A metal string A is suspended from a rigid support and its free end is attached to a block of mass

Question

A metal string A is suspended from a rigid support and its free end is attached to a block of mass $m$ . Second block having mass $2 m$ is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings A to B is 2 and the ratio of their Young’s moduli $(Y_{A} / Y_{B})$ is 0.5. The ratio of elongations in A to B is ______. [2026]

Smart Achievers · Accepted Answer

(4)

$Δ l = \frac{T l}{Y A}$

$Δ l_{1} = \frac{3 m g l}{Y A}$

$Δ l_{2} = \frac{2 m g (l / 2)}{2 Y A} = \frac{m g l}{2 Y A}$

$\frac{Δ l_{1}}{Δ l_{2}} = 6$

Solution

1 1

2 4

3 8

4 6

Ans.
(4)

$Δ l = \frac{T l}{Y A}$

$Δ l_{1} = \frac{3 m g l}{Y A}$

$Δ l_{2} = \frac{2 m g (l / 2)}{2 Y A} = \frac{m g l}{2 Y A}$

$\frac{Δ l_{1}}{Δ l_{2}} = 6$

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Solution

1 1

2 4

3 8

4 6

Ans. (4) Δl=TlYA Δl1=3mglYA Δl2=2mg(l/2)2YA=mgl2YA Δl1Δl2=6

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Ans.
(4)

$Δ l = \frac{T l}{Y A}$

$Δ l_{1} = \frac{3 m g l}{Y A}$

$Δ l_{2} = \frac{2 m g (l / 2)}{2 Y A} = \frac{m g l}{2 Y A}$

$\frac{Δ l_{1}}{Δ l_{2}} = 6$