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Question: Answered & Verified by Expert
A wooden cylinder of mass 20 g and area of cross-section 1 cm 2 , having a piece of lead of mass 60g attached to its bottom, floats in water. The cylinder is depressed and then released. The frequency of oscillations is  N π s 1. Find the value of N. [Neglect the volume of water displaced by the lead piece, take g=9.8m/s2, density of water ρw=1 g cm-3] If your answer is K, mark 4K as answer after rounding off to nearest integer.
PhysicsOscillationsJEE Main
Solution:
1699 Upvotes Verified Answer
The correct answer is: 7

Suppose that the loaded wooden block sinks upto a height h. Then,
weight of water displaced by the block
= weight of the block with lead
If a is area of cross-section of the block and σ, the density of water, then
a h σ g = M + m g
or h = M + m a σ
= 2 0 + 6 0 1 × 1 = 8 0 cm density of water = 1gcm - 3
Now for small displacement ' x' from equilibrium
σga( x+h )( M+m )g=-( M+m )A
σgax=-( M+m )A
A=- σga M+m x
f= 1 2π g h
= 1 2π 980 80   = 1.75 π s 1
 


 

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