Question

When a copper sphere is heated, percentage change is :

• A. Maximum in radius
• B. Maximum in volume
• C. Maximum in density
• D. Equal in radius, volume and density

Answer 1

The correct option is B Maximum in volume

Let $R$ be the radius of sphere, $V$ its volume and $\rho$ its density. Then,

$\Delta R=R\alpha \Delta \theta$

and percentage change in radius

$\frac{\Delta R}{R}\times 100=100\alpha \Delta \theta$ ....(i)

Now, $\Delta V=\gamma v\Delta \theta =3\alpha v\Delta \theta$ $\{\because \gamma =3\alpha \}$

$\therefore$ Percentage increase in volume

$=\frac{\Delta V}{V}\times 100=300\alpha \Delta \theta$ ...(ii)

Again, ${\rho }^{{}^{\prime }}=\frac{\rho }{1+\gamma \Delta \theta }=\frac{\rho }{1+3\alpha \Delta \theta }$

$\therefore \Delta \rho =\rho -{\rho }^{{}^{\prime }}$

$=\rho [1-\frac{1}{1+3\alpha \Delta \theta }]=\frac{\left(3\alpha \Delta \theta \right)\rho }{1+3\alpha \Delta \theta }$

Percentage change in density

$=\frac{\Delta \rho }{\rho }\times 100=\frac{300\alpha \Delta \theta }{1+3\alpha \Delta \theta }$ ...(iii)

From equations (i), (ii) and (iii), we see that the percentage change is maximum in volume.