Question

A uniform solid brass sphere is rotating with angular speed $\omega_{0}$ about a diameter. If its temperature is now increased by $100^{\circ} C$. What will be its new angular speed. (Given $\alpha_{B}=2.0 \times 10^{-5}$ per $^{\circ} C$ )

• A. $1.1\, \omega_{0}$
• B. $1.01\, \omega_{0}$
• C. $0.996\, \omega_{0}$
• D. $0.824\, \omega_{0}$

Answer 1

Answer: (C)

We use sphere volume
$V=\frac{4}{3} \pi R^{3}$ ...(1)
when temperature is increased by $100^{\circ} C$
$V^{\prime} =\frac{4}{3} \pi R^{\prime 3}$
$V(1+\gamma \Delta T) =\frac{4}{3} \pi R^{\prime 3}$
$V\left[1+\left(8 \times 10^{-5}\right)(100)\right] =\frac{4}{3} \pi R^{\prime 3}$
$V(1.008)=\frac{4}{3} \pi R^{\prime 3}$ ...(2)
Dividing (1)by(2),we get
$\frac{1}{1.008} =\frac{R^{3}}{R^{\prime 3}}$
$R^{3} =1.008\, R^{3}$
$R^{\prime} =1.00266\, R$
Now, $I \omega =\text { constant }$
$\frac{2}{5} M R^{2} \omega_{0} =\frac{2}{5} M R^{\prime 2} \omega$
$R^{2} \omega_{0} =1.0054 R^{2} \omega$
$\omega =0.996\, \omega_{0}$