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Question Number 50808 by Tinkutara last updated on 20/Dec/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 21/Dec/18

	$a t t e m p a r a t u r e t M . I i s$ $I = \frac{1}{2} (π R^{2} h ρ) R^{2} + {M d}^{2} a t t e m p a r a t u r e t$ $d = d i s t a n c e b e t w e e n a x i s o f c y l i n d e r a n d a x i s$ $o f r o t a t i o n . . .$ $I = \frac{π ρ}{2} (R^{4} h)$ $\frac{d I}{d t} = \frac{π ρ}{2} (R^{4} \frac{d h}{d t} + 4 {h R}^{3} \frac{d R}{d t}) + (\frac{d M}{d t}) d^{2}$ $d M = 0 n o c h a n g e o f m a s s w i t h t e m p a r a t u r e$ $r i s e$ $α = \frac{d l}{l d t} = \frac{c h a n g e o f l e n g t h}{o r i g i n a l l e n g t h \times c h a n g e o f t e m p a r s t u r e}$ $s o$ $\frac{d I}{d t} = \frac{π ρ}{2} (R^{4} h \times \frac{d h}{h d t} + 4 {h R}^{4} \times \frac{d R}{R d t})$ $d I = \frac{π ρ R^{4} h}{2} α d t + 4 \times \frac{π ρ {h R}^{4}}{2} \times (- α) d t [d R = - v e]$ $d I = I α d t - 4 \times I α d t = - 3 I α d t$ $s i n c e v a l u e o f h i n c r e a s e s s o v a l u e o f R d e c r e a s e s$ $h e n c e M . I g e t d e c r e a s e s . . .$ $p l s c h e c k i s i t r i g h t . . .$
	Commented by Tinkutara last updated on 22/Dec/18
	No sir your answer is wrong.
	Answered by ajfour last updated on 22/Dec/18

	$I = \frac{{m r}^{2}}{2}$ $△ I = \frac{m}{2} (2 α r^{2}) △ t$ $\Rightarrow △ I = 2 α I △ t [o p t i o n (c)] .$
	Commented by Tinkutara last updated on 23/Dec/18
	Thanks Sir!
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