The pressure applied from all directions on a cube is \(\mathrm{P}\). How much its temperature should be raised to maintain the orginal volume ? The volume elasticity, of the cube is \(\beta\) and the coefficient of volume expansion is \(\alpha\). (A) \([\mathrm{P} / \alpha \beta]\) (B) \([\mathrm{P\alpha} / \beta]\) (C) \([\beta \mathrm{p} / \alpha]\) (D) \([\alpha \beta / \mathrm{p}]\)

The volume expansion formula relates the change in volume of an object with respect to its original volume and the change in temperature. It is given by: \[\Delta V = V_0 \cdot \alpha \cdot \Delta T\] We are considering a situation where the change in volume is zero (i.e., ΔV = 0), so the formula becomes: \[0 = V_0 \cdot \alpha \cdot \Delta T\]

Next we consider the work done due to the pressure applied on the cube. The work done by the pressure is given by: \[W = -\mathrm{P} \cdot \Delta V\] As ΔV = 0 (no change in volume), the work done by the pressure will also be zero, meaning that the cube's volume remains constant despite the change in temperature.

The relationship between volume elasticity, pressure, and change in volume is given by the following formula: \[\mathrm{P} = - \beta \cdot \frac{\Delta V}{V_0}\] Since ΔV = 0, we can rewrite this relationship as: \[\mathrm{P} = 0\] Therefore, the pressure applied on the cube is not causing any change in volume.

Now, using the relationships we've established, we'll solve for ΔT in the volume expansion formula: \[0 = V_0 \cdot \alpha \cdot \Delta T\] As V0 ≠ 0 (there is some volume) and α ≠ 0 ( there is some coefficient of volume expansion), we can divide both sides by V₀α to get: \[\Delta T = \frac{0}{\alpha \cdot V_0}\] Since we established that \(\mathrm{P} = 0\), we can write ΔT as follows: \[\Delta T = \frac{\mathrm{P}}{\alpha \cdot V_0}\] Now, let's replace V₀ in the ΔT equation using the volume elasticity relationship: \[\frac{\Delta V}{V_0} = - \frac{\mathrm{P}}{\beta}\] \[V_0 = - \frac{\Delta V \cdot \beta}{\mathrm{P}}\] Substitute the value of V₀ into the ΔT formula: \[\Delta T = \frac{\mathrm{P}}{\alpha} \cdot \frac{-\Delta V \cdot \beta}{\mathrm{P}}\] Simplify the expression and cancel out common factors: \[\Delta T = \frac{-\Delta V \cdot \beta}{\alpha}\] As we know, the change in volume (ΔV) is zero, so the expression becomes: \[\Delta T = 0\] Therefore, there is no change in temperature required to maintain the original volume of the cube, as it is not affected by the pressure applied on it. This does not match any of the given answer choices, so none of them are correct.