Mechanical Properties of Fluids JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Mechanical Properties of Fluids, free to read — no sign-in needed. The full chapter has 41 questions; sign in to attempt the remaining 36 in the exam simulator.

1. Q1JEE Advanced Adv 2024 (Paper 2)
   A table tennis ball has radius $(3 / 2) \times 10^{-2} \mathrm{~m}$ and mass $(22 / 7) \times 10^{-3} \mathrm{~kg}$. It is slowly pushed down into a swimming pool to a depth of $d=0.7 \mathrm{~m}$ 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: $\pi=22 / 7, g=10 \mathrm{~m} \mathrm{~s}^{-2}$, density of water $=1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, viscosity of water $=1 \times 10^{-3} \mathrm{~Pa}$-s.]
   1. A.The work done in pushing the ball to the depth $d$ is $0.077 \mathrm{~J}$.
   2. B.If we neglect the viscous force in water, then the speed $v=7 \mathrm{~m} / \mathrm{s}$.
   3. C.If we neglect the viscous force in water, then the height $H=1.4 \mathrm{~m}$.
   4. D.The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.
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   Answer: A,B,D

   Work done in pushing the ball $$W=(v \rho g) d-(v \sigma g) d$$ Where, $\rho \rightarrow$ Density of water $\sigma \rightarrow$ Density of ball $$\begin{aligned} & \Rightarrow W=\frac{4}{3} \pi R^3 \times 10 \times 0.7\left[1000-\frac{3}{4} \times \frac{10^{-3}}{R^3}\right] \\ & W=0.077 \mathrm{~J} \end{aligned}$$ [1 is correct] $\Rightarrow$ When ball is released at bottom same work (i.e. $0.077 \mathrm{~J}$ ) is done on ball. $$\begin{aligned} & \therefore \frac{1}{2} m v^2=0.077 \\ & v=\sqrt{\frac{0.077 \times 2}{\frac{22}{7} \times 10^{-3}}} \\ & =7 \mathrm{~m} / \mathrm{s} \end{aligned}$$ [2 is correct] $$\Rightarrow \text { also, } H=\frac{v^2}{2 g}=\frac{7 \times 7}{2 \times 10}=2.45 \mathrm{~m}$$ [3 is incorrect] $\Rightarrow$ Net force $F_{\text {net }}=v \sigma g-v \sigma g=0.11 \mathrm{~N}$ Also, viscous force is maximum when $v=7 \mathrm{~m} / \mathrm{s}$ $$\begin{aligned} & \therefore\left(F_v\right)_{\max }=6 \pi \eta r v \\ & =6 \times \frac{22}{7} \times 10^{-3}\left(\frac{3}{2} \times 10^{-2}\right) \times 7 \\ & =18 \times 11 \times 10^{-5} \mathrm{~N} \end{aligned}$$ Now, $$\frac{F_{\text {net }}}{\left(F_v\right)_{\max }}=\frac{500}{9}$$ [4 is correct]
2. Q2JEE Advanced Adv 2020 (Paper 2)
   A hot air balloon is carrying some passengers, and a few sandbags of mass $1kg$ each so that its total mass is $480kg$. Its effective volume giving the balloon its buoyancy is $V$. The balloon is floating at an equilibrium height of $100m$. When $N$ number of sandbags are thrown out, the balloon rises to a new equilibrium height close to $150m$ with its volume $V$ remain unchanged. If the variation of the density of air with height $h$ from the ground is $\rho \left(h\right)={\rho }_{0}{e}^{-\dfrac{h}{{h}_{0}}},$ where ${\rho }_{0}=1.25kg{m}^{-3}$ and ${h}_{0}=6000m,$ the value of $N$ is _________.
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   Answer: 4

   Initially $mg={F}_{b}$ $480\times g=V\left({\rho }_{0}{e}^{-\dfrac{100}{6000}}\right)g$ ....(1) Finally $(480-m)g=V\left({\rho }_{0}{e}^{-\dfrac{150}{6000}}\right)g$ .....(2) dividing $\dfrac{480-m}{480}={e}^{-\dfrac{1}{40}+\dfrac{1}{60}}$ $m=480\left(1-{e}^{-1/120}\right)$ Hence ${e}^{-1/120}\approx 1-\dfrac{1}{120}$ $1-{e}^{-1/110}=\dfrac{1}{120}$ $m=4kg$ Hence, four sandbags are thrown.
3. Q3JEE Advanced Adv 2018 (Paper 1)
   A uniform capillary tube of inner radius $r$ is dipped vertically into a beaker filled with water. The water rises to a height $h$ in the capillary tube above the water surface in the beaker. The surface tension of water is $\sigma$ . The angle of contact between water and the wall of the capillary tube is $\theta$ . Ignore the mass of water in the meniscus. Which of the following statements is (are) true?
   1. A.For a given material of the capillary tube, $h$ decreases with increase in $r$
   2. B.For a given material of the capillary tube, $h$ is independent of $\sigma$
   3. C.If this experiment is performed in a lift going up with a constant acceleration, then $h$ decreases
   4. D.$h$ is proportional to contact angle $\theta$
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   Answer: A,C

   $\dfrac{2\sigma }{R}=\rho gh$ $\left[R\rightarrow Radiusofmeniscus\right]$ $h=\dfrac{2\sigma }{R\rho g}$ $R=\dfrac{r}{\cos ⁡\theta }$ [r→radius of capillary; $\theta \rightarrow$ contact angle] $h=\dfrac{2\sigma \cos ⁡\theta }{r\rho g}$ $(A)$ For given material, $\theta =constant$ $h\propto \dfrac{1}{r}$ $(B)$ h depends on $\sigma$ $(C)$ If lift is going up with constant acceleration, ${g}_{eff}=\left(g+a\right)$ $h=\dfrac{2\sigma \cos ⁡\theta }{r\rho \left(g+a\right)}$ It means h decreases $(D)$ h is proportional to $\cos ⁡\theta ,Not\theta$
4. Q4JEE Advanced Adv 2017 (Paper 1)
   A drop of liquid of radius $R={10}^{-2}m$ having surface tension $S=\dfrac{0.1}{4\pi }N{m}^{-1}$ divides itself into K identical drops. In the process the total change in the surface energy $\Delta U={10}^{-3}J$ . If $K={10}^{\alpha }$ then the value of $\alpha$ is
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   Answer: 6

   By mass conservation, $\rho .\dfrac{4}{3}\pi {R}^{3}=\rho .K.\dfrac{4}{3}\pi {r}^{3}$ $\Rightarrow R={K}^{\dfrac{1}{3}}r$ $∴\Delta U=T\Delta A=T\left(K.4\pi {r}^{2}-4\pi {R}^{2}\right)$ $=T\left(K.4\pi {R}^{2}{K}^{-\dfrac{2}{3}}-4\pi {R}^{2}\right)$ $\Delta U=4\pi {R}^{2}T\left[{K}^{\dfrac{1}{3}}-1\right]$ Putting the value’s $\Rightarrow {10}^{-3}=\dfrac{{10}^{-1}}{4\pi }\times 4\pi \times {10}^{-4}\left[{K}^{\dfrac{1}{3}}-1\right]$ $100={K}^{\dfrac{1}{3}}-1$ $\Rightarrow {K}^{\dfrac{1}{3}}\cong 100={10}^{2}$ Given that $K={10}^{\alpha }\Rightarrow ∴{10}^{\dfrac{\alpha }{3}}={10}^{2}$ $\Rightarrow \dfrac{\alpha }{3}=2$ $\Rightarrow \alpha =6$
5. Q5JEE Advanced Adv 2016 (Paper 1)
   Consider two solid spheres P and Q each of density 8 gm $c{m}^{-3}$ and diameters 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density 0.8 gm $c{m}^{-3}$ and viscosity $\eta =3$ poiseulles. Sphere Q is dropped into a liquid of density 1.6 gm $c{m}^{-3}$and viscosity $\eta =2$ poiseulles. the ratio of the terminal velocities of P and Q is.
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   Answer: 3

   ${V}_{T}\propto \dfrac{{r}^{2}\left[{d}_{m}-{d}_{L}\right]}{n}$ $\dfrac{{V}_{TP}}{{V}_{TQ}}={\left(\dfrac{{r}_{P}}{{r}_{Q}}\right)}^{2}\times \dfrac{{n}_{{L}_{2}}}{{n}_{{L}_{1}}}\times \left[\dfrac{{d}_{m}-{d}_{{L}_{1}}}{{d}_{m}-{d}_{{L}_{2}}}\right]$ $\dfrac{{V}_{TP}}{{V}_{TQ}}={\left(\dfrac{2}{1}\right)}^{2}\times \dfrac{2}{3}\times \left[\dfrac{8-0.8}{8-1.6}\right]$ $\dfrac{{V}_{TP}}{{V}_{TQ}}=3$