Question

Horizontal force $F$ to keep the cylinder in static equilibrium, if it is placed on a smooth horizontal plane, is

• A. $7.2N$
• B. $10N$
• C. $15.5N$
• D. $20.4N$

Answer 1

Answer: (A)

$7.2N$

Since area of a hole is very small in comparison to base area $A$ of the cylinder, velocity of liquid inside the cylinder is negligible. Let velocity of efflux be $v$ and atmospheric pressure be $P_0$.

Consider two points A (inside the cylinder) and B (must outside the hole) in the same horizontal line as shown in the figure in question.

Pressure at A: $P_A=P_0+h{\rho }_{2}g+(h-y){\rho }_{1}g$

Pressure at A: $P_B=P_0$

Applying Bernoulli's theorem at points A and B

$P_A+\frac{1}{2}{\rho }_2{v}_2^2+({\rho }_{2}gh+{\rho }_{1}g(h-y\left)\right)=P_B+\frac{1}{2}{\rho }_1{v}_1^2+({\rho }_{2}gh+{\rho }_{1}g(h-y\left)\right)$ .....(1)

Substituting $P_A$ and $P_B$ in eqn(1) we get,

$P_0+h{\rho }_{2}g+(h-y){\rho }_{1}g+\frac{1}{2}{\rho }_2{v}_2^2+({\rho }_{2}gh+{\rho }_{1}g(h-y\left)\right)=P_0+\frac{1}{2}{\rho }_1{v}_1^2+({\rho }_{2}gh+{\rho }_{1}g(h-y\left)\right)$ ....(2)

Given,

$A=0.5m^2$

$a=5c{m}^2=25\times {10}^{-4}{m}^2$

From continuity equation,

$A{v}_2=a{v}_1$

$\therefore \frac{v_2}{v_1}=\frac{a}{A}=\frac{25\times {10}^{-4}}{0.5}=50\times {10}^{-4}$

Given $h=60cm=0.6m$

$y=20cm=0.2m$

${\rho }_1=900kg{m}^{-3}$

${\rho }_2=600kg{m}^{-3}$

$P_0+h{\rho }_{2}g+(h-y){\rho }_{1}g+\frac{1}{2}{\rho }_2{v}_2^2=P_0+\frac{1}{2}{\rho }_1{v}_1^2$

As given in the problem, we ignore $v_2$ since a is very small.

$\therefore P_0+h{\rho }_{2}g+(h-y){\rho }_{1}g=P_0+\frac{1}{2}{\rho }_1{v}_1^2$

$\therefore h{\rho }_{2}g+(h-y){\rho }_{1}g=\frac{1}{2}{\rho }_1{v}_1^2$

Substituting the values, $0.6\times 600\times 10+(0.6-0.2)\times 900\times 10=\frac{1}{2}\times 900\times v_1^2$

$\Rightarrow v_1=4m{s}^{-1}$

When the cylinder is on the smooth horizontal plane, force $F$ required to keep the cylinder stationary equals horizontal thrust exerted by fluid jet. But that is equal to mass flowing per second change in velocity of this mass.

$\therefore av{\rho }_1(v-0)=a{v}^2{\rho }_1=5\times {10}^{-4}{4}^2\times 900=7.2N$