In an experiment to calculate the coefficient | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The density of the ball $ho$ is given by $ho = \frac{m}{V} = \frac{\pi/2}{\frac{4}{3}\pi(0.5)^3} = \frac{3}{8 	imes 0.125} = 3 \ gm/cc$.The t

Vedclass · Accepted Answer

$9.8$

Vedclass · Answer

(B) The density of the ball $ho$ is given by $ho = \frac{m}{V} = \frac{\pi/2}{\frac{4}{3}\pi(0.5)^3} = \frac{3}{8 	imes 0.125} = 3 \ gm/cc$.The terminal velocity $V_T$ is $V_T = \frac{	ext{distance}}{	ext{time}} = \frac{50 \ cm}{5 \ s} = 10 \ cm/s$.According to Stoke's Law, the terminal velocity is $V_T = \frac{2}{9} \frac{r^2(ho - \sigma)g}{\eta}$, where $\sigma = 1.2 \ gm/cc$ is the density of the liquid and $g = 980 \ cm/s^2$.Rearranging for viscosity $\eta$: $\eta = \frac{2}{9} \frac{r^2(ho - \sigma)g}{V_T}$.Substituting the values: $\eta = \frac{2}{9} 	imes \frac{(0.5)^2 	imes (3 - 1.2) 	imes 980}{10}$.$\eta = \frac{2}{9} 	imes \frac{0.25 	imes 1.8 	imes 980}{10} = \frac{2}{9} 	imes \frac{0.45 	imes 980}{10} = \frac{2}{9} 	imes 44.1 = 9.8 \ 	ext{poise}$.

Similar Questions

$A$ ball of density $\rho$ is released from deep inside a viscous liquid of density $2\rho$. It will move up:

An iron sphere of mass $20 \times 10^{-3} \ kg$ falls through a viscous liquid with terminal velocity $0.5 \ ms^{-1}$. The terminal velocity (in $ms^{-1}$) of another iron sphere of mass $54 \times 10^{-2} \ kg$ is (in $.5$)

Two drops of equal radius are falling with a terminal velocity of $5 \, cm/s$. If they coalesce to form a single larger drop,what will be the terminal velocity of the new drop?

Which of the following is the incorrect graph for a sphere falling in a viscous liquid? (Given at $t = 0$,velocity $v = 0$ and displacement $x = 0$.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module