Question

A sonometer wire under suitable tension having specific gravity \( Q \), vibrates with frequency \( n \) in air. If the load is completely immersed in water the frequency of vibration of the wire will become

• A. \( \left( \frac{Q - 1}{n Q} \right)^{1/2} \)
• B. \( \left( \frac{Q}{Q - 1} \right)^{1/2} n \)
• C. \( \left( \frac{n Q}{Q - 1} \right)^{1/2} \)
• D. \( n \left( \frac{Q - 1}{Q} \right)^{1/2} \)

Hint:

When an object is immersed in a fluid, its effective mass decreases due to buoyant force, which in turn affects the frequency of vibration.

Answer 1

The Correct Option is D

Step 1: Effect of immersion in water.
When a sonometer wire is immersed in water, the effective weight of the wire decreases due to buoyancy. The tension in the wire remains the same, but the frequency of vibration changes.
Step 2: Frequency relation.
The frequency of vibration is inversely proportional to the square root of the effective mass, which depends on the specific gravity \( Q \). The frequency in water will be given by: \[ f_{\text{water}} = n \left( \frac{Q - 1}{Q} \right)^{1/2} \]
Step 3: Conclusion.
The frequency of vibration of the wire when immersed in water will become \( n \left( \frac{Q - 1}{Q} \right)^{1/2} \), so the correct answer is (D).