91Ó°ÊÓ

• The density of rock is, $\left(P_{density}\right)$=1200kg/m³
• The tension in the string when the rock is in air is, $\left(T_{air}\right)$(=28.0N.
• The density of the liquid is, $\left(P_{liquid}\right)$ =750kg/m^3.
• The rock is immersed totally in the liquid.

Buoyant force occurs when a body is submerged in water and the fluid exerts an upward draw on it. Archimedes' principle asserts that the buoyant force of an object is equal to its weight in fluid when it is displaced by a fluid.

First diagram shows forces acting on rock when it is in air, tension is acting upwards and weight of rock is acting downwards. As rock is suspended and at same time at rest, the net force acting is zero.

$\begin{array}{r}\mathrm{Î\pounds ¹ó}={\mathrm{T}}_{\text{air}}−\mathrm{W}=0\\ {\mathrm{T}}_{\text{air}}=\mathrm{W}\\ {\mathrm{T}}_{\text{air}}=\mathrm{mg}\\ \mathrm{Putting}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{equation},\\ \mathrm{m}=\frac{28.0\mathrm{N}}{9.80\mathrm{N}/\mathrm{kg}}\\ =\frac{20}{7}\mathrm{kg}\end{array}$

Second diagram shows force acting on rock when it is immersed in liquid, tension is acting upwards, weight of rock is acting downwards and buoyant force is acting upwards. Here total force will be zero.

$\begin{array}{c}∑\mathrm{F}={\mathrm{T}}_{\text{liquid}}−\mathrm{W}+\mathrm{B}=0\\ {\mathrm{T}}_{\text{liquid}}−{\mathrm{T}}_{\text{air}}+{\mathrm{ÒÏ}}_{\text{liquid}}{\mathrm{V}}_{\text{rock}}\mathrm{g}=0\\ {\mathrm{T}}_{\text{liquid}}−{\mathrm{T}}_{\text{air}}+{\mathrm{ÒÏ}}_{\text{liquid}}\frac{{\mathrm{m}}_{\text{rock}}}{{\mathrm{ÒÏ}}_{\text{rock}}}\mathrm{g}=0\\ \mathrm{Putting}\mathrm{values}\mathrm{in}\mathrm{equation}\mathrm{we}\mathrm{get},\\ {\mathrm{T}}_{\text{liquid}}=28−\left(750\right)\left(\frac{\frac{20}{7}}{1200}\right)(9.80)\\ {\mathrm{T}}_{\text{liquid}}=10.5\mathrm{N}\end{array}$