Question

A plank is floating in a liquid as shown. Fraction \[f\] of its volume is immersed. Choose the correct options.


A. If the system is taken to a place where atmospheric pressure is more, '\[f\] ' will increase.
B. In the above condition \[f\] will remain unchanged.
C. If the temperature is increased and expansion of only liquid is considered \[f\] will increase.
D. If the temperature is increased and expansion of only plank is considered \[f\] will decrease.

Answer 1

Hint:The downforce generated by the object is equal to the upthrust. The force depends on many factors. By using the formula of equilibrium of a plank, we will find the relation between density and temperature.

Formula used
For equilibrium of the plank,
\[Vdg = (f_{V})\rho_{ L}g\]
Here, V is the volume of the plank.
d is the density of the plank
g is acceleration of gravity.
${f_V}$ is a buoyant force.
\[\rho_{ L}\] is the density of the liquid.
g is acceleration of gravity.

Complete step by step solution:
For equilibrium of the plank,
\[Vdg = (fV)\rho Lg\]
When the liquid expansion is taken into account, density drops as the temperature rises. Therefore, as \[l\] drops, \[f\] increases. Density \[d\] of plank falls when plank expansion is taken into account. Therefore, \[f\] declines.

Hence, if the temperature is increased and expansion of only liquid is considered \[f\] will increase, and if the temperature is increased and expansion of only plank is considered \[f\] will decrease.

Hence the correct options are C and D.

Additional information: A particle is considered to be in equilibrium when the net external force exerted on it is zero. We can state that the particle is either at rest or in uniform motion in this scenario by applying Newton's First Law of Motion.

Note: Students are often confused with the relation between density and temperature. If the temperature of a liquid increases then the density of the liquid decreases. If the temperature of a liquid decreases then the density of the liquid increases.