Question

A block of wood weighs 12 kg and has a relative density 0.6. It is to be in water with 0.9 of its volume immersed. What weight of a metal is needed if the metal is on the top of wood ? [Relative density of metal=14]
A. 2kg
B. 4kg
C. 6kg
D. 8kg

Answer 1

Hint:-The density of a body is defined as the ratio of mass and volume of the body. The relative density of the anybody is equal to the ratio of density of the body to the density of a standard liquid which is water having density$\rho = 1000\dfrac{{kg}}{{{m^3}}}$.

Formula used:
The formula of density of anybody is given by,
$\rho = \dfrac{M}{V}$
Where M is the mass of the body, V is the volume of the body and $\rho $ is the density of the body.

Complete step-by-step solution
It is given in the problem that a block of wood of mass 12kg having relative density 0.6 and 0.9 of the part of the wooden block is immersed in the water and we need to find the value of weight that is required in order to dip 0.9 of the volume of the wood into the water.
Let a metal of mass ‘m’ is taken and put onto the top of the wooden block then the volume of the metal is equal to,
$ \Rightarrow \rho = \dfrac{M}{V}$
$ \Rightarrow {V_m} = \dfrac{m}{{14}}$………eq. (1)
Volume of the wooden block can be also calculated in the same way,
$ \Rightarrow \rho = \dfrac{M}{V}$
$ \Rightarrow {V_w} = \dfrac{{12}}{{0 \cdot 6}}$………eq. (2)
The total volume of the wooden block and metal block is given by,
${V_t} = {V_m} + {V_w}$
Replace the value of volume of wooden block and volume of metal mass.
$ \Rightarrow {V_t} = {V_m} + {V_w}$
$ \Rightarrow {V_t} = \dfrac{m}{{14}} + \dfrac{{12}}{{0 \cdot 6}}$
$ \Rightarrow {V_t} = \dfrac{m}{{14}} + 20$………eq. (3)
The volume displaced will be of wood so the value of the buoyant force on the wooden block is given by,
$ \Right Arrow \left( {m + 12} \right)g = g\left( {0 \cdot 9 \times 20} \right)$
$ \Rightarrow m = 6kg$.
The mass of the metal is equal to$m = 6kg$. The correct answer for this problem is option C.

Note:- Whenever a body is merged into a liquid then the liquid applies a buoyant force on the body which is in upwards direction and in the equilibrium condition the weight of the body is equal to the buoyant force on the body.