Q 11-10-010NEETJEE MainEAMCET 1983 (Engineering)Easy
A metal rod having a linear coefficient of expansion $2 \times 10^{-5}\ °\text{C}^{-1}$ has a length of $1$ m at $25°$C. The temperature at which it is shortened by $1$ mm is
Answer: (C) $-25°$C
Concept. $\Delta L = L\alpha\Delta T$; a shortening means a negative $\Delta T$.
Step 1. Temperature change.
$$\Delta T = \frac{\Delta L}{L\alpha} = \frac{-10^{-3}}{1 \times 2 \times 10^{-5}} = -50°\text{C}$$
Step 2. New temperature. $25 - 50 = -25°$C.
Answer. $-25°$C (Option C).
Solution by Sreeraj P, M.Sc Physics