Mechanics

A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 150 J . What are its translational and rotational kinetic energies respectively?

Feb 23, 2024 PCM TUTORIALS 2 Comment

Rolling of a solid disc on horizontal

Rolling , specifically a pure rolling is a classic example of combined translational and rotational motion. While talking about the rolling of a solid disc over horizontal ground, actually we are discussing a simple case of rolling. Pure rolling refers rolling without slipping. Thus, there is no presence of kinetic friction between point of contact and horizontal surface. Kinetic friction keeps acting until pure rolling is achieved. Static friction may or may not be present in the motion, it depends on on factor whether a force is present throughout the motion or not.
For permanent pure rolling(without slipping), we have conditions i.e. $v=\;R\omega$ and $a=\;R\alpha$
These are relations between angular speed and linear speed (angular acceleration and tangential acceleration) defines the situation of rolling without slipping.

Rolling of a solid disc on horizontal - A Case of Pure rolling

Here, we are considering an example which incorporates the concept of rolling of a solid disc on horizontal – a case of Pure rolling and its both kinetic energies- translational and rotational. Also, we can use these conditions (of rolling) to relate both kinetics energies of combined translational and rotational motion.

Example - A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 150 J . Its translational and rotational kinetic energies respectively are

(A) 50 J , 100 J

(B)100 J , 50J

(C) 75 J, 75J

(D) 125J , 25J

Solution

As we know that, for pure rolling(without slipping) $v=\;R\omega$

$\frac{K.E._R}{K.E_T}=\frac{{\displaystyle\frac12}I\;\omega^2}{{\displaystyle\frac12}m\;v^2}$

$\frac{K.E._R}{K.E_T}=\;\frac{\displaystyle\frac12\left(mK^2\right)\;\omega^2}{\displaystyle\frac12m\;v^2}$

$\frac{K.E._R}{K.E_T} =\frac{\displaystyle K^2\;\omega^2}{\displaystyle v^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle{(R\omega)}^2}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle R^2\;\omega^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;}{\displaystyle R^2\;}$ .......(1)

Now, $I=\;\frac12m\;R^2=m\;K^2\; where\; K \;is \;radius\; of\; Gyration$

$\Rightarrow\frac{K^2}{R^2}=\frac12$ .......(2)

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac{K^2}{R^2}=\frac12$ [from (1) and (2)]

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac12$

$\Rightarrow\ K.E_T=\;\left(\frac2{1+2}\right)\;K.E._{total}$

$\Rightarrow\ K.E_T=\;\left(\frac2{3}\right)\;*150$

$\Rightarrow\ K.E_T= 2* 50 = 100J$

Similarly, $K.E_R=\;\left(\frac1{1+2}\right)\;K.E._{total}$

$\Rightarrow K.E_R=\;\left(\frac1{1+2}\right)\;*150$

$\Rightarrow K.E_R=\;\left(\frac1{3}\right)\;*150$

$\Rightarrow K.E_R=1*50=50J$

(B) is correct option

Conclusion: -

From above discussion , we can easily understand that the basic knowledge of moment of inertia, kinetic energies and condition of rolling (without slipping) helps us to solve problems based on such concepts. For more question visit the page Physics-XI Mechanics

Mechanics

10 Important Problems for IIT-JEE and NEET

Feb 13, 2024 PCM TUTORIALS 0 Comment

10 Important Problems for IIT-JEE and NEET

Problems 1 of "10 Important Problems for IIT-JEE and NEET"

Q1.A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 150 J . Its translational and rotational kinetic energies respectively are

(A) 50 J , 100 J

(B)100 J , 50J

(C) 75 J, 75J

(D) 125J , 25J

Solution

As we know that, for pure rolling(without slipping) $v=\;R\omega$

$\frac{K.E._R}{K.E_T}=\frac{{\displaystyle\frac12}I\;\omega^2}{{\displaystyle\frac12}m\;v^2}$

$\frac{K.E._R}{K.E_T}=\;\frac{\displaystyle\frac12\left(mK^2\right)\;\omega^2}{\displaystyle\frac12m\;v^2}$

$\frac{K.E._R}{K.E_T} =\frac{\displaystyle K^2\;\omega^2}{\displaystyle v^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle{(R\omega)}^2}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle R^2\;\omega^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;}{\displaystyle R^2\;}$ .......(1)

Now, $I=\;\frac12m\;R^2=m\;K^2\; where\; K \;is \;radius\; of\; Gyration$

$\Rightarrow\frac{K^2}{R^2}=\frac12$ .......(2)

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac{K^2}{R^2}=\frac12$ [from (1) and (2)]

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac12$

$\Rightarrow\ K.E_T=\;\left(\frac2{1+2}\right)\;K.E._{total}$

$\Rightarrow\ K.E_T=\;\left(\frac2{3}\right)\;*150$

$\Rightarrow\ K.E_T= 2* 50 = 100J$

Similarly, $K.E_R=\;\left(\frac1{1+2}\right)\;K.E._{total}$

$\Rightarrow K.E_R=\;\left(\frac1{1+2}\right)\;*150$

$\Rightarrow K.E_R=\;\left(\frac1{3}\right)\;*150$

$\Rightarrow K.E_R=1*50=50J$

(B) is correct option

Problems 2 of "10 Important Problems for IIT-JEE and NEET"

Q2.For given system of two blocks, find the maximum value of force F which can be applied on the system as shown in figure so that both blocks move together. [Given coefficient of static friction between both blocks = 0.3]

(A) 3N

(B) 6N

(C) 9N

(D) 12N

Solution

Free body diagram of block of upper block is

We can see that the block 1 Kg (upper block) is moved by friction force. If block moves together (without slipping) it means the frictional force is static friction. Therefore,

$f_{s,max}=\;\mu_sN$

$f_{s,max}=\;\mu_s mg$

$f_{s,max}=\;\ 0.3 * 1 * 10 = 3N$

Thus, maximum acceleration that can be produced in upper block without slipping is

$a_{max}=\frac{f_{s,max}}m=\;\frac31=3ms^{-2}$

For system of blocks , to move together the maxmium force can be

$F_{max}=\;M_{sys,net\;}\ast\;\;a_{max}$

$F_{max}=(1+2) * 3= 9N$

(C) is correct option

Problems 3 of "10 Important Problems for IIT-JEE and NEET"

Q3.What is the maximum compression in the spring, if the lower block is shifted to rightward with acceleration a. (Given that all surfaces are smooth as shown).

(A) $\frac{ma}k$

(B) $\frac{2ma}k$

(C) $\frac{mak}2$

(D) $\frac{ma}{2k}$

Solution

Let us assume that the upper block of mass 'm' displaces by distance 'x' towards leftward with velocity 'v' (w.r.t to lower block) when lower block is shifted towards rightward with acceleration a.

We can solve this problem by two method (1) Force Method (2) Energy Method from non-inertial frame of refence (Lower plank/block) .

Force Method

Acc. to Newton's 2nd law of motion,

$F_{net}=F_{pseu}-F_{spring}$

$F_{net}=ma- kx$

$\frac{\operatorname dv}{\operatorname dt}=ma\;-\;kx$

$v\frac{\operatorname dv}{\operatorname dx}=ma\;-\;kx$

$v\;dv=\left(ma\;-\;kx\right)\;dx$

Integrating both sides

$\int_{v_i}^{v_f}\;vdv=\int_{x_i}^{x_f}\left(ma\;-\;kx\right)\;dx$

$\int_{0}^{0}\;vdv=\int_{0}^{x_o}\left(ma\;-\;kx\right)\;dx$

$\left[\frac{v^2}2\right]_0^0=\;ma\;\left[x\right]_0^{x_0}-k\left[\frac{x^2}2\right]_0^{x_o}$

$0-0=\;ma\;\left(x_o-0\right)-\;\frac12k\;\left(x_o^2-0\right)$

$0=\;ma x_o - \;\frac12k\ x_o^2$

$ma x_o = \frac12k\ x_o^2$

$2ma= k x_{o}$

$x_o=\frac{2ma}k$

(B) is correct option

let $a_1$ be the acceleration of upper block w.r.t the lower block

FBD of upper block from lower block reference frame

Boundary conditions for integration:

Initially, the upper block is at rest when x=0, therefore $v_{i}=0$

When the spring has maximum compression $x_{o}$ , again the block becomes stationary w.r.t. lower block.

So, $v_{f}=0$ , $when x=x_{o}$

Alternate Method

We can also solve this problem by Energy Method from non-inertial frame of refence (Lower plank/block) . We will apply Work-Energy theorem from lower block frame of reference.

Energy Method

We can apply Work-Energy theorem for initial and final position of upper block w.r.t. lower block

Since, at initial position the upper block is at rest, therefore $v_{i}=0$

When the spring has maximum compression $x_{o}$ , again the block becomes stationary w.r.t. lower block. So, also $v_{f}=0$

Acc. to Work-Energy Theorem,

$F_{net}=ma- kx$

$W.D_{net \;by\;all\;forces}\;=\triangle K.E.\;$

$W.D._{pseudo}\;\;+\;W.D._{spring}\;=\;\frac12m\;v_f^2-\frac12m\;v_i^2$

$W.D._{pseudo}\;\;+\;W.D._{spring}\;=\;\frac12m\;0^2-\frac12m\;0^2$

$\Rightarrow\;ma\;x_o-\;\frac12k\;x_o^2=0$

$\Rightarrow\;ma\;x_o=\;\frac12k\;x_o^2$

$\Rightarrow\;ma\;=\;\frac12k\;x_o$

$\Rightarrow\;x_o=\frac{2ma}k$

(B) is correct option

For more mechanics problems visit our question bank page

We must know that the work done by conservative forces is

$\;W.D._{cons}\;=-\triangle U$

$\;\Rightarrow\;W.D._{spring}\;=-\triangle U$

$\;\Rightarrow W.D._{spring}\;=-\;\frac12k\;x_{o}^2$

And, $W.D._{pseu}={\overrightarrow{F\;}}_{pseu}.\overrightarrow{x_o}$

$\Rightarrow W.D._{pseu}=F_{pseu}\;x_{o\;}\cos0^o$

$\Rightarrow W.D._{pseu}=ma\;x_{o\;}$

Problems 4 of "10 Important Problems for IIT-JEE and NEET"

Q4. A spaceship is launched into a circular orbit close to the earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth =6400 km , g= 9.8 $ms^{-2}$

(A) $3.2 km/s$

(B) $11.2 km/s$

(C) $1.5 km/s$

(D) $8 km/s$

Solution

As we know that the speed of the spaceship in a circular orbit close to the earth's surface , $v_o=\;\sqrt{gR}$

Ans also, escape speed from earth's surface , $v_e=\;\sqrt{2gR}$

Additional velocity required to make spaceship escape,

$v_{additional}= v_{e}- v_o$ = $\;\sqrt{2gR}-\sqrt{gR}$ = $\left(\sqrt2-1\right)\;\sqrt{gR}$

$v_{additional}=0.414 * 7.9= 3.27 km/s$

(A) is correct option

Problems 5 of "10 Important Problems for IIT-JEE and NEET"

Q5. Moment of inertia of a uniform rod of length L and mass M, about an axis passing through L/3 from one end and perpendicular to its length is

(A) $\frac2{27}M\;L^2$

(B) $\frac1{6}M\;L^2$

(C) $\frac1{12}M\;L^2$

(D) $\frac19M\;L^2$

Solution

According to Parallel axis theorem ,

$I_{parallel\;\;}=\;I_{COM}\;+\;M_{total\;}\;d^2$

Here, d= perpendicular distance of parallel axis from the axis passing through C.O.M.

$\;I_{COM}\;$ = Moment of Inertia of rod passing through C.O.M. (perpendicular to rod)

$\Rightarrow\;I_{parallel}=\frac1{12}M\;L^2+\;M\;\left(\frac L6\right)^2$

$\Rightarrow\;I_{parallel}=\;ML^2\;\left[\frac1{12}+\frac1{36}\right]$

$\Rightarrow\;I_{parallel}=\frac19M\;L^2$

(D) is correct option

We have , $I_{com}=\frac1{12}ML^2$

$And \;total\;mass\;of\;rod,\;M_{total}=M$

$d=\frac L2-\frac L3=\frac L6$

Problems 6 of "10 Important Problems for IIT-JEE and NEET"

Q6.F-x equation of a body of mass 4 kg in S.H.M. is $F+\;16x\;=0$

Here, F is newton and x in meter. The time period of oscillations will be

(A) $6.28\; sec$

(B) $3.14\; sec$

(C) $1.57\; sec$

(D) $1 \; sec$

Solution

Given that, $F+\;16x\;=0$

Also, $m= 4kg$

$F=\;-16x$ .......(1)

On comparing (1) and (2)

$-kx= -16x$

$\Rightarrow\; k= 16$

Time period of S.H.M. is $T=2\pi\sqrt{\frac mk}$

$\Rightarrow T=2\pi\sqrt{\frac4{16}}$

$\Rightarrow T=2\pi\sqrt{\frac14}$

$\Rightarrow T=\frac{2\pi}2=\pi$

$\Rightarrow T=\pi=\;3.14\;sec$

(B) is correct option

We already know, the equation of S.H.M. can be written as

$F = -kx$ ......(2)

Problems 7 of "10 Important Problems for IIT-JEE and NEET"

Q7.A ball of mass m moving at a sped v makes a head on collision with an identical ball at rest. If the kinetic energy of the balls after collision is $\frac34th$ of the original. Then the coefficient of restitution is

(A) $e=\frac1{\sqrt2}$

(B) $e=\frac12$

(C) $e=\frac{\sqrt2}3$

(D) $e=\frac13$

Solution

Given that, mass of each ball $= m$

According to law of conservation of momentum

$m_1 u_1 +m_2 u_2 = m_1 v_1 + m_2 v_2$

$m v +m *0 = m v_1 + m v_2$

$m v = m (v_1 + v_2)$

$v = (v_1 + v_2)$

$v_1 + v_2 = v$ .....(1)

Given that , $K.E._{net,f\;}=\frac34K.E._{net,i\;}$

Therefore, $\frac12\;m_1\;v_1^2\;+\;\frac12\;m_2\;v_2^2\;=\;\frac34\left[\frac12\;m_1\;u_1^2\;+\;\frac12\;m_2\;u_2^2\;\right]$

$\frac12\;m\;\left(\frac{v(1-e)}2\right)^2\;+\;\frac12\;m\;\left(\frac{v(1+e)}2\right)^2\;=\;\frac34\left[\frac12\;m\;v^2\;+\;\frac12\;m\;{\ast\;0}^2\;\right]$

$\frac12\;\;\frac{v^2\;\left(1+e\right)^2}4\;+\;\frac12\;\;\frac{v^2\;\left(1-e\right)^2}4\;=\;\frac34\left(\frac12\;v^2\;\right)$

$\Rightarrow\left(1+e\right)^2+\left(1-e\right)^2=3$

$\Rightarrow1+\;e^2+\;2e\;+\;1\;+\;e^2-2e\;= 3$

$\Rightarrow\;2\;e^2=\;1$

$\Rightarrow\;e^2=\;\frac12$

$\Rightarrow\;e=\;\frac1{\sqrt2}$

(A) is correct option

The formula of coefficient of restitution is :

$e=\;\frac{vel. of sepration}{vel. of approach}$

$e=\;\frac{v_2-v_1}{u_1-u_2}$

$e=\;\frac{v_2-v_1}{v-0}$

$v_2- v_1= ev$ ....(2)

On adding (1) and (2), we get

$2 v_2 = v +ev$

$v_2=\;\frac{v(1+e)}2$

On subtracting (2) from (1), we get

$2 v_1 = v - ev$

$v_1=\;\frac{v(1-e)}2$

Problems 8 of "10 Important Problems for IIT-JEE and NEET"

Q8.The density of a rod AB increases linearly from A to B. Its mid-point is O and centre of mass is at C. Four axes pass through A, B, O and C , all perpendicular to the length of rod. The moments of inertia of the rod about these axes are $I_{A}$ , $I_{B}$ , $I_{O}$ and $I_{C}$ respectively. Then

(A) $I_{A}> I_{B}$

(B) $I_{C} < I_{B}$

(C) $I_{O}> I_{C}$

(D) All of these

Solution

Problems 9 of "10 Important Problems for IIT-JEE and NEET"

Q9.A simple harmonic oscillation has an amplitude A and time period T. The time required to travel from $x=A\;to\;x=\frac A2$ is

(A) $\frac T4$

(B) $\frac T5$

(C) $\frac T6$

(D) $\frac T{12}$

Solution

Problems 10 of "10 Important Problems for IIT-JEE and NEET"

Q10. A geostationary satellite is orbiting the earth at a height of 6R above the surface of earth where R is the radius of earth. The time period of another satellite at a distance of 3.5 R from the centre of earth is

(A) $16.96 hours$

(B) $8.48 hours$

(C) $9.25 hours$

(D) $15.76 hours$

Solution

We know that , orbital speed of satellite is

$v_o=\;\sqrt{\frac{GM}r}\;$

$where\;r\;is\;dis\tan ce\;from\;centre\;of\;planet/earth$

$T\;=\frac{dis\tan ce\;}{speed}=\;\frac{2\pi\;r}{v_o}$

$T\;=\;\frac{2\pi\;r}{\sqrt{\displaystyle\frac{GM}r}}$

$T\;=\;2\pi\;\sqrt{\frac{r^3}{GM}}$

$T\;=\;2\pi\;\sqrt{\frac{r^3}{GM}}\\T^2=\;4\;\pi^2\;\frac{r^3}{GM}$

$\Rightarrow T^2\;\;\propto\;r^3$

This is known as Kepler's Law of period for planetary motion

$r= R+ h$ where R is radius of planet/earth and h is height of satellite from surface of earth

$\Rightarrow r_1 = R+ 6R=7R \;and \;r_2= 3.5R$

$\Rightarrow T\;\propto\;r^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{r_2}{r_1}\right)^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{3.5R}{7R}\right)^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{1}{2}\right)^\frac32$

$\frac{T_2}{T_1}=\frac1{2\sqrt2}$

$T_2 =T_1 * \frac1{2\sqrt2}$

$T_2 =24 * \frac1{2\sqrt2}$

$T_2 =12* \frac1{\sqrt2}$

$T_2 =12* 0.707= 8.48 \;hours$

(B) is correct option

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

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 -

Category Mechanics

Rolling of a solid disc on horizontal

Rolling of a solid disc on horizontal - A Case of Pure rolling

Example - A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 150 J . Its translational and rotational kinetic energies respectively are

(A) 50 J , 100 J

(B)100 J , 50J

(C) 75 J, 75J

(D) 125J , 25J

Solution

As we know that, for pure rolling(without slipping) v=\;R\omega

\frac{K.E._R}{K.E_T}=\frac{{\displaystyle\frac12}I\;\omega^2}{{\displaystyle\frac12}m\;v^2}

\frac{K.E._R}{K.E_T}=\;\frac{\displaystyle\frac12\left(mK^2\right)\;\omega^2}{\displaystyle\frac12m\;v^2}

\frac{K.E._R}{K.E_T} =\frac{\displaystyle K^2\;\omega^2}{\displaystyle v^2}

\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle{(R\omega)}^2}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle R^2\;\omega^2}

\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;}{\displaystyle R^2\;}.......(1)

Now, I=\;\frac12m\;R^2=m\;K^2\; where\; K \;is \;radius\; of\; Gyration

\Rightarrow\frac{K^2}{R^2}=\frac12.......(2)

\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac{K^2}{R^2}=\frac12 [from (1) and (2)]

\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac12

\Rightarrow\ K.E_T=\;\left(\frac2{1+2}\right)\;K.E._{total}

\Rightarrow\ K.E_T=\;\left(\frac2{3}\right)\;*150

\Rightarrow\ K.E_T= 2* 50 = 100J

Similarly, K.E_R=\;\left(\frac1{1+2}\right)\;K.E._{total}

\Rightarrow K.E_R=\;\left(\frac1{1+2}\right)\;*150

\Rightarrow K.E_R=\;\left(\frac1{3}\right)\;*150

\Rightarrow K.E_R=1*50=50J

(B) is correct option

Conclusion: -

10 Important Problems for IIT-JEE and NEET

Problems 1 of "10 Important Problems for IIT-JEE and NEET"

Q1.A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 150 J . Its translational and rotational kinetic energies respectively are

(A) 50 J , 100 J

(B)100 J , 50J

(C) 75 J, 75J

(D) 125J , 25J

Solution

As we know that, for pure rolling(without slipping) v=\;R\omega

\frac{K.E._R}{K.E_T}=\frac{{\displaystyle\frac12}I\;\omega^2}{{\displaystyle\frac12}m\;v^2}

\frac{K.E._R}{K.E_T}=\;\frac{\displaystyle\frac12\left(mK^2\right)\;\omega^2}{\displaystyle\frac12m\;v^2}

\frac{K.E._R}{K.E_T} =\frac{\displaystyle K^2\;\omega^2}{\displaystyle v^2}

\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle{(R\omega)}^2}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle R^2\;\omega^2}

\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;}{\displaystyle R^2\;}.......(1)

Now, I=\;\frac12m\;R^2=m\;K^2\; where\; K \;is \;radius\; of\; Gyration

\Rightarrow\frac{K^2}{R^2}=\frac12.......(2)

\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac{K^2}{R^2}=\frac12 [from (1) and (2)]

\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac12

\Rightarrow\ K.E_T=\;\left(\frac2{1+2}\right)\;K.E._{total}

\Rightarrow\ K.E_T=\;\left(\frac2{3}\right)\;*150

\Rightarrow\ K.E_T= 2* 50 = 100J

Similarly, K.E_R=\;\left(\frac1{1+2}\right)\;K.E._{total}

\Rightarrow K.E_R=\;\left(\frac1{1+2}\right)\;*150

\Rightarrow K.E_R=\;\left(\frac1{3}\right)\;*150

\Rightarrow K.E_R=1*50=50J

(B) is correct option

Problems 2 of "10 Important Problems for IIT-JEE and NEET"

Q2.For given system of two blocks, find the maximum value of force F which can be applied on the system as shown in figure so that both blocks move together. [Given coefficient of static friction between both blocks = 0.3]

(A) 3N

(B) 6N

(C) 9N

(D) 12N

Solution

Free body diagram of block of upper block is

We can see that the block 1 Kg (upper block) is moved by friction force. If block moves together (without slipping) it means the frictional force is static friction. Therefore,

f_{s,max}=\;\mu_sN

f_{s,max}=\;\mu_s mg

f_{s,max}=\;\ 0.3 * 1 * 10 = 3N

f_{s,max}=\;\ 0.3 * 1 * 10 = 3N

Thus, maximum acceleration that can be produced in upper block without slipping is

a_{max}=\frac{f_{s,max}}m=\;\frac31=3ms^{-2}

For system of blocks , to move together the maxmium force can be

F_{max}=\;M_{sys,net\;}\ast\;\;a_{max}

F_{max}=(1+2) * 3= 9N

(C) is correct option

Problems 3 of "10 Important Problems for IIT-JEE and NEET"

Q3.What is the maximum compression in the spring, if the lower block is shifted to rightward with acceleration a. (Given that all surfaces are smooth as shown).

(A) \frac{ma}k

(B) \frac{2ma}k

(C) \frac{mak}2

(D) \frac{ma}{2k}

Solution

As we know that, for pure rolling(without slipping) $v=\;R\omega$

$\frac{K.E._R}{K.E_T}=\frac{{\displaystyle\frac12}I\;\omega^2}{{\displaystyle\frac12}m\;v^2}$

$\frac{K.E._R}{K.E_T}=\;\frac{\displaystyle\frac12\left(mK^2\right)\;\omega^2}{\displaystyle\frac12m\;v^2}$

$\frac{K.E._R}{K.E_T} =\frac{\displaystyle K^2\;\omega^2}{\displaystyle v^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle{(R\omega)}^2}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle R^2\;\omega^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;}{\displaystyle R^2\;}$ .......(1)

Now, $I=\;\frac12m\;R^2=m\;K^2\; where\; K \;is \;radius\; of\; Gyration$

$\Rightarrow\frac{K^2}{R^2}=\frac12$ .......(2)

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac{K^2}{R^2}=\frac12$ [from (1) and (2)]

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac12$

$\Rightarrow\ K.E_T=\;\left(\frac2{1+2}\right)\;K.E._{total}$

$\Rightarrow\ K.E_T=\;\left(\frac2{3}\right)\;*150$

$\Rightarrow\ K.E_T= 2* 50 = 100J$

Similarly, $K.E_R=\;\left(\frac1{1+2}\right)\;K.E._{total}$

$\Rightarrow K.E_R=\;\left(\frac1{1+2}\right)\;*150$

$\Rightarrow K.E_R=\;\left(\frac1{3}\right)\;*150$

$\Rightarrow K.E_R=1*50=50J$

As we know that, for pure rolling(without slipping) $v=\;R\omega$

$\frac{K.E._R}{K.E_T}=\frac{{\displaystyle\frac12}I\;\omega^2}{{\displaystyle\frac12}m\;v^2}$

$\frac{K.E._R}{K.E_T}=\;\frac{\displaystyle\frac12\left(mK^2\right)\;\omega^2}{\displaystyle\frac12m\;v^2}$

$\frac{K.E._R}{K.E_T} =\frac{\displaystyle K^2\;\omega^2}{\displaystyle v^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle{(R\omega)}^2}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle R^2\;\omega^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;}{\displaystyle R^2\;}$ .......(1)

Now, $I=\;\frac12m\;R^2=m\;K^2\; where\; K \;is \;radius\; of\; Gyration$

$\Rightarrow\frac{K^2}{R^2}=\frac12$ .......(2)

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac{K^2}{R^2}=\frac12$ [from (1) and (2)]

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac12$

$\Rightarrow\ K.E_T=\;\left(\frac2{1+2}\right)\;K.E._{total}$

$\Rightarrow\ K.E_T=\;\left(\frac2{3}\right)\;*150$

$\Rightarrow\ K.E_T= 2* 50 = 100J$

Similarly, $K.E_R=\;\left(\frac1{1+2}\right)\;K.E._{total}$

$\Rightarrow K.E_R=\;\left(\frac1{1+2}\right)\;*150$

$\Rightarrow K.E_R=\;\left(\frac1{3}\right)\;*150$

$\Rightarrow K.E_R=1*50=50J$

$f_{s,max}=\;\mu_sN$

$f_{s,max}=\;\mu_s mg$

$f_{s,max}=\;\ 0.3 * 1 * 10 = 3N$

$f_{s,max}=\;\ 0.3 * 1 * 10 = 3N$

$a_{max}=\frac{f_{s,max}}m=\;\frac31=3ms^{-2}$

$F_{max}=\;M_{sys,net\;}\ast\;\;a_{max}$

$F_{max}=(1+2) * 3= 9N$

(A) $\frac{ma}k$

(B) $\frac{2ma}k$

(C) $\frac{mak}2$

(D) $\frac{ma}{2k}$

$F_{net}=F_{pseu}-F_{spring}$

$F_{net}=ma- kx$

$\frac{\operatorname dv}{\operatorname dt}=ma\;-\;kx$

$v\frac{\operatorname dv}{\operatorname dx}=ma\;-\;kx$

$v\;dv=\left(ma\;-\;kx\right)\;dx$

$\int_{v_i}^{v_f}\;vdv=\int_{x_i}^{x_f}\left(ma\;-\;kx\right)\;dx$

$\int_{0}^{0}\;vdv=\int_{0}^{x_o}\left(ma\;-\;kx\right)\;dx$

$\left[\frac{v^2}2\right]_0^0=\;ma\;\left[x\right]_0^{x_0}-k\left[\frac{x^2}2\right]_0^{x_o}$

$0-0=\;ma\;\left(x_o-0\right)-\;\frac12k\;\left(x_o^2-0\right)$

$0=\;ma x_o - \;\frac12k\ x_o^2$

$ma x_o = \frac12k\ x_o^2$

$2ma= k x_{o}$

$x_o=\frac{2ma}k$

let $a_1$ be the acceleration of upper block w.r.t the lower block

Initially, the upper block is at rest when x=0, therefore $v_{i}=0$

When the spring has maximum compression $x_{o}$ , again the block becomes stationary w.r.t. lower block.

So, $v_{f}=0$ , $when x=x_{o}$

Since, at initial position the upper block is at rest, therefore $v_{i}=0$

When the spring has maximum compression $x_{o}$ , again the block becomes stationary w.r.t. lower block. So, also $v_{f}=0$

$F_{net}=ma- kx$

$W.D_{net \;by\;all\;forces}\;=\triangle K.E.\;$

$W.D._{pseudo}\;\;+\;W.D._{spring}\;=\;\frac12m\;v_f^2-\frac12m\;v_i^2$

$W.D._{pseudo}\;\;+\;W.D._{spring}\;=\;\frac12m\;v_f^2-\frac12m\;v_i^2$

$W.D._{pseudo}\;\;+\;W.D._{spring}\;=\;\frac12m\;0^2-\frac12m\;0^2$

$\Rightarrow\;ma\;x_o-\;\frac12k\;x_o^2=0$

$\Rightarrow\;ma\;x_o=\;\frac12k\;x_o^2$

$\Rightarrow\;ma\;=\;\frac12k\;x_o$

$\Rightarrow\;x_o=\frac{2ma}k$

$\;W.D._{cons}\;=-\triangle U$

$\;\Rightarrow\;W.D._{spring}\;=-\triangle U$

$\;\Rightarrow W.D._{spring}\;=-\;\frac12k\;x_{o}^2$

And, $W.D._{pseu}={\overrightarrow{F\;}}_{pseu}.\overrightarrow{x_o}$

$\Rightarrow W.D._{pseu}=F_{pseu}\;x_{o\;}\cos0^o$

$\Rightarrow W.D._{pseu}=ma\;x_{o\;}$

Q4. A spaceship is launched into a circular orbit close to the earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth =6400 km , g= 9.8 $ms^{-2}$