MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 5

A stone of mass

$1\;kg$ is tied to the end of a string of

$1\;m$ long. It is whirled in a vertical circle. If the velocity of the stone at the top be

$4\;m/s$ . What is the tension in the string?

0%
$\;6\;N$
0%
$\;16\;N$
0%
$\;5\;N$
0%
$\;10\;N$

Explanation

Given :

$v =4$

$m/s$

$m =1 kg$

$r = 1$ m

Using circular motion equation :

$\dfrac{mv^2}{r} =T + mg$

$\implies$

$T =\dfrac{mv^2}{r} -mg$

$\therefore$

$T =\dfrac{1 \times 4^2}{1} -1 \times 10$

$\implies T = 6$

$N$

Figures

$I, II, III\ and\ IV$ depicts variation of force with time. In which situation impulse will be maximum?

0%
$I \&II$
0%
$III \& I$
0%
$III \& IV$
0%
$Only\ IV$

Explanation

$I$ . Area is

$0.25$

$II$ . Area is

$0.15$

$III$ . Area is

$0.5$

$IV$ . Area is

$0.5$
Since area under the curve (

$\Delta P$ is impulse ),

$III$ and

$IV$ have maximum impulse.

A particle of mass

$'m'$ describes a circle of radius

$(r)$ . The centripetal acceleration of the particle is

$\displaystyle\frac{4}{r^2}$ . The momentum of the particle:

0%
$\;\displaystyle\frac{2m}{r}$
0%
$\;\displaystyle\frac{2m}{\sqrt{r}}$
0%
$\;\displaystyle\frac{4m}{r}$
0%
$\;\displaystyle\frac{4m}{\sqrt{r}}$

Explanation

Centripetal acceleration of the particle,

$a = \dfrac{v^2}{r}$

$\therefore$

$\dfrac{4}{r^2}= \dfrac{v^2}{r}$

$\implies v = \dfrac{2}{\sqrt{r}}$

Thus the momentum of the particle,

$P = mv = \dfrac{2m}{\sqrt{r}}$

A mass tied to a string moves in a vertical circle with a uniform speed of

$5\;m/s$ as shown. At the point

$P$ the string breaks. The mass will reach a height above

$P$ of nearly:

$(g=10\ m/s^{2})$

0%
$\;1\;m$
0%
$\;0.5\;m$
0%
$\;1.75\;m$
0%
$\;1.25\;m$

Explanation

Given :

$u =5 m/s$

$a = - g = -10 m/s^2$

When the string breaks at point P, from there onwards the mass has the free falling motion.

At the maximum height attained, the velocity of the mass is zero i.e

$v = 0 m/s$

Using

$v^2 - u^2 = 2aS$

$\therefore$

$0 - 5^2 = 2 (-10) h$

$\implies h = 1.25$ m

In a vertical circle of radius

$(r)$ , at what point in its path a particle may have tension equal to zero:

0%
highest point
0%
lowest point
0%
at any point
0%
at a point horizontal from the centre of radius

Explanation

Let the tension in the string at the highest point be

$T$

Minimum speed required by the particle at the highest point to complete the vertical circular motion is

$\sqrt{gr}$

$\therefore$

$\dfrac{mv^2}{r} =T + mg$

$\dfrac{m (gr)}{r} = T+mg$

$\implies T =0$

Thus tension can be zero at the highest point.

A stone attached to one end of a string is whirled in a vertical circle. The tension in the string is maximum when

0%
the string is horizontal
0%
the string is vertical with the stone at highest position
0%
the string is vertical with the stone at the lowest position
0%
the string makes an angle of $45^{\circ}$ with the vertical

Explanation

$Answer:-$ C

When the stone is at the bottom position forces acting on stone are:

Upward force tension (T), Downward forces are weight (mg) and also

$\dfrac{mv^2}{R}$ centrifugal force

So balancing forces we get :

$T=mg+\dfrac{mv^2}{R}$

And minimum tension is obtained at highest point which will be:

$T=mg-$

$\dfrac{mv^2}{R}$

A particle is moving in a vertical circle, the tension in the string when passing through two position at angle

$30^{\circ}$ and

$60^{\circ}$ from vertical from lowest position are

$T_1$ and

$T_2$ respectively, then

0%
$T_1 = T_2$
0%
$T_1 > T_2$
0%
$T_1 < T_2$
0%
$T_1\ge T_2$

Explanation

In equilibrium,

$T=mv^2/r+mg\cos\theta$
Since,

$\theta$ increases

$\cos\theta$ and

$v$ both will decrease.
Hence, for

$\theta=60^o$ , the tension will be less.
Thus,

$T_1>T_2$ .

A pendulum bob has a speed

$3\;m/s$ while passing through its lowest position, length of the pendulum is

$0.5\;m$ then its speed when it makes an angle of

$60^{\circ}$ with the vertical is

0%
$\;2\;m/s$
0%
$\;1\;m/s$
0%
$\;4\;m/s$
0%
$\;3\;m/s$

Explanation

$\textbf{Step 1: Apply energy conservation}$

$\Delta ABD$ ,

$AB = AD \cos \theta$

$\Rightarrow AB = l \cos \dfrac{\pi}{3} = \dfrac{l}{2}$

$\Rightarrow h = l - \dfrac{l}{2} = \dfrac{l}{2}$

As there are no dissipative force energy will be conserved.

$\Rightarrow (E)_i = E_f$

$\Rightarrow K_1 + U_1 = K_2 + U_2$

Taking point

$C$ is zero potential level

$\Rightarrow \dfrac{1}{2} mu^2 = \dfrac{1}{2} mv^2 + (mgh)$

$\textbf{Step 2: Calculations}$

$\Rightarrow v = \sqrt{u^2 - 2gh}$

Putting given values

$v = \sqrt{(3)^2(m/s)^2 - 2 \times 9.8(m/s^2) \times \dfrac{0.5}{2}(m)} = 2 m/s$

Hence

$A$ is the correct option.

2115111_280800_ans_d10d9bce5ded4e0a9afbe5fe1264a62f.png

In the given figure a block of weight

$10 N$ is resting on a horizontal surface. The coefficient of static friction between the block and the surface is

$\mu_s = 0.4$ . A force of

$3.5 N$ will keep the block in uniform motion, once it has been set in motion. A horizontal force of

$3 N$ is applied to the block, then the block will

0%
move over the surface with constant velocity.
0%
move having accelerated motion over the surface.
0%
will not move.
0%
first it will move with a constant velocity for some time and then it will have an accelerated motion.

Explanation

$f_s^{max}=\mu N = 0.4\times 10N = 4N$
The applied force is less than max

$f_s^{max}$ , so the block would not move.

A bob hangs from a rigid support by an inextensible string of length

$\ell$ . If it is displaced through a distance

$\ell$ (from the lowest position) keeping the string straight, & then released. The speed of the bob at the lowest position is:

0%
$\sqrt{g\ell}$
0%
$\sqrt{3g\ell}$
0%
$\sqrt{2g\ell}$
0%
$\sqrt{5g\ell}$

Explanation

Distance=

$l$
change in height=

$lcos60 = l/2$

From energy conservation,
PE at max height= KE at lowest point=

$mgl/2= mv^2/2$

$v=\sqrt{gl}$

A metal block is resting on a rough wooden surface. A horizontal force applied to the block is increased uniformly. Which of the following curves correctly represents velocity of the block ?

Explanation

The force on the block starts increasing uniformly which states that the acceleration is increasing uniformly. The block starts moving after some time when the force is greater than the friction acting into play. The

$velocity$

$-$

$time$ graph for a uniformly accelerated body is a curve and for this case it is the curve starting after time

$t$

A ball of mass 50 gm is dropped from a height h = 10 m. It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for At =0.01 sec, the impulse of the impact force is
(take 9 = 10

$m/s^2$ )

0%
1.3 N - s
0%
1.06 N - s
0%
1300 N - s
0%
105 N - s

Explanation

Given :

$m = 0.05$ kg

$h = 10$ m

Kinetic energy of ball just before collision

$E_i = mgh = 0.05\times 10\times 10 = 5$ J

Thus initial momentum of the ball

$P_i = \sqrt{2mE_i} = \sqrt{2(0.05)(5)} =0.707$ N-s (downwards)

Kinetic energy of ball just after collision

$E_f = \dfrac{E_i}{4} =\dfrac{5}{4} = 1.25$ J

Thus final momentum of the ball

$P_f = \sqrt{2mE_f} = \sqrt{2(0.05)(1.25)} =0.353$ N-s (upwards)

$\therefore$ Impulse

$I = \Delta P = P_f- (-P_i) = 0.353 + 0.707 = 1.06$ N-s

A block of mass

$2 kg$ is placed on the floor. The coefficient of static friction is

$0.4$ . Force of

$2.8N$ is applied on the block. The force of friction between the block and the floor is (Taken

$g = 10m/s^2$ )

0%
2.8 N
0%
8 N
0%
2.0 N
0%
zero

Explanation

The maximum static frictional force is

$\mu mg=0.4\times2\times10=8N$ , which is more than the applied force.

So, when a force of

$2.8N$ is applied, static frictional force is also

$2.8$ N, as it is a self-adjusting force.

A particle slides from rest from the topmost point of a vertical circle of radius (

$r$ ) along a smooth chord making an angle

$(\theta)$ with the vertical the time of decent is

0%
Least for $\theta=0^{\circ}$
0%
Maximum for $\theta=0^{\circ}$
0%
Least for $\theta=45^{\circ}$
0%
Independent of $\theta$

Explanation

Force on particle along the cord = $mg\cos { \theta }$

Distance travelled by the particle = $\quad \dfrac { d }{ \cos { \theta } }$ , where $d$ is the diameter or the vertical circle.

$s = \dfrac { a{ t }^{ 2 } }{ 2 }$

$\Rightarrow\ t =\sqrt { \dfrac { 2s }{ a } }$ .

Hence, $\ t$ is independent of $\theta$ .

A body of mass M is kept on a rough horizontal surface (friction coefficient =

$\mu$ ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on A is F where

0%
$F=Mg$
0%
$F=\mu Mg$
0%
$Mg\leq F\leq Mg\sqrt {1+\mu^2}$
0%
$Mg\geq F\geq Mg\sqrt {1-\mu^2}$

Explanation

Force applied by any surface is resultant of friction force and Normal force.
0

$\le$ Friction force (f)

$\le$

$\mu \times N$
where, N is Normal force applied by surface.
Normal force

$=$ Mg (perpendicular to surface)

$\therefore$ Resultant force (F)

$=$

$\sqrt { { \left( N \right) }^{ 2 }+{ f }^{ 2 } }$

${ F }_{ min }$

$=$ N

$=$ Mg when f

$=$ 0

${ F }_{ max }$

$=$ =

$\sqrt { { N }^{ 2 }+{ f }^{ 2 } }$ when f =

$\mu$ mg

$=$

$\sqrt { { \left( Mg \right) }^{ 2 }+\left( \mu Mg \right) ^{ 2 } }$

$=$

$Mg\sqrt { 1+{ \mu }^{ 2 } }$

$\therefore$

$Mg\quad \leq F\quad \leq Mg\sqrt { 1+{ \mu }^{ 2 } }$

A rod of weight

$w$ is supported by two parallel knife edges

$A$ &

$B$ and is in equilibrium in a horizontal position. The knives are at a distance

$d$ from each other. The center of mass of the rod is at a distance

$x$ from

$A.$ The normal reactions at

$A$ and

$B$ will be

0%
$N_A \, =\, 2w (1\, -\, x/d),\, N_B\, =\, wx/d$
0%
$N_A\, =\, w\, (1\, -\, x/d),\, N_B\, =\, wx/d$
0%
$N_A\, =\, 2w(1\, -\, x/d), N_B\, =\, 2wx/d$
0%
$N_A\, =\, w(2\, -\, x/d), N_B\, =\, wx/d$

Explanation

Rod is in the equilibrium,

$\sum F_Y$ =0

$\Rightarrow N_A + N_B = w$ ......1

Taking moments about point A,

Moments clockwise = Moments Anticlockwise

$w\times x = N_B \times d$

$\therefore N_B= \dfrac{wx}{d}$ .......2

Solving equations 1 and 2,

$N_A= w(1-\dfrac{x}{d})$ ,

$N_B= \dfrac{wx}{d}$

A hammer weighing 3 kg, moving with a velocity of 10 m/s, strikes against the head of a spike and drives it into a block of wood. If the hammer comes to rest in 0.025 s, the impulse associated with the spike will be :

0%
30 Ns
0%
-30 Ns
0%
15 Ns
0%
-15 Ns

Explanation

Initial momentum of the hammer

$P_i = mu = 3\times 10 = 30 \ kg \ m/s$

Final momentum of the hammer

$P_f = 0$

Change in momentum of hammer

$\Delta P = P_f - P_i = 0-30 = -30 \ kg \ m/s$

Impulse acting on the hammer

$I_{hammer} = \Delta P = -30 \ Ns$

Impulse acting on the spike

$I_{spike} = -I_{hammer} = 30 \ N \ s$

A bead is free to slide down a smooth wire tightly stretched between points

$A$ and

$B$ on a verticle circle of radius

$R$ . If the bead starts from rest at '

$A$ ', the highest point on the circle, its velocity when it arrives at

$B$ is

0%
$2\sqrt{gR}$
0%
$2\sqrt{gR} \cos{\theta}$
0%
$\displaystyle \dfrac{2\sqrt{R}}{g}$
0%
None of the above

Explanation

$a= g\cos{\theta}\\AB = 2R\cos{\theta}\Rightarrow v^2 =u^2 + 2as \\v^2 = 0 + 2 (g\cos{\theta}) 2R\cos{\theta}\Rightarrow v^2 = 4gR \cos ^{2 }{\theta}\\ \Rightarrow v = 2\sqrt{ gR} \cos{\theta}$

Physical independence of force is a consequence of _____.

0%
first law of motion
0%
second law of motion
0%
third law of motion
0%
all ofthese laws

The change in momentum of a vehicle weighing 2000 kg when its speed increases from 36 km/h to 72 km/h uniformly is :

0%
15000 kg.m/s
0%
4000 kg.m/s
0%
20000 kg.m/s
0%
7200 kg.m/s

Explanation

Given :

$m = 2000$ kg

$u = 36$ km/h

$= 36 \times \dfrac{5}{18} = 10$ m/s

$v= 72$ km/h

$= 72 \times \dfrac{5}{18} = 20$ m/s

Change in momentum

$\Delta P = m(v-u) = 2000\times (20-10) =20000$ kg.m/s

A stationary ball weighing 0.25 kg acquire a speed of

$10\:m/s$ when hit by a hockey stick. The impulse imparted to the ball is:

0%
$0.25\:N\times s$
0%
$2.5\:N\times s$
0%
$2\:N\times s$
0%
$0.5\:N\times s$

Explanation

Given :

$m = 0.25$ kg

Initial velocity of the ball

$u = 0$ m/s

Final velocity of the ball

$v = 10$ m/s

Impulse imparted

$I = \Delta P = mv-mu =0.25\times 10 - 0 = 2.5$

$N\times s$

A body of mass 'm' and velocity 'u' strikes a wall and rebounds with velocity 'v'. The change in momentum is:

0%
$m(v-u)$
0%
$m(u-v)$
0%
$-m(v+u)$
0%
none of these

Explanation

Given :

$\vec{u} = u\hat{x}$

$\vec{v} = -v\hat{x}$

Thus change in momentum

$\Delta \vec{P} = m(\vec{v} - \vec{u})$

$\therefore$

$\Delta \vec{P} = m(-v - u) = -m(u+v)$

$\hat{x}$

A stationary ball weighing 0.25 kg acquires a speed of 10 m/s when hit by a hockey stick. The impulse imparted to the ball is :

0%
2.5 Ns
0%
2.0Ns
0%
1.5 Ns
0%
0.5Ns

Explanation

impulse is change in momentum so initial momentum is zero and final momentum is

$2.5 Ns$ .

so impulse is

$2.5Ns$ which is change in momentum .

so the answer is A.

A ball weighing

$10 g$ hits a hard surface vertically with a speed of

$5 {m}/{sec}$ and rebounds with the same speed. The ball remains in contact with the surface for

${1}/{100} sec$ . The average force exerted by the surface on the ball is

0%
$100 N$
0%
$10 N$
0%
$1 N$
0%
$0.1 N$

Explanation

Given,

$m=10g=0.01kg$

$v=5m/s$

$t=\dfrac{1}{100} sec$

Average force,

$F_{avg}=\dfrac{2mv}{t}$

$F_{}=\dfrac{2\times 0.01\times 5}{1/100}=10N$

The correct option is B.

A boat at rest has two persons of unequal weights seated at either end. If they walk across the boat and interchange their positions,

0%
The boat will move in the direction of walking of the lighter person
0%
The boat will move in the direction of walking of the heavier person
0%
The boat will not move at all
0%
The boat will oscillate about a mean position and stop once the persons have moved to the new positions
0%
None of these

Two forces act on either side of rigid body of negligible mass suspended by string as shown in the figure. If R is the resultant force the tension of the string T = _____ gwt.

0%
$26$
0%
$41$
0%
$82$
0%
$16$

Explanation

Since the rod is in equilibrium, so net force acting on the rod is zero. Thus tension in the string balances the

$32 \ gwt$ and

$50 \ gwt$ .

$\therefore$

$T = 32+50 = 82 \ gwt$

Every action has reaction. These action-reaction are

0%
Equal, opposite and acting on the same body
0%
Equal, in the same direction and acting on same body
0%
Equal, opposite and acting on different bodies
0%
Equal, in same direction and acting on different bodies.

The effect of an impulse force on the body is measured only in terms of its

0%
Force
0%
Moment of momentum
0%
Impulse
0%
None of these

Explanation

The relation between the impulsive force and impulse produced by it is

$F\times t =$ Impulse

The effect of an impulse force of the body is measured in terms of its impulse.

Friction is a/an

0%
self-adjusting force
0%
necessary evil
0%
important force in daily life
0%
all the above

A rubber of mass

$200$ g hits a wall normally with a velocity of

$\displaystyle { 10\ ms }^{ -1 }$ and bounces back with a velocity of

$\displaystyle { 8\ ms }^{ -1 }$ . The impulse is _____ Ns.

0%
$-0.5$
0%
$+0.5$
0%
$-3.6$
0%
$+3.6$

Explanation

Initial momentum

$P_i = mu = 0.2\times (10) = 2 \ kg \ m/s$

Final impulse

$P_f = mv = 0.2\times (-8) = -1.6\ kg \ m/s$

Impulse

$I = P_f - P_i$

$\implies \ I = -1.6-2 = -3.6 \ N \ s$

If the momentum of a body is doubled, the kinetic energy is

0%
Halved
0%
Unchanged
0%
Doubled
0%
Becomes 4 times

Explanation

The correct answer is option (D).

Hint: Relate the formula of momentum and kinetic energy of any body.

Step 1: Find the relation between momentum and kinetic energy.

By formula,

$p=mv$ (where,

$p$ is the momentum of the body)

$v=\dfrac{p}{m}$

Also,

$K.E.=\dfrac{1}{2}mv^2$

Substituting the value of

$v$ in the above equation, we get,

$K.E.=\dfrac{1}{2}m{(\dfrac{p}{m}})^2$

$K.E.=\dfrac{1}{2}m\dfrac{p^2}{m^2}$

$K.E.=\dfrac{p^2}{2m}$

Hence,

$K.E.\propto p^2$

Step 2: Calculate the final kinetic energy.

From the above relation, we can write,

$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{p_2^2}$

According to the question,

$p_2=2p_1$

$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{(2p_2)^2}$

$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{4p_1^2}$

$K.E._2=4K.E._1$

Hence, the kinetic energy becomes four times.

The correct answer is option (D).

A disc of mass 10 g is kept floating horizontally by throwing 10 marbles per second against it from below. The marbles strike the disc normally and rebound downwards with the same speed. If the mass of each marble is 5 g, the velocity with which the marbles are striking the disc is

$\displaystyle \left( g=9.8{ m }/{ { s }^{ 2 } } \right)$

0%
$\displaystyle 0.98m/s$
0%
$\displaystyle 9.8m/s$
0%
$\displaystyle 1.96m/s$
0%
$\displaystyle 19.6m/s$

Explanation

Given that:

Mass of disc,

$M=10g=0.010kg$

Mass of each marble,

$m=5g=0.005kg$

Suppose the number of marbles striking the disc per second=n and velocity with which marbles strike disc is v.

Then Weight of disc acting downward

$=Mg$

Change in velocity of marbles when they rebound

$=v-(-v)=2v$

Therefore the change in momentum of each marble when it strikes the disc=

$m * 2v=2mv$

and total momentum imparted per second to the disc

$=2mnv$ = The force exerted by the marbles in the upward direction

The disc will remain at rest if the net force acting on it is zero provided it is initially at rest.

Therefore for the disc to remain at rest,

Weight of disc=Upward force on it

$= Mg=2mnv$

$v=\dfrac{Mg}{2mn}=\dfrac{(0.010 *9.8)}{(2*0.005*10)}$

$=0.98 m/s$

A ball is moving in a circular path of radius

$5m$ . If tangential acceleration at any instant is

$10\ m/s^2$ and the net acceleration makes an angle

$30^o$ with the centripetal acceleration, then the instantaneous speed is.

0%
$50\sqrt{3}m/s$
0%
$9.3\ m/s$
0%
$6.6\ m/s$
0%
$5.4\ m/s$

Explanation

$\tan\theta=\displaystyle\frac{a_t}{a_c}=\frac{a_t}{v^2/r}$

$\Rightarrow\displaystyle v^2=\frac{a_t r}{\tan\theta}$

$=\displaystyle\frac{10\times 5}{\tan 30^o}=\frac{10\times 5}{1/\sqrt{3}}$

$v^2=50\times \sqrt 3$

$=50\times 1.732=86.6$

$v=\sqrt{86.6}=9.3m/s$

An aircraft executes a horizontal loop with a speed of

$150$ m/s with its wings banked at an angle of

$\displaystyle { 12 }^{ o }$ . The radius of the loop is

$\displaystyle \left( g=10{ m }/{ { s }^{ 2 } } \right)$

0%
10.6 km
0%
9.6 km
0%
7.4 km
0%
5.8 km

Explanation

Using the relation for the radius (r) of loop.

$\displaystyle \tan { \theta } =\frac { { v }^{ 2 } }{ rg }$

or, $\displaystyle \tan { { 12 }^{ o } } =\frac { { \left( 150 \right) }^{ 2 } }{ r\times 10 }$

or, $\displaystyle r=\frac { 2250 }{ 0.2125 } =10.6\times { 10 }^{ 3 }m=10.6\quad km$

A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is

$20\ N$ , mass of the stone in

$kg$ is:
(

$\displaystyle g=10{ ms }^{ -2 }$ )

0%
$1.0$
0%
$1.5$
0%
$0.5$
0%
$0.75$

Explanation

Let the mass of the stone be

$m$ , and the angular speed with which it rotates be

$\omega$ .

Thus at the bottom most point,

$T_b-mg=m\omega^2 l$

And at the top most point,

$T_t+mg=m\omega ^2 l$

$\implies T_b-T_t=2mg=20$

$\implies mg=10$

$\implies m=1\ kg$

In vertical circular motion, the ratio of kinetic energy to potential energy at the horizontal position is ____________.

0%
$5:2$
0%
$2:1$
0%
$3:2$
0%
$2:3$

Explanation

We assume that the object in motion just manages to complete the vertical circle. In that case,

$v_{top} = \sqrt{gR}$
Total energy at the topmost point

$=\dfrac12mgR + 2mgR = \dfrac52mgR$

At the horizontal position

$PE = mgR$

Total energy

$= \dfrac52mgR$

$\therefore KE = (\dfrac52-1) = \dfrac32mgR$

$\therefore \dfrac{KE}{PE} = \dfrac32$

A stone tied to a string is rotated in a vertical plane. If mass of the stone is

$m$ , the length of the string is

$r$ and the linear speed of the stone is

$v$ ,When the stone is at its lowest point, then the tension in the string will be :

0%
$\displaystyle \frac { m{ v }^{ 2 } }{ r } +mg$
0%
$\displaystyle \frac { m{ v }^{ 2 } }{ r } -mg$
0%
$\displaystyle \frac { mv }{ r }$
0%
$\displaystyle mg$

Explanation

At the lowest point, as shown in the figure both mg and centrifugal force

$\dfrac { m{ v }^{ 2 } }{ r } \\$ will act in the same direction so,

$T=mg+\dfrac { m{ v }^{ 2 } }{ r } \\$

If the string were to break at point II, what would be the path of the ball?

0%
0%
0%
0%
0%
None of these

Explanation

When the bob is swinging in vertical circle the centripetal force and the pseudo force varies. Centripetal force causes the tension in the string.

Let centripetal force be

$C$ , pseudo force be

$P$ and weight of bob be

$W$ .

At position 2,

$C = W$

If the string were to break at this position, the only external force acting on the bob is

$W$ (downward). Hence the bob would move vertically down.

Therefore option D is correct.

$50 kg$ acrobat is swinging on a rope with a length of

$15 m$ from one horizontal platform to another. Both platforms are at equal height.
If the maximum tension the rope can support is

$1200 N$ , which of the following answer best represents the maximum velocity the acrobat can reach without breaking the rope?

0%
$25 {m}/{s}$
0%
$19 {m}/{s}$
0%
$20 {m}/{s}$
0%
$15 {m}/{s}$
0%
$5 {m}/{s}$

Explanation

Given :

$T =1200$ N

$m = 50$ kg

$L =15$ m

Let the maximum velocity of the acrobat be

$v$ .

From figure,

$\dfrac{mv^2}{L} =T - mg$

$\therefore$

$\dfrac{50 \times v^2}{15} =1200 - 50 \times 9.8$

$\implies v =\sqrt{213} \approx 15$

$m/s$

Which of the following statements is true?

0%
The tension in the string is greater at points III than at point I
0%
The tension in the string is greater at points I than at point III
0%
The tension at point I is equal to the tension at point III
0%
The tension in the string is greatest at point II
0%
The tension in the string is same at points I, II and III

Explanation

When the bob is swinging in vertical circle the centripetal force and the pseudo force varies. Centripetal force causes the tension in the string.

Let centripetal force be

$C$ , pseudo force be

$P$ and weight of bob be

$W$ .

At position 1,

$C = P - W$

At position 2,

$C = W$

At position 3,

$C = P + W$

As C at position 3 is greatest, the tension in the string is greatest at this position.

Therefore option A is correct.

A ball of mass

$m$ is attached with the string of length

$R$ , rotating in circular motion, with instantaneous velocity

$v$ and centripetal acceleration

$a$ .

Calculate the centripetal acceleration of the ball if the length of the string is doubled?

0%
${a}/{4}$
0%
${a}/{2}$
0%
$a$
0%
$2a$
0%
$4a$

Explanation

Centripetal acceleration

$a = \dfrac{v^2}{R}$

Now the length of the string is doubled keeping the instantaneous velocity constant i.e

$R' = 2R$

$\therefore$ New centripetal acceleration

$a' = \dfrac{v^2}{R'} = \dfrac{v^2}{2R} =\dfrac{a}{2}$

A uniform rod of mass 1 kg is hanging from a thread attached at the midpoint of the rod. A block of mass

$m$ =3kg hangs at the left end of the rod and block of mass

$M$ hangs at the right side at 80 cm from block

$m$ . If system is in equilibrium. Calculate

$M$ .

0%
4 kg
0%
5 kg
0%
6 kg
0%
8 kg
0%
9 kg

Explanation

Given :

$m = 3 kg$

As the rod is in equilibrium, thus net torque about the point O is zero i.e

$\tau_o = 0$

$\therefore$

$mg (0.5 ) -Mg (0.3) = 0$

$\implies$

$M = \dfrac{m(0.5)}{0.3} = \dfrac{3 \times 0.5}{0.3} = 5 kg$

The given rod is uniform and has a mass m. Find the tension in the string.

0%
$\dfrac { \left( mg\sin { \theta } \right) }{ 2 }$
0%
$\dfrac { mg\sin { } }{ 2 }$
0%
$\dfrac { \left( mg\cos { \theta } \right) }{ 2 }$
0%
$\dfrac { mg }{ 2 }$
0%
$mg$

Explanation

Let the tension in the string be

$T$ and

$L$ be the length of the rod.

As the rod is in equilibrium, thus the torque about point O is zero i.e

$\tau_o = 0$

$\therefore$

$T (L sin\theta) - mg \times \dfrac{L}{2} sin\theta = 0$

$\implies T =\dfrac{mg}{2}$

$110 kg$ panda is riding on a

$3.0 m$ long swing whose mass can be considered negligible. The highest point of its arc occurs when the swing makes a

${20}^{o}$ angle with the vertical.
What is the magnitude of the total tension in the ropes of the swing at that point?

0%
$0N$
0%
$103 N$
0%
$369 N$
0%
$1013 N$
0%
$1078 N$

Explanation

Given :

$m =110$ kg

Tension in the string

$T = mg$

$cos20^o = 110 \times 9.8 \times 0.94 = 1013$ N

A uniform bar is lying on a flat horizontal table. The bar is acted upon by

${F}_{1}$ and

${F}_{2}$ ,besides the gravitational and normal forces (which cancel),

${F}_{1}$ and

${F}_{2}$ are parallel to the surface of the table. If the net force on the rod is zero, then which one of the following statement is true?

0%
The net torque on the bar must also be zero
0%
The bar can accelerate translationally if ${F}_{1}$ and ${F}_{2}$ are not applied at the same point
0%
The net torque will be zero if ${F}_{1}$ and ${F}_{2}$ are applied at the same point
0%
The bar cannot accelerate translationally or rotationally
0%
None of the above

Two objects rest on a seesaw. The first object has a mass of 3 kg and rests 10 m from the pivot. The other rests 1 m from the pivot. What is the mass of the second object if the seesaw is in equilibrium?

0%
0.3 kg
0%
3 kg
0%
10 kg
0%
30 kg
0%
50 kg

Explanation

Torque is given by,

$Torque,\quad \tau =\vec { r } \times \vec { F } =rF\sin { \theta } \\ Since\ \vec { r } \ perpendicular\quad to\ \vec { F } ,\quad \theta ={ 90 }^{ \circ }\\ So,\quad \tau =rF\\ For\quad the\quad seesaw\quad to\quad be\quad in\quad equilibrium,\quad \\ { r }_{ 1 }{ F }_{ 1 }={ r }_{ 2 }{ F }_{ 2 }\\ { r }_{ 1 }{ m }_{ 1 }g={ r }_{ 2 }{ m }_{ 2 }g\\ 1\times { m }_{ 1 }=10\times 3\\ { \quad \quad m }_{ 1 }=30kg$

The catcher prepares to receive a pitch from the pitcher. As the ball reaches and makes contact with his glove, the catcher pulls his hand backward. This action reduces the impact of the ball on the catcher's hand because:

0%
The energy absorbed by his hand is reduced
0%
The momentum of the pitch is reduced
0%
The time of impact is increased
0%
The time of impact is reduced
0%
The force exerted on his hand remains the same

Explanation

We know that,

impulse = change in momentum

$F\times t=p_{2}-p_{1}$

$p_{2}$ is zero , because final velocity of the ball becomes zero , so RHS of this equation is constant (

$=p_{1}$ ), when catcher pulls his hands backward, he increases the time of impact

$t$ and due to this impact of ball decreases.

A person is moving on a rough horizontal surface towards east. Then the direction of friction force is towards

0%
west
0%
east
0%
north
0%
cant say depends on the situation.

A plane travelling at

$300 {m}/{s}$ banks its wings to enter into a horizontal circular turn. The circular path has a radius of

$2.7 km$ .
Which of these values best represents the angle of the wings relative to the horizontal if the plane experiences no change in its altitude during the turn?

0%
${8}^{o}$
0%
${16}^{o}$
0%
${20}^{o}$
0%
${25}^{o}$
0%
${32}^{o}$

Explanation

In the flight of a plane the angle of banking with vertical is given by ,

$\tan \theta=v^{2}/rg$

given

$v=300m/s , r=2.7km=2700m$

therefore

$\tan\theta=\frac{300^{2}}{2700\times9.8}=3.4$

$\tan\theta=\tan74$

$\theta=74^{o}$

so the angle with horizontal will be ,

$\theta'=90-74=16^{o}$

A ball of mass 100 g is moving with a velocity of 10 m

$s^{-1}$ . The force of the blow by the bat acts for 0.01 seconds. What is the force exerted on the ball by the bat if the ball retraces its path with same speed?

0%
100 N
0%
200 N
0%
300 N
0%
400 N

Explanation

Initial velocity of ball

$u = -10 \ m/s$

Initial momentum

$P_i = mu = 0.1\times (-10) = -1 \ kg \ m/s$

Final velocity of ball

$v = 10 \ m/s$

Final momentum

$P_f = mv = 0.1\times 10 = 1 \ kg \ m/s$

Time of action

$t = 0.01 \ s$

Using

$F\times t = P_f - P_i$

$F\times 0.01 = 1-(-1) = 2$

$\implies \ F = 200 \ N$

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.