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The Young's modulus of a rubber string 8...

The Young's modulus of a rubber string 8 cm long and density `1.5kg//m^(3)` is `5xx10^(8)N//m^(2)` is suspended on the ceiling in a room. The increase in length due to its own weight will be-

A

`9.6 xx 10^(-5) m`

B

`9.6 xx 10^(-3) m`

C

`9.6 m`

D

`9.6 xx 10^(-7) m`

Text Solution

Verified by Experts

The correct Answer is:
D
`Y = (MgL)/(pir^(2)l) :. L = (MgL)/(pir^(2)Y)`
Mass of the string `(M) = V xx d = pir^(2)Ld`
Since the weight acts through the midpoint of the string,
the length is considered as `(L)/(2)`.
`:. l = ((MgL)/(2))/(pir^(2)Y) = (pir^(2)Ldg xx L)/(2pir^(2)Y) = (L^(2)dg)/(2Y)`
`l = (L^(2)dg)/(2Y) = (8^(2) xx 1.5 xx 10)/(2 xx 5 xx 10^(8))`
`= (64 xx 1.5)/(10^(8))`
`= 96 xx 10^(-8) = 9.6 xx 10^(-7) m`
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