Question 101 of 280

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 $1 \mathrm{gm} / \mathrm{cm}^3$.) The unknown mass is ________ kg.

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Answer: 3

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 $\left(\mathrm{F}_{\mathrm{B}}\right)=1000 \times \frac{10^{-3}}{2} \times 10=5 \mathrm{~N}$
F.B.D. of blocks
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 $=\rho \times 10^{-3}$
$=3000 \times 10^{-3}=3 \mathrm{~kg}$