Question

A sphere of brass released in a long liquid column attains a terminal speed ${v_0}$ . If the terminal speed attained by sphere of marble of the same radius and released in the same liquid in $n{v_0}$ , then the value of $n$ will be –
(Given: The specific gravities of brass, marbles and the liquid are $8.5$, $2.5$ and $0.8$ respectively).
A. $\dfrac{5}{{17}}$
B. $\dfrac{{17}}{{77}}$
C. $\dfrac{{11}}{{31}}$
D. $\dfrac{{17}}{5}$

Answer 1

The radius for both the spheres are the same hence the radius for the equation of terminal velocity will be constant. $\eta$ will be constant as it satisfies the same liquid.Terminal velocity is the highest velocity that can be attained by an object when it falls through the air. It happens when the sum of the dragged force and buoyancy is equal to the downward force of gravity acting on the body. The object holds zero acceleration since the net force acting is zero.

Formula used:
The formula for terminal velocity is given by,
${v_t} = \dfrac{2}{9}{r^2}\left( {\dfrac{{\rho - {\rho _l}}}{\eta }} \right)g$
Here,
$r$ is the radius of the sphere.
$\rho$ is the density of the object.
${\rho _l}$ is the density of the liquid.
$\eta$ is the coefficient of viscosity
$g$ is the acceleration due to gravity.

Complete step by step answer:
Given,
Specific gravity of brass, ${\rho _b} = 8.5$
Specific gravity of marble, ${\rho _m} = 2.5$
Specific gravity of liquid, ${\rho _l} = 0.8$

Terminal velocity is given by,
${v_t} = \dfrac{2}{9}{r^2}\left( {\dfrac{{\rho - {\rho _l}}}{\eta }} \right)g$…… (1)
Here, we can see that the terms $r$, $\eta$, and $g$ are constant because the radius for both the spheres are the same.

Therefore equation (1) can be rewritten as,
${v_t} \propto \left( {\rho - {\rho _l}} \right)$ …… (2)
In case of brass,
${v_1} \propto \left( {{\rho _b} - {\rho _l}} \right)$ …… (3)
In case of marble,
${v_2} \propto \left( {{\rho _m} - {\rho _l}} \right)$ …… (4)

According to the question,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{v_0}}}{{n{v_0}}}$
Substitute the values of ${v_1}$ and ${v_2}$ from the equations (3) and (4) in the above equation.
$\dfrac{1}{n} = \dfrac{{{\rho _b} - {\rho _l}}}{{{\rho _m} - {\rho _l}}}$…… (5)

Now, substitute ${\rho _b} = 8.5$, ${\rho _m} = 2.5$, and ${\rho _l} = 0.8$ in equation (5)
$\dfrac{1}{n} = \dfrac{{8.5 - 0.8}}{{2.5 - 0.8}}$
$\Rightarrow \dfrac{1}{n} = \dfrac{{7.7}}{{1.7}} \\\ \Rightarrow \dfrac{1}{n} = \dfrac{{77}}{{17}} \\\ \therefore n = \dfrac{{17}}{{77}}$

Hence, the value of $n$ is $\dfrac{{17}}{{77}}$ .The correct option B is correct.

Note: In this question we are asked to calculate the value of $n$ . Remember that both the brass sphere and the marble sphere are of the same radius. Only the terms associated with the specific gravity remains varying. Hence, do not get confused to which factors thermal velocity is proportional to and evaluate the value of $n$ .