Mechanical Properties of Fluids JEE Main previous year questions with solutions

4 solved JEE Main questions on Mechanical Properties of Fluids, free to read — no sign-in needed. The full chapter has 188 questions; sign in to attempt the remaining 184 in the exam simulator.

1. Q1JEE Main 2026 (02 Apr, Shift 1)Archmidies Principle
   A tub is filled with water and a wooden cube $10$ cm $\times$ $10$ cm $\times$ $10$ cm is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by $3.87$ cm. The mass of the metal coin is _______ gram. (Take water density as $1$ g/cm$^3$ and density of wood as $0.4$ g/cm$^3$)
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   Answer: 387

   Let $A$ be the base area of the wooden cube. $A = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2$ Let $\Delta x$ be the increase in the submerged depth of the cube when the coin is placed on it. $\Delta x = 3.87 \text{ cm}$ The additional volume of water displaced due to the weight of the metal coin is $\Delta V = A \times \Delta x$. $\Delta V = 100 \times 3.87 = 387 \text{ cm}^3$ By Archimedes' principle, the weight of the metal coin is balanced by the additional buoyant force, which is equal to the weight of the additional water displaced. $m g = \Delta V \rho_w g$ $m = \Delta V \rho_w$ Substituting the values of $\Delta V$ and the density of water $\rho_w = 1 \text{ g/cm}^3$: $m = 387 \text{ cm}^3 \times 1 \text{ g/cm}^3 = 387 \text{ g}$ Answer: $387$
2. Q2JEE Main 2025 (22 Jan, Shift 2)Pascal's Law
   A tube of length 1 m is filled completely with an ideal liquid of mass 2 M , and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is $\sqrt{\frac{\mathrm{F}}{\alpha \mathrm{M}}}$ in SI unit. The value of $\alpha$ is __________.
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   Answer: 1

   $\begin{aligned} & F=m \omega^2 r_{\mathrm{cm}} \\ & r_{\mathrm{cm}}=\frac{1}{2} \mathrm{~m} \\ & \omega=\sqrt{\frac{2 F}{m}} \\ & m=2 M \\ & \omega=\sqrt{\frac{F}{M}} \\ & \alpha=1\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 1)Hydrodynamics
   Correct Bernoulli's equation is (symbols have their usual meaning) :
   1. A.$P+m g h+\frac{1}{2} m v^2=$ constant
   2. B.$P+\rho g h+\frac{1}{2} \rho v^2=$ constant
   3. C.$P+\rho g h+\rho v^2=$ constant
   4. D.$P+\frac{1}{2} \rho g h+\frac{1}{2} \rho v^2=$ constant
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   Answer: (B)

   $\mathrm{P}+\rho \mathrm{gh}+\frac{1}{2} \rho \mathrm{V}^2=$ constant
4. Q4JEE Main 2023 (01 Feb, Shift 2)Efflux of liquid
   The surface of water in a water tank of cross section area $750{cm}^{2}$ on the top of a house is $hm$. above the tap level. The speed of water coming out through the tap of cross section area $500{mm}^{2}$ is $30cm{s}^{-1}$. At that instant, $\dfrac{dh}{dt}$ is $x\times {10}^{-3}m{s}^{-1}$. The value of $x$ will be ______.
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   Answer: 2

   Using continuity equation ${A}_{1}{v}_{1}={A}_{2}{v}_{2}$ We have, $750\times {10}^{-4}{v}_{1}=500\times {10}^{-6}\times 0.3$ ${v}_{1}=\dfrac{500\times 3\times {10}^{-3}}{750}$ $=2\times {10}^{-3}m{s}^{-1}$ Thus, $\dfrac{dh}{dt}=-2\times {10}^{-3}m{s}^{-1}$ Hence, the value of $x=2$.