JEE Advanced Viscosity & Terminal Velocity PYQ with Solutions

Q 1 :

Consider two solid spheres P and Q each of density $8 {gm cm}^{- 3}$ and diameters 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density $0.8 {gm cm}^{- 3}$ and viscosity $η = 3$ poiseulles. Sphere Q is dropped into a liquid of density $1.6 {gm cm}^{- 3}$ and viscosity $η = 2$ poiseulles. The ratio of the terminal velocities of P and Q is [2016]

(3)

$As we know, terminal velocity$

$V_{T} = \frac{2 r^{2}}{9 η} (ρ - σ) g$

$\frac{V_{P}}{V_{Q}} = \frac{\frac{2 r_{1}^{2} (σ - ρ_{1}) g}{9 η_{1}}}{\frac{2 r_{2}^{2} (σ - ρ_{2}) g}{9 η_{2}}}$

$= \frac{r_{1}^{2} (σ - ρ_{1})}{r_{2}^{2} (σ - ρ_{2})} \times \frac{η_{2}}{η_{1}}$

$= \frac{1^{2}}{{(0.5)}^{2}} [\frac{8 - 0.8}{8 - 1.6}] \times \frac{2}{3} = 3$

Q 2 :

A table tennis ball has radius $(\frac{3}{2}) \times 10^{- 2} m$ and mass $(\frac{22}{7}) \times 10^{- 3} kg .$ It is slowly pushed down into a swimming pool to a depth of $d = 0.7 m$ below the water surface and then released from rest. It emerges from the water surface at speed $v$ , without getting wet, and rises up to a height $H$ . Which of the following option(s) is(are) correct?

[Given: $π = \frac{22}{7}, g = 10$ ${m s}^{- 2}$ , density of water = $1 \times 10^{3} {kg m}^{- 3}$ , viscosity of water = $1 \times 10^{- 3} Pa-s .$ ] [2024]

The work done in pushing the ball to the depth $d$ is $0.077 J$
If we neglect the viscous force in water, then the speed $v = 7 m/s$ .
If we neglect the viscous force in water, then the height $H = 1.4 m$ .
The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $\frac{500}{9} .$

1

2

3

4

Select one or more options

(1, 2, 4)

$(1)$ $Work done in pushing the ball to the depth d = 0.7 m$

$W_{all} = k_{f} - k_{i} = 0$

$w_{g} + w_{B} + w_{v} + w_{ext} = 0$

$m g d - ρ_{w} \cdot v \cdot g d - 6 π η r v d + w_{ext} = 0$

$w_{ext} = ρ_{w} \cdot v \cdot g d - m g d$

$= (1000 \times \frac{4}{3} \times \frac{22}{7} \times {(\frac{3}{2} \times 10^{- 2})}^{3} - \frac{22}{7} \times 10^{- 3}) g d$

$w_{ext} = \frac{22}{7} \times 10^{- 3} [\frac{9}{2} - 1] \times 10 \times 0.7 = \frac{22}{7} \times 10^{- 3} \times \frac{7}{2} \times 7$

$w_{ext} = 77 \times 10^{- 3} J = 0.077 J$

$so option (1) is correct.$

$(2) When ball is released at bottom same work i.e. 0.077 J$ $is done on ball$

$∴$ $\frac{1}{2} m v^{2} = 0.077$ $∴$ $v = \sqrt{\frac{2 \times 0.077}{\frac{22}{7} \times 10^{- 3}}} = 7 m/s$

$So option (2) is correct$

$(3) Height H = \frac{v^{2}}{2 g} = \frac{49}{20} = 2.45 m$

$so option (3) is incorrect$

$Net force F_{net} = v d g - v ρ g = 0.11 N$

$And viscous force is maximum when v = 7 m/s$

$∴$ ${(F_{v})}_{\max} = 6 π η r v = 6 \times \frac{22}{7} \times 10^{- 3} \times (\frac{3}{2} \times 10^{- 2}) \times 7$

$= 18 \times 11 \times 10^{- 5} N$

$∴$ $\frac{F_{net}}{{(F_{v})}_{\max}} = \frac{0.11}{18 \times 11 \times 10^{- 5}} = \frac{500}{9}$

$so, option (4) is correct$

Q 3 :

Two spheres P and Q of equal radii have densities $ρ_{1}$ and $ρ_{2}$ , respectively. The spheres are connected by a massless string and placed in liquids $L_{1}$ and $L_{2}$ of densities $σ_{1}$ and $σ_{2}$ and viscosities $η_{1}$ and $η_{2}$ , respectively. They float in equilibrium with the sphere P in $L_{1}$ and sphere Q in $L_{2}$ and the string being taut (see figure). If sphere P alone in $L_{2}$ has terminal velocity ${\vec{V}}_{P}$ and Q alone in $L_{1}$ has terminal velocity ${\vec{V}}_{Q}$ , then [2015]

[IMAGE 398]

$\frac{| {\vec{V}}_{P} |}{| {\vec{V}}_{Q} |} = \frac{η_{1}}{η_{2}}$
$\frac{| {\vec{V}}_{P} |}{| {\vec{V}}_{Q} |} = \frac{η_{2}}{η_{1}}$
${\vec{V}}_{P} \cdot {\vec{V}}_{Q} > 0$
${\vec{V}}_{P} \cdot {\vec{V}}_{Q} < 0$

1

2

3

4

Select one or more options

(1, 4)

[IMAGE 399]

$Since string is taut, ρ_{1} < σ_{1} and ρ_{2} < σ_{2}$

$For floating, net weight of system = net upthrust$

$(ρ_{1} + ρ_{2}) V g = (σ_{1} + σ_{2}) V g$

$Upward terminal velocity$

$V_{P} = \frac{2 r^{2}}{9 η_{2}} (σ_{2} - ρ_{1}) g$

$Where r is radius of sphere.$

$Downward terminal velocity$

$V_{Q} = \frac{2 r^{2} (ρ_{2} - σ_{1}) g}{9 η_{1}}$

$∴$ $| \frac{V_{P}}{V_{Q}} |$ $= \frac{η_{1}}{η_{2}}$

$(∵ ρ_{1} - σ_{2} = σ_{1} - ρ_{2})$

$Again V_{P} \cdot V_{Q} < 0 i.e., negative as V_{P} and V_{Q} are opposite to each other.$

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