JEE Thermal Expansion Questions

Let the original length of the strip at temperature $$T_1$$ be $$L_0$$. For a solid with constant linear expansion coefficient $$\alpha$$, the change in length for a temperature change $$\Delta T$$ is

$$\Delta L = L \,\alpha\,\Delta T \quad -(1)$$

Elongation from $$T_1$$ to $$T_2$$
The temperature rise is $$T_2-T_1 = \Delta T$$. Using $$(1)$$ with the initial length $$L_0$$, we have

$$\Delta L_1 = L_0 \,\alpha\,\Delta T \quad -(2)$$

After this heating, the new length at $$T_2$$ becomes

$$L_{T_2} = L_0 + \Delta L_1 = L_0\bigl(1+\alpha\Delta T\bigr) \quad -(3)$$

Elongation from $$T_2$$ to $$T_3$$
We are told $$T_3+T_1 = 2T_2$$. Re-arranging, $$T_3 - T_2 = T_2 - T_1 = \Delta T \quad -(4)$$ Thus the second heating also raises the temperature by $$\Delta T$$.

Now use $$(1)$$ again, but the initial length for this stage is $$L_{T_2}$$ from $$(3)$$:

$$\Delta L_2 = L_{T_2}\,\alpha\,(T_3-T_2) = L_0\bigl(1+\alpha\Delta T\bigr)\alpha\Delta T \quad -(5)$$

Compare $$(5)$$ with $$(2)$$ (note that $$\Delta L_1 = L_0\alpha\Delta T$$):

$$\Delta L_2 = \Delta L_1\bigl(1+\alpha\Delta T\bigr)$$

Hence

$$\boxed{\;\Delta L_2 = \Delta L_1[\,1+\alpha\Delta T\,]\;}$$

Therefore the correct option is
Option D which is: $$\Delta L_1[1 + \alpha\Delta T]$$.

We need to find the temperature to which a brass wire must be cooled to increase its tension from $$T$$ to $$1.4T$$.

Since when a wire held between rigid supports is cooled it tries to contract but cannot, a thermal stress develops, producing a tension given by $$T = YA\alpha\Delta\theta$$, where $$Y$$ = Young's modulus, $$A$$ = cross-sectional area, $$\alpha$$ = coefficient of linear expansion, and $$\Delta\theta$$ = temperature change.

Initially the wire is at 27°C and is cooled to -43°C, so $$\Delta\theta_1 = 27 - (-43) = 70°C$$; substituting into the tension expression gives $$T = YA\alpha \times 70$$.

For the increased tension $$1.4T$$ we therefore have $$1.4T = YA\alpha \times \Delta\theta_2$$.

Dividing these equations yields $$\frac{1.4T}{T} = \frac{\Delta\theta_2}{70}$$ and hence $$\Delta\theta_2 = 1.4 \times 70 = 98°C$$.

This means the wire must be cooled from 27°C by 98°C, giving $$\theta = 27 - 98 = -71°C$$. Therefore, the required temperature is Option 4: -71°C.

For the bimetallic strip to maintain a constant length difference (20 cm), both metals must expand by the same amount when the temperature changes.

The change in length for each material is:

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

For the difference to remain constant, the expansions must be equal:

$$\Delta L_{\text{brass}} = \Delta L_{\text{iron}}$$

$$L_{\text{brass}} \times \alpha_{\text{brass}} \times \Delta T = L_{\text{iron}} \times \alpha_{\text{iron}} \times \Delta T$$

Cancelling $$\Delta T$$ from both sides:

$$L_{\text{brass}} \times \alpha_{\text{brass}} = L_{\text{iron}} \times \alpha_{\text{iron}}$$

$$40 \times 1.8 \times 10^{-5} = L_{\text{iron}} \times 1.2 \times 10^{-5}$$

$$L_{\text{iron}} = \dfrac{40 \times 1.8}{1.2} = \dfrac{72}{1.2} = 60 \text{ cm}$$

We can verify: The difference $$L_{\text{iron}} - L_{\text{brass}} = 60 - 40 = 20 \text{ cm}$$, which is the required scale length.

Therefore, the length of iron is $$\boxed{60}$$ cm.

A gold ring of diameter 6.230 cm needs to be heated to fit on a wooden bangle of diameter 6.241 cm. Room temperature is 27°C. $$\alpha_L = 1.4 \times 10^{-5}$$ K$$^{-1}$$.

$$L = L_0(1 + \alpha_L \Delta T)$$

where $$L_0 = 6.230$$ cm is the initial diameter and $$L = 6.241$$ cm is the required diameter.

$$6.241 = 6.230(1 + 1.4 \times 10^{-5} \times \Delta T)$$

$$\frac{6.241}{6.230} = 1 + 1.4 \times 10^{-5} \times \Delta T$$

$$\frac{6.241 - 6.230}{6.230} = 1.4 \times 10^{-5} \times \Delta T$$

$$\frac{0.011}{6.230} = 1.4 \times 10^{-5} \times \Delta T$$

$$1.765 \times 10^{-3} = 1.4 \times 10^{-5} \times \Delta T$$

$$\Delta T = \frac{1.765 \times 10^{-3}}{1.4 \times 10^{-5}} = \frac{1765}{14} \approx 125.7 \text{ °C}$$

$$T = 27 + 125.7 = 152.7 \text{ °C}$$

Hence, the correct answer is Option D: 152.7°C.

We have a cubic metal box with each side of length $$a$$ at room temperature $$T$$. The coefficient of linear expansion of the metal sheet is $$\alpha$$. The temperature is increased by $$\Delta T$$.

The initial volume of the cube is $$V_0 = a^3$$.

When the temperature increases by $$\Delta T$$, each side of the cube expands. The new length of each side is $$a' = a(1 + \alpha \Delta T)$$.

The new volume is $$V = (a')^3 = a^3(1 + \alpha \Delta T)^3$$.

Since $$\alpha \Delta T$$ is very small (the problem states "small temperature $$\Delta T$$"), we can use the binomial approximation $$(1 + x)^3 \approx 1 + 3x$$ for small $$x$$.

So $$V \approx a^3(1 + 3\alpha \Delta T) = a^3 + 3a^3\alpha \Delta T$$.

The increase in volume is $$\Delta V = V - V_0 = 3a^3\alpha \Delta T$$.

This result is consistent with the general formula for volume expansion: $$\Delta V = V_0 \gamma \Delta T$$, where the coefficient of volume expansion $$\gamma = 3\alpha$$ for isotropic materials.

Hence, the correct answer is Option B.

We are told that when the temperature of a metal wire rises from $$0^{\circ}\text C$$ to $$10^{\circ}\text C$$ its length increases by $$0.02\%$$. First, let us convert this percentage increase in length into a fractional form because all expansion formulae use fractional changes.

$$0.02\% = \frac{0.02}{100}=0.0002$$

So the fractional change in length is

$$\frac{\Delta L}{L}=0.0002$$

The formula for linear (one-dimensional) thermal expansion is first stated:

$$\frac{\Delta L}{L}=\alpha\,\Delta T$$

where $$\alpha$$ is the coefficient of linear expansion and $$\Delta T$$ is the rise in temperature. We substitute the known values:

$$0.0002=\alpha\,(10^{\circ}\text C)$$

Solving for $$\alpha$$, we divide both sides by $$10$$:

$$\alpha=\frac{0.0002}{10}=2\times10^{-5}\;{}^{\circ}\text C^{-1}$$

Now we turn to the volume change. For an isotropic solid, the relation between the coefficient of volume expansion $$\beta$$ and the linear coefficient $$\alpha$$ is stated:

$$\beta = 3\alpha$$

Therefore, substituting $$\alpha = 2\times10^{-5}\;{}^{\circ}\text C^{-1}$$, we get

$$\beta = 3 \times 2\times10^{-5}=6\times10^{-5}\;{}^{\circ}\text C^{-1}$$

The fractional change in volume is then

$$\frac{\Delta V}{V}=\beta\,\Delta T = (6\times10^{-5})(10)=6\times10^{-4}$$

To convert this fractional change to a percentage, we multiply by $$100$$:

$$6\times10^{-4}\times100 = 0.06\%$$

Thus, the volume increases by $$0.06\%$$ when the temperature rises from $$0^{\circ}\text C$$ to $$10^{\circ}\text C$$.

Because the mass of the wire does not change, density $$\rho$$ varies inversely with volume. The relation is written as

$$\rho = \frac{M}{V}\quad\Longrightarrow\quad \frac{\Delta\rho}{\rho} = -\frac{\Delta V}{V}$$

So the fractional change in density is the negative of the fractional change in volume. Hence

$$\frac{\Delta\rho}{\rho}= -6\times10^{-4}$$

In percentage terms this is

$$-6\times10^{-4}\times100 = -0.06\%$$

The negative sign merely tells us the density decreases; the magnitude of the percentage change is $$0.06\%$$.

Hence, the correct answer is Option A.