JEE Main20258 Apr 2025Evening ShiftPhysicsRotational MotionActual
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 o
Correct answer
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Step-by-step solution
Given, volume of block = (10 10⁻² )^3=10⁻³ ~m ^3 Let density of block = kg / m ^3 mass of block = 10⁻³ ~kg Buoyant Force ( F _ B )=1000 10⁻³ 2 10=5 ~N F.B.D. of blocks Balancing torque about point O , we get aligned & mg (2 10⁻² )- F _ B (2 10⁻² )=0.2 ~g (25 10⁻² ) & 10⁻³ 10 2-10=50 & =3000 ~kg / m ^3 aligned Hence, mass of block = 10⁻³ =3000 10⁻³=3 ~kg