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

Ler $R$ be the radius of sphere, $V$ be its volume and $\rho$ be its density.
Then, from the definition of linear expansion we get,
$\Delta R=R\alpha \Delta \theta$
Percentage increase in radius is given by
$\frac{\Delta R}{R}\times 100=100\alpha \Delta \theta .......\left(1\right)$
From the definition of volume expansion we get ,
$\Delta V=V\gamma \Delta \theta$
Percentage increase in volume is given by
$\frac{\Delta V}{V}\times 100=300\alpha \Delta \theta ......\left(2\right)(\because \gamma =3\alpha )$
Variation of density with temperature is given by
${\rho }^{\prime }=\frac{\rho }{1+\gamma \Delta \theta }=\frac{\rho }{1+3\alpha \Delta \theta }$
$\therefore \Delta \rho =\rho -{\rho }^{\prime }=\rho [1-\frac{1}{1+3\alpha \Delta \theta }]=\frac{\left(3\alpha \Delta \theta \right)\rho }{1+3\alpha \Delta \theta }$
Percentage change in density
$\frac{\Delta \rho }{\rho }\times 100=\frac{300\alpha \Delta \theta }{1+3\alpha \Delta \theta }.......\left(3\right)$
From $\left(1\right)$ , $\left(2\right)$ and $\left(3\right)$ we can say that , Percentage change is maximum in volume.
Thus, option (b) is the correct answer.