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(A) First,calculate the average time $t_{	ext{avg}}$ taken by the ball to sink: $t_{	ext{avg}} = \frac{2.75 + 2.65 + 2.70}{3} = 2.7 \,s$.The termina

Vedclass · Accepted Answer

$0.14 \,Pa \cdot s$

Vedclass · Answer

(A) First,calculate the average time $t_{	ext{avg}}$ taken by the ball to sink: $t_{	ext{avg}} = \frac{2.75 + 2.65 + 2.70}{3} = 2.7 \,s$.The terminal velocity $v_t$ is given by $v_t = \frac{h}{t_{	ext{avg}}} = \frac{0.9}{2.7} = \frac{1}{3} \,m/s$.According to Stokes' Law,the terminal velocity is $v_t = \frac{2r^2g(ho_s - ho_l)}{9\eta}$,where $\eta$ is the coefficient of viscosity.Rearranging for $\eta$: $\eta = \frac{2r^2g(ho_s - ho_l)}{9v_t}$.Given $r = \sqrt{3} 	imes 10^{-3} \,m$,so $r^2 = 3 	imes 10^{-6} \,m^2$.Given $(ho_s - ho_l) = 7000 \,kg/m^3$ and $g = 10 \,m/s^2$.Substituting the values: $\eta = \frac{2 	imes (3 	imes 10^{-6}) 	imes 10 	imes 7000}{9 	imes (1/3)} = \frac{6 	imes 10^{-5} 	imes 7000}{3} = 2 	imes 10^{-5} 	imes 7000 = 0.14 \,Pa \cdot s$.

Trial No.	Time taken by the ball to fall by $h$ (in second)
$1$.	$2.75$
$2$.	$2.65$
$3$.	$2.70$

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