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NVAMediumJEE 2025Torque & Angular Momentum

JEE Physics 2025 Question with Solution

A cube having a side of 10 cm10 \, \text{cm} with unknown mass and 200 gm200 \, \text{gm} mass were hung at two ends of an uniform rigid rod of 27 cm27 \, \text{cm} long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm200 \, \text{gm} weight as 25 cm25 \, \text{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 the unknown mass is more than that of the water, the mass did not absorb water and water density is 1 gm/cm31 \, \text{gm/cm}^3.) The unknown mass is _____ kg\text{kg}.

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: side of cube = 10 cm10 \, \text{cm}, rod length = 27 cm27 \, \text{cm}, distance from wedge to 200 gm200 \, \text{gm} mass = 25 cm25 \, \text{cm}. Half of the cube is submerged in water of density 1 gm/cm31 \, \text{gm/cm}^3.

Find: the unknown mass.

Distance from the wedge to the unknown mass:

27−25=2 cm27 - 25 = 2 \, \text{cm}

Volume of the cube:

V=103=1000 cm3V = 10^3 = 1000 \, \text{cm}^3

Half-submerged volume:

Vsub=500 cm3V_{\text{sub}} = 500 \, \text{cm}^3

Buoyant force corresponds to displaced water mass:

ρwaterVsub=1 gm/cm3×500 cm3=500 gm\rho_{\text{water}} V_{\text{sub}} = 1 \, \text{gm/cm}^3 \times 500 \, \text{cm}^3 = 500 \, \text{gm}

Let the unknown mass be MM grams. At balance, torques about the wedge are equal:

200×g×25=(M×g−500×g)×2200 \times g \times 25 = (M \times g - 500 \times g) \times 2

Cancelling gg:

200×25=(M−500)×2200 \times 25 = (M - 500) \times 2

Now simplify:

5000=2M−10005000 = 2M - 1000 2M=60002M = 6000 M=3000 gm=3 kgM = 3000 \, \text{gm} = 3 \, \text{kg}

Therefore, the unknown mass is 3 kg3 \, \text{kg}.

Torque Equation Directly

Given: the lever arm of the 200 gm200 \, \text{gm} mass is 25 cm25 \, \text{cm}, and the lever arm of the unknown mass is 2 cm2 \, \text{cm}. Half the cube is submerged.

Find: the unknown mass.

Using the direct torque balance from the extracted working:

25×0.2×g=2×(m−ρv)×g25 \times 0.2 \times g = 2 \times (m - \rho v) \times g

So,

m−ρv=2.5 kgm - \rho v = 2.5 \, \text{kg}

Now,

ρv=1×10−3 kgcm3×103 cm32=12 kg\rho v = \frac{1 \times 10^{-3} \, \text{kg}}{\text{cm}^3} \times \frac{10^3 \, \text{cm}^3}{2} = \frac{1}{2} \, \text{kg}

Hence,

m=2.5+0.5=3 kgm = 2.5 + 0.5 = 3 \, \text{kg}

Therefore, the unknown mass is 3 kg3 \, \text{kg}.

Common mistakes

  • Using the full cube volume for buoyancy is incorrect because only half of the cube is inside water. Use displaced volume 500 cm3500 \, \text{cm}^3, not 1000 cm31000 \, \text{cm}^3.

  • Taking the wrong lever arm for the unknown mass is a common error. The wedge is 25 cm25 \, \text{cm} from the 200 gm200 \, \text{gm} mass, so the other side is 27−25=2 cm27 - 25 = 2 \, \text{cm}.

  • Adding the buoyant force to the unknown weight is wrong. Buoyancy acts upward, so it reduces the effective downward force to (M−500)g(M - 500)g in gram-force form.

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