Question

A certain block weights $ 15\;N $ in the air. It weighs $ 12\;N $ when immersed in water. When immersed in another liquid, it weighs $ 13\;N $ . The relative density of the blocks and the other liquid are respectively.
(A) $ 5,\dfrac{2}{3} $
(B) $ \dfrac{2}{3},5 $
(C) $ \dfrac{4}{5},5 $
(D) $ 5,\dfrac{4}{5} $

Answer 1

Hint: To solve this question we will use the concept of buoyant force to find the actual force on the block when it will immerse in the water. Then using the density formula we will find the relative density of the block in air and the water. Using the same method we will find the relative density of the block in water and unknown liquid.

Formula used:
Force acting due to the gravitational acceleration
 $ \Rightarrow F = mg $
Where $ m $ is mass and $ g $ is the gravitational acceleration.
Density formula
 $ \Rightarrow \rho = \dfrac{m}{V} $
Where $ V $ is the volume.

Complete Step-by-step solution:
The force acting on the block is given as $ 15\;N $ when it is in air, which can be given as
 $ \Rightarrow F = mg = 15\;N $ ………. $ (1) $
Assume that the density of the block is given as $ {\rho _b} $ when in air, hence
 $ \Rightarrow {\rho _b} = \dfrac{m}{V} $
 $ \Rightarrow m = {\rho _b} \times V $
Substituting it in the equation $ (1) $ , hence
 $ \Rightarrow {\rho _b} \times V = 15\;N $ …….. $ (2) $
Now when the block is immersed in water then the buoyant force $ {F_b} $ acts on it which is balanced by the force acting downwards $ m\;g $ , hence
 $ \Rightarrow 15\;N = 12\;N - {F_b} $
 $ \Rightarrow {F_b} = 3\;N $
If $ {\rho _w} $ is the density of the water then the force on the block when immersed in water is given as
 $ \Rightarrow {\rho _w} \times V = 5\;N $
Now evaluating the relative density of the block and water is given as
 $ \Rightarrow {\left( {{R_\rho }} \right)_b} = \dfrac{{{\rho _b}}}{{{\rho _w}}} $
 $ \Rightarrow {\left( {{R_\rho }} \right)_b} = \dfrac{{\dfrac{{15N}}{V}}}{{\dfrac{{3N}}{V}}} $
Hence the relative density of the block is given as
 $ \Rightarrow {\left( {{R_\rho }} \right)_b} = 5\; $
Now when the block is immersed in other liquid then the buoyant force $ {F_b}^\prime $ acts on it which is balanced by the force acting downwards $ m\;g $ , hence
 $ \Rightarrow 15\;N = 13\;N - {F_b}\prime $
 $ \Rightarrow {F_b}\prime = 2\;N $
If $ {\rho _L} $ is the density of the other liquid then the force on the block when immersed in other liquid is given as
 $ \Rightarrow {\rho _L} \times V = 2\;N $
Now evaluating the relative density of the block and in other liquid when it is removed from water which is given as
 $ \Rightarrow {\left( {{R_\rho }} \right)_L} = \dfrac{{{\rho _L}}}{{{\rho _w}}} $
 $ \Rightarrow {\left( {{R_\rho }} \right)_L} = \dfrac{{\dfrac{{2N}}{V}}}{{\dfrac{{3N}}{V}}} $
Hence the relative density of the block is given as
 $ \therefore {\left( {{R_\rho }} \right)_L} = \dfrac{2}{3} $
Hence the option (A) is the correct answer.

Note:
Here we have used the formula of the relative density of the substance between the two mediums given. We have to be ensured that the forces should be properly balanced and the proper sign should be mentioned while doing the calculation.