The ratio of the coefficient of linear expansion of two metals âˆ1/âˆ[sub]2[/sub] is 3:4. If, when heated through the same temperature change, the ratio of the increase in length of the two metals, e1/e[sub]2[/sub] is 1:2, the ratio of the original lengths l1/l[sub]2[/sub] is
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Correct Answer: Option C
Explanation:
Ratio of their linear expansion = âˆ1/âˆ[sub]2[/sub]= 3:4.
When heated to the same temperature range, the ratio of their increase in length e1/e[sub]2[/sub] = 1:2
But the increase in length of
1 = e1 = âˆ1l1Δθ and increase i length of
2 = e2 = âˆ2l2Δθ
=> 6l1 = 4l2
Ratio of their linear expansion = âˆ1/âˆ[sub]2[/sub]= 3:4.
When heated to the same temperature range, the ratio of their increase in length e1/e[sub]2[/sub] = 1:2
But the increase in length of
1 = e1 = âˆ1l1Δθ and increase i length of
2 = e2 = âˆ2l2Δθ
| e2 | âˆ2l2Δθ | âˆ2l2 | 2 |
| âˆ2 | 4 | 4 | l2 | 2 |
=> 6l1 = 4l2
| l2 | 6 | 3 |