# Top 5 Fluid Mechanics problems for IIT-JEE NEET

## Q1. from "Top 5 Fluid Mechanics problems for IIT-JEE NEET"

## Q1.The tension in a string holding a solid block below the surface of a liquid (of density greater than that of solid) as shown in the figure is T_o when the system is at rest. What will be the tension in the string if the system has upward acceleration a.

`T_o\frac ag`

`T_o\left(1-\frac ag\right)`

`T_o\left(1+\frac ag\right)`

`T_o\left(\frac ag-1\right) `

## Q1.The tension in a string holding a solid block below the surface of a liquid (of density greater than that of solid) as shown in the figure is T_o (To) when the system is at rest. What will be the tension in the string if the system has upward acceleration a.

## (A) T_o\frac ag

## (B) T_o\left(1-\frac ag\right)

## (C) T_o\left(1+\frac ag\right)

## (D) T_o\left(\frac ag-1\right)

## Solution

#### At Equilibrium (Rest), Free body diagram of block

###### T_0+mg=F_{up}

###### T_o+V\rho_Sg=\;\;V\rho_Lg\\

###### Vg(\;\rho_L-\;\rho_S)\;=T_o

###### \rho_L-\;\rho_S=\frac{T_o}{Vg}\;-----(1)

#### Free body diagram of block when it moves upwards.

## When system moves up with acceleration 'a' then effective weight of fluid displaced also changes which is called apparent weight (of fluid displaced).

## F_{up}^I=V\;\rho_L\;g_{eff}

## Since, the system is moving up with acceleration 'a' .Thus, \\g_{eff}=\;g+a-----(2)

## As 'a' be the acceleration of system , Use Free body diagram of the moving block

###### F_{net}=F_{up}^I-mg-T

## Therefore, ma=V\;\rho_L\;g_{eff}-mg-T

## m(\;a\;+g)=V\;\rho_L\;g_{eff}\;-T

## V\;\rho_S\;(\;a\;+g)=V\;\rho_L\;g_{eff}\;-T

## Now use (2), V\;\rho_S\;(\;g+a)=V\;\rho_L\;(g+a)\;-T

## V\;\rho_S\;(\;g+a)=V\;\rho_L\;(g+a)\;-T

## T=\;V(g+a)\;(\;\rho_L-\;\rho_S)

## Using (1), T=\;V(g+a)\;({\textstyle\frac{T_o}{V\;g}})\;

## T=\;\;T_o\left(1+\frac ag\right)\\

## Ans. (C) is correct option

## Q2. from "Top 5 Fluid Mechanics problems for IIT-JEE NEET"

## Q2. A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

g\left(\frac\sigma\rho-1\right)\;downwards

g\left(\frac\sigma\rho-1\right)\;upwards

g\left(\frac\sigma\rho\right)\;downwards

g\left(\frac\sigma\rho\right)\;upwards

## Q2.A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

## (A) g\left(\frac\sigma\rho-1\right)\;downwards

## (B) g\left(\frac\sigma\rho-1\right)\;upwards

## (C) g\left(\frac\sigma\rho\right)\;downwards

## (D) g\left(\frac\sigma\rho\right)\;upwards

## Solution

#### Let V = Volume of body

#### \rho = density of body

#### \sigma = density of lake water

#### Thus, m=\;\rho\;V

#### As the density of liquid is more than that of body( σ > ρ), Therefore the body has tendency to float on the liquid because of larger upwards force in comparison to that of weight of body. Hence, net force will be upwards.

#### W know that F_{up}=V\;\sigma g\;

#### According to Newton's 2nd law of motion, we can write

#### F_{net}= F_{up}-mg (upwards)

#### F_{net}= V\sigma g -V \rho g

#### ma = V\sigma g -V \rho g

#### V\rho a = V\sigma g -V \rho g

#### a=\frac{V(\sigma-\rho)g}{V\rho}

#### a=\frac{(\sigma-\rho)g}{\rho}

#### a=g\left(\frac\sigma\rho-1\right)\;upwards

#### (B) is correct option

## Q3. from "Top 5 Fluid Mechanics problems for IIT-JEE NEET"

## Q3.A partially immersed solid block of \rho_s is floating in a liquid of density \rho_L as shown in figure. If beaker container moves up with positive acceleration 'a' , what is correct statement about the block?

It sinks more inside the liquid

It sinks lesser inside the liquid

It neither sinks more nor lesser inside the liquid

Can not be predicted as data is insufficient