If the temperature of a uniform steel rod of mass 12kg and length 1m is increased by 20∘C, its moment of inertia about its perpendicular bisector increases by - [α=11×10−6/∘C]
A
2.2×10−4kg-m2
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B
4.4×10−4kg-m2
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C
1.1×10−4kg-m2
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D
6.6×10−4kg-m2
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Solution
The correct option is B4.4×10−4kg-m2 Given:
Change in temperature, ΔT=20∘C
Co-efficient of linear expansion of steel rod, α=11×10−6/∘C
Length of rod, L=1m
Mass of rod, m=12kg
We know that change in moment of inertia with change in temperature is given by ΔI=2αIΔT ⇒ΔI=2×11×10−6×12×1212×20 [Irod=mL212 about perpendicular bisector] ⇒ΔI=4.4×10−4kg-m2