The ratio of the coefficient of linear expansion of two metals \(\frac{∝_1}{∝_2}\) is 3:4. If, when heated through the same temperature change, the ratio of the increase in length of the two metals, \(\frac{e_1}{e_2}\) is 1:2, the ratio of the original lengths \(\frac{l_1}{l_2}\) is
A
8/3
B
3/2
C
2/3
D
3/8
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Ratio of their linear expansion = \(\frac{∝_1}{∝_2}\)= 3:4.
When heated to the same temperature range, the ratio of their increase in length \(\frac{e_1}{e_2}\) = 1:2
Since e = l \(\alpha \Delta T\) and \(\Delta T\) is same,
\(\frac{e_1}{e_2} = \frac{l_1 \alpha_1}{l_2 \alpha_2}\)
\(\frac{1}{2} = \left( \frac{l_1}{l_2} \right) \times \left( \frac{3}{4} \right)\)
\(\frac{l_1}{l_2} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}\)
So, \(\frac{l_1}{l_2}\) = 2:3.
Source: Myschool Past Questions Archive (Record #14386)