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A weight is attached to the free end of a sonometer wire. It gives resonance at a length of 40 cm when resonanced with a tuning fork of frequency 512 Hz. The weight is then immersed wholly in water. The new length for resonance is:

a) 50 cm
b) 60 cm
c) 30 cm
d) 20 cm

User Bdoubleu
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1 Answer

3 votes

Final answer:

Option B is correct. The new length for resonance after immersing the weight wholly in water is 50 cm, when weight is attached to the free end of a sonometer wire. It gives resonance at a length of 40 cm when resonanced with a tuning fork of frequency 512 Hz.

Step-by-step explanation:

To find the new length for resonance after immersing the weight wholly in water, we need to understand the concept of resonance in a sonometer wire.

When a tuning fork is struck and brought near the wire, the wire vibrates at the same frequency as the tuning fork.

The length at which resonance occurs is related to the frequency of the tuning fork.

Given that the resonance occurs at a length of 40 cm when resonated with a tuning fork of frequency 512 Hz, we can use the formula:

Resonance length = (n/4) * wavelength

where n is an integer and wavelength is the distance between consecutive nodes.

Since the weight is then immersed wholly in water, the length for resonance increases.

Using the same formula, we find that the new length for resonance is 50 cm, which means option B is correct.

User Pedro Vale
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7.9k points