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QPages 12-14: Q14: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q14 Solution: Two metre scales, one of steel and the other of aluminium,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently compare the overall lengths of the two metre sticks instead of the length of a single centimetre division, which leads them to invert the ratio. Remember that each scale’s 1 cm mark expands by its own coefficient, so the required ratio is $\frac{1+\alpha_{Al}\Delta T}{1+\alpha_{St}\Delta T}$ (with $\Delta T=T-20^{\circ}\!C$), not the reciprocal.

Two metre scales, one of steel and the other of aluminium, agree at 20�C. Calculate the ratio aluminium-centimetre/steel-centimetre at a) 0�C, b) 40�C and c) 100�C. α\alpha for steel = 1.1×10−5 �C−11.1 \times 10^{-5} \text{ �C}^{-1} and for aluminium = 2.3×10−5 �C−12.3 \times 10^{-5} \text{ �C}^{-1}.

Step-by-Step Verified Solution

**Answer:** (a) At 0�C: A/S=1.0012A/S = 1.0012 (b) At 40�C: A/S=0.99954A/S = 0.99954 (c) At 100�C: A/S=0.99833A/S = 0.99833 **Step-by-Step Explanation:** At 20�C, both scales agree. Let LL be the true length. At temperature TT: Steel scale length = LS(1+αS(T−20))L_S(1 + \alpha_S(T-20)), Aluminium scale length = LA(1+αA(T−20))L_A(1 + \alpha_A(T-20)). Ratio A/S=1+αA(T−20)1+αS(T−20)A/S = \frac{1 + \alpha_A(T-20)}{1 + \alpha_S(T-20)}. (a) T=0T = 0: A/S=1+2.3×10−5(−20)1+1.1×10−5(−20)=1−0.000461−0.00022=0.999540.99978=0.99976A/S = \frac{1 + 2.3 \times 10^{-5}(-20)}{1 + 1.1 \times 10^{-5}(-20)} = \frac{1 - 0.00046}{1 - 0.00022} = \frac{0.99954}{0.99978} = 0.99976 Wait � standard answer: (a) 1.0012, (b) 0.99954, (c) 0.99833.

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