The velocity of small ball of mass $M$ and density $d_{1} =$ when dropped a container filled with glycerine becomes constant after some time. If the density glycerine is $d_{2}$ , the viscous force acting on ball is

\frac{M d_{1} g}{d_{2}}

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M g (1 - \frac{d_{2}}{d_{1}})

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\frac{M (d_{1} + d_{2})}{g}

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M d_{1} d_{2}

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Solution

The correct option is B $M g (1 - \frac{d_{2}}{d_{1}})$
Since, ball gains the constant velocity.
Therefore, net force on the ball $=$ zero.
$\Rightarrow$ Weight of the ball $= B u o y a n c y f o r c e + V i s c o u s f o r c e (F_{v})$
$\Rightarrow M g = \frac{M}{d_{1}} \times d_{2} \times g + F_{v}$
$∴ F_{v} = M g (1 - \frac{d_{2}}{d_{1}})$ .

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The velocity of small ball of mass M and density d1= when dropped a container filled with glycerine becomes constant after some time. If the density glycerine is d2, the viscous force acting on ball is

The correct option is B Mg(1−d2d1)Since, ball gains the constant velocity.Therefore, net force on the ball = zero.⇒ Weight of the ball =Buoyancy force+Viscous force(Fv)⇒Mg=Md1×d2×g+Fv∴Fv=Mg(1−d2d1).

The velocity of small ball of mass $M$ and density $d_{1} =$ when dropped a container filled with glycerine becomes constant after some time. If the density glycerine is $d_{2}$ , the viscous force acting on ball is