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Q.

A dense liquid is completely filled in a closed large container and kept at rest on a horizontal surface. A steel ball is moving downward with a constant speed of 0.5 cm/sec. Find the momentum of dense liquid in C.G.S unit, if its density is $3 g m / c m^{3}$ . Given that volume is steel ball of $1.5 c m^{3}$

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answer is 2.25.

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Detailed Solution

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In time dt shift in centre of mass of system (ball + liquid)
$d S_{C M} = \frac{m_{1} d s_{1} + m_{2} d s_{2}}{M} = \frac{(ρ_{b} V) v d t - (ρ_{l} V) v d t}{M}$ …..(i)
Momentum of system $M . \frac{d S_{c m}}{d t}$ ….(ii)
$= (ρ_{0} V) v - (V ρ_{l}) v$
Therefore momentum of liquid $= (V ρ_{l}) v$

$= 2.25 g m / c m^{3}$

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