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An external presseure P is applied on a ...

An external presseure P is applied on a cube at 273K hence it compresses equally from all sides `alpha` is coefficient of linear expansion & K is bulk modulus of material. To bring cube to its original size by heating the temperature rise must be

A

P/3`alpha`K

B

P/`alpha`K

C

P/2`alpha`K

D

P/4`alpha`K

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To solve the problem, we need to determine the temperature rise required to return a cube to its original size after it has been compressed by an external pressure \( P \). We will use the concepts of bulk modulus and linear expansion. ### Step-by-Step Solution: 1. **Understanding Bulk Modulus**: The bulk modulus \( K \) is defined as: \[ K = -\frac{\Delta P}{\frac{\Delta V}{V}} \] where \( \Delta P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V \) is the original volume. 2. **Relating Pressure and Volume Change**: Rearranging the formula gives: \[ \Delta V = -\frac{\Delta P \cdot V}{K} \] Here, \( \Delta P = P \) (the applied pressure). 3. **Volume Change in Terms of Linear Expansion**: The change in volume \( \Delta V \) can also be expressed in terms of the coefficient of linear expansion \( \alpha \) and the change in temperature \( \Delta \theta \): \[ \Delta V = \gamma \cdot V \cdot \Delta \theta \] where \( \gamma = 3\alpha \) for a cube (since volume is proportional to the cube of the linear dimensions). 4. **Setting the Equations Equal**: From the two expressions for \( \Delta V \): \[ -\frac{P \cdot V}{K} = 3\alpha \cdot V \cdot \Delta \theta \] We can cancel \( V \) from both sides (assuming \( V \neq 0 \)): \[ -\frac{P}{K} = 3\alpha \Delta \theta \] 5. **Solving for Temperature Change**: Rearranging the equation to solve for \( \Delta \theta \): \[ \Delta \theta = -\frac{P}{3\alpha K} \] Since we are interested in the magnitude of the temperature rise, we can drop the negative sign: \[ \Delta \theta = \frac{P}{3\alpha K} \] ### Final Answer: The temperature rise required to bring the cube back to its original size is: \[ \Delta \theta = \frac{P}{3\alpha K} \]
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