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Assertion If a wire is stretched, only h...

Assertion If a wire is stretched, only half of the work done in stretching the wire remains stored as elastic potential energy.
Reason Potential energy stored in the wire is 1/2 (stress)`xx` (strain)

A

If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

B

If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

C

If Assertion is true but Reason is false.

D

If both Assertion and Reason are false.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze both the assertion and the reason provided. ### Step 1: Understanding the Assertion The assertion states that if a wire is stretched, only half of the work done in stretching the wire remains stored as elastic potential energy. - When a wire is stretched, work is done on the wire. The work done (W) is given by the formula: \[ W = \frac{1}{2} k x^2 \] where \( k \) is the spring constant and \( x \) is the extension of the wire. - The elastic potential energy (U) stored in the wire is also given by: \[ U = \frac{1}{2} k x^2 \] - Therefore, it can be concluded that the work done on the wire is equal to the elastic potential energy stored in the wire. ### Step 2: Understanding the Reason The reason states that the potential energy stored in the wire is \( \frac{1}{2} (\text{stress}) \times (\text{strain}) \). - The potential energy per unit volume (u) is given by: \[ u = \frac{1}{2} \times \text{stress} \times \text{strain} \] - However, the total potential energy (U) stored in the wire would be: \[ U = u \times V = \frac{1}{2} \times \text{stress} \times \text{strain} \times V \] where \( V \) is the volume of the wire. - The reason is incorrect because it does not specify that the potential energy is per unit volume. ### Conclusion - The assertion is true: only half of the work done remains as elastic potential energy. - The reason is false because it does not correctly represent the relationship of potential energy stored in the wire. ### Final Answer The correct answer is that the assertion is true, but the reason is false.

To solve the question, we need to analyze both the assertion and the reason provided. ### Step 1: Understanding the Assertion The assertion states that if a wire is stretched, only half of the work done in stretching the wire remains stored as elastic potential energy. - When a wire is stretched, work is done on the wire. The work done (W) is given by the formula: \[ W = \frac{1}{2} k x^2 ...
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