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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2025 · 8 April · Evening Shift · Physics · Rotational Motion

A cube having a side of 10 cm with unknown mass and 200 gm mass...

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 ________ kg.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 cmA 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

Correct answer

0

Step-by-step solution

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
Given, volume of block =(10×10−2)3=10−3 m3=\left(10 \times 10^{-2}\right)^3=10^{-3} \mathrm{~m}^3
Let density of block =ρkg/m3=\rho \mathrm{kg} / \mathrm{m}^3
mass of block =ρ×10−3 kg=\rho \times 10^{-3} \mathrm{~kg}
Buoyant Force \left(\mathrm{F}_{\mathrm{B}}\right)=1000 \times \frac{10^{-3}}{2} \times 10=5 \mathrm{~N}<br/>F.B.D.ofblocks<imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeemain/O3CBl3mxXbDKlW68I9ix5DVQGd13qW6YN3aeXVNPPb4.original.fullsize.png"/><br/>BalancingtorqueaboutpointO,weget<br/><br/>F.B.D. of blocks<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/O3CBl3mxXbDKlW68I9ix5DVQGd13qW6YN3aeXVNPPb4.original.fullsize.png"/><br/>Balancing torque about point O , we get<br/><br/>mg⁡(2×10−2)−FB(2×10−2)=0.2 g(25×10−2)ρ×10−3×10×2−10=50ρ=3000 kg/m3<br/>\begin{aligned}<br/>& \operatorname{mg}\left(2 \times 10^{-2}\right)-\mathrm{F}_{\mathrm{B}}\left(2 \times 10^{-2}\right)=0.2 \mathrm{~g}\left(25 \times 10^{-2}\right) \\ & \rho \times 10^{-3} \times 10 \times 2-10=50 \\ & \rho=3000 \mathrm{~kg} / \mathrm{m}^3<br/>\end{aligned}<br/>Hence,massofblock<br/>Hence, mass of block=\rho \times 10^{-3}<br/><br/>=3000 \times 10^{-3}=3 \mathrm{~kg}
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