A block of mass 1 kg and density 0.8 $g/cm^3$ is held stationary with the help of a string as shown in figure. The block is completely immersed into the water filled in tank.If the tank is accelerating vertically upwards with an acceleration a = 2 $m/s^2.$

(i) Find the tension in the string.

(A) 3 N

(B) 2 N

(C) 3.5 N

(D) 5.5 N

Solution

Let $V$ = Volume of body

$\rho$ = density of block = 0.8 $\frac{g}{cm^3}$ = $800\frac{kg}{m^3}$

$\sigma$ = density of water =1000 $\frac{kg}{m^3}$

We know that, the mass of block= $m=\;\rho\;V$

$\rho=\frac mV$

$V=\frac m\rho$

We know that $F_{up}=V\sigma(g+a)\;$

$\Rightarrow F_{up}=\frac m\rho\sigma(g+a)\;$

$\Rightarrow F_{up}=\frac1{800}\ast1000\ast(10+2)=15N$

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

$F_{net}= F_{up}-T- mg$

$ma= F_{up} -T- mg$

$12= 15-T-110$

$2= 15 - T- 10$

$T= 15 - 10-2=3N$

(A) is correct option

Q5. from "Top 5 Fluid Mechanics problems for IIT-JEE NEET"

Q5.A wooden ball of density $900 kg / m^{3}$ is immersed in a liquid of density $1200 kg / m^{3}$ to a depth of 126 cm and the released. The height h above the surface of which the ball rises will be

Click here to unlock this option (A)

40 cm

Click here to unlock this option (B)

42 cm

Click here to unlock this option (C)

45 cm

Click here to unlock this option (D)

50 cm

Choose correct options among the following options.

A wooden ball of density $900 kg / m^{3}$ is immersed in a liquid of density $1200 kg / m^{3}$ to a depth of 126 cm and the released. The height h above the surface of which the ball rises will be

(A) 40 cm

(B) 42 cm

(C) 45 cm

(D) 50 cm

Solution

Let $V$ = Volume of ball

Given that, $\rho$ = density of ball= $900\frac{kg}{m^3}$

$\sigma$ = density of liquid =1200 $\frac{kg}{m^3}$

As, the density of ball = $\rho=\frac mV$

$m=V\rho$

We know that $F_{up}=V\sigma g\;$

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

$F_{net}= F_{up}-mg$

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

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

$a=\;\frac{V\sigma g-V\rho g}{V\rho}$

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

$a=\;\frac{(1200-900)g}{900}$

$a=\;\frac{(300)g}{900}$

$a=\;\frac{g}{3}$

Now, u=0 (because the ball is released from rest)

When ball reaches to surface, S= 126 cm = 1.26 m. Also, let us assume that it attains velocity v at top surface.

According to Kinematics Equation of U.A.M., $v^2-u^2=2aS$

$\Rightarrow v^2-0^2=2\left(\frac g3\right)\;\left(1.26\right)$

$\Rightarrow v^2=0.84 g .....(1)$

h= height the ball attains when it rises up with velocity v from top surafce

Since , the ball rises up vertically under the effect of gravity, so it would come to rest momentarily when it attains height 'h'

$v^{'}=0$

$A =-g$

(-ve sign appears as the diraction of acceleration is against motion)

Again,apply $V^2-U^2=2AS$

$\Rightarrow\;v'^2-v^2=2(-g) h$

$\Rightarrow\;0^2-v^2=2(-g) h$

$\Rightarrow\ v^2=2gh ....(2)$

From (1) and (2), we get,

$0.84g = 2gh$

$h=\frac{0.84g}{2g}$

$h=0.42m=42cm$

(B) is correct option

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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.

Choose correct options among the following options.

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).

Acc. to Archimedes Principle , Upthrust = Weight (effective) of liquid/fluid displaced by object

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

According to Newton's 2nd Law of Motion

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)

Choose correct options among the following options.

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?

Choose correct options among the following options.

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 ?

Solution

Let V_{in} = Volume of block immersed in the liquid

For Equilibrium of rest, see FBD as shown aside

F_{up}=V_{in}\;\rho_Lg\;=\;mg

V_{in}\;\rho_Lg\;=\;V\;\rho_S\;g\;

\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}

Therefore, fraction of volume inside the liquid is-

\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}.....(1)

When block is moving up , the FBD is shown in figure

\;\\F_{up}^I=V_{in}^I\;\rho_L\;g_{eff}\\

F_{up}^I=V_{in}^I\;\rho_L\;(g+a).....(2)

According to Newton's 2nd Law of Motion

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

ma =F_{up}^I-mg

F_{up}^I=m(g+a)

F_{up}^I=V\;\rho_S\;(g+a)\\....(3)

From (2) and (3), V_{in}^I\;\rho_L\;(g+a)=V\;\rho_S\;(g+a)

\;\frac{V_{in}^I}V\;=\frac{\rho_S}{\rho_L}

\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}=\;\frac{V_{in}^I}V\;

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.

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

$T_0+mg=F_{up}$

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

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

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

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

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

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

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

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

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

Let $V$ = Volume of body

$\rho$ = density of body

$\sigma$ = density of lake water

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

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

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

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

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

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

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

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

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

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?

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 ?

Let $V_{in}$ = Volume of block immersed in the liquid

$F_{up}=V_{in}\;\rho_Lg\;=\;mg$

$V_{in}\;\rho_Lg\;=\;V\;\rho_S\;g\;$

$\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}$

$\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}.....(1)$

$\;\\F_{up}^I=V_{in}^I\;\rho_L\;g_{eff}\\$

$F_{up}^I=V_{in}^I\;\rho_L\;(g+a).....(2)$

$F_{net}=F_{up}^I-mg$

$ma =F_{up}^I-mg$

$F_{up}^I=m(g+a)$

$F_{up}^I=V\;\rho_S\;(g+a)\\....(3)$

From (2) and (3), $V_{in}^I\;\rho_L\;(g+a)=V\;\rho_S\;(g+a)$

$\;\frac{V_{in}^I}V\;=\frac{\rho_S}{\rho_L}$

$\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}=\;\frac{V_{in}^I}V\;$

Hence, fraction of volume immersed is again equal to $\frac{\rho_S}{\rho_L}$ . It means immersed volume remains same .

Q4.A block of mass 1 kg and density 0.8 $g/cm^3$ is held stationary with the help of a string as shown in figure. The block is completely immersed into the water filled in tank.If the tank is accelerating vertically upwards with an acceleration a = 2 $m/s^2$ . Find the tension in the string.

A block of mass 1 kg and density 0.8 $g/cm^3$ is held stationary with the help of a string as shown in figure. The block is completely immersed into the water filled in tank.If the tank is accelerating vertically upwards with an acceleration a = 2 $m/s^2.$

Let $V$ = Volume of body

$\rho$ = density of block = 0.8 $\frac{g}{cm^3}$ = $800\frac{kg}{m^3}$

$\sigma$ = density of water =1000 $\frac{kg}{m^3}$

We know that, the mass of block= $m=\;\rho\;V$

$\rho=\frac mV$

$V=\frac m\rho$

We know that $F_{up}=V\sigma(g+a)\;$

$\Rightarrow F_{up}=\frac m\rho\sigma(g+a)\;$

$\Rightarrow F_{up}=\frac1{800}\ast1000\ast(10+2)=15N$

$F_{net}= F_{up}-T- mg$

$ma= F_{up} -T- mg$

$12= 15-T-110$

$2= 15 - T- 10$

$T= 15 - 10-2=3N$

Q5.A wooden ball of density $900 kg / m^{3}$ is immersed in a liquid of density $1200 kg / m^{3}$ to a depth of 126 cm and the released. The height h above the surface of which the ball rises will be

A wooden ball of density $900 kg / m^{3}$ is immersed in a liquid of density $1200 kg / m^{3}$ to a depth of 126 cm and the released. The height h above the surface of which the ball rises will be

Let $V$ = Volume of ball

Given that, $\rho$ = density of ball= $900\frac{kg}{m^3}$

$\sigma$ = density of liquid =1200 $\frac{kg}{m^3}$

As, the density of ball = $\rho=\frac mV$

$m=V\rho$

We know that $F_{up}=V\sigma g\;$

$F_{net}= F_{up}-mg$

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