Physics · Fluid Mechanics

JEE Main 2025 — 8 April, Evening Shift — Question 68

A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 cm . Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water. (Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is 1gm/cm31 \mathrm{gm} / \mathrm{cm}^{3}.) The unknown mass is \qquad kg .

Question figure

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

25×0.2×g=2×(m−ρ×v)g25 \times 0.2 \times g=2 \times(m-\rho \times v) g m−ρv=2.5 kgm-\rho v=2.5 \mathrm{~kg}

ρv=1×10−3 kg cm3×103 cm32=12 kg\rho v=\frac{1 \times 10^{-3} \mathrm{~kg}}{\mathrm{~cm}^{3}} \times \frac{10^{3} \mathrm{~cm}^{3}}{2}=\frac{1}{2} \mathrm{~kg} m=3 kgm=3 \mathrm{~kg}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Fluid Mechanics
Topic
Properties of Fluids & Hydrostatic Pressure
A cube having a side of 10 cm with unknown mass and 200 gm mass were… | JEE Main 2025 PYQ with Solution · DhiX AI