s modulus. Young’s modulus $$Y$$ of a material is defined by the formula

$$Y \;=\; \dfrac{\text{stress}}{\text{strain}}.$$

Here the axial stress caused by the applied force $$F$$ is simply the force divided by the area of cross-section:

$$\text{stress} \;=\; \dfrac{F}{A}.$$

Substituting this stress into the defining equation of Young’s modulus, we get the mechanical strain:

$$\text{strain}_{\text{mech}} \;=\; \dfrac{\text{stress}}{Y} \;=\; \dfrac{F/A}{Y} \;=\; \dfrac{F}{A\,Y}.$$

Because the applied force is compressive, it attempts to shorten the rod. Hence this mechanical strain is negative with respect to the original length. To indicate the sign explicitly, we may write it as $$-\dfrac{F}{A\,Y}.$$

Adding the two strains and imposing the zero-net-strain condition. The net strain is the algebraic sum of the thermal strain and the mechanical strain. Setting their sum to zero gives

$$\text{thermal strain} \;+\; \text{mechanical strain} \;=\; 0,$$

$$\alpha \,\Delta T \;-\; \dfrac{F}{A\,Y} \;=\; 0.$$

Now we solve this equation step by step for the unknown force $$F$$. First, move the mechanical-strain term to the right-hand side:

$$\alpha \,\Delta T \;=\; \dfrac{F}{A\,Y}.$$

Next, multiply both sides by $$A\,Y$$ to isolate $$F$$:

$$F \;=\; A\,Y \, \alpha \,\Delta T.$$

Thus the compressive force necessary to counteract the expansion produced by the temperature rise and leave the rod’s length unchanged is

$$F \;=\; A Y \alpha \Delta T.$$

We compare this result with the given options. It matches Option B.

Hence, the correct answer is Option B.

We have to compare the mechanical stress on the leg before and after the man becomes a giant. In mechanics, stress is defined first:

$$\sigma=\frac{F}{A}$$

where $$\sigma$$ is the stress, $$F$$ is the force acting on the area (here the weight of the body that the legs must support), and $$A$$ is the area over which that force is distributed (the cross-sectional area of the legs).

The problem states that all linear dimensions (height, width, depth, diameter of the leg, etc.) are enlarged by a common factor of $$9$$. Let us call this scaling factor $$k$$, so here $$k = 9$$.

Step 1: How does the volume change?

The volume of any three-dimensional object is proportional to the cube of a characteristic length. Because every linear dimension grows by $$k$$, the new volume $$V'$$ becomes

$$V' = k^3 V = 9^3 V = 729\,V.$$

Step 2: How does the mass (and hence weight) change?

The density is given to remain the same. Mass is density times volume, so if volume increases by the factor $$729$$, mass increases by exactly the same factor:

$$m' = 729\,m.$$

The weight is the gravitational force $$F = mg$$. Therefore, the new weight $$F'$$ is

$$F' = m' g = 729\,m g = 729\,F.$$

Step 3: How does the supporting area of the leg change?

The cross-section of the leg is a two-dimensional quantity, so it scales with the square of the linear factor:

$$A' = k^2 A = 9^2 A = 81\,A.$$

Step 4: Compute the new stress.

Using the stress formula on the enlarged man, we have

$$\sigma' = \frac{F'}{A'} = \frac{729\,F}{81\,A}.$$

We now separate the numerical factor:

$$\sigma' = \frac{729}{81}\,\frac{F}{A} = 9\,\sigma.$$

So the stress in the giant’s legs is nine times the original stress.

Hence, the correct answer is Option B.

We know that any material, if free to expand, would increase its length by the thermal relation

$$\Delta L = \alpha L \Delta T,$$

where $$\alpha$$ is the coefficient of linear expansion, $$L$$ is the original length and $$\Delta T$$ is the rise in temperature. Here, however, the rail is rigidly fixed, so no actual change in length is allowed. This restriction produces a compressive mechanical strain exactly equal in magnitude (but opposite in sense) to the free thermal strain.

So the strain locked inside the rail is

$$\text{strain} = \alpha \Delta T.$$

Young’s modulus $$Y$$ links stress and strain through the formula

$$\text{stress} = Y \times \text{strain}.$$

Substituting the expression for strain, we obtain

$$\text{stress} = Y \, \alpha \, \Delta T.$$

The numerical values given are

$$Y = 2 \times 10^{11}\ \text{N m}^{-2}, \qquad \alpha = 1.2 \times 10^{-5}\ \text{K}^{-1}, \qquad \Delta T = 10^\circ\text{C}.$$

Putting these into the stress formula, we have

$$\text{stress} = \left(2 \times 10^{11}\right)\left(1.2 \times 10^{-5}\right)(10) = 2 \times 1.2 \times 10^{11} \times 10^{-5} \times 10.$$

Now, $$1.2 \times 10^{-5} \times 10 = 1.2 \times 10^{-4},$$ so

$$\text{stress} = 2 \times 1.2 \times 10^{11} \times 10^{-4} = 2.4 \times 10^{7}\ \text{N m}^{-2}.$$

Force is obtained from stress through the relation

$$\text{force} = \text{stress} \times \text{area}.$$

The cross-sectional area is 40 cm$$^2$$. Converting this to square metres:

$$40\ \text{cm}^2 = 40 \times 10^{-4}\ \text{m}^2 = 4.0 \times 10^{-3}\ \text{m}^2.$$

Substituting the values,

$$\text{force} = (2.4 \times 10^{7}) \times (4.0 \times 10^{-3}) = 2.4 \times 4.0 \times 10^{7} \times 10^{-3}.$$

Since $$10^{7} \times 10^{-3} = 10^{4},$$ we get

$$\text{force} = 9.6 \times 10^{4}\ \text{N}.$$

Rounding to one significant figure, this is

$$\text{force} \approx 1 \times 10^{5}\ \text{N}.$$

Hence, the correct answer is Option D.

We have a conical wire whose upper, thinner end of radius $$R$$ is rigidly fixed and whose lower, thicker end of radius $$3R$$ supports a mass $$M$$. Because the mass hangs at the very end, every cross-section of the wire carries the same tensile force, namely the weight $$Mg$$ of the mass. The weight of the wire itself is not mentioned, so we neglect it.

The extension of a small element will be obtained from the basic definition of Young’s modulus. First we state the formula:

For a small element of length $$dx$$, cross-sectional area $$A$$ and tension $$F$$, the incremental extension is

$$d\ell=\frac{F}{YA}\;dx.$$

To find the total extension we integrate this expression along the full length of the wire, taking into account that the area varies with position.

Choose the coordinate $$x$$ measured from the fixed upper end (so $$x=0$$ at the top and $$x=L$$ at the bottom). Because the radius increases linearly from $$R$$ to $$3R$$, the radius at a distance $$x$$ from the top is

$$r(x)=R+\left(\frac{3R-R}{L}\right)x=R+\frac{2R}{L}x=R\left(1+\frac{2x}{L}\right).$$

Hence the cross-sectional area is

$$A(x)=\pi r(x)^2=\pi R^2\left(1+\frac{2x}{L}\right)^2.$$

The tensile force in every element is simply $$F=Mg$$. Substituting these expressions into the incremental extension formula, we get

$$d\ell=\frac{Mg}{Y\;\pi R^2\left(1+\dfrac{2x}{L}\right)^2}\;dx.$$

Integrating from the top ($$x=0$$) to the bottom ($$x=L$$) gives the total extension $$\Delta L$$:

$$\Delta L=\int_{0}^{L}\frac{Mg}{Y\;\pi R^2\left(1+\dfrac{2x}{L}\right)^2}\;dx.$$

Now we perform the integral. Put

$$u=1+\frac{2x}{L}\quad\Rightarrow\quad du=\frac{2}{L}dx\quad\Rightarrow\quad dx=\frac{L}{2}\,du.$$

When $$x=0$$, $$u=1$$; when $$x=L$$, $$u=3$$. Substituting, we obtain

$$\Delta L=\frac{Mg}{Y\pi R^2}\int_{u=1}^{3}\frac{L}{2}\,\frac{du}{u^{2}}$$ $$=\frac{MgL}{2Y\pi R^2}\int_{1}^{3}u^{-2}\,du.$$

We evaluate the integral:

$$\int u^{-2}\,du=-\frac{1}{u}\quad\text{so}\quad\int_{1}^{3}u^{-2}\,du=-\frac{1}{3}+1=\frac{2}{3}.$$

Substituting this result gives

$$\Delta L=\frac{MgL}{2Y\pi R^2}\cdot\frac{2}{3}=\frac{MgL}{3Y\pi R^2}.$$

The equilibrium (extended) length $$L_{\text{eq}}$$ is therefore

$$L_{\text{eq}}=L+\Delta L=L\left(1+\frac{Mg}{3Y\pi R^{2}}\right).$$

This expression exactly matches Option C.

Hence, the correct answer is Option C.

$$\Delta V = \pi a^2 b - \pi (a + \Delta a)^2 b$$
$$\Delta V = \pi b [a^2 - (a^2 + 2a\Delta a + (\Delta a)^2)]$$

$$\Delta V \approx -2\pi a b \Delta a$$

$$\frac{\Delta V}{V} = \frac{-2\pi a b \Delta a}{\pi a^2 b} = -\frac{2 \Delta a}{a}$$

The Bulk Modulus ($$B$$) relates pressure to volumetric strain:

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

$$P = -B \left( -\frac{2 \Delta a}{a} \right) = \frac{2 B \Delta a}{a}$$

This pressure $$P$$ acts as the normal stress on the inner surface of the bottle neck.

The total Normal Force ($$N$$) exerted by the cork on the bottle is the pressure multiplied by the contact area (lateral surface area of the cylinder):

$$N = P \times (2\pi a b)$$

$$N = \left( \frac{2 B \Delta a}{a} \right) \times 2\pi a b = 4\pi B b \Delta a$$

The force ($$F$$) needed to push the cork must overcome the frictional force ($$f = \mu N$$):

$$F = \mu (4\pi B b \Delta a)$$

$$F = (4\pi \mu B b) \Delta a$$

Since it takes significantly less effort to twist or slide a metal block than it does to stretch it or compress its entire volume, the energy required scales accordingly. Therefore, the resistance to changing shape is the lowest, while the resistance to changing volume is the highest, giving the order: Shear moduli < Young's moduli < bulk moduli

The steel ruptures when the shear stress reaches $$3.5 \times 10^8$$ N m$$^{-2}$$. To punch a hole, we need to apply a force that causes this shear stress over the area being sheared.

When punching a hole of diameter 1 cm in a sheet of thickness 0.3 cm, the sheared area is the lateral surface area of the cylindrical hole. This area is calculated as the circumference of the hole multiplied by the thickness of the sheet.

First, convert all measurements to meters for consistency. The diameter is 1 cm, which is 0.01 m, so the radius $$r$$ is half of that: $$r = \frac{0.01}{2} = 0.005$$ m. The thickness $$t$$ is 0.3 cm, which is 0.003 m.

The circumference of the hole is $$2 \pi r$$. Therefore, the sheared area $$A$$ is:

$$A = 2 \pi r t$$

Substituting the values:

$$A = 2 \times \pi \times 0.005 \times 0.003$$

First, multiply the numerical parts:

$$0.005 \times 0.003 = 0.000015$$

Then:

$$A = 2 \times \pi \times 0.000015 = 0.00003 \pi$$

Using $$\pi \approx 3.1416$$,

$$A \approx 0.00003 \times 3.1416 = 0.000094248 \text{ m}^2$$

In scientific notation, $$A \approx 9.4248 \times 10^{-5}$$ m$$^2$$.

The shear stress $$\tau$$ is related to the force $$F$$ and area $$A$$ by:

$$\tau = \frac{F}{A}$$

At rupture, $$\tau = 3.5 \times 10^8$$ N m$$^{-2}$$. Solving for $$F$$:

$$F = \tau \times A$$

Substituting the values:

$$F = (3.5 \times 10^8) \times (9.4248 \times 10^{-5})$$

First, multiply the coefficients:

$$3.5 \times 9.4248 = 32.9868$$

Then multiply the powers of 10:

$$10^8 \times 10^{-5} = 10^{3}$$

So:

$$F \approx 32.9868 \times 10^3 = 32986.8 \text{ N}$$

Which is $$3.29868 \times 10^4$$ N. Rounding to two significant figures, as in the options, gives approximately $$3.3 \times 10^4$$ N.

Alternatively, using exact values:

$$A = 2 \pi r t = 2 \times \pi \times 0.005 \times 0.003 = 2 \times \pi \times 1.5 \times 10^{-5} = 3\pi \times 10^{-5} \text{ m}^2$$

Then:

$$F = (3.5 \times 10^8) \times (3\pi \times 10^{-5}) = 3.5 \times 3 \times \pi \times 10^{8-5} = 10.5 \pi \times 10^3 = 10500\pi \text{ N}$$

With $$\pi \approx 3.1416$$,

$$F \approx 10500 \times 3.1416 = 32986.8 \text{ N} \approx 3.3 \times 10^4 \text{ N}$$

Comparing with the options, $$3.3 \times 10^4$$ N corresponds to option C.

Hence, the correct answer is Option C.

Tension in Brass wire ($$F_b$$): It supports only the bottom mass $$2M$$, hence, $$F_b = 2Mg$$
Tension in Steel wire ($$F_s$$): It supports both masses $$M$$ and $$2M$$, hence, $$F_s = Mg + 2Mg = 3Mg$$

$$\frac{F_s}{F_b} = \frac{3Mg}{2Mg} = \frac{3}{2}$$

$$\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$$

$$\frac{\Delta L_s}{\Delta L_b} = \left( \frac{F_s}{F_b} \right) \times \left( \frac{L_s}{L_b} \right) \times \left( \frac{r_b}{r_s} \right)^2 \times \left( \frac{Y_b}{Y_s} \right)$$

$$\frac{\Delta L_s}{\Delta L_b} = \left( \frac{3}{2} \right) \times (a) \times \left( \frac{1}{b} \right)^2 \times \left( \frac{1}{c} \right)$$

$$\frac{\Delta L_s}{\Delta L_b} = \frac{3a}{2b^2c}$$

We have a composite wire made of copper and steel joined end to end. The copper wire is 1.0 m long and the steel wire is 0.5 m long, both with the same cross-sectional area. The composite wire is stretched by a load that causes the copper wire to extend by 1 mm. We need to find the total extension of the composite wire.

Given:

• Length of copper wire, $$ L_c = 1.0 \text{m} $$
• Length of steel wire, $$ L_s = 0.5 \text{m} $$
• Extension of copper wire, $$ \Delta L_c = 1 \text{mm} = 0.001 \text{m} $$
• Young's modulus of copper, $$ Y_c = 1.0 \times 10^{11} \text{Nm}^{-2} $$
• Young's modulus of steel, $$ Y_s = 2.0 \times 10^{11} \text{Nm}^{-2} $$

Since the wires are joined end to end, the same force $$ F $$ acts on both wires. The cross-sectional area $$ A $$ is the same for both.

Using Hooke's law for the copper wire:

$$ Y_c = \frac{F \cdot L_c}{A \cdot \Delta L_c} $$

Solving for the force $$ F $$:

$$ F = \frac{Y_c \cdot A \cdot \Delta L_c}{L_c} $$

Substituting the values:

$$ F = \frac{(1.0 \times 10^{11}) \cdot A \cdot (0.001)}{1.0} $$

$$ F = (1.0 \times 10^{11}) \cdot A \cdot 0.001 $$

$$ F = 1.0 \times 10^{11} \times 10^{-3} \cdot A $$

$$ F = 1.0 \times 10^{8} A \text{ Newtons} $$

Now, for the steel wire, the same force $$ F $$ is applied. Using Hooke's law:

$$ Y_s = \frac{F \cdot L_s}{A \cdot \Delta L_s} $$

Solving for the extension of steel $$ \Delta L_s $$:

$$ \Delta L_s = \frac{F \cdot L_s}{A \cdot Y_s} $$

Substituting the values:

$$ \Delta L_s = \frac{(1.0 \times 10^{8} A) \cdot (0.5)}{A \cdot (2.0 \times 10^{11})} $$

The $$ A $$ cancels out:

$$ \Delta L_s = \frac{1.0 \times 10^{8} \cdot 0.5}{2.0 \times 10^{11}} $$

$$ \Delta L_s = \frac{5.0 \times 10^{7}}{2.0 \times 10^{11}} $$

$$ \Delta L_s = \frac{5.0}{2.0} \times 10^{7-11} $$

$$ \Delta L_s = 2.5 \times 10^{-4} \text{m} $$

Converting to millimeters (since 1 m = 1000 mm):

$$ \Delta L_s = 2.5 \times 10^{-4} \times 1000 = 0.25 \text{mm} $$

The total extension of the composite wire is the sum of the extensions of the copper and steel parts:

$$ \Delta L_{\text{total}} = \Delta L_c + \Delta L_s = 1 \text{mm} + 0.25 \text{mm} = 1.25 \text{mm} $$

Hence, the correct answer is Option D.

A uniform wire is subjected to a longitudinal tensile stress, and we are given Young's modulus $$Y = 2 \times 10^{11}$$ N/m² and stress $$\sigma = 5 \times 10^7$$ N/m². The overall volume change is 0.02%, which means $$\frac{\Delta V}{V} = \frac{0.02}{100} = 0.0002$$. We need to find the fractional decrease in the radius, which is $$\left| \frac{\Delta r}{r} \right|$$.

When a wire is stretched, the volume change relates to Poisson's ratio ($$\nu$$) and the longitudinal strain. The formula for volume change is:

$$\frac{\Delta V}{V} = (1 - 2\nu) \cdot \epsilon_{\text{long}}$$

where $$\epsilon_{\text{long}}$$ is the longitudinal strain. Using Hooke's law, $$\sigma = Y \cdot \epsilon_{\text{long}}$$, so $$\epsilon_{\text{long}} = \frac{\sigma}{Y}$$.

First, calculate $$\epsilon_{\text{long}}$$:

$$\epsilon_{\text{long}} = \frac{\sigma}{Y} = \frac{5 \times 10^7}{2 \times 10^{11}} = \frac{5}{2} \times 10^{7-11} = 2.5 \times 10^{-4}$$

Now substitute into the volume change formula:

$$\frac{\Delta V}{V} = (1 - 2\nu) \cdot \epsilon_{\text{long}}$$

$$0.0002 = (1 - 2\nu) \cdot (2.5 \times 10^{-4})$$

Solve for $$1 - 2\nu$$:

$$1 - 2\nu = \frac{0.0002}{2.5 \times 10^{-4}} = \frac{2 \times 10^{-4}}{2.5 \times 10^{-4}} = \frac{2}{2.5} = 0.8$$

Now solve for $$\nu$$:

$$1 - 2\nu = 0.8$$

$$-2\nu = 0.8 - 1 = -0.2$$

$$\nu = \frac{-0.2}{-2} = 0.1$$

Poisson's ratio $$\nu$$ is 0.1. The fractional decrease in radius is given by the magnitude of the lateral strain. Poisson's ratio relates lateral strain ($$\epsilon_{\text{lat}}$$) to longitudinal strain:

$$\nu = -\frac{\epsilon_{\text{lat}}}{\epsilon_{\text{long}}}$$

So,

$$\epsilon_{\text{lat}} = -\nu \cdot \epsilon_{\text{long}}$$

The fractional change in radius is $$\frac{\Delta r}{r} = \epsilon_{\text{lat}}$$, so the fractional decrease is $$\left| \frac{\Delta r}{r} \right| = \nu \cdot \epsilon_{\text{long}}$$.

Substitute the values:

$$\left| \frac{\Delta r}{r} \right| = \nu \cdot \epsilon_{\text{long}} = 0.1 \cdot (2.5 \times 10^{-4}) = 0.25 \times 10^{-4}$$

This matches option C. Therefore, the fractional decrease in the radius is $$0.25 \times 10^{-4}$$.

Hence, the correct answer is Option C.

For a uniform wire of length $$L$$, cross-sectional area $$A$$ and Young’s modulus $$Y$$, the elongation $$\Delta L$$ produced by a stretching force (load) $$F$$ is

$$\Delta L=\frac{F\,L}{Y\,A}\qquad -(1)$$

Equation $$-(1)$$ can be rearranged to write the load in terms of the elongation:

$$F=\frac{Y\,A}{L}\,\Delta L\qquad -(2)$$

In a load-elongation graph, the slope is

$$\text{slope}=\frac{F}{\Delta L}=\frac{Y\,A}{L}\qquad -(3)$$

For wires made of the same material ($$Y$$ common) and having the same length ($$L$$ common), the slope is directly proportional to the cross-sectional area $$A$$:

$$\text{slope}\;\propto\;A\qquad -(4)$$

Therefore:

• A larger slope corresponds to a larger area (thicker wire).
• A smaller slope corresponds to a smaller area (thinner wire).

Among the four straight lines shown, line $$OA$$ has the smallest slope (it rises the least in load for a given elongation). Hence, $$OA$$ represents the wire with the smallest cross-sectional area, i.e., the thinnest wire.

Option A which is: $$OA$$

Breaking of a wire occurs when the tensile stress in the material reaches its ultimate (breaking) stress. Breaking stress is a material property; it does not depend on the length of the wire but only on the nature of the material and its cross-sectional area.

For the original steel wire:
maximum load $$W_{\max} = 100 \text{ kgf}$$ (where $$1 \text{ kgf}=g \text{ newton}$$).
Corresponding breaking stress is
$$\sigma_{\text{break}} = \frac{W_{\max} \, g}{A},$$
where $$A$$ is the wire’s cross-sectional area.

If the wire is cut into two equal halves, the length becomes $$\tfrac{1}{2}$$ of the original, but the cross-sectional area $$A$$ remains unchanged. Under any load $$W$$ the tensile stress in either half is still $$\sigma = \dfrac{W\,g}{A}$$, the same formula as before.

Because the breaking stress of steel is unchanged and the area is unchanged, the maximum load that produces this critical stress is also unchanged. Hence each half can sustain the same load of $$100 \text{ kgf}$$ before breaking.

Therefore, each part can sustain a weight of 100 kg.

Option C which is: $$100 \text{ kg}$$

The metal ring is free to expand and contract but its length is fixed by the wooden wheel once it has cooled down.
Initial circumference of the wheel is $$2\pi R$$; the ring is manufactured a little shorter (length $$L$$).
On heating, it expands by $$\Delta L = \alpha L\,\Delta T$$ and just slips over the wheel, so while cooling it contracts by exactly the same amount but is prevented from shortening. Hence the ring is left in uniform tensile strain

Strain in the ring, $$\varepsilon = \alpha\,\Delta T$$

Young’s modulus relates stress and strain: $$\sigma = Y\,\varepsilon = Y\,\alpha\,\Delta T$$

Tension developed in the ring (force along the strip) is therefore
$$T = \sigma S = S\,Y\,\alpha\,\Delta T$$

To find the force with which the two semicircular wooden parts press against each other, draw a free-body diagram of one semicircle. At every point of contact the ring pulls tangentially with magnitude $$T$$. Take radial components and integrate over the half-circumference:

The resultant of tangential tensions around a semicircle equals $$2T$$ and is directed along the diameter joining the two semicircles. (Mathematically, $$\displaystyle \vec F = \int_{-\pi/2}^{\pi/2} T\,\hat t\,d\theta = 2T\,\hat x$$.)

Hence the compressive force exerted by one half of the wheel on the other is
$$F = 2T = 2\,S\,Y\,\alpha\,\Delta T$$

Option D which is: $$2SY\alpha\Delta T$$

The percentage strain is obtained from the definition

$$\text{strain} \; \varepsilon = \frac{\text{stress}}{\text{Young’s modulus}}$$

Hence we first find the longitudinal stress produced in the rod.

Step 1: Cross-sectional area
Radius $$r = 10\ \text{mm} = 0.01\ \text{m}$$
$$A = \pi r^{2} = \pi (0.01)^{2} = \pi \times 10^{-4}\ \text{m}^{2}$$

Step 2: Stress
Applied force $$F = 100\ \text{kN} = 100 \times 10^{3}\ \text{N}$$
$$\sigma = \frac{F}{A} = \frac{100 \times 10^{3}}{\pi \times 10^{-4}} \approx 3.18 \times 10^{8}\ \text{N\,m}^{-2}$$

Step 3: Strain
Young’s modulus $$Y = 2 \times 10^{11}\ \text{N\,m}^{-2}$$
$$\varepsilon = \frac{\sigma}{Y} = \frac{3.18 \times 10^{8}}{2 \times 10^{11}} = 1.59 \times 10^{-3}$$

Step 4: Percentage strain
$$\varepsilon(\%) = \varepsilon \times 100 = 1.59 \times 10^{-3} \times 100 = 0.159\% \approx 0.16\%$$

Therefore, the percentage strain is about 0.16 %.
Option A which is: 0.16 %.

Let the original lengths of the two wires be $$L_1$$ and $$L_2$$ and let their common Young’s modulus be $$Y$$.

Step 1: Relate the lengths using equal volumes.
Both wires are made of the same material and have the same volume $$V$$.

For wire 1: $$V = A\,L_1$$
For wire 2: $$V = 3A\,L_2$$

Equating the two expressions for $$V$$ gives
$$A\,L_1 = 3A\,L_2 \;\;\Longrightarrow\;\; L_2 = \frac{L_1}{3}$$

Step 2: Write the extension formula for each wire.
For a wire of length $$L$$ and cross-sectional area $$A_c$$, the increase in length produced by a force $$F$$ is
$$\Delta x = \frac{F\,L}{A_c\,Y} \quad -(1)$$

Wire 1 (area $$A$$, force $$F$$):
$$\Delta x = \frac{F\,L_1}{A\,Y} \quad -(2)$$

Wire 2 (area $$3A$$, unknown force $$F_2$$):
$$\Delta x = \frac{F_2\,L_2}{3A\,Y} \quad -(3)$$

Step 3: Impose the condition of equal elongations.
Set the right-hand sides of (2) and (3) equal and substitute $$L_2 = L_1/3$$:

$$\frac{F\,L_1}{A\,Y} = \frac{F_2\,(L_1/3)}{3A\,Y}$$

Simplifying yields
$$\frac{F\,L_1}{A\,Y} = \frac{F_2\,L_1}{9A\,Y}$$
$$\Longrightarrow\; F = \frac{F_2}{9}$$
$$\Longrightarrow\; F_2 = 9F$$

Conclusion
To produce the same extension $$\Delta x$$, wire 2 must be pulled with a force nine times larger than $$F$$.

Option D which is: $$9F$$

Case 1: Wire fixed at one support and loaded with $$W$$

Let the original length of the wire be $$L$$, its cross-sectional area be $$A$$, and the Young's modulus of its material be $$Y$$

In this setup, the tension $$T$$ in the wire is uniform and equal to the applied load $$W$$.

According to Hooke's Law, the initial elongation $$\ell$$ is given by:

$$\ell = \frac{T L}{A Y} = \frac{W L}{A Y}$$

Case 2: Wire over a smooth pulley with weights $$W$$ at both ends

$$T' = W$$.

Even though the wire is bent over the pulley, the total length subjected to this tension is still $$L$$. Since the tension $$T'$$ is exactly the same as in Case 1 ($$T' = W$$), the total elongation $$\Delta L$$ depends entirely on this same tension acting over the entire length $$L$$:

$$\Delta L = \frac{T' L}{A Y} = \frac{W L}{A Y}$$

Comparing this with our equation from Case 1, we clearly see that:

$$\Delta L = \ell$$ (option B)

The elastic potential energy stored per unit volume (also known as energy density, $$u$$) in a stretched wire is given by the standard formula:

$$u = \frac{1}{2} \times \text{Stress} \times \text{Strain}$$

By definition, Young's modulus ($$Y$$) is the ratio of longitudinal stress to longitudinal strain:

$$Y = \frac{\text{Stress}}{\text{Strain}}$$

Rearranging this formula to solve for Strain gives:

$$\text{Strain} = \frac{\text{Stress}}{Y}$$

We are given that the stress is denoted by $$S$$. Substituting $$S$$ into the strain equation yields:

$$\text{Strain} = \frac{S}{Y}$$

Now, substitute the expressions for Stress ($$S$$) and Strain ($$\frac{S}{Y}$$) back into the initial energy density formula:

$$u = \frac{1}{2} \times S \times \left( \frac{S}{Y} \right)$$

Simplifying the expression:

$$u = \frac{S^2}{2Y}$$

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{W / A}{\Delta l / l} = \frac{W \cdot l}{A \cdot \Delta l}$$

From the given linear graph, picking a clear coordinates point $$(W, \Delta l)$$:

Load ($$W$$) = $$20\text{ N}$$, Extension ($$\Delta l$$) = $$2 \times 10^{-4}\text{ m}$$

$$Y = \frac{20 \times 1}{10^{-5} \times (2 \times 10^{-4})}$$

$$Y = \frac{20}{2 \times 10^{-9}} = 10 \times 10^{9} = 1.0 \times 10^{10}\text{ N/m}^2$$

The extension of a uniform string under a steady load is obtained from
$$\Delta L = \dfrac{F\,L}{A\,Y}$$
where $$F$$ is the tension (equal to the load’s weight), $$L$$ is the original length, $$A$$ is the cross-sectional area and $$Y$$ is Young’s modulus.

Data for the two strings
String A: $$L_1 = 0.314\ \text{m},\quad Y_1 = 2 \times 10^{10}\ \text{N m}^{-2}$$
String B: $$L_2 = 2L_1 = 0.628\ \text{m},\quad Y_2 = 2Y_1 = 4 \times 10^{10}\ \text{N m}^{-2}$$
Common radius: $$r = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m}$$

Cross-sectional area of either string:
$$A = \pi r^{2} = \pi\,(0.2 \times 10^{-3})^{2}$$ $$A = \pi \times 0.04 \times 10^{-6} = 4\pi \times 10^{-8}\ \text{m}^{2}$$

Load attached at the lower end:
$$m = 0.8\ \text{kg},\quad g = 10\ \text{m s}^{-2}$$ $$F = mg = 0.8 \times 10 = 8\ \text{N}$$

Extension of String A
$$\Delta L_1 = \dfrac{F\,L_1}{A\,Y_1} = \dfrac{8 \times 0.314}{(4\pi \times 10^{-8})\,(2 \times 10^{10})}$$ Denominator: $$4\pi \times 10^{-8} \times 2 \times 10^{10} = 8\pi \times 10^{2} = 800\pi$$
Hence
$$\Delta L_1 = \dfrac{2.512}{800\pi}\ \text{m} \approx 0.000999\ \text{m} = 0.999\ \text{mm}$$

Extension of String B
$$\Delta L_2 = \dfrac{F\,L_2}{A\,Y_2} = \dfrac{8 \times 0.628}{(4\pi \times 10^{-8})\,(4 \times 10^{10})}$$ Denominator: $$4\pi \times 10^{-8} \times 4 \times 10^{10} = 16\pi \times 10^{2} = 1600\pi$$
So
$$\Delta L_2 = \dfrac{5.024}{1600\pi}\ \text{m} \approx 0.000999\ \text{m} = 0.999\ \text{mm}$$

Total extension of the series combination
$$\Delta L_{\text{total}} = \Delta L_1 + \Delta L_2 \approx 0.999\ \text{mm} + 0.999\ \text{mm} \approx 1.998\ \text{mm}$$

Rounded to the nearest millimetre, the net change in length is $$2\ \text{mm}$$.

Option B which is: $$2$$

$$ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} $$

Rearranging for elongation ($$ \Delta L $$):

$$ \Delta L = \frac{FL}{AY} $$

According to the problem, both wires are joined together and kept under the same tension,have same area, same elongation:

$$ F_A = F_B = F $$

$$ A_A = A_B = A $$

$$ \Delta L_A = \Delta L_B $$

Substitute the expression for elongation for both wires:

$$ \frac{F \cdot L_A}{A \cdot Y_A} = \frac{F \cdot L_B}{A \cdot Y_B} $$

$$ \frac{L_A}{Y_A} = \frac{L_B}{Y_B} $$

We need to find the length of wire B ($$ L_B $$). Rearranging the equation gives:

$$ L_B = L_A \times \frac{Y_B}{Y_A} $$

$$ \frac{Y_A}{Y_B} = \frac{20}{11} \implies \frac{Y_B}{Y_A} = \frac{11}{20} $$

$$ L_A = 2.2\text{ m} $$

Substitute these values into the equation for $$ L_B $$:

$$ L_B = 2.2 \times \frac{11}{20} $$

$$ L_B = \frac{24.2}{20} $$

$$ L_B = 1.21\text{ m} $$

Two liquids with the same densities and surface tensions $$T_1$$ and $$T_2$$ rise to heights $$h_1$$ and $$h_2$$, respectively, in capillary tubes of radii $$r_1 > r_2$$.

We first recall the capillary rise formula:

$$h = \frac{2T\cos\theta}{\rho g r}$$

Assuming that both liquids have the same contact angle (so that $$\cos\theta = 1$$) and the same density, we can rearrange to find

$$T = \frac{h\rho g r}{2}$$, which shows that $$T \propto hr$$.

Because the surface tensions are equal ($$T_1 = T_2$$), it follows that

$$h_1 r_1 = h_2 r_2$$, and since $$r_1 > r_2$$, we have $$h_1 < h_2$$.

This conclusion matches Option C, which states $$h_1 < h_2$$ and $$T_1 = T_2$$.

The correct answer is Option C.

The extension in a wire under a load is related to Young’s modulus $$Y$$ by the formula
$$Y = \frac{F\,L}{A\,\Delta L}$$
where
  $$F$$ = stretching force,
  $$L$$ = original length of the wire,
  $$A$$ = cross-sectional area, and
  $$\Delta L$$ = extension produced.

First, find each quantity from the data given:

Force
The load is a mass of $$50\ \text{kg}$$, so the force is its weight:
$$F = m g = 50 \times 3\pi = 150\pi\ \text{N}$$

Original length
$$L = 3\ \text{m}$$

Extension
$$\Delta L = 0.1\ \text{mm} = 0.1 \times 10^{-3}\ \text{m} = 1 \times 10^{-4}\ \text{m}$$

Cross-sectional area
The wire has radius $$r = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m}$$.
Area $$A = \pi r^{2} = \pi \left(3 \times 10^{-3}\right)^{2} = \pi \times 9 \times 10^{-6} = 9\pi \times 10^{-6}\ \text{m}^{2}$$

Substitute into the formula for $$Y$$
$$Y = \frac{150\pi \times 3}{9\pi \times 10^{-6} \times 1 \times 10^{-4}}$$

Simplify step by step:

• Multiply force and length:
$$150\pi \times 3 = 450\pi$$

• Multiply area and extension:
$$9\pi \times 10^{-6} \times 1 \times 10^{-4} = 9\pi \times 10^{-10}$$

• Divide:
$$Y = \frac{450\pi}{9\pi \times 10^{-10}} = \frac{450}{9} \times 10^{10} = 50 \times 10^{10}\ \text{N m}^{-2}$$

• Write in the required power-of-ten form:
$$50 \times 10^{10} = 5 \times 10^{11}\ \text{N m}^{-2}$$

Hence the Young’s modulus is $$P \times 10^{11}\ \text{N m}^{-2}$$ with $$P = 5$$.

Therefore, the correct option is Option A.

The cylindrical rod is subjected to a horizontal (shear) force at its upper end while the lower end is fixed. Such a loading produces a uniform shear stress over the circular cross-section.

Step 1 : Shear stress on the cross-section
Shear stress $$\tau$$ is force per unit area: $$\tau = \dfrac{F}{A}$$

The cross-sectional area is $$A = \pi r^{2}$$.
Given radius $$r = 4\text{ cm} = 0.04\text{ m}$$, so
$$A = \pi (0.04)^{2} = \pi \times 0.0016 = 0.0016\,\pi\;\text{m}^{2}$$.

With the applied force $$F = 10^{5}\text{ N}$$,
$$\tau = \dfrac{10^{5}}{0.0016\,\pi} = \dfrac{10^{5}\times625}{\pi} = \dfrac{6.25\times10^{7}}{\pi}\;\text{N m}^{-2}$$ $$-(1)$$

Step 2 : Shear strain produced
For an elastic material, shear strain $$\gamma$$ is related to shear stress by the shear modulus $$G$$:
$$\gamma = \dfrac{\tau}{G}$$.

Here $$G = 10^{10}\;\text{N m}^{-2}$$. Substitute $$\tau$$ from $$(1)$$:
$$\gamma = \dfrac{6.25\times10^{7}/\pi}{10^{10}} = \dfrac{6.25}{\pi\times10^{3}} = \dfrac{0.00625}{\pi}$$.

Step 3 : Relation between shear strain and angular displacement
For a small angular displacement of the top relative to the fixed bottom, shear strain equals the angle in radians:
$$\theta \approx \gamma$$.

Therefore,
$$\theta = \dfrac{0.00625}{\pi} = \dfrac{1}{160\,\pi}\;\text{radian}$$.

Result
Angular displacement $$\theta = \dfrac{1}{160\pi}\;\text{rad}$$.

Hence, the correct option is Option A.

For a metallic wire stretched by a tensile force, the extension (elongation) produced is given by the Young’s modulus relation: $$\Delta L = \dfrac{F\,L}{A\,Y}$$ where
$$F$$ = applied force, $$L$$ = original length, $$A$$ = cross-sectional area, $$Y$$ = Young’s modulus.

The two wires A and B are made of the same material, so $$Y$$ is identical for both. The same force $$F$$ is applied to both wires, so $$F$$ also cancels when we take a ratio.

Hence the ratio of their extensions is determined only by their lengths and cross-sectional areas: $$\dfrac{\Delta L_A}{\Delta L_B} \;=\; \dfrac{F\,L_A / (A_A\,Y)}{F\,L_B / (A_B\,Y)} = \dfrac{L_A}{L_B}\;\dfrac{A_B}{A_A}$$

Step 1: Length ratio
Given $$\dfrac{L_A}{L_B} = \dfrac{1}{3}$$, so $$L_A : L_B = 1 : 3$$.

Step 2: Area ratio
For circular wires, the area $$A = \dfrac{\pi d^{2}}{4}$$, hence $$A \propto d^{2}$$. Given $$\dfrac{d_A}{d_B} = 2$$, so $$\left(\dfrac{d_A}{d_B}\right)^2 = 2^{2} = 4$$ leading to $$\dfrac{A_A}{A_B} = 4$$. Therefore $$\dfrac{A_B}{A_A} = \dfrac{1}{4}$$.

Step 3: Extension ratio
Substituting the two ratios into the expression for $$\dfrac{\Delta L_A}{\Delta L_B}$$: $$\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{1}{3}\right)\left(\dfrac{1}{4}\right) = \dfrac{1}{12}$$.

Thus $$\Delta L_A : \Delta L_B = 1 : 12$$.

The required ratio of elongations is 1 : 12, which corresponds to Option B.

Young’s modulus of the steel wire is $$E = 2.0 \times 10^{11} \,\text{N m}^{-2}$$.
Poisson ratio is $$\nu = 0.2$$.
Given transverse (lateral) strain, $$\varepsilon_t = 10^{-3}$$.

For a rod under uniaxial tension, Poisson ratio is defined as
$$\nu = - \dfrac{\varepsilon_t}{\varepsilon_l}$$, where $$\varepsilon_l$$ is the longitudinal (axial) strain.

Therefore,
$$\varepsilon_l = -\,\dfrac{\varepsilon_t}{\nu} = -\,\dfrac{10^{-3}}{0.2} = -\,5 \times 10^{-3}$$.
(The negative sign only indicates that the lateral and longitudinal strains are opposite in sense; for energy we need the magnitude.)
So, $$|\varepsilon_l| = 5 \times 10^{-3}$$.

The corresponding longitudinal stress is obtained from Hooke’s law:
$$\sigma = E\,\varepsilon_l = 2.0 \times 10^{11} \times 5 \times 10^{-3}\; \text{N m}^{-2}$$.
$$\sigma = 1.0 \times 10^{9}\; \text{N m}^{-2}$$.

Strain-energy density (elastic potential energy per unit volume) for uniaxial loading is
$$u = \dfrac{1}{2}\,\sigma\,\varepsilon_l$$.

Substituting the values:
$$u = \dfrac{1}{2}\,(1.0 \times 10^{9})\,(5 \times 10^{-3})$$
$$u = 0.5 \times 5 \times 10^{6}$$
$$u = 2.5 \times 10^{6}\;\text{J m}^{-3}$$.

Expressing the result as $$\times 10^{5}$$ J m$$^{-3}$$:
$$u = 25 \times 10^{5}\;\text{J m}^{-3}$$.

Hence, the elastic potential energy density of the wire is 25 × 105 J m-3.

For a slab under a tangential (shearing) force $$F$$,
 • the shearing (tangential) stress is $$\tau = \dfrac{F}{A}$$, where $$A$$ is the area of the face on which the force acts.
 • the shearing strain is $$\gamma = \tan\theta \approx \theta$$ for small angles.
 • shear modulus (rigidity modulus) is defined by $$G = \dfrac{\text{shear stress}}{\text{shear strain}} = \dfrac{\tau}{\gamma}.$$

Both slabs have square cross-section of side $$l$$ and thickness $$d$$. The force $$F$$ is applied on a narrow face of each slab; that face has dimensions $$l \times d$$, so its area is $$A = l\,d$$.

Thickness $$d_1$$, narrow-face area $$A_1 = l\,d_1$$. Shear stress $$\tau_1 = \dfrac{F}{A_1} = \dfrac{F}{l\,d_1}$$. Let the small angle of deformation be $$\theta_1$$, so shear strain $$\gamma_1 = \theta_1$$.

Hence $$G_1 = \dfrac{\tau_1}{\gamma_1} = \dfrac{F}{l\,d_1\,\theta_1} \quad -(1)$$.

Thickness $$d_2 = 2d_1$$, narrow-face area $$A_2 = l\,d_2 = l\,(2d_1) = 2l\,d_1$$. Shear stress $$\tau_2 = \dfrac{F}{A_2} = \dfrac{F}{2l\,d_1} = \dfrac{\tau_1}{2}$$. Given angle $$\theta_2 = 2\theta_1$$, so shear strain $$\gamma_2 = \theta_2 = 2\theta_1$$.

Therefore $$G_2 = \dfrac{\tau_2}{\gamma_2} = \dfrac{\tau_1/2}{2\theta_1} = \dfrac{\tau_1}{4\theta_1} \quad -(2)$$.

Divide $$(2)$$ by $$(1)$$:

$$\dfrac{G_2}{G_1} = \dfrac{\tau_1/(4\theta_1)}{F/(l\,d_1\,\theta_1)} = \dfrac{F}{4l\,d_1\,\theta_1}\times\dfrac{l\,d_1\,\theta_1}{F} = \dfrac{1}{4}.$$

Hence $$G_2 = \dfrac{G_1}{4}$$.

Given $$G_1 = 4 \times 10^9 \text{ N m}^{-2}$$,

$$G_2 = \dfrac{4 \times 10^9}{4} = 1 \times 10^9 \text{ N m}^{-2}.$$

The required value is $$x = 1$$.

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

We first calculate the pressure inside the bubble using

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

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

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

Substituting the numerical values gives

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

$$= 2000 + 190$$

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

The answer is 2190.

For a spring with natural length $$L_0$$ and spring constant $$k$$:

Under tension 3N: $$a = L_0 + 3/k$$ ... (1)

Under tension 2N: $$b = L_0 + 2/k$$ ... (2)

From (1) - (2): $$a - b = 1/k$$, so $$k = 1/(a-b)$$

From (2): $$L_0 = b - 2(a-b) = 3b - 2a$$

For length $$3a - 2b$$: $$T = k(L - L_0) = \frac{(3a-2b)-(3b-2a)}{a-b} = \frac{5a-5b}{a-b} = 5$$ N

The answer is 5.

$$a  = \frac{m_R g}{m_P + m_Q + m_R}$$

$$a = \frac{3 \times 10}{3 + 3 + 3} = \frac{30}{9} = \frac{10}{3} \text{ m s}^{-2}$$

Wire $$B$$ is responsible for accelerating blocks $$P$$ and $$Q$$ at the rate calculated above.

$$T_B = (m_P + m_Q) \cdot a$$

$$T_B = (3 + 3) \times \frac{10}{3} = 6 \times \frac{10}{3} = 20 \text{ N}$$

Strain is the ratio of stress to Young's modulus ($$Y$$). Stress is tension divided by the cross-sectional area ($$S$$).

$$\epsilon = \frac{\text{Stress}}{Y} = \frac{T_B}{S \cdot Y}$$

$$\epsilon = \frac{20}{(5 \times 10^{-7}) \times (2 \times 10^{11})}$$

$$\epsilon = \frac{20}{10 \times 10^4} = \frac{20}{10^5} = 2 \times 10^{-4}$$

The value is 2.

Find the ratio of longitudinal strain of the upper wire to the lower wire.

The upper wire supports both loads (2 kg + 1 kg), while the lower wire supports only the bottom load (1 kg).

Force on upper wire: $$F_1 = (2 + 1) \times g = 3 \times 10 = 30$$ N.

Force on lower wire: $$F_2 = 1 \times g = 1 \times 10 = 10$$ N.

Strain $$\epsilon = \frac{\Delta L}{L} = \frac{F}{YA}$$ (from Young's modulus $$Y = \frac{F/A}{\Delta L/L}$$).

Since both wires are similar (same $$Y$$ and $$A$$): $$\epsilon \propto F$$.

$$ \frac{\epsilon_1}{\epsilon_2} = \frac{F_1}{F_2} = \frac{30}{10} = 3 $$

The answer is 3.

$$\Delta P = \rho g h$$, $$\frac{\Delta V}{V} = \frac{\Delta P}{B}$$.

$$0.02\% = 0.0002 = \frac{\rho g h}{B} = \frac{10^3 \times 10 \times h}{9 \times 10^8}$$.

$$h = \frac{0.0002 \times 9 \times 10^8}{10^4} = \frac{1.8 \times 10^5}{10^4} = 18$$ m.

The answer is $$\boxed{18}$$.

Masses$$:m₁=2kg,m₂=4kg$$

For Atwood machine:

Acceleration:
$$a=\frac{(m₂−m₁)g}{(m₁+m₂)}$$
= $$\frac{\left((4−2)\times10\right)}{(2+4)}$$

$$=\frac{20}{6}=\frac{10}{3}$$

Tension:
$$T=m₁(g+a)$$
$$=2(10+10/3)$$
=$$2\times(40/3)$$
$$=\frac{80}{3}N$$

Now strain:

$$strain=stress/Y$$

$$Stress=T/A$$

Area of wire:
$$A=\pi r^2=\pi(4\times10⁻⁵)^2$$
$$=\pi\times16\times10⁻^1⁰$$
$$=16\pi\times10⁻^1⁰$$

So:

$$strain=T/(AY)$$

$$=(80/3)/[(16\pi\times10⁻^1⁰)(2\times10^{11})]$$

Simplify denominator:

$$16\times2=32$$
$$10⁻^1⁰\times10^{11}=10$$

$$Sodeno\min ator=32\pi\times10=320\pi$$

Now:

$$strain=(80/3)/(320\pi)$$
$$=80/(960\pi)$$
$$=1/(12\pi)$$

So given:

$$strain=1/(α\pi)$$

$$⇒α=12$$

Final answer:

$$12$$

Two metallic wires $$P$$ and $$Q$$ have the same volume and are made of the same material (so they have the same Young's modulus $$Y$$). Their cross-sectional areas are in the ratio $$A_P : A_Q = 4 : 1$$.

Since both wires have the same volume:

$$V = A_P \times L_P = A_Q \times L_Q$$

$$\Rightarrow \frac{L_P}{L_Q} = \frac{A_Q}{A_P} = \frac{1}{4}$$

Using Young's modulus formula:

$$Y = \frac{F \cdot L}{A \cdot \Delta l}$$

Rearranging for force to produce extension $$\Delta l$$:

$$F = \frac{Y \cdot A \cdot \Delta l}{L}$$

For wire P:

$$F_1 = \frac{Y \cdot A_P \cdot \Delta l}{L_P}$$

For wire Q (same extension $$\Delta l$$):

$$F_2 = \frac{Y \cdot A_Q \cdot \Delta l}{L_Q}$$

Taking the ratio:

$$\frac{F_1}{F_2} = \frac{A_P}{A_Q} \times \frac{L_Q}{L_P}$$

Since $$L = V/A$$, we have $$L_Q/L_P = A_P/A_Q$$:

$$\frac{F_1}{F_2} = \frac{A_P}{A_Q} \times \frac{A_P}{A_Q} = \left(\frac{A_P}{A_Q}\right)^2 = \left(\frac{4}{1}\right)^2 = 16$$

The answer is $$\frac{F_1}{F_2} = 16$$.

Given: $$F = 200$$ N, $$Y = 1 \times 10^{11}$$ N/m², $$L = 2$$ m, $$A = 2$$ cm² = $$2 \times 10^{-4}$$ m².

When two persons pull a wire from both ends, each exerting 200 N, the tension in the wire is 200 N (not 400 N, since the forces are in opposite directions maintaining equilibrium).

Using Young's modulus formula: $$Y = \frac{FL}{A\Delta L}$$

$$\Delta L = \frac{FL}{AY} = \frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}} = \frac{400}{2 \times 10^{7}} = 2 \times 10^{-5} \text{ m} = 20 \text{ }\mu\text{m}$$

The answer is $$\boxed{20}$$ μm.

Find how surface energy changes when 1000 small droplets coalesce into one big drop.

We relate the radii by conserving the total volume: $$1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$ which gives $$R^3 = 1000r^3$$ and hence $$R = 10r$$.

The total surface area of the 1000 small drops is $$A_{\text{small}} = 1000 \times 4\pi r^2$$, while the surface area of the single large drop is $$A_{\text{big}} = 4\pi R^2 = 4\pi(10r)^2 = 400\pi r^2$$.

Since the surface energy is given by $$E = T \times A$$ with constant surface tension $$T$$, the ratio of the energies is $$\frac{E_{\text{big}}}{E_{\text{small}}} = \frac{A_{\text{big}}}{A_{\text{small}}} = \frac{400\pi r^2}{4000\pi r^2} = \frac{1}{10}$$.

Therefore, the surface energy becomes $$\frac{1}{10}$$th of its original value.

The correct answer is Option D: $$\frac{1}{10}$$th.

We need to find how the elongation changes when both the force and radius are halved.

From Young's modulus we have $$Y = \frac{F/A}{\Delta l/L} = \frac{FL}{\pi r^2 \Delta l}$$ which gives $$\Delta l = \frac{FL}{\pi r^2 Y}.$$

When the force is halved ($$F' = F/2$$) and the radius is halved ($$r' = r/2$$), while $$L$$ and $$Y$$ remain the same, the new elongation becomes $$\Delta l' = \frac{F'L}{\pi (r')^2 Y} = \frac{(F/2)L}{\pi (r/2)^2 Y} = \frac{(F/2)L}{\pi r^2 Y / 4} = \frac{4FL}{2\pi r^2 Y} = \frac{2FL}{\pi r^2 Y} = 2\Delta l.$$

The new elongation is twice the original; halving the radius (which quarters the cross-sectional area) has a larger effect than halving the force.

The correct answer is Option (4): 2 times.

Assertion (A): Elasticity is the property by which a body tends to regain its original shape when external force is removed. True.

Reason (R): The restoring force depends on inter-atomic and inter-molecular forces. True. The elastic behavior of solids is indeed due to these forces.

R correctly explains A — the ability to regain shape (elasticity) is due to the restoring forces arising from interatomic bonds.

Both (A) and (R) are true and (R) is the correct explanation of (A). The answer corresponds to Option (3).

Match each definition carefully.

(A) “A force that restores an elastic body of unit area to its original state”

Force per unit area developed internally to restore original shape is definition of stress.

So

A→(III)

(B) “Two equal and opposite forces parallel to opposite faces”

Parallel tangential forces produce shear deformation, associated with shear modulus.

So

B→(IV)

(C) “Forces perpendicular everywhere to the surface per unit area same everywhere”

This describes uniform normal pressure acting in all directions, associated with bulk modulus.

So

C→(I)

(D) “Two equal and opposite forces perpendicular to opposite faces”

Normal tensile/compressive stress produces longitudinal strain, associated with Young’s modulus.

So

D→(II)

Young's modulus is a material property that depends only on the nature of the material, not on the dimensions of the wire.

By definition, Young's modulus is: $$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$$

While stress and strain depend on the dimensions (length $$L$$ and cross-sectional area $$A$$), the ratio $$Y$$ is an intrinsic property of the material itself. Therefore, if the length is doubled and the cross-sectional area is halved, the Young's modulus remains unchanged at $$Y$$.

Option Y

We are asked to find the extension of a vertically hanging uniform rod caused solely by its own weight. The data given are:

Weight of the rod $$W = 10\;{\rm kg\,m\,s^{-2}} = 10\;{\rm N}$$
Length $$L = 20\;{\rm cm} = 0.20\;{\rm m}$$
Cross-sectional area $$A = 100\;{\rm cm^2}$$

Because $$1\;{\rm cm} = 10^{-2}\;{\rm m},$$ we have

$$A = 100\;(10^{-2}\,{\rm m})^{2} = 100 \times 10^{-4}\;{\rm m^2} = 10^{-2}\;{\rm m^2} = 0.01\;{\rm m^2}.$$

Young’s modulus of the material is given as
$$Y = 2 \times 10^{11}\;{\rm N\,m^{-2}}.$$

Let us denote by $$x$$ the distance from the fixed support down the rod. At a section situated at this distance, the portion of the rod lying below that section has a length $$(L - x)$$ and therefore weighs

$$P(x) = \frac{W}{L}\;(L - x).$$

The factor $$\dfrac{W}{L}$$ is simply the weight per unit length, so we first evaluate it:

$$\frac{W}{L} = \frac{10\;{\rm N}}{0.20\;{\rm m}} = 50\;{\rm N\,m^{-1}}.$$

Hence the tensile force acting at a distance $$x$$ from the top is

$$P(x) = 50\,(L - x)\;{\rm N}.$$

For a small element of thickness $$dx$$ situated at that point, the corresponding elongation $$d\ell$$ is obtained from Hooke’s (Young’s) law in the form

$$d\ell = \frac{P(x)}{A\,Y}\;dx.$$

Substituting $$P(x)$$, we get

$$d\ell = \frac{50\,(L - x)}{A\,Y}\;dx.$$

To find the total extension of the rod, we integrate $$d\ell$$ from the top end $$x = 0$$ to the bottom end $$x = L$$:

$$\Delta L = \int_{0}^{L} d\ell = \int_{0}^{L} \frac{50\,(L - x)}{A\,Y}\;dx.$$

The constants $$\dfrac{50}{A\,Y}$$ can be taken outside the integral:

$$\Delta L = \frac{50}{A\,Y}\int_{0}^{L}(L - x)\;dx.$$

We now evaluate the integral:

$$\int_{0}^{L}(L - x)\;dx = \bigg[Lx - \frac{x^{2}}{2}\bigg]_{0}^{L} = \left(L\cdot L - \frac{L^{2}}{2}\right) = \frac{L^{2}}{2}.$$

Therefore

$$\Delta L = \frac{50}{A\,Y}\,\frac{L^{2}}{2} = \frac{50\,L^{2}}{2\,A\,Y}.$$

Substituting the numerical values now:

$$\Delta L = \frac{50\,(0.20\,{\rm m})^{2}}{2 \times 0.01\,{\rm m^{2}} \times 2 \times 10^{11}\,{\rm N\,m^{-2}}}.$$

We first compute $$L^{2}$$:

$$L^{2} = (0.20)^{2} = 0.04\;{\rm m^{2}}.$$

Multiplying by 50:

$$50 \times 0.04 = 2.0.$$

Now the denominator:

$$2 \times 0.01 \times 2 \times 10^{11} = 0.02 \times 2 \times 10^{11} = 0.04 \times 10^{11} = 4 \times 10^{9}.$$

Finally,

$$\Delta L = \frac{2.0}{4 \times 10^{9}} = 0.5 \times 10^{-9}\;{\rm m} = 5 \times 10^{-10}\;{\rm m}.$$

Hence, the correct answer is Option A.

We have a simple pendulum whose bob is suspended by a uniform metallic wire of cross-sectional area $$A$$. Let the actual (already slightly stretched) length of this wire be $$L$$ when only the original bob is hanging. With this length the measured time period is

$$T = 2\pi \sqrt{\dfrac{L}{g}}\;.$$

Now an additional mass $$M$$ is attached to the same bob. The extra load increases the tension in the wire by the amount

$$F = Mg\;.$$

Because the wire obeys Hooke’s law, this extra force produces an extra elongation $$\Delta L$$. For a wire of natural length $$L$$, Young’s modulus $$Y$$ is defined by the relation

$$Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{F/A}{\Delta L/L} = \dfrac{F\,L}{A\,\Delta L}\;.$$

Re-arranging, the extra extension is

$$\Delta L = \dfrac{F\,L}{A\,Y} = \dfrac{Mg\,L}{A\,Y}\;.$$

Because of this elongation, the new effective length of the pendulum becomes $$L + \Delta L$$ and the new time period is therefore

$$T_M = 2\pi \sqrt{\dfrac{L + \Delta L}{g}}\;.$$

Squaring the expressions for both time periods, we get

$$T^{2} = 4\pi^{2}\dfrac{L}{g}\;,\qquad T_M^{2} = 4\pi^{2}\dfrac{L + \Delta L}{g}\;.$$

Subtract the first equation from the second:

$$T_M^{2} - T^{2} = 4\pi^{2}\dfrac{(L + \Delta L) - L}{g} = 4\pi^{2}\dfrac{\Delta L}{g}\;.$$

Hence the extra elongation is also expressed as

$$\Delta L = \dfrac{g}{4\pi^{2}}\bigl(T_M^{2} - T^{2}\bigr)\;.$$

We now have two expressions for $$\Delta L$$, so we equate them:

$$\dfrac{Mg\,L}{A\,Y} = \dfrac{g}{4\pi^{2}}\bigl(T_M^{2} - T^{2}\bigr)\;.$$

The factor $$g$$ cancels from both sides, leaving

$$\dfrac{M\,L}{A\,Y} = \dfrac{1}{4\pi^{2}}\bigl(T_M^{2} - T^{2}\bigr)\;.$$

But from the first squared relation we already have $$L = \dfrac{g\,T^{2}}{4\pi^{2}}$$. Substitute this value of $$L$$ into the left-hand side:

$$\dfrac{M}{A\,Y}\left(\dfrac{g\,T^{2}}{4\pi^{2}}\right) = \dfrac{1}{4\pi^{2}}\bigl(T_M^{2} - T^{2}\bigr)\;.$$

Multiply both sides by $$4\pi^{2}$$ to clear the denominator:

$$\dfrac{M\,g\,T^{2}}{A\,Y} = T_M^{2} - T^{2}\;.$$

Solve for $$\dfrac{1}{Y}$$:

$$\dfrac{1}{Y} = \dfrac{T_M^{2} - T^{2}}{M\,g\,T^{2}}\,A = \left[\dfrac{T_M^{2}}{T^{2}} - 1\right]\dfrac{A}{M\,g}\;.$$

Writing the ratio compactly,

$$\dfrac{1}{Y} = \left[\left(\dfrac{T_M}{T}\right)^{2} - 1\right]\dfrac{A}{M\,g}\;.$$

This matches Option B in the given list.

Hence, the correct answer is Option B.

We are told that a steel wire is heated through a temperature rise of $$\Delta T = 100\;^{\circ}\mathrm{C}$$, but its length must remain unchanged. Normally, heating would make the wire expand. The thermal (free) linear strain that tries to develop is given by the well-known relation

$$ \text{Thermal strain} = \alpha \,\Delta T, $$

where $$\alpha$$ is the coefficient of linear thermal expansion. For steel we have $$\alpha = 1.1 \times 10^{-5}\;\mathrm{K^{-1}}$$. Substituting the given temperature rise, the thermal strain would be

$$ \alpha \,\Delta T = \left(1.1 \times 10^{-5}\right)\times 100 = 1.1 \times 10^{-3}. $$

To keep the length actually constant, an equal and opposite mechanical (compressive) strain must be produced. According to Hooke’s law for a linear elastic solid, the mechanical strain produced by an applied stress $$\sigma$$ is

$$ \text{Mechanical strain} = \frac{\sigma}{Y}, $$

where $$Y = 2 \times 10^{11}\;\mathrm{N\,m^{-2}}$$ is Young’s modulus for steel. Setting the mechanical strain equal in magnitude to the unwanted thermal strain, we write

$$ \frac{\sigma}{Y} = \alpha \,\Delta T. $$

Solving for the required stress $$\sigma$$ (which will be equal to the necessary external pressure $$P$$ because stress and pressure have the same dimensions), we obtain

$$ \sigma = Y \, \alpha \,\Delta T. $$

Now we substitute all the numerical values step by step:

$$ \sigma = \left(2 \times 10^{11}\;\mathrm{N\,m^{-2}}\right) \left(1.1 \times 10^{-5}\right) \left(100\right). $$

First multiply the powers of ten:

$$ 10^{11}\times 10^{-5}\times 100 = 10^{11}\times 10^{-5}\times 10^{2} = 10^{11}\times 10^{-3} = 10^{8}. $$

Next multiply the numerical coefficients:

$$ 2 \times 1.1 = 2.2. $$

Combining these two results, the stress (and therefore the pressure) required is

$$ \sigma = 2.2 \times 10^{8}\;\mathrm{N\,m^{-2}} = 2.2 \times 10^{8}\;\mathrm{Pa}. $$

This matches exactly the value in Option A.

Hence, the correct answer is Option A.

Both strings have the same cross-sectional area, call it $$A$$. The masses of the strings themselves are negligible, so the only forces are the weights of the attached masses.

For string B
Only the mass $$2M$$ is hanging from its lower end, so the tension throughout string B is
$$F_B = 2Mg$$

For string A
String A has to support both masses: the mass $$M$$ fixed to its lower end and the mass $$2M$$ hanging further below. Hence the tension in string A is
$$F_A = (M + 2M)g = 3Mg$$

The elongation of a string of length $$L$$ under tension $$F$$ is given by
$$\Delta L = \frac{F\,L}{A\,Y}$$
where $$Y$$ is Young’s modulus.

Let the original lengths be $$L_A$$ and $$L_B$$. The data give
$$\frac{L_A}{L_B} = 2 \quad\Rightarrow\quad L_A = 2L_B$$
and
$$\frac{Y_A}{Y_B} = 0.5 \quad\Rightarrow\quad Y_B = 2Y_A$$

Now write the ratio of elongations:

$$\frac{\Delta L_A}{\Delta L_B} = \frac{\dfrac{F_A L_A}{A Y_A}} {\dfrac{F_B L_B}{A Y_B}} = \frac{F_A L_A Y_B}{F_B L_B Y_A}$$

Insert the known values:

$$\frac{\Delta L_A}{\Delta L_B} = \frac{(3Mg)\,(2L_B)\,(2Y_A)}{(2Mg)\,(L_B)\,(Y_A)} = \frac{3 \times 2 \times 2}{2} = 6$$

Therefore, the ratio of elongations is $$\mathbf{6}$$.

Option D which is: $$6$$

Young’s modulus:

Y = (stress) / (strain)

$$Y=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$$

Now if:

radius → 2r ⇒ area $$A\propto r^2$$ becomes 4A
length → 2L

Substitute in formula:

$$Y'=\frac{F(2L)}{(4A)(\Delta L')}$$

But extension $$ΔL′$$ also changes such that:

$$\Delta L'=\frac{F(2L)}{Y(4A)}=\frac{1}{2}\Delta L$$

Putting back:

Y′=Y

Breaking stress = $$12\times10^8N/m^2$$

Areas:
$$upper=0.008cm^2=8\times10^{-7}m^2$$
$$lower=0.004cm^2=4\times10^{-7}m^2$$

step 1: max tension each wire can bear

upper wire:

$$T_u=σA=12\times10^8\times8\times10^{-7}=960N$$

lower wire:

$$T_l=12\times10^8\times4\times10^{-7}=480N$$

step 2: tensions in wires

let mass added = mmm

lower wire carries:

$$T_l=(10+m)g=(10+m)\times10$$

=100+10m

limit:

$$100+10m\le480\Rightarrow m\le38$$

upper wire carries total mass:

$$30+10+m=40+m$$

$$T_u=(40+m)\times10=400+10m$$

limit:

$$400+10m\le960\Rightarrow m\le56$$

step 3: final

smaller limit governs → m=38

We need to find the fractional compression of water at a depth of 2.5 km below sea level.

The fractional change in volume due to pressure is given by:

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

where $$P$$ is the pressure at the depth and $$B$$ is the bulk modulus.

Calculate the pressure at depth $$h = 2.5 \, \text{km} = 2500 \, \text{m}$$:

$$P = \rho g h = 10^3 \times 10 \times 2500 = 2.5 \times 10^7 \, \text{N/m}^2$$

Calculate the fractional compression:

$$\frac{\Delta V}{V} = \frac{P}{B} = \frac{2.5 \times 10^7}{2 \times 10^9} = 1.25 \times 10^{-2} = 0.0125$$

Express as a percentage:

$$\frac{\Delta V}{V} = 0.0125 \times 100\% = 1.25\%$$

The correct answer is Option (1): 1.25%.

When the mass $$m$$ is suspended, the vertical components of the tension ($$T$$) in the two halves of the wire balance the weight: $$2T \sin \theta = mg$$

For small angles ($$\theta = 1/100$$ rad), we use the approximation $$\sin \theta \approx \theta$$

$$2T\theta = mg \implies T = \frac{mg}{2\theta}$$

The original length of half the wire is $$L$$. The new length $$L'$$ is $$L' = \frac{L}{\cos \theta}$$

$$\text{Strain} = \frac{L' - L}{L} = \frac{1}{\cos \theta} - 1$$

Using the small angle approximation $$\cos \theta \approx 1 - \frac{\theta^2}{2}$$:

$$\text{Strain} \approx \left(1 + \frac{\theta^2}{2}\right) - 1 = \frac{\theta^2}{2}$$

$$Y = \frac{T / A}{\theta^2 / 2} = \frac{2T}{A\theta^2}$$

$$Y = \frac{2(mg / 2\theta)}{A\theta^2} = \frac{mg}{A\theta^3}$$

$$A = \frac{mg}{Y\theta^3}$$

$$A = \frac{2 \times 10}{(2 \times 10^{11}) \times (10^{-2})^3}$$

$$A = \frac{20}{2 \times 10^{11} \times 10^{-6}} = \frac{10}{10^5} = 1 \times 10^{-4} \text{ m}^2$$

The answer is 1.

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\Delta L/L} = \frac{TL}{A\Delta L}$$

$$\implies \Delta L = \frac{TL}{AY}$$

$$\ell_1 - L = \frac{T_1L}{AY} \implies \frac{L}{AY} = \frac{\ell_1 - L}{T_1} \quad \text{--- (1)}$$

$$\ell_2 - L = \frac{T_2L}{AY} \implies \frac{L}{AY} = \frac{\ell_2 - L}{T_2} \quad \text{--- (2)}$$

$$\frac{\ell_1 - L}{T_1} = \frac{\ell_2 - L}{T_2}$$ $$\implies T_2\ell_1 - T_2L = T_1\ell_2 - T_1L$$

$$\implies T_1L - T_2L = T_1\ell_2 - T_2\ell_1$$ $$\implies L(T_1 - T_2) = T_1\ell_2 - T_2\ell_1$$

$$\implies L = \frac{T_1\ell_2 - T_2\ell_1}{T_1 - T_2}$$

We start with the definition of bulk modulus. For an isotropic body under a uniform (hydrostatic) pressure $$P$$, the bulk modulus $$B$$ is defined as

$$B \;=\; -\,\dfrac{P}{\dfrac{\Delta V}{V}}$$

Here $$\dfrac{\Delta V}{V}$$ is the volumetric (volume) strain and the minus sign accounts for the fact that an applied pressure (a compressive stress) makes the volume decrease (so $$\Delta V$$ is negative).

Rearranging the above formula to isolate the volume strain, we obtain

$$\dfrac{\Delta V}{V} \;=\; -\,\dfrac{P}{B}$$

Now we substitute the given numerical values. The hydrostatic pressure is $$P = 4 \text{ GPa} = 4 \times 10^{9}\ {\rm Pa}$$, and the bulk modulus is $$B = 8 \times 10^{10}\ {\rm Pa}$$. Therefore,

$$\dfrac{\Delta V}{V} \;=\; -\,\dfrac{4 \times 10^{9}}{8 \times 10^{10}}$$

Carrying out the division step by step, first divide the coefficients:

$$\dfrac{4}{8} = 0.5$$

Next handle the powers of ten:

$$\dfrac{10^{9}}{10^{10}} = 10^{-1} = 0.1$$

Multiplying the two intermediate results,

$$0.5 \times 0.1 = 0.05$$

Remembering the negative sign, we obtain

$$\dfrac{\Delta V}{V} = -\,0.05$$

This number means the volume decreases by 5 %. To convert this volume strain into linear (length) strain for a cube we use a geometrical relation. For a cube with side length $$l$$, the volume is $$V = l^{3}$$. If the side changes by a small amount $$\Delta l$$, we differentiate:

$$V = l^{3} \;\;\Longrightarrow\;\; \dfrac{dV}{V} = 3\,\dfrac{dl}{l}$$

In infinitesimal form, $$dV/V$$ is the volume strain and $$dl/l$$ is the linear strain. Re-writing this for finite but small changes, we approximate

$$\dfrac{\Delta V}{V} \approx 3\,\dfrac{\Delta l}{l}$$

Hence the linear strain is one-third of the volume strain:

$$\dfrac{\Delta l}{l} = \dfrac{1}{3}\,\dfrac{\Delta V}{V}$$

Substituting the value we already found,

$$\dfrac{\Delta l}{l} = \dfrac{1}{3}\,(-0.05)$$

Dividing by 3,

$$\dfrac{\Delta l}{l} = -0.016666\dots$$

To express this as a percentage change in length we multiply the absolute value by 100 %:

$$\bigl|\text{percentage change}\bigr| = 0.016666\dots \times 100\% = 1.6666\dots \%$$

Rounding to the number of significant figures implied by the data, we state

$$\text{percentage change in length} \approx 1.67\%$$

Hence, the correct answer is Option D.

We begin by recalling that when the temperature of a free rod rises by $$\Delta T$$, its natural tendency is to increase its length. The linear thermal expansion formula is first stated:

$$\Delta L_{\text{thermal}} = \alpha L \Delta T,$$

where $$\alpha$$ is the coefficient of linear expansion, $$L$$ is the original length and $$\Delta T$$ is the rise in temperature.

This change of length produces a thermal strain (fractional change in length) given by

$$\varepsilon_{\text{thermal}} = \frac{\Delta L_{\text{thermal}}}{L} = \alpha \Delta T.$$

In the problem, however, a compressive force $$F$$ is applied at both ends so that the rod is not allowed to change its length at all. Therefore, the total strain in the rod must be zero:

$$\varepsilon_{\text{total}} = \varepsilon_{\text{thermal}} + \varepsilon_{\text{mechanical}} = 0.$$

We have already written $$\varepsilon_{\text{thermal}} = \alpha \Delta T.$$ Next, we write the mechanical strain caused by the external compressive force. Young’s modulus $$Y$$ is defined by the relation

$$Y = \frac{\text{stress}}{\text{strain}} \quad \Longrightarrow \quad \text{strain} = \frac{\text{stress}}{Y}.$$

The compressive stress produced by the force $$F$$ over the cross-sectional area $$A$$ is

$$\sigma = \frac{F}{A}.$$

Because the force is compressive, the corresponding mechanical strain is also compressive (negative):

$$\varepsilon_{\text{mechanical}} = -\frac{\sigma}{Y} = -\frac{F}{A\,Y}.$$

Now we impose the condition that the rod’s length does not change, i.e. $$\varepsilon_{\text{total}} = 0$$. Substituting the expressions for the two strains, we get

$$\alpha \Delta T \;+\; \left(-\frac{F}{A\,Y}\right) = 0.$$

Simplifying,

$$\alpha \Delta T = \frac{F}{A\,Y}.$$

We now solve for Young’s modulus $$Y$$ by cross-multiplying:

$$Y = \frac{F}{A\,\alpha \,\Delta T}.$$

This expression matches Option C in the given list.

Hence, the correct answer is Option C.

The maximum stress is the maximum force per unit area. Therefore, the maximum tension ($$ T_{max} $$) is:

$$ T_{max} = \sigma_{max} \times A $$

Where the cross-sectional area $$ A = \pi r^2 $$.

Substitute the values:

$$ T_{max} = (4 \times 10^8) \times \left(3.14 \times (4 \times 10^{-3})^2\right) $$

$$ T_{max} = (4 \times 10^8) \times (3.14 \times 16 \times 10^{-6}) $$

$$ T_{max} = 64 \times 3.14 \times 10^2 $$

$$ T_{max} = 200.96 \times 100 $$

$$ T_{max} = 20096\text{ N} $$

When the lift is accelerating upwards with a maximum acceleration $$ a $$, the net upward force is $$ T_{max} - mg $$.

According to the equation of motion:

$$ T_{max} - mg = ma $$

$$ T_{max} = m(g + a) $$

Substitute the known values into the equation:

$$ 20096 = 1600(10 + a) $$

Solve for $$ a $$:

$$ \frac{20096}{1600} = 10 + a $$

$$ 12.56 = 10 + a $$

$$ a = 12.56 - 10 $$

$$ a = 2.56\text{ m/s}^2 $$

Young's modulus, $$Y = \frac{\text{Stress}}{\text{Strain}}$$

In the given plot, strain is on the $$y$$-axis and stress is on the $$x$$-axis.

$$\text{Slope} = \frac{\text{Strain}}{\text{Stress}} = \frac{1}{Y}$$

Therefore, the material with the least slope will have the largest Young's modulus.

Observing the plot, material C has the minimum slope. Therefore, it possesses the largest Young's modulus.

Using Stokes' law: $$v_t = \frac{2r^2(\rho_s - \rho_l)g}{9\eta}$$

r = 1 mm = 0.1 cm, ρ_s = 10.5 g/cm³, ρ_l = 1.5 g/cm³, η = 10 P = 10 g/(cm·s), g = 1000 cm/s².

$$v_t = \frac{2 \times 0.01 \times 9 \times 1000}{9 \times 10} = \frac{180}{90} =2.0$$ cm/s

The answer is Option 2: 2.0 cm/s.

Wire A: length 6.0 cm, area $$3.0 \times 10^{-5}$$ m$$^2$$. Wire B: length 5.4 cm, area $$4.5 \times 10^{-5}$$ m$$^2$$. Same elongation under same load. Find $$Y_A/Y_B = x:3$$.

Apply Hooke's law for each wire.

$$\Delta L = \frac{FL}{AY}$$

Same $$\Delta L$$ and same $$F$$:

$$\frac{FL_A}{A_A Y_A} = \frac{FL_B}{A_B Y_B}$$

$$\frac{L_A}{A_A Y_A} = \frac{L_B}{A_B Y_B}$$

$$\frac{Y_A}{Y_B} = \frac{L_A \cdot A_B}{L_B \cdot A_A} = \frac{6.0 \times 4.5 \times 10^{-5}}{5.4 \times 3.0 \times 10^{-5}} = \frac{27}{16.2} = \frac{5}{3}$$

So $$Y_A:Y_B = 5:3$$, meaning $$x = 5$$.

The correct answer is Option D: 5.