# A square lamina is rotating about any axis perpendicular to its plane . The temperature is increased by $\;\bigtriangleup t\;$ , its moment of inertia $\;(I_{0})\;$ will increase by - (side of square lamina =a )

$(a)\;2I_{0} \alpha \bigtriangleup t\qquad(b)\;I_{0} \alpha \bigtriangleup t\qquad(c)\;Depends\;on\;the\;value\;of\;'a'\qquad(d)\;None\;of\;these$

Answer : $2I_{0} \alpha \bigtriangleup t$
Explanation :
$I_{0}=\large\frac{ma^2}{12}$
$a_{T}=a (1+\alpha \bigtriangleup t)$
$a_{T}^{2}=a (1+\alpha \bigtriangleup t)^2$
$\approx a^2 (1+2 \alpha \bigtriangleup t)$
$= {\bigtriangleup t \to 0}$
$I_{T}=\large\frac{ma^2(1+2 \alpha \bigtriangleup t)}{12}$
$I_{T}=I_{0} (1+2 \alpha \bigtriangleup t)$
Therefore , $\bigtriangleup I=2I_{0} \alpha \bigtriangleup t$