91Ó°ÊÓ

a) Given:

Drag force experienced by sphere in liquid $F_{drag}=bv$

Radius of the sphere $=R$

$b=6\mathrm{Ï€}nR$

Initial speed of sphere $v_i=v_o$$v_i=v_o$

We need to determine the velocity of sphere with Time . As there is no other force except drag force in $x$direction hence its velocity will be affected by drag force only. Also the drag force is opposite to the direction of motion, hence it will lead to decrease in speed with time.

To find speed variations with time we proceed as follow:

$ma=-F_{drag}$

$a=\frac{-F_{drag}}{m}$

$F_{drag}=6\mathrm{π}\mathrm{η}Rv$

$⟹a=\frac{6\mathrm{π}\mathrm{η}Rv}{m}$

Using equation of motion:

$v_f=v_i+a\mathrm{Δ}t$

$v_f=v_o-\frac{6\mathrm{π}\mathrm{η}Rv}{m}t$

$\text{Hence}v\left(t\right)=v_o-\frac{6\mathrm{π}\mathrm{η}Rv}{m}t$

b) Given

The coefficient of viscosity$\mathrm{η}=1×{10}^{-}3\mathrm{Ns}/{\mathrm{m}}^2$

Diameter $d=4\mathrm{cm}$

$R=2\mathrm{cm}=0.02\mathrm{m}$

Mass $m=33\mathrm{g}=0.033\mathrm{Kg}$

To determine the time taken by sphere to reach $50\%$of initial speed.

Using the result of (a)

$v_f=v_o-\frac{6\mathrm{π}\mathrm{η}Rv}{m}t$

$0.5v_o=v_o-\frac{6×\mathrm{π}×1.0×{10}^{-3}×0.02×0.5v}{0.033}t$

$f0.5\backslash \mathrm{not}p=f\frac{6×\mathrm{π}×1.0×{10}^{-3}×0.02×0.5\backslash \mathrm{not}x}{0.033}t$

$t=87.57\mathrm{s}=1.46\min$