Bernoulli's Principle and Viscosity — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Bernoulli's Principle and Viscosity MCQs with step-by-step solutions (6 questions). Part of Mechanical Properties of Fluids. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Bernoulli's Principle and Viscosity · medium · numerical
The velocity of a small ball of mass $M$ and density $d$ when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $d/2$, then the viscous force acting on the ball will be
A. $\dfrac{Mg}{2}$ ✓ Correct
B. $Mg$
C. $\dfrac{3}{2}Mg$
D. $2Mg$
Solution: At constant (terminal) velocity, viscous force $=$ weight $-$ buoyant force $= Vd_1 g - Vd_2 g$, with $d_1 = d$ and $d_2 = d/2$. So viscous force $= Vdg - V\dfrac{d}{2}g = \dfrac{Vdg}{2} = \dfrac{Mg}{2}$ (since $M = d\times V$).
Q2 — Bernoulli's Principle and Viscosity · medium · numerical
Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_1$ and $\rho_2$ ($\rho_1 = 8\rho_2$) and have radii of $1$ mm and $2$ mm, respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1\rho_2$. The ratio of their terminal velocities would be
A. $\dfrac{79}{72}$
B. $\dfrac{19}{36}$
C. $\dfrac{39}{72}$
D. $\dfrac{79}{36}$ ✓ Correct
Solution: Terminal velocity $v_t = \dfrac{2(\rho-\sigma)r^2 g}{9\eta}$, so $\dfrac{v_{t_1}}{v_{t_2}} = \dfrac{(8\rho_2-\sigma)}{(\rho_2-\sigma)}\left(\dfrac{r_1}{r_2}\right)^2$. With $\sigma = 0.1\rho_2$, $r_1 = 1$ mm and $r_2 = 2$ mm, $= \dfrac{7.9\rho_2}{0.9\rho_2}\left(\dfrac{1}{2}\right)^2 = \dfrac{79}{36}$.
Q3 — Bernoulli's Principle and Viscosity · medium · numerical
A small hole of area of cross-section $2$ mm$^2$ is present near the bottom of a fully filled open tank of height $2$ m. Taking $g = 10$ m/s$^2$, the rate of flow of water through the open hole would be nearly
A. $8.9\times10^{-6}$ m$^3$/s
B. $2.23\times10^{-6}$ m$^3$/s
C. $6.4\times10^{-6}$ m$^3$/s
D. $12.6\times10^{-6}$ m$^3$/s ✓ Correct
Solution: Rate $R = a\times v$ with $v = \sqrt{2gh}$, so $R = a\sqrt{2gh}$. With $a = 2\times10^{-6}$ m$^2$ and $h = 2$ m, $R = 2\times10^{-6}\times\sqrt{2\times10\times2} = 2\times10^{-6}\times6.32 = 12.64\times10^{-6} \approx 12.6\times10^{-6}$ m$^3$/s.
Q4 — Bernoulli's Principle and Viscosity · medium · numerical
A small sphere of radius $r$ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to
A. $r^5$ ✓ Correct
B. $r^2$
C. $r^3$
D. $r^4$
Solution: Rate of heat produced $\dfrac{dQ}{dt} = F\times v_T = 6\pi\eta r v_T\times v_T = 6\pi\eta r v_T^2$. Since $v_T = \dfrac{2}{9}\dfrac{r^2(\rho-\sigma)}{\eta}g$, $v_T \propto r^2$. Hence $\dfrac{dQ}{dt} \propto r\cdot(r^2)^2 \propto r^5$.
Q5 — Bernoulli's Principle and Viscosity · medium · numerical
A wind with speed $40$ m/s blows parallel to the roof of a house. The area of the roof is $250$ m$^2$. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be ($\rho_{air} = 1.2$ kg/m$^3$)
A. $4.8\times10^5$ N, downwards
B. $4.8\times10^5$ N, upwards
C. $2.4\times10^5$ N, upwards ✓ Correct
D. $2.4\times10^5$ N, downwards
Solution: From Bernoulli's theorem, $p_1 - p_2 = \dfrac{1}{2}\rho(v_2^2 - v_1^2) = \dfrac{1}{2}\times1.2(40^2 - 0) = 960$ N/m$^2$. Force $= (p_1 - p_2)A = 960\times250 = 24\times10^4 = 2.4\times10^5$ N. As the pressure inside is greater than outside, the force acts upwards.
Q6 — Bernoulli's Principle and Viscosity · medium · numerical
The cylindrical tube of a spray pump has radius $R$, one end of which has $n$ fine holes, each of radius $r$. If the speed of the liquid in the tube is $v$, the speed of the ejection of the liquid through the holes is
A. $\dfrac{vR^2}{n^2r^2}$
B. $\dfrac{vR^2}{nr^2}$ ✓ Correct
C. $\dfrac{vR^2}{n^3r^2}$
D. $\dfrac{v^2R}{nr}$
Solution: By the equation of continuity, $Av=$ constant, so the volume flow rate in the tube equals the total volume flow rate out of the holes: $\pi R^2 v = n\pi r^2 v'$. $\Rightarrow v' = \dfrac{R^2 v}{nr^2}$. Hence the speed of ejection through the holes is $\dfrac{vR^2}{nr^2}$.