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Byju's Answer

Standard XII

Physics

Compressibility

The system sh...

Question

The system shown in the adjacent figure is in equilibrium. The mass of the container with liquid is M, density of liquid in the container is $ρ$ and volume of the block is V. If the container is now displaced downwards through a distance $x_{0}$ and released such that the block remains well inside the liquid, then during subsequent motion.

time period of SHM of the container will be

2 π \sqrt{\frac{M}{k}}

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time period of SHM of the container will be

2 π \sqrt{\frac{(M + p v)}{k}}

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amplitude of SHM of the container will be greater than

x_{0}

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container will not perform SHM

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Solution

The correct option is B time period of SHM of the container will be $2 π \sqrt{\frac{(M + p v)}{k}}$
$M g - k (x + Δ x) + ρ (V) (g - a) = M a$
$W h e r e M g + p V g = k Δ x$
$\Rightarrow (M + ρ V) a = - K x$
$O r a = - \frac{k x}{(M + ρ V)} o r ω = - \sqrt{\frac{k}{M + ρ V}}$
$\Rightarrow T = 2 π \sqrt{\frac{(M + ρ V)}{k}}$

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The correct option is B time period of SHM of the container will be 2π√(M+pv)kMg−k(x+Δx)+ρ(V)(g−a)=Ma Where Mg+pVg=kΔx ⇒(M+ρV)a=−Kx Or a=−kx(M+ρV) or ω=−√kM+ρV ⇒T=2π√(M+ρV)k