A metal ball initially at a pressure of 10^5 | vedclass

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(B) The metal ball tends to expand due to heating, but since its volume is kept constant, thermal stress is developed within the material.The bulk mod

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$6 	imes 10^8 \ Pa$

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(B) The metal ball tends to expand due to heating, but since its volume is kept constant, thermal stress is developed within the material.The bulk modulus $B$ is defined as $B = \frac{\Delta P}{\Delta V / V}$, where $\Delta P$ is the change in pressure.The fractional change in volume due to thermal expansion is given by $\frac{\Delta V}{V} = \gamma \Delta T$, where $\gamma$ is the coefficient of volume expansion and $\Delta T$ is the change in temperature.For a solid, $\gamma = 3\alpha$, where $\alpha$ is the coefficient of linear expansion.Given: $\alpha = 10^{-5} \ {^{\circ} C}^{-1}$, $\Delta T = 127^{\circ} C - 20^{\circ} C = 107^{\circ} C$, and $B = 2 	imes 10^{11} \ N \ m^{-2}$.Substituting these into the formula for pressure change: $\Delta P = B 	imes (3\alpha \Delta T)$.$\Delta P = (2 	imes 10^{11}) 	imes (3 	imes 10^{-5} 	imes 107) = 6 	imes 10^6 	imes 107 = 6.42 	imes 10^8 \ Pa$.The final pressure $P_f = P_i + \Delta P = 10^5 \ Pa + 6.42 	imes 10^8 \ Pa \approx 6.42 	imes 10^8 \ Pa$.Rounding to the nearest given option, the pressure becomes approximately $6 	imes 10^8 \ Pa$.

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