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When the temperature of a rod increases ...

When the temperature of a rod increases from t to `r+Delta t`, its moment of inertia increases from I to `I+Delta I`. If `alpha` is the value of `Delta I//I` is

A

`2 alpha Delta t`

B

`alpha Delta t`

C

`(alpha Delta t)/(2)`

D

`(Delta t)/(alpha)`

Text Solution

Verified by Experts

The correct Answer is:
A
Initial moment of inertia -
`I_(0) = (ML^2)/(12)`
Final moment of inertia.
`I = (M)/(12) (L + Delta L )^(2) = (M)/(12)[L^2 + 2L Delta L + Delta L^(2)]`
`therefore Delta I = I- I_(0) = (M/12) 2 L Delta L (Delta L^(2) rarr " neglectred")`
`(Delta I)/(I) = 2 (Delta L)/(L ) rArr Delta L = L alpha Delta t`
`therefore (Delta L )/(L) = alpha Delta t,`
`therefore (Delta I)/(I) = 2 alpha Delta t.`
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