A small steel ball is dropped into a long cyl | vedclass

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

Question

$\mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma}$

$\left(\rho \frac{4}{3} \pi \mathrm{r}^3\right) \mathrm{g}-\left(\rho_{\m

Vedclass · Accepted Answer

Vedclass · Answer

$\mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma}$

$\left(ho \frac{4}{3} \pi \mathrm{r}^3ight) \mathrm{g}-\left(ho_{\mathrm{L}} \frac{4}{3} \pi \mathrm{r}^3ight) \mathrm{g}-6 \pi \eta \mathrm{r}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$

$	ext { Let } \frac{4}{3 \mathrm{~m}} \pi \mathrm{R}^3 \mathrm{~g}\left(ho-ho_{\mathrm{L}}ight)=\mathrm{K}_1 	ext { and } \frac{6 \pi \eta \mathrm{r}}{\mathrm{m}}=\mathrm{K}_2$

$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}$

$\int_0^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}}=\int_0^{\mathrm{t}} \mathrm{dt}$

$-\frac{1}{\mathrm{~K}_2} \ln \left[\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}ight]_0^{\mathrm{v}}=\mathrm{t}$

$\ln \left(\frac{\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}}{\mathrm{K}_1}ight)=-\mathrm{K}_2 \mathrm{t}$

$\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}=\mathrm{K}_1 \mathrm{e}^{-\mathrm{K}_2 \mathrm{t}} $

$\mathrm{v}=\frac{\mathrm{K}_1}{\mathrm{~K}_2}\left[1-\mathrm{e}^{-\mathrm{K}_2 \mathrm{t}}ight]$

A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?

Similar Questions

A small spherical solid ball is dropped from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the

The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to

A ball of mass $m$ and radius $ r $ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m' $ such that $m > m'$. The terminal velocity is proportional to

The terminal velocity of a copper ball of radius $5\,mm$ falling through a tank of oil at room temperature is $10\,cm\,s ^{-1}$. If the viscosity of oil at room temperature is $0.9\,kg\,m ^{-1} s ^{-1}$, the viscous drag force is :

Velocity of water in a river is