JEE Physics Mechanical Properties Of Fluids MCQs Set 28

Answer: (b) \( 3\times10^{-6} \) m

Explanation:
1. \( v=\frac{2r^2\rho g}{9\eta}\Rightarrow r^2=\frac{9\eta v}{2\rho g}=\frac{9\times3.6\times10^{-5}\times5\times10^{-4}}{2\times900\times10}=9\times10^{-12} \).
2. So \( r=3\times10^{-6} \) m.

Teacher's Note:
1. Terminal velocity is proportional to r squared.
2. Neglect the density of air relative to oil.

Answer: (d) directly proportional to the square of the radius of the sphere

Explanation:
1. \( v_T=\frac{2r^2(\rho-\sigma)g}{9\eta}\propto r^2 \).

Teacher's Note:
1. Weight and buoyancy scale as r cubed, while Stokes drag scales as r times v.
2. Balancing them gives v proportional to r squared.

Answer: (b) there will be no terminal velocity

Explanation:
1. In vacuum there is no viscous drag, so the sphere keeps accelerating with g.
2. A terminal velocity needs a retarding force that grows with speed.

Teacher's Note:
1. Terminal velocity exists only when drag balances the driving force.
2. In vacuum the speed keeps increasing.

Answer: (d) 10 cm/s

Explanation:
1. Effective weight \( w \) equals the drag at 10 cm/s: \( w=6\pi\eta r(10) \).
2. Moving up, drag acts downward: \( 2w=w+6\pi\eta rv' \) so \( v'=10 \) cm/s.

Teacher's Note:
1. Drag always opposes the motion, so when the ball moves up both weight and drag act downward.
2. Net upward force then equals w.

Answer: (b) \( r^5 \)

Explanation:
1. Rate of heat production \( =F\,v=6\pi\eta rv\cdot v\propto rv^2 \).
2. With \( v\propto r^2 \): \( \propto r\cdot r^4=r^5 \).

Teacher's Note:
1. At terminal velocity the viscous force does work at the rate F times v.
2. Both F and v depend on r.

Answer: (d) 3

Explanation:
1. \( v_T\propto\frac{r^2(\rho-\sigma)}{\eta} \).
2. \( v_P\propto\frac{(0.5)^2\times7.2}{3}=0.6 \) and \( v_Q\propto\frac{(0.25)^2\times6.4}{2}=0.2 \).
3. Ratio = 3.

Teacher's Note:
1. Use the radius, not the diameter, in the formula.
2. Subtract the liquid density from the sphere density in each case.

Answer: (b) 6.30 m

Explanation:
1. Initial speed \( v_0=\frac{J}{m}=\frac{1.0}{0.4}=2.5 \) m/s.
2. \( x=\int_0^\tau v_0e^{-t/\tau}dt=v_0\tau(1-e^{-1})=2.5\times4\times0.63=6.30 \) m.

Teacher's Note:
1. The impulse gives the initial speed by J = m v0.
2. Integrate the exponential velocity over time.

Answer: (d) 0.1 m/s

Explanation:
1. \( v_T\propto(\rho-\sigma) \) for equal radius.
2. \( \frac{v_{Ag}}{v_{Au}}=\frac{10.5-1.5}{19.5-1.5}=\frac{9}{18}=\frac12 \), so \( v_{Ag}=0.1 \) m/s.

Teacher's Note:
1. Same size and liquid means only the density difference matters.
2. Subtract the liquid density from each metal density.

Answer: (b) \( 6.25\times10^{-4} \) cm/s

Explanation:
1. \( v_T\propto\frac{\rho-\sigma}{\eta} \).
2. \( \frac{v_g}{v_w}=\frac{(7.8-1.2)/13.2}{(7.8-1.0)/(8.5\times10^{-4})}=\frac{0.5}{8000} \).
3. \( v_g=10\times\frac{0.5}{8000}=6.25\times10^{-4} \) cm/s.

Teacher's Note:
1. Take the ratio of the two terminal velocity formulas; radius cancels.
2. Water density is 1 g/cm³.

Answer: (c) 20 cm/s

Explanation:
1. Volume conserved: \( R^3=8r^3 \), so \( R=2r \).
2. \( v_T\propto r^2 \) so \( v'=4\times5=20 \) cm/s.

Teacher's Note:
1. Terminal velocity scales as the square of the radius.
2. Eight drops give a radius doubled.

Answer: (a) 0.12 N s/m²

Explanation:
1. Gap \( y=5 \) mm, speed of the surface \( v=\omega R_1=2\pi\times0.15 \) m/s.
2. Torque \( \tau=\eta\frac{\omega R_1}{y}\cdot(2\pi R_1L)\cdot R_1=\frac{2\pi\eta\omega R_1^3L}{y} \).
3. \( \eta=\frac{0.98\times0.005}{2\pi\times2\pi\times(0.15)^3\times0.3}\approx0.12 \) N s/m².

Teacher's Note:
1. Shear stress times area gives the force, and force times radius gives the torque.
2. Convert 60 rpm to 2 pi rad/s.

Answer: (b) \( \tfrac12\pi R^2v_0 \)

Explanation:
1. \( Q=\int_0^Rv_0\left(1-\frac{r^2}{R^2}\right)2\pi r\,dr=2\pi v_0\left(\frac{R^2}{2}-\frac{R^2}{4}\right)=\frac{\pi R^2v_0}{2} \).

Teacher's Note:
1. Integrate over thin annular rings of area 2 pi r dr.
2. The average speed is half the axial speed.

Answer: (c) \( \frac{4\eta\ell v_0}{R^2} \)

Explanation:
1. Shear stress at the wall \( =\eta\left|\frac{dv}{dr}\right|_R=\frac{2\eta v_0}{R} \).
2. Friction force on the wall \( =\frac{2\eta v_0}{R}\times2\pi R\ell=4\pi\eta\ell v_0 \).
3. This equals the net pressure force: \( \Delta p\,\pi R^2=4\pi\eta\ell v_0 \), so \( \Delta p=\frac{4\eta\ell v_0}{R^2} \).

Teacher's Note:
1. In steady flow the pressure force on the fluid balances the viscous force at the wall.
2. Differentiate the velocity profile to get the velocity gradient at r = R.