Question

If the length of a string is \(P\) when the tension in it is 6 N and its length is \(Q\) when the tension in it is 8 N, then the original length of the string is:

• A. \( 3P + 4Q \)
• B. \( 3P - 4Q \)
• C. \( 4P + 3Q \)
• D. \( 4P - 3Q \)

Hint:

Use the relation \( L = L_0(1 + \frac{T}{Y}) \) for elongation due to tension.
- Solve for \( L_0 \) using the given tensions and stretched lengths.

Answer 1

The Correct Option is D

The length of a string under tension follows the relationship: \[ L = L_0 \left(1 + \frac{T}{Y}\right) \] where \( L \) is the stretched length, \( L_0 \) is the original length, \( T \) is the tension, and \( Y \) is Young's modulus. From the given data: \[ P = L_0 \left(1 + \frac{6}{Y}\right) \] \[ Q = L_0 \left(1 + \frac{8}{Y}\right) \] Dividing the two equations: \[ \frac{Q}{P} = \frac{1 + \frac{8}{Y}}{1 + \frac{6}{Y}} \] Rearranging and solving for \( L_0 \), we derive: \[ L_0 = 4P - 3Q \] Thus, the original length of the string is \( \boxed{4P - 3Q} \).