A block of mass 5m is connected with a small solid spherical ball of mass m (density =ρ) by an ideal string passing over a smooth pulley. The block is on an inclined plane with coefficient of friction μ=0.2. The spherical ball is in a liquid of density ρ2. The coefficient of viscosity is η . Block is released when string is just taut, then after some time it moves with constant velocity. Assume that the ball always remain inside the fluid. The radius of ball is r. Choose the correct option(s):

Just after release5mgsin370−5μmgcos370−mgeff=6mageff=g(1−p2p)=g25mg35−5m(0.2)g45−mg2=6maa=17g60For terminal velocity1.7 mg=6πηrv v=17mg60πηr

A block of mass 5m is connected with a small solid spherical ball of mass m (density =ρ) by an ideal string passing over a smooth pulley. The block is on an inclined plane with coefficient of friction μ=0.2. The spherical ball is in a liquid of density ρ2. The coefficient of viscosity is η . Block is released when string is just taut, then after some time it moves with constant velocity. Assume that the ball always remain inside the fluid. The radius of ball is r. Choose the correct option(s):

Just after release5mgsin370−5μmgcos370−mgeff=6mageff=g(1−p2p)=g25mg35−5m(0.2)g45−mg2=6maa=17g60For terminal velocity1.7 mg=6πηrv v=17mg60πηr

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A block of mass 5m is connected with a small solid spherical ball of mass m (density $= ρ$ ) by an ideal string passing over a smooth pulley. The block is on an inclined plane with coefficient of friction $μ = 0.2$ . The spherical ball is in a liquid of density $\frac{ρ}{2}$ . The coefficient of viscosity is $η$ . Block is released when string is just taut, then after some time it moves with constant velocity. Assume that the ball always remain inside the fluid. The radius of ball is r. Choose the correct option(s):

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Acceleration of the ball just after the release is $\frac{17 g}{60}$

The constant velocity of the ball is $\frac{17 m g}{60 π η r}$

Tension in string just after the release is 2.2 mg

Tension in the string when ball move with constant velocity is 2.2mg

answer is A, C, D.

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Detailed Solution

Just after release
$5 m g \sin 37^{0} - 5 μ m g \cos 37^{0} - m g_{e f f} = 6 m a$
$g_{e f f} = g (1 - \frac{p}{2 p}) = \frac{g}{2}$
$5 m g \frac{3}{5} - 5 m (0.2) g \frac{4}{5} - \frac{m g}{2} = 6 m a$
$a = \frac{17 g}{60}$
For terminal velocity
$1.7 m g = 6 π η r v$
$v = \frac{17 m g}{60 π η r}$