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Oscillations question

2006 · Q167
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Oscillations question

2006 · Q167

NEETPhysicsOscillationsMCQ+4 / −1
A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ\rhoρ. If it is given a small vertical displacement from equilibrium it undergoes with a time period T, then
  1. A
    T∝1mT \propto {1 \over {\sqrt m }}T∝m​1​
  2. B
    T∝ρT \propto \sqrt \rhoT∝ρ​
  3. C
    T∝1AT \propto {1 \over {\sqrt A }}T∝A​1​
  4. D
    T∝1ρT \propto {1 \over \rho }T∝ρ1​
View written solutionFree

Correct answer: C

Let l be the length of block immersed in liquid as shown in the figure. When the block is floating,

AIPMT 2006 Physics - Oscillations Question 33 English Explanation


∴\therefore∴ mg = Alρ\rho ρm

If the block is given vertical displacement y then the effective restoring force is

F=−[A(l+y)ρg−mg]F = - \left[ {A\left( {l + y} \right)\rho g - mg} \right]F=−[A(l+y)ρg−mg]

=−[A(l+y)ρg−Alρg]=−Alρgy = - \left[ {A\left( {l + y} \right)\rho g - Al\rho g} \right] = - Al\rho gy=−[A(l+y)ρg−Alρg]=−Alρgy

Restoring force = −[Alρg]y - \left[ {Al\rho g} \right]y−[Alρg]y. As this F is directed towards its equilibrium position of block, so if the block is left free, it will execute simple harmonic motion.

Here inertia factor = mass of block = m
Spring factor = Aρg{A\rho g}Aρg

∴\therefore∴ Time period = T=2πmAρgT = 2\pi \sqrt {{m \over {A\rho g}}} T=2πAρgm​​   i.e. T∝1AT \propto {1 \over {\sqrt A }}T∝A​1​

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