While maintaining a constant volume, a steel ball initially at a...

The fractional change in volume due to a temperature change $\Delta T$ is given by

$\dfrac{\Delta V}{V} = \gamma \Delta T = 3\alpha \Delta T$

To maintain a constant volume, the increase in pressure must counteract this thermal expansion. The required change in pressure $\Delta P$ is related to the bulk modulus $B$ by

$\Delta P = B \dfrac{\Delta V}{V} = B (3\alpha) \Delta T$

Given values are:

$B = 1.6 \times 10^{11} \text{ N m}^{-2}$

\alpha = 12 \times 10^{-6} ^{}^{{}^{\circ}}\textC^{-1}

\Delta T = 120^{}^{{}^{\circ}}\text{C} - 20^{}^{{}^{\circ}}\text{C} = 100^{}^{{}^{\circ}}\text{C}

Substituting these values:

$\Delta P = (1.6 \times 10^{11}) \times (3 \times 12 \times 10^{-6}) \times 100$

$\Delta P = 1.6 \times 10^{11} \times 36 \times 10^{-4}$

$\Delta P = 57.6 \times 10^{7} \text{ Pa} = 5.76 \times 10^{8} \text{ Pa}$

The final pressure inside the ball is the sum of the initial pressure and the change in pressure:

$P = P_{initial} + \Delta P$

$P = 1.0 \times 10^{5} + 5.76 \times 10^{8} \approx 5.8 \times 10^{8} \text{ Pa}$