The ratio of the coefficient of linear expansion of two metals ∝1/∝2 is 3:4. If, when heated through the same temperature change, the ratio of the increase in length of the two metals, e1/e2 is 1:2, the ratio of the original lengths l1/l2 is
- 8/3
- 3/2
-
2/3
- 3/8
The correct answer is: C
Explanation
Ratio of their linear expansion = ∝1/∝2= 3:4.When heated to the same temperature range, the ratio of their increase in length e1/e2 = 1:2
But the increase in length of
1 = e1 = ∝1l1Δθ and increase i length of
2 = e2 = ∝2l2Δθ
| => | e1 | = | ∝1l1Δθ | = | ∝1l1 | = | 1 |
| e2 | ∝2l2Δθ | ∝2l2 | 2 |
| But | ∝1 | = | 3 | , ∴ | 3 | x | l1 | = | 1 |
| ∝2 | 4 | 4 | l2 | 2 |
=> 6l1 = 4l2
| ∴ | l1 | = | 4 | = | 2 | OR 2 : 3 |
| l2 | 6 | 3 |