When a copper sphere is heated, percentage change is

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, $\quad \Delta V=\gamma v \Delta \theta=3 \alpha v \Delta \theta \quad\{\because \gamma=3 \alpha\}$ $\therefore$ Percentage increase in volume $=\frac{\Delta V}{V} \times 100=300 \alpha \Delta \theta$ ...(ii) Again, $\quad \rho^{\prime}=\frac{\rho}{1+\gamma \Delta \theta}=\frac{\rho}{1+3 \alpha \Delta \theta}$ $\therefore \quad \Delta \rho=\rho-\rho^{\prime}$ $=\rho\left[1-\frac{1}{1+3 \alpha \Delta \theta}\right]=\frac{(3 \alpha \Delta \theta) \rho}{1+3 \alpha \Delta \theta}$ Percentage change in density $=\frac{\Delta \rho}{\rho} \times 100=\frac{300 \alpha \Delta \theta}{1+3 \alpha \Delta \theta}$ ...(iii) From Eqs. (i), (ii) and (iii), we see that the percentage change is maximum in volume.