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If the temperature of metal sphere increases by $\;\bigtriangleup t\;$ , the fractional change in its moment of inertia is $\;(\alpha \;= Coefficient\; of\; linear \;expansion )$

$(a)\;2 \alpha \bigtriangleup t\qquad(b)\;\alpha \bigtriangleup t\qquad(c)\;2 \alpha\qquad(d)\;2 \bigtriangleup t$

1 Answer

Answer : $\;2 \alpha \bigtriangleup t$
Explanation :
$I_{0} = \large\frac{2}{5} mr_{0}^2$
$I^{'} = \large\frac{2}{5} m\; [r_{0}^2 (1+\alpha \bigtriangleup t)^2]$
$= \large\frac{2}{5} mr_{0}^2\;[1 +2 \alpha \bigtriangleup t]$
$\bigtriangleup I = \large\frac{2}{5} mr_{0}^2 \;(2 \alpha \bigtriangleup t)$
$\large\frac{ \bigtriangleup I}{I_{0}} =2 \alpha \bigtriangleup t$
answered Apr 8, 2014 by yamini.v

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