JEE MAIN - Physics (2025 - 8th April Evening Shift - No. 25)
A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of a 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 1 gm/cm3.)
The unknown mass is _____ kg.
(Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is 1 gm/cm3.)
The unknown mass is _____ kg.
Answer
3
Explanation
_8th_April_Evening_Shift_en_25_1.png)
Given, volume of block $=\left(10 \times 10^{-2}\right)^3=10^{-3} \mathrm{~m}^3$
Let density of block $=\rho \mathrm{kg} / \mathrm{m}^3$
mass of block $=\rho \times 10^{-3} \mathrm{~kg}$
Buoyant Force $\left(F_B\right)=1000 \times \frac{10^{-3}}{2} \times 10=5 \mathrm{~N}$
F.B.D. of blocks
_8th_April_Evening_Shift_en_25_2.png)
Balancing torque about point O , we get
$$\begin{aligned} & \operatorname{mg}\left(2 \times 10^{-2}\right)-\mathrm{F}_{\mathrm{B}}\left(2 \times 10^{-2}\right)=0.2 \mathrm{~g}\left(25 \times 10^{-2}\right) \\ & \rho \times 10^{-3} \times 10 \times 2-10=50 \\ & \rho=3000 \mathrm{~kg} / \mathrm{m}^3 \end{aligned}$$
Hence, mass of block $=\rho \times 10^{-3}$
$$=3000 \times 10^{-3}=3 \mathrm{~kg}$$
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