JEE MAIN - Physics (2006 - No. 23)
If the terminal speed of a sphere of gold (density $$ = 19.5\,\,kg/{m^3}$$) is $$0.2$$ $$m/s$$ in a viscous liquid (density $$ = 1.5\,\,kg/{m^3}$$, find the terminal speed of a sphere of silver (density $$ = 10.5\,\,kg/{m^3}$$) of the same size in the same liquid
$$0.4$$ $$m/s$$
$$0.133$$ $$m/s$$
$$0.1$$ $$m/s$$
$$0.2$$ $$m/s$$
Explanation
Let Terminal velocity = vt
Upward viscous force = downward weight of sphere
$$ \Rightarrow 6\pi \eta r{v_t} = \left( {{4 \over 3}\pi {r^3}} \right)\left( {\rho - \sigma } \right)g$$
$$ \Rightarrow {v_t} = {{2{r^2}\left( {\rho - \sigma } \right)g} \over {9\eta }}$$ ........ (1)
where, $$\rho $$ = density of substance of a body
$$\sigma $$ = density of liquid
Now let the terminal velocity of gold = vg and silver = vs.
From equation (1), we can write
$${{{v_g}} \over {{v_s}}} = {{{\rho _g} - \sigma } \over {{\rho _s} - \sigma }}$$ $$ = {{19.5 - 1.5} \over {10.5 - 1.5}}$$ = $${{18} \over 9}$$ $$ = {2 \over 1}$$
$$\therefore$$ $${v_s} = {{{v_g}} \over 2}$$ = $${{0.2} \over 2}$$ = 0.1
Upward viscous force = downward weight of sphere
$$ \Rightarrow 6\pi \eta r{v_t} = \left( {{4 \over 3}\pi {r^3}} \right)\left( {\rho - \sigma } \right)g$$
$$ \Rightarrow {v_t} = {{2{r^2}\left( {\rho - \sigma } \right)g} \over {9\eta }}$$ ........ (1)
where, $$\rho $$ = density of substance of a body
$$\sigma $$ = density of liquid
Now let the terminal velocity of gold = vg and silver = vs.
From equation (1), we can write
$${{{v_g}} \over {{v_s}}} = {{{\rho _g} - \sigma } \over {{\rho _s} - \sigma }}$$ $$ = {{19.5 - 1.5} \over {10.5 - 1.5}}$$ = $${{18} \over 9}$$ $$ = {2 \over 1}$$
$$\therefore$$ $${v_s} = {{{v_g}} \over 2}$$ = $${{0.2} \over 2}$$ = 0.1
Comments (0)
