Four smooth steel balls of equal mass at rest are free to move along a straight line without friction. The first ball is given a velocity of $0.4 \mathrm{~m} / \mathrm{s}$. It collides head on with the second elastically, the second one similarly with the third and so on. The velocity of the last ball is
✓
$0.4 \mathrm{~m} / \mathrm{s}$
B
$0.2 \mathrm{~m} / \mathrm{s}$
C
$0.1 \mathrm{~m} / \mathrm{s}$
D
$0.05 \mathrm{~m} / \mathrm{s}$
Answer
Correct option: A.
$0.4 \mathrm{~m} / \mathrm{s}$
In head on elastic collision velocity get interchanged (if masses of particle are equal). i.e. the last ball will move with the velocity of first ball i.e $0.4 \mathrm{~m} /\mathrm{s}$
Consider elastic collision of a particle of mass $m$ moving with a velocity $u$ with another particle of the same mass at rest. After the collision the projectile and the struck particle move in directions making angles $\theta_1$ and $\theta_2$ respectively with the initial direction of motion. The sum of the angles. $\theta_1+\theta_2$, is
A
$45^{\circ}$
✓
$90^{\circ}$
C
$135^{\circ}$
D
$180^{\circ}$
Answer
Correct option: B.
$90^{\circ}$
If the masses are equal and target is at rest and after collision both masses moves in different direction.
Then angle between direction of velocity will be $90^{\circ}$,ifcollision is elastic.
An object of mass $3 \mathrm{~m}$ splits into three equal fragments. Two fragments have velocities $\hat{v j}$ and $\hat{v i}$. The velocity of the third fragment is
A drop of mercury of radius $2 \mathrm{~mm}$ is split into $8$ identical droplets. Find the increase in surface energy. (Surface tension of mercury is $0.465 \mathrm{~J} / \mathrm{m}^2$ )
✓
$23.4 \mu \mathrm{J}$
B
$18.5 \mu \mathrm{J}$
C
$26.8 \mu \mathrm{J}$
D
$16.8 \mu \mathrm{J}$
Answer
Correct option: A.
$23.4 \mu \mathrm{J}$
Increase in surface energy or work done in splitting a big drop
$=4 \pi R^2 T\left(n^{1 / 3}-1\right)$
$ \Rightarrow W=4 \pi \times\left(2 \times 10^{-3}\right)^2 \times 0.465\left(8^{1 / 3}-1\right)=23.4 \mu J$
Water rises in a vertical capillary tube upto a height of $2.0 \mathrm{~cm}$. If the tube is inclined at an angle of $60^{\circ}$ with the vertical, then upto what length the water will rise in the tube
Two bodies of masses $1 \mathrm{~kg}$ and $5 \mathrm{~kg}$ are dropped gently from the top of a tower. At a point $20 \mathrm{~cm}$ from the ground, both the bodies will have the same
A
Momentum
B
Kinetic energy
✓
Velocity
D
Total energy
Answer
Correct option: C.
Velocity
(c) Velocity of fall is independent of the mass of the falling body.
Two identical cylindrical vessels with their bases at same level each contains a liquid of density $\rho$. The height of the liquid in one vessel is $h_1$ and that in the other vessel is $h_2$. The area of either base is$A$. The work done by gravity in equalizing the levels when the two vessels are connected, is
A
$\left(h_1-h_2\right) g \rho$
B
$\left(h_1-h_2\right) g A \rho$
C
$\frac{1}{2}\left(h_1-h_2\right)^2 g A \rho$
✓
$\frac{1}{4}\left(h_1-h_2\right)^2 g A \rho$
Answer
Correct option: D.
$\frac{1}{4}\left(h_1-h_2\right)^2 g A \rho$
$=\frac{1}{4} \rho g A\left(h_1 \sim h_2\right)^2$
A body of mass $10 \mathrm{~kg}$ is dropped to the ground from a height of $10$ metres. The work done by the gravitational force is $\left(g=9.8 \mathrm{~m} / \mathrm{sec}^2\right)$
A
$-490$ Joules
B
$+490$ Joules
C
$-980$ Joules
✓
$+980$ Joules
Answer
Correct option: D.
$+980$ Joules
As the body moves in the direction of force therefore work done by gravitationalforcewill be positive.
$W=F S=m g h=10 \times 9.8 \times 10=980 J$
Two masses $m_A$ and $m_B$ moving with velocities $v_A$ and $v_B$ in opposite directions collide elastically. After that the masses $m_A$ and $m_B$ move with velocity $v_B$ and $v_A$ respectively. The ratio $\left(m_A / m_B\right)$ is
✓
1
B
$\frac{v_A-v_B}{v_A+v_B}$
C
$\left(m_A+m_B\right) / m_A$
D
$v_A / v_B$
Answer
Correct option: A.
1
(a) Since bodies exchange their velocities, hence their masses are equal so that $\frac{m_A}{m_B}=1$
A space craft of mass $M$ is moving with velocity $V$ and suddenly explodes into two pieces. A part of it of mass $m$ becomes at rest, then the velocity of other part will be
✓
$\frac{M V}{M-m}$
B
$\frac{M V}{M+m}$
C
$\frac{m V}{M-m}$
D
$\frac{(M+m) V}{m}$
Answer
Correct option: A.
$\frac{M V}{M-m}$
After explosion $m$ mass comes at rest and let Rest $(M-m)$ mass moves with velocity$v$.By the law of conservation of momentum
$M V=(M-m) v$$\Rightarrow v=\frac{M V}{(M-m)}$
A steel ball of radius $2 \mathrm{~cm}$ is at rest on a frictionless surface. Another ball of radius $4 \mathrm{~cm}$ moving at a velocity of $81 \mathrm{~cm} / \mathrm{sec}$ collides elastically with first ball. After collision the smaller ball moves with speed of
A body of mass $50 \mathrm{~kg}$ is projected vertically upwards with velocity of $100 \mathrm{~m} / \mathrm{sec} .5$ seconds after this body breaks into $20 \mathrm{~kg}$ and $30 \mathrm{~kg}$. If $20 \mathrm{~kg}$ piece travels upwards with $150 \mathrm{~m} / \mathrm{sec}$, then the velocity of other block will be
✓
$15 \mathrm{~m} / \mathrm{sec}$ downwards
B
$15 \mathrm{~m} / \mathrm{sec}$ upwards
C
$51 \mathrm{~m} / \mathrm{sec}$ downwards
D
$51 \mathrm{~m} / \mathrm{sec}$ upwards
Answer
Correct option: A.
$15 \mathrm{~m} / \mathrm{sec}$ downwards
(a) Velocity of $50 \mathrm{~kg}$.mass after $5\mathrm{sec}$of projection $v=ugt=100-9.8\times 5=51 \mathrm{~m} / \mathrm{s}$
At this instant momentum of body is in upward direction $P_{\text{initial }}=50 \times 51=2550 \mathrm{~kg}-\mathrm{m} / \mathrm{s}$
After breaking $20 \mathrm{~kg}$ piece travels upwards with $150 \mathrm{~m} / \mathrm{s}$
let the speed of $30\mathrm{~kg}$ mass is $V$
$P_{\text {final }}=20 \times 150+30 \times V$
By the law of conservation of momentum $P_{\text {initial }}=P_{\text {final }} $
$\Rightarrow2550=20\times150+30\times V \Rightarrow V=-15 \mathrm{~m} / \mathrm{s}$
i.e. it moves in downward direction.
A capillary tube of radius $R$ is immersed in water and water rises in it to a height $H$. Mass of water in the capillary tube is $M$. If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be
A
$M$
✓
$2 M$
C
$M / 2$
D
$4 M$
Answer
Correct option: B.
$2 M$
(b) Mass of liquid in capillary tube$M=\pi R^2 H \times \rho \therefore M \propto R^2 \times\left(\frac{1}{R}\right)(\text { As } H \propto 1 / R)$$\therefore \mathrm{M} \propto R$. If radius becomes double then mass will becomes twice.
A sphere of mass $m$ moving with a constant velocity $u$ hits another stationary sphere of the same mass. If $e$ is the coefficient of restitution, then the ratio of the velocity of two spheres after collision will be
Radius of a capillary is $2 \times 10^{-3} \mathrm{~m}$. A liquid of weight $6.28 \times 10^{-4} N$ may remain in the capillary then the surface tension of liquid will be
A tennis ball is released from height $h$ above ground level. If the ball makes inelastic collision with the ground, to what height will it rise after third collision
✓
$h e^6$
B
$e^2 h$
C
$e^3 h$
D
None of these
Answer
Correct option: A.
$h e^6$
(a) $\quad h_n=h e^{2 n}$ after third collision $h_3=h e^6 \quad$ [as $n=3$ ]
A particle of mass $m$ moving with velocity $V_0$ strikes a simple pendulum of mass $m$ and sticks to it. The maximum height attained by the pendulum will be
A big ball of mass $M$, moving with velocity $u$ strikes a small ball of mass $m$, which is at rest. Finally small ball obtains velocity $u$ and big ball $v$. Then what is the value of $v$
If pressure at half the depth of a lake is equal to $2 / 3$ pressure at the bottom of the lake then what is the depth of the lake
A
$10 \mathrm{~m}$
✓
$20 \mathrm{~m}$
C
$60 \mathrm{~m}$
D
$30 \mathrm{~m}$
Answer
Correct option: B.
$20 \mathrm{~m}$
(b) Pressure at half the depth $=P_0+\frac{h}{2} d g$Pressure at the bottom $=P_0+h d g$According to given condition$\begin{aligned}& P_0+\frac{h}{2} d g=\frac{2}{3}\left(P_0+h d g\right) \\& \Rightarrow 3 P_0+\frac{3 h}{2} d g=2 P_0+2 h d g \\& \Rightarrow h=\frac{2 P_0}{d g}=\frac{2 \times 10^5}{10^3 \times 10}=20 \mathrm{~m}\end{aligned}$
A car of mass $1250 \mathrm{~kg}$ is moving at $30 \mathrm{~m} / \mathrm{s}$. Its engine delivers $30 \mathrm{~kW}$ while resistive force due to surface is $750 N$. What max acceleration can be given in the car
A
$\frac{1}{3} \mathrm{~m} / \mathrm{s}^2$
B
$\frac{1}{4} m / s^2$
✓
$\frac{1}{5} \mathrm{~m} / \mathrm{s}^2$
D
$\frac{1}{6} m / \mathrm{s}^2$
Answer
Correct option: C.
$\frac{1}{5} \mathrm{~m} / \mathrm{s}^2$
Force produced by the engine $F=\frac{P}{v}=\frac{30 \times 10^3}{30}=10 \mathrm{~N}$
$\text { Acceleration }=\frac{\text{Forward force by engine}}{\text{resistiveforce}}{\text { mass of car }}$
$ =\frac{1000-750}{1250}=\frac{250}{1250}=\frac{1}{5}\mathrm{~m}\mathrm{s}^2$
A bomb of $12 \mathrm{~kg}$ divides in two parts whose ratio of masses is $1: 3$. If kinetic energy of smaller part is $216 \mathrm{~J}$, then momentum of bigger part in $\mathrm{kgm} / \mathrm{sec}$ will be
✓
$36$
B
$72$
C
$108$
D
Data is incomplete
Answer
Correct option: A.
$36$
The bomb of mass $12$ kg divides into two masses $m$ and $m$ then $m_1+m_2=12\dots(i)$
and $\frac{m_1}{m_2}=\frac{1}{3}\dots(ii)$
by solving we get $m_1=3 kg$ and $m_2=9 kg$
Kinetic energy of smaller part $=\frac{1}{2} m_1 v_1^2=216 J$
$\therefore v_1^2=\frac{216 \times 2}{3} \Rightarrow v_1=12 m / s$
So its momentum $=m_1 v_1=3 \times 12=36 kg- m / s$
As both parts possess same momentum therefore momentum of each part is $36 kg- m / s$
By inserting a capillary tube upto a depth $/$ in water, the water rises to a height $h$. If the lower end of the capillary is closed inside water and the capillary is taken out and closed end opened, to what height the water will remain in the tube
A
Zero
B
$l+h$
C
$2 h$
✓
$h$
Answer
Correct option: D.
$h$
(d) The water rises to height $h$ due to capilarity.
If the increase in the kinetic energy of a body is $22 \%$, then the increase in the momentum will be
A
$22 \%$
B
$44 \%$
✓
$10 \%$
D
$300 \%$
Answer
Correct option: C.
$10 \%$
$P=\sqrt{2 m E}$. If $m$ is constant then
$\frac{P_2}{P_1}=\sqrt{\frac{E_2}{E_1}}=\sqrt{\frac{1.22 E}{E}} \Rightarrow \frac{P_2}{P_1}=\sqrt{1.22}=1.1$
$\Rightarrow P_2=1.1 P_1 \Rightarrow P_2=P_1+0.1 P_1=P_1+10 \% \text { of } P_1$
So the momentumwill increase by $10 \%$
An air bubble in a water tank rises from the bottom to the top. Which of the following statements are true
Bubble rises upwards because pressure at the bottom is less than that at the top.
Bubble rises upwards because pressure at the bottom is greater than that at the top.
As the bubble rises, its size increases
As the bubble rises, its size decreases
✓
$2$ and $3$
B
$1$ and $4$
C
$1$ and $3$
D
$1$ and $2$
Answer
Correct option: A.
$2$ and $3$
$P_{\text {Bottom }}>P_{\text {Surface }}$. So bubble rises upward. At constant temperature $V \propto \frac{1}{P} ($Boyle's law$)$ Since as the bubble rises upward, pressure decreases, then from above law volume of bubble will increase $i.e.$ its size increases.
