The correct answer is (d) \(\frac{∂∆V}{∂z}=-m_v dxdydz \frac{∂∆σ’}{∂t}\)
Best explanation: Since the decrease in volume is given by the equation,
∆v=-mv V0 ∆σ’ where V0=dxdydz,
differentiating the equation, we get,
\(\frac{∂∆V}{∂z}=-m_v dxdydz \frac{∂∆σ’}{∂t}\) where, mv is the coefficient of volume change.