Mechanics

A capillary tube of radius r is immersed in water and water rises in it to a height h . The mass of water in the capillary tube is 4g . Another capillary tube of radius 2 r is immersed in water. The mass of water that will rise in this tube is:

Feb 11, 2024 PCM TUTORIALS 0 Comment

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 4g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

(A) 2g

(B) 2.5g

(C) 8g

(D) 16g

Solution

The rise of water in the capillary tube is

$h=\frac{2T\;\cos\theta}{\rho rg}$

$h=\frac{2T\;\cos\theta}{\left({\displaystyle\frac mV}\right)rg}$

$h=\frac{2T\;V\:\cos\theta}{mrg}$

$h=\frac{2T\;\left(\pi\;r^2\;h\right)\;\cos\theta}{mrg}$

$1 =\frac{2T\;\left(\pi\;r\;\right)\;\cos\theta}{m\;g}$

$m=\left(\frac{2\pi T\cos\theta}g\right)\;r$

$m\;\propto\;r$

$\frac{m_1}{m_2}=\frac{r_1}{r_2}$

$\frac{4}{m_2}=\frac{r}{2r}$

$m_2=\;\left(\frac{2r}r\right)\ast\;4$

$\Rightarrow m_2=\;2\ast\;4=8\;$

$\Rightarrow m_2=8g\;$

(C) option is correct

Mechanics

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

Feb 7, 2024 PCM TUTORIALS 0 Comment

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 wooden ball= $900\frac{kg}{m^3}$

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

As, the density of wooden 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 wooden ball attains when it rises up with velocity v from top surafce

Since , the wooden 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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A block of mass 1 kg and density 0.8 [latex]g/cm^3[/latex] 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[latex]m/s^2[/latex]. Find the tension in the string.

Feb 6, 2024 PCM TUTORIALS 0 Comment

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

(ii) Determine the acceleration produced in the of block when the string is cut off.

(A) $3 ms^{-2}$ downwards

(B) $3 ms^{-2}$ upwards

(C) $2 ms^{-2}$ upwards

(D) $5 ms^{-2}$ upwards

When string is cut off then T=0 and the block will move upwards due to upwards force as density of block is lesser than that of water.

We know that, the acceleration = $a^{'}=\frac{F_{net}}m$

$a^{'}=\frac{F_{up}-mg}m=\;\frac{15-1\ast10}1=\frac{15-10}1=5\;ms^{-2}$

(D) is correct option

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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)

Feb 6, 2024 PCM TUTORIALS 0 Comment

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

Mechanics

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 [latex]T_o[/latex] (To) when the system is at rest. What will be the tension in the string if the system has upward acceleration a.

Feb 4, 2024 PCM TUTORIALS 2 Comment

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

Therefore, the tension in a string holding a solid block is

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

Ans. (C) is correct option

A partially immersed solid block of density [latex]\rho_s[/latex] is floating in a liquid of density [latex]\rho_L[/latex] as shown in figure. If beaker container moves up with positive acceleration ‘a’ , what is correct statement about the block?

Feb 3, 2024 PCM TUTORIALS 0 Comment

Top 5 Fluid Mechanics problems for IIT-JEE NEET

Q.A partially immersed solid block of density $\rho_s$ 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 ?

(A) It sinks more

(B) It sinks lesser

(C) It neither sinks more nor lesser

(D) Cannot be predicted as data is insufficient

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

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

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

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

Ans. (C) is correct option

Actually, this question is all about the Archimedes Principle which focus on the concept of Upthrust/Buoyant force acting on completely or partially immersed solid block or any other solid/hollow object. It helps in calculating the loss of weight when any solid substance is partially or completely immersed into the liquid. The liquid container may be stationary or moving with constant velocity/ variable velocity.

When completely or partially immersed solid block lies in stationary beaker container :-

When beaker container is stationary then , according to Archimedes Principle , it loses its weight to equal to the weight of liquid displaced by the object (like completely or partially immersed solid block) . Therefore,

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

When completely or partially immersed solid block lies in beaker container moving up with constant acceleration :-

When beaker container is moving up with uniform acceleration, then the loss of weight of object is now equal to apparent weight of liquid displaced by it.Thus, buoyant force/ upthrust changes so as to balance the increased effective weight of solid block when it observed from frame of reference of accelerated beaker.

$F_{up}^I=F_{pseu}+W$

$F_{up}^I=ma\;+\;mg$

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

$F_{up}^I=mg_{eff}$ where $g_{eff}=g+a$

Therefore, we can write

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

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

Conclusion-

From the above discussion, we understand that the volume immersed does not change when the container accelerates uniformly upwards or downwards along vertical. Immersed part remains same. Block neither sinks more nor lesser. This problems is best combine of Newton’s laws of motion, Archimedes’s Principle and Law of Flotation. This is how we reach to the conclusion that the knowledge of Free Body Diagram and Pseudo force is necessary to get into problems solving process. In fact the basic idea of pseudo force can be summarized into one line that a problem which is solved by application Newton’s 2nd law (from ground frame) gets converted into problem of Equilibrium of rest (from non-inertial frame of reference) by introducing concept of pseudo force.
Pseudo force can be written as:

${\overrightarrow{F\;\;}}_{pseu}=-M_{object}\;\;{\overrightarrow{a\;\;}}_{frame}$
The negative sign shows that the peudo force is taken opposite to direction of acceleration of non-inertial frame.
Law of flotation simply gives us the condition under which an object floats partially or completely immersed inside the liquid. It suggest that
(i) if the density of solid is lesser than density of liquid then it floats partially immersed in the liquid
(ii) if the density of solid is equal to the density of liquid then it floats completely immersed in the liquid
(iii) if the density of solid is more than the density of liquid then it sinks completely unitl it settles down to bottom of container.
It is very interesting to note that this law also helps to find out fraction of immersed volume of object inside the liquid. We have to understand the relationships between different concepts of mechanics to solve such type of fluid mechanics problems.

Mechanics

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

Conculsion -

Tag Fluid Mechanics

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 4g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

(A) 2g

(B) 2.5g

(C) 8g

(D) 16g

Solution

h=\frac{2T\;\cos\theta}{\rho rg}

h=\frac{2T\;\cos\theta}{\left({\displaystyle\frac mV}\right)rg}

h=\frac{2T\;V\:\cos\theta}{mrg}

h=\frac{2T\;\left(\pi\;r^2\;h\right)\;\cos\theta}{mrg}

1 =\frac{2T\;\left(\pi\;r\;\right)\;\cos\theta}{m\;g}

m=\left(\frac{2\pi T\cos\theta}g\right)\;r

m\;\propto\;r

\frac{m_1}{m_2}=\frac{r_1}{r_2}

\frac{4}{m_2}=\frac{r}{2r}

m_2=\;\left(\frac{2r}r\right)\ast\;4

\Rightarrow m_2=\;2\ast\;4=8\;

\Rightarrow m_2=8g\;

(C) option is correct

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 wooden ball= 900\frac{kg}{m^3}

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

As, the density of wooden 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 wooden ball attains when it rises up with velocity v from top surafce

Since , the wooden 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

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

1*2= 15-T-1*10

$h=\frac{2T\;\cos\theta}{\rho rg}$

$h=\frac{2T\;\cos\theta}{\left({\displaystyle\frac mV}\right)rg}$

$h=\frac{2T\;V\:\cos\theta}{mrg}$

$h=\frac{2T\;\left(\pi\;r^2\;h\right)\;\cos\theta}{mrg}$

$1 =\frac{2T\;\left(\pi\;r\;\right)\;\cos\theta}{m\;g}$

$m=\left(\frac{2\pi T\cos\theta}g\right)\;r$

$m\;\propto\;r$

$\frac{m_1}{m_2}=\frac{r_1}{r_2}$

$\frac{4}{m_2}=\frac{r}{2r}$

$m_2=\;\left(\frac{2r}r\right)\ast\;4$

$\Rightarrow m_2=\;2\ast\;4=8\;$

$\Rightarrow m_2=8g\;$

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 wooden ball= $900\frac{kg}{m^3}$

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

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

$m=V\rho$

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

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

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

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

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

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

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

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

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

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

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

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

$v^{'}=0$

$A =-g$

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

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

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

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

$0.84g = 2gh$

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

$h=0.42m=42cm$

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

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

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

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

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

$12= 15-T-110$

$2= 15 - T- 10$

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

(A) $3 ms^{-2}$ downwards

(B) $3 ms^{-2}$ upwards

(C) $2 ms^{-2}$ upwards

(D) $5 ms^{-2}$ upwards

We know that, the acceleration = $a^{'}=\frac{F_{net}}m$

$a^{'}=\frac{F_{up}-mg}m=\;\frac{15-1\ast10}1=\frac{15-10}1=5\;ms^{-2}$

(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)

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

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

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

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

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

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

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