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A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork, is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30 cm. The relative density of the stone is
  • a)
    16/9
  • b)
    16/7
  • c)
    16/5
  • d)
    16/3
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A stone hangs from the free end of a sonometer wire whose vibrating le...
 where M = mass of stone. If ρ is the density of the stone and V its volume, then m = ρV.
When the stone is wholly immersed in water of density ρ', the effective weight of the stone
Given l = 40 cm and l' = 30 cm.
Also v = v', which gives
Hence the correct choice is (2).
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Community Answer
A stone hangs from the free end of a sonometer wire whose vibrating le...
Given data:
Vibrating length of sonometer wire when tuned to the tuning fork = 40 cm
Resonant length of sonometer wire when the stone is wholly immersed in water = 30 cm

To find: Relative density of the stone

Let's consider the following points to solve the problem:

1. Resonance in a sonometer wire:
When a tuning fork is struck and placed near the opening of a sonometer wire, it sets the wire into resonance. The length of the wire that vibrates maximally is called the vibrating length.

2. Resonance in a sonometer wire with a stone immersed in water:
When the stone is wholly immersed in water, it changes the effective density of the wire. This results in a change in the resonant length of the wire.

3. Relationship between resonant length and relative density:
The resonant length of a sonometer wire is inversely proportional to the square root of the effective density of the wire. Mathematically, we can represent this relationship as:
L1/L2 = √(ρ2/ρ1)
where L1 and L2 are the resonant lengths in air and water respectively, and ρ1 and ρ2 are the densities of the wire in air and water respectively.

Calculation:
Given that L1 = 40 cm and L2 = 30 cm.

Using the above formula, we can rewrite it as:
30/40 = √(ρ2/ρ1)
Simplifying further, we get:
(3/4)^2 = ρ2/ρ1
9/16 = ρ2/ρ1

Comparing the equation with the given options, we find that option 'B' is the correct answer:
Relative density of the stone = 16/7

Therefore, the relative density of the stone is 16/7.
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A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork, is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30 cm. The relative density of the stone isa)16/9b)16/7c)16/5d)16/3Correct answer is option 'B'. Can you explain this answer?
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A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork, is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30 cm. The relative density of the stone isa)16/9b)16/7c)16/5d)16/3Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork, is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30 cm. The relative density of the stone isa)16/9b)16/7c)16/5d)16/3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork, is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30 cm. The relative density of the stone isa)16/9b)16/7c)16/5d)16/3Correct answer is option 'B'. Can you explain this answer?.
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