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Thermal Properties of Matter

JEE Main level, single correct, medium

13 of 126

Question 13 of 126

The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$ resulting in increase of its length by $\Delta L_1$ . The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta L_2$ . Given $T_3+T_1=2T_2$ and $T_2-T_1=\Delta T$ , the value of $\Delta L_2$ is ______.

Show answer and solution

Answer: (D) $\Delta L_1[1+\alpha\Delta T]$

Let the length of the metal strip at temperature

T_1

L_1

.

When the temperature is increased from

T_1

T_2

, the new length

L_2

is given by:

L_2 = L_1(1 + \alpha \Delta T)

where

\Delta T = T_2 - T_1

.

The increase in length is:

\Delta L_1 = L_2 - L_1 = L_1 \alpha \Delta T

When the temperature is further increased from

T_2

T_3

, the new length

L_3

is:

L_3 = L_2(1 + \alpha \Delta T')

where

\Delta T' = T_3 - T_2

.

Given

T_3 + T_1 = 2T_2

, we have

T_3 - T_2 = T_2 - T_1 = \Delta T

.

Thus,

\Delta T' = \Delta T

.

The increase in length for this second temperature change is:

\Delta L_2 = L_3 - L_2 = L_2 \alpha \Delta T

Substituting

L_2 = L_1(1 + \alpha \Delta T)

into the equation for

\Delta L_2

\Delta L_2 = L_1(1 + \alpha \Delta T) \alpha \Delta T

\Delta L_2 = (L_1 \alpha \Delta T)(1 + \alpha \Delta T)

Since

\Delta L_1 = L_1 \alpha \Delta T

, substituting this gives:

\Delta L_2 = \Delta L_1(1 + \alpha \Delta T)

Answer:

\Delta L_1[1+\alpha\Delta T]

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JEE Main 2025 Apr 2 Shift 1 question paper

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