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a) block B will oscillate but not simple harmonically
b) block B will oscillate simple harmonically
c) the system will remain in equilibrium
d) none of these

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Let $m_A=2m$

$m_B=3m$

density of liquid $=\rho$

density of block $B= 2 \rho$

The acceleration of the system , when block B is inside the liquid, will be given by,

net force acting on the system.

$a_1 = \large\frac{m_Ag-(m_Bg-upthrust \;on\;B)}{m_A+m_B}$

$\qquad= \large\frac{2mg-\bigg[3mg -\Large\frac{3m}{2 \rho}\normalsize \rho g\bigg]}{5m}$

$\qquad= \large\frac{g}{10}$

$\qquad= Constant$

But when the block is outside the liquid.

the acceleration of the system is in opposite direction.

$a_2= \large\frac{m_B g- m_A g}{m_A+m_B}$

$\qquad= \large\frac{3mg-2mg}{5m}$

$\qquad=\large\frac{g}{5}$

$\qquad$= constant

Since $a_1$ and $a_2$ are constants, motion is periodic but not simple harmonic.

Hence a is the correct answer

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