Question

metal sphere of radius r and specific heat S
is rotated about an axis passing through its
centre at a speed of n rotations per second. It
is stopped and 50% of its energy is used in
increasing its temperature, then the raise in
temperature of the sphere is
1)2/5(pi^2*n^2*r^2÷s)
2)1/10(pi^2*n^2÷r^2*s)
3)7/8(pi^2*r^2*n^2*s)
4)5/14(pi*r*n÷s)​

Answer 1

Answer:-

dT= 5s2π ^2 r ^2 n^ 2

Step-by-step explanation:

moment of inertia of a sphere:-

$I= \frac{5}{2} {mr}^{2}$

$w = 2\pi \: n \: rad/sec \: \\ KE= \frac{1}{2} {w}^{2} I \\ \ = > \: \frac{1}{2} 2\pi \frac{2}{5} m {r}^{2}$

$= > \frac{4}{5} m {\pi}^{2} {n}^{2} {r}^{2}$

$half \: of \: this \: energy \: is \: \: converted \: to \: heat \: energy =$

$dQ = \frac{4}{5} m {\pi}^{2} {r}^{2} {n}^{2} \times \frac{1}{2}$

Dt= msdQ

dT= ms2/5mπ^2 ssar^2 n^2

dT= 5s2π ^2 r ^2 n^ 2

hope it helps you ♥️♥️

sorry for error in last

lines but hope you

understand ☺️☺️