Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

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 )

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

Can you answer this question?

1 Answer

0 votes
Answer : $2I_{0} \alpha \bigtriangleup t$
Explanation :
$a_{T}=a (1+\alpha \bigtriangleup t)$
$a_{T}^{2}=a (1+\alpha \bigtriangleup t)^2$
$\approx a^2 (1+2 \alpha \bigtriangleup t)$
$= {\bigtriangleup t \to 0}$
$I_{T}=\large\frac{ma^2(1+2 \alpha \bigtriangleup t)}{12}$
$I_{T}=I_{0} (1+2 \alpha \bigtriangleup t)$
Therefore , $\bigtriangleup I=2I_{0} \alpha \bigtriangleup t$
answered Mar 13, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App