Physics - Elasticity Question with Solution | TestHub

Given:

Diameter $D=20$ cm, so radius $R=10$ cm.

Initial temperature $T_1=0^{\circ }$C

Final temperature $T_2=100^{\circ }$C

Coefficient of linear expansion ${\alpha }_{\mathrm{Al}}=23\times 10^{-6}{/}^{\circ }$C

1. Calculate the initial volume $V_0$ of the sphere:

$$V_0=\frac{4}{3}\pi R^3=\frac{4}{3}\pi (10cm{)}^3=\frac{4000}{3}\pi c{m}^3$$

2. Calculate the change in temperature $\Delta T$:

$$\Delta T=T_2-T_1=100^{\circ }C-0^{\circ }C=100^{\circ }C$$

3. Calculate the coefficient of volumetric expansion $\gamma$:

$$\gamma =3{\alpha }_{\mathrm{Al}}=3\times 23\times 10^{-6}{/}^{\circ }C=69\times 10^{-6}{/}^{\circ }C$$

4. Calculate the change in volume $\Delta V$:

$$\Delta V=V_0\gamma \Delta T=\left(\frac{4000}{3}\pi c{m}^3\right)(69\times 10^{-6}{/}^{\circ }C)(100^{\circ }C)$$

$$\Delta V=\frac{4000}{3}\pi \times 69\times 10^{-4}c{m}^3$$

$$\Delta V=4000\pi \times 23\times 10^{-4}c{m}^3$$

$$\Delta V=4\pi \times 23\times 10^{-1}c{m}^3$$

$$\Delta V=92\pi \times 10^{-1}c{m}^3$$

$$\Delta V=9.2\pi c{m}^3$$

Using $\pi \approx 3.14159$:

$$\Delta V\approx 9.2\times 3.14159c{m}^3\approx 28.902c{m}^3$$

Therefore, the volume changes by approximately $28.9c{m}^3$.

The final answer is $\boxed{28.9cc}$.