MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 8

A boy is swinging in a swing. If he stands the time period will

0%
First decreases, then increase
0%
Decrease
0%
Increase
0%
Remain same

Explanation

As the child stand up the effective length of pendulum decreases due to the reason that center of gravity rises up.

According to

$T=2\pi\sqrt{\left(\dfrac{l}{g}\right)}$

Thus, the time peroid decreases

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

${30}^{o}$ and

${60}^{o}$ from vertical 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

$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ r } \\ T=mgcos\theta +\dfrac { m{ v }^{ 2 } }{ r } \\ for\quad \theta =30\\ { T }_{ 1 }=mgcos30+\dfrac { m{ v }^{ 2 } }{ r } \\ for\quad \theta =60\\ { T }_{ 2 }=mgcos60+\dfrac { m{ v }^{ 2 } }{ r } \\ { T }_{ 1 }>{ T }_{ 2 }$

If force F =(500 - 100)t, then impulse as a function of time will be:-

0%
$500t - 50 t^2$
0%
$50 t - 10$
0%
$50 - t ^2$
0%
$100 t^2$

Explanation

We know that force is given as

$F=\dfrac{dP}{dt}$

so impulse

$dP$ is given as

$dP=Fdt$

putting value of

$F$ and integrating both sides we get

$\int dP=\int (500-100t)dt$

$P=500t-50t^2+c$ ,

$c$ is an integration constant.

as the

$P$ is here the

$momentum$ supplied or abstracted so

$c=0$

A hammer of mass 1 kg moving with a speed of 6 m/s strikers a wall and comes to rest in 0.1 s, calculate Retarding force that stops the hammer.

0%
60 N
0%
50 N
0%
40 N
0%
30 N

Explanation

$m=1kg$

$u = 6m/s ,v=0 ,t=0.15$

impulse =

$f\times t = m(v-u) = 1(0-6) = -6Ns$

retarding force stops the hammer

$F = \dfrac{I}{T} = \dfrac{6}{0.1} = 60N$

A boy is swinging on a swing such that his lowest and highest positions are at heights of

$2m$ and

$4.5m$ respectively. His velocity at the lowest position is:

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$2.5{ms}^{-1}$
0%
$7{ms}^{-1}$
0%
$14{ms}^{-1}$
0%
$20{ms}^{-1}$

Explanation

The height of the boy falls with the swing,

$h=4.5-2=2.5m$

Therefore,

The velocity at the lowest position

$=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times2.5}=\sqrt{49}$

$v=7 m/s$

In the figure, a 4.0 kg ball is on the end of a 1.6 m rope that is fixed at O. The ball is held at point A, with the rope horizontal is given an initial downward velocity. The ball moves theough three qyarters of a circle with no friction and arrives at B, with the rope barely under tension. Thee initial velocity of the ball, at point A, is closest to

0%
4.0 m/s.
0%
5.6 m/s.
0%
6.2 m/s
0%
6.8 m/s

Explanation

The motion of the ball is always perpendicular to the tension in the Rope. Therefore, the tension does no work on the system and the total mechanical energy of the ball is conserved.

$E_A = E_B$

$mgR + \dfrac{1}{2} mv_A^2 = mg(2R) + \dfrac{1}{2} mv_B^2$

$v_A^2 = 2gR +v_B^2$

Newton's second law for the ball at point B

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

$T \approx 0$

$v_B = gR$

$v_A = 3gR$

= 3(9.8)(1.6)

= 6.9 m/s

Since the velocity is near to 6.8m/s.

Hence (D) option is correct answer

A stone of mass

$0.2kg$ is tied to one end of a thread of length

$0.1m$ whirled in a vertical circle. When the stone is at the lowest point of circle, tension in thread is

$52N$ , then velocity of the stone will be:

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$4m/s$
0%
$5m/s$
0%
$6m/s$
0%
$7m/s$

Explanation

$T-mg=\dfrac { { mv }^{ 2 } }{ r } \\ T=\dfrac { { mv }^{ 2 } }{ r } +mg\\ 52=0.2(10)\quad +\dfrac { 0.2({ v }^{ 2 }) }{ 0.1 } \\ 2{ v }^{ 2 }=52-2=50\\ \qquad v=5m/sec$

A stone of mass

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

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

$4m/s$ . What is the tension in the string (at that instant)?

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

Explanation

$T+mg=\dfrac { { mv }^{ 2 } }{ r } \\ T=\dfrac { { mv }^{ 2 } }{ r } -mg\\ \dfrac { 1\times { 4 }^{ 2 } }{ 1 } -1(10)\\ =16-10\\ =6N$

One end of a string of length

$1.0m$ is tied to a body of mass

$0.5kg$ . It is whirled in a vertical circle with angular velocity

$4 rad/s$ . The tension in the string when body is at the lower most point of its motion is equal to [Take

$g=10m/s^2$ ].

0%
$3N$
0%
$5N$
0%
$8N$
0%
$13N$

Explanation

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

or,

$T - mg = m \omega ^2 r$

$\therefore T = mg + m\omega^2 r$

$= (0.5 \times 10) + (0.5 \times 4^2 \times 1)$

$= 5+8 = 13N$

980092_1075445_ans_2ef8b06d2afb41ff81c6ea7d6737324d.png

A body of mass $1 kg$ tied to one end of string is revolved in a horizontal circle of radius $0.1 m$ with a speed of $3 evolution/sec$ , assuming the effect of gravity is negligible, then linear velocity, acceleration and tension in the string be :

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$1.88 m/s, 35.5 m/s^2, 35.5 N$
0%
$2.88 m/s, 45.5 m/s^2, 45.5 N$
0%
$3.88 m/s, 55.5 m/s^2, 55.5 N$
0%
None

Explanation

$3rev/sec\implies 3\times 2 \pi \ rad/sec$

$\omega=6\pi\ radian/sec$

$v=r\omega=0.1\times 6\pi$

$=1.88m/s$

$T=m\omega^{2}r=1\times 36\times \pi^{2}\times 0.1$

$=35.5N$

$a_{t}=\alpha r$
and

$a=0$

$\therefore a_{t}=0$

$a_{c}=\omega^{2}r$

$=36\times \pi^{2}\times 0.1$

$a_{c}=35.5N$

A small sphere of mass m suspended by a thread is first taken aside so that the thread forms the right angle with the vertical and then released, then:

The total acceleration of the sphere and the thread tension as a function of

$\theta$ , the angle of deflection of the thread from the vertical will be

0%
$g\sqrt{1+3cos^2} \theta, T=3 mg\,cos\,\theta$
0%
$g \, cos\,\theta, T=3 mg \, cos\,\theta$
0%
$g\sqrt{1+3sin^2} \theta, T=5 mg\,cos\,\theta$
0%
$g \, sin\,\theta, T=5 mg \, cos\,\theta$

The angle

$\theta$ between the thread and the vertical at the moment when the total acceleration vector of the sphere is directed horizontally will be

0%
$cos\, \theta =\dfrac{1}{\sqrt{3}}$
0%
$cos\, \theta =\dfrac{1}{3}$
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$sin\, \theta =\dfrac{1}{\sqrt{3}}$
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$sin\, \theta =\dfrac{1}{\sqrt{2}}$

A bullet of mass

$50\ g$ travelling at

$500\ m/s$ penetrates

$100\ cm$ into a wooden block. Find the average force exerted on the block.

0%
$0.625\times 10^{4}\ N$
0%
$62.5\times 10^{4}\ N$
0%
$6.25\times 10^{4}\ N$
0%
$2.65\times 10^{4}\ N$

Explanation

Mass, $m = 0.05\ kg$
Initial velocity, $u = 500\ m/s$
Distance covered, $S = 100 cm = 1\ m$

Final velocity $v=0$

Now, from the equation of motion

${{v}^{2}}-{{u}^{2}}=2as$

$0-{{\left( 500 \right)}^{2}}=2\times a\times 1$

$a=-125000\,m/{{s}^{2}}$

Now, the average force exerted on the block

$F=ma$

$F=0.05\times 125000$

$F=6.25\times {{10}^{3}}$

$F=0.625\times {{10}^{4}}\,N$

Hence, the average force is $0.625\times {{10}^{4}}\,N$

An object is moving on a plane surface uniform velocity

$10 ms^{-1}$ in presence of a force

$10 N$ . The frictional force between the object and the surface is:

0%
ZERO
0%
-10 N
0%
10 N
0%
100 N

A car is moving in a circular path with a uniform speed

$v$ . When the car rotates through an angle

$\theta$ , the magnitude of change in its velocity is

0%
$\triangle v = v\sin \left (\dfrac {\theta}{2}\right )$
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$\triangle v = v\sin \theta$
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$\triangle v = 2v\sin \left (\dfrac {\theta}{2}\right )$
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$\triangle v = 2v\sin \theta$

Which of the following groups of forces could be in equilibrium?

0%
$3 N, 4 N, 5 N$
0%
$4 N, 5 N, 10 N$
0%
$30 N, 40 N, 80 N$
0%
$1 N, 3 N, 5 N$

Explanation

For the equilibrium of force, the resultant of two smaller is equal and opposite to third one.

$C^2=A^2+B^2$

Lets see,

$5^2=3^2+4^2$

$25=25$

$10^2=5^2+4^2$

$100{\neq} 41$

$80^2=40^2+30^2$

$6400\neq2500$

$5^2=1^2+3^2$

$25\neq10$

From the above discussion,

The correct option is A.

The frictional force is?

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Self adjustable
0%
Not self adjustable
0%
Scalar quantity
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Equal to the limiting force

A ball of mas

$150 g$ , moving with an acceleration

$20 m/s^2$ , is hit by a force which acts on it for

$1$ sec. The impulsive force is:-

0%
$3 Ns$
0%
$1 Ns$
0%
$05 Ns$
0%
$1.2 Ns$

Explanation

$force = mass x acceleration$

$F = ma$

$F = 150/1000 x 20$

$F = 3N$

$Impulse = Force \times Time$

$Impulse = 3 \times 1 => 3 N s$

Hence,

option $A$ is correct answer.

An inclined track ends in a circular loop of diameter

$D$ . From what height on the track a particles should be released so that it completes that loop in the vertical plane?

0%
$\dfrac {5D}{2}$
0%
$\dfrac {2D}{5}$
0%
$\dfrac {5D}{4}$
0%
$\dfrac {4D}{5}$

A stone, tied at the end of a string

$80$ cm long, is whirled in a horizontal circle with a constant speed. If the stone makes

$14$ revolutions in

$25$ sec, what is the magnitude of acceleration of the stone.

0%
$680$ cm $/s^2$
0%
$720$ cm $/s^2$
0%
$860$ cm $/s^2$
0%
$990$ cm $/s^2$

Explanation

Given length of string = 80 cm and angular velocity

$\omega$ = 2

$\pi$ f=

$\dfrac { 88 }{ 25 }$

And we know centripetal acceleration =

$r{ \omega }^{ 2 }$

$80\times { (\frac { 88 }{ 25 } ) }^{ 2 }$

=991.232 cm/s

=990 cm/s

A stone of mass

$1\ kg$ is tied to a string

$4\ m$ long and is rotated at constant speed of

$40\ ms^{-1}$ in a vertical circle. The ration of the tension at the top and the bottom is

0%
11:12
0%
39:41
0%
41:39
0%
12:11

A field gun of mass 150

$kg$ fires a shell of mass 1 kg with velocity of 150 m/s. Calculate the velocity of the recoil of the gun.

0%
1 m/sec
0%
3 m/sec
0%
1.5 m/sec
0%
5 m/sec

A ball of weight

$0.1$ kg coming with a speed

$30$ m/s strikes a bat and returns in opposite direction with speed of

$40$ m/s, then impulse is

0%
$-0.1 \times 40 -0.1 \times 30 Ns$
0%
$0.1 \times 40 -0.1 \times (-30) Ns$
0%
$0.1 \times 40 +0.1 \times (-30) Ns$
0%
$0.1 \times 40 -0.1 \times 20 Ns$

Explanation

Given,

$m=0.1kg$

$u=30m/s$

$v=-40m/s$

Impulse = change in momentum

$mv_{2}−mv_{1}= 0.1\times 40−0.1\times(−30)$

The correct option is B

Velocity of a particle of mass

$2\ kg$ varies with time

$t$ according to the equation

$\vec {v} = (2t\hat {i} + 4\hat {j}) m/s$ . Here

$t$ is in seconds. Find the impulse imparted to the particle in the time interval from

$t = 0$ to

$t = 2s$ .

0%
$8\hat {i} N-s$
0%
$10\hat {i} N-s$
0%
$12\hat {i} N-s$
0%
$16\hat {i} N-s$

Explanation

velocity vector

$=v=(2t\hat i+4\hat j)\ m/s$

since acceleration is the rate of change in velocity,

$\therefore a=\dfrac{dv}{dt}=\dfrac{d(2t\hat i+4\hat j)\ m/s}{dt}=2\hat i\ m/s^2$

$\therefore$ particle has constant acceleration. This means particle is acts upon by constant force.

$\therefore F=ma=2\ kg\times (2\hat i)\ m/s^2=4\hat i\ N$

Since a constant force acts on the particle the impulse imparts to the particle is given by :-

Impulse

$=\hat i \Delta t=(4\hat i N)\times (2s-0s)=8\hat i\ N\ s$

$\therefore$ option

$A$ is correct.

Two balls of same mass are dropped from the same height h , on to the floor . the first ball bounces to a height h/4, after the collection & the second ball to a height h/the impulse applied by the first & second ball on the floor are

${I_1}$ and

${I_2}$ respectively . then

0%
$5{I_1} = 6{I_2}$
0%
$6{I_1} = 5{I_2}$
0%
$3{I_1} = 2{I_2}$
0%
$2{I_1} = {I_2}$

Explanation

When the ball is dropped from height h its velocity

${v}_{1}=\sqrt{2gh}$

Now velocity of the ball after the bounce, where height

$h=\dfrac{h}{4}$ ,velocity

${v}_{1a}=\sqrt{2g\dfrac{h}{4}}$

Now impulse

${I}_{1}={F}_{1}t=mat=m({v}_{1}-{v}_{1a})=m(\sqrt { 2gh } -\sqrt { 2g\dfrac { h }{ 4 } } )=m({ v }_{ 1 }-\dfrac { 1 }{ 2 } { v }_{ 1 })=\dfrac { 1 }{ 2 } m{ v }_{ 1 }$

Similarly for the ball when reaches a height of

$\dfrac{h}{4}$ after bounce,velocity

${v}_{2a}=\sqrt{2g\dfrac{h}{16}}$ and velocity of the ball before

bounce

${v}_{2}=\sqrt{2gh}$

Similarly the impulse of the second ball

${I}_{2}=\dfrac{3}{4}{v}_{2}m$

Given mass of both the ball are same.

We can compare both the impulses as

$\dfrac{4{I}_{2}}{3{v}_{2}}=\dfrac{2{I}_{2}}{{v}_{1}}$

Now as both

${v}_{2}={v}_{1}=\sqrt{2gh}$

$3{I}_{1}=2{I}_{2}$

A rubber ball of mass 10 g falls from a height of

$80 cm$ and rebounds to half of the original height. Impulse exerted by the ball on the floor is

0%
0.068 N s
0%
6.8 N s
0%
0.0065 N s
0%
0.68 N s

A force of 100 dynes acts on mass of 5 gm for 10 sec. The velocity produced is:

0%
2 cm/sec
0%
20 cm/sec
0%
200 cm/sec
0%
2000 cm/sec

Explanation

Given that,

Force $F=100\,dyn$

Mass $m=5\,g$

Time $t=10\,s$

Initial velocity $u=0$

Now, the acceleration is

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{100\times {{10}^{-5}}}{5\times {{10}^{-3}}}$

$a=0.20\,m/{{s}^{2}}$

Now, from equation of motion

$v=u+at$

$v=0+0.20\times 10$

$v=2\,m/s$

$v=200\,cm/s$

Hence, the velocity produced is $200\ cm/s$

Two particles A and B equal masses are respectively tied to the centre and one end of a string, whose other end 0 is fixed. Both the particles always revolve in concentric circles of centre O. The ratio of tensions in both the parts of the string will be

0%
3:2
0%
2:3
0%
1:2
0%
1:1

Read statement-1 and statement-2 and choose the correct option.
Statement-1: In some situations, friction facilitates the motion and in some situations, it opposes the motion.
Statement-2: Friction force always opposes the relative motion between two surfaces in contact.
Choose the correct option.

0%
Only statement -1 is correct
0%
Only statement-2 is correct
0%
Both statement-1 and statement-2 are correct
0%
Neither statement-1 nor statement-2 is correct

A 40 N block is supported by two ropes. One rope is horizontal and the other makes an angle of ${30^ \circ }$ with the ceiling. The tension in the rope attached to the ceiling is approximately:

0%
80 N
0%
40 N
0%
34.6 N
0%
46.2 N

Explanation

$x-$ direction:

$T_{1}=T_{2}\cos 30^{o}\rightarrow (i)$

$y-$ direction

$T_{2}\sin 30^{o}=40$

$T_{2}=80\ N$

Option

$A$

1382638_1132916_ans_8f3143df88084721a7421c7d8216ccd2.png

Circular flexible current loop of radius R carrying current I is placed in an inward magnetic field B. If we spin the loop with angular speed

$/omega$ , then tension in string

0%
Is zero
0%
is more than iBR
0%
is less than iBR
0%
Does not depend on rotation

If a car is to travel with a speed

$v$ along the frictionless, banked circular track of radius

$r$ , the required angle of banking, so that the car does skids is ?

0%
$\theta=\tan^{-1}(\dfrac {v^{2}}{rg})$
0%
$\theta=\tan^{-1}(\dfrac {v}{rg})$
0%
$\theta=\tan^{-1}(\dfrac {r^{2}}{rg})$
0%
$\theta < \tan^{-1}(\dfrac {v^{2}}{rg})$

The disc is at rest at the top of a rough inclined plane. It rolls without slipping. At the bottom of inclined plane there is a vertical groove if radius

$R$ . In order to loop the groove, the minimum height of incline required is:

0%
$\dfrac{15 R}{4}$
0%
$\dfrac{9 R}{4}$
0%
$\dfrac{5 R}{2}$
0%
$\dfrac{7 R}{5}$

Explanation

Hence, option

$(A)$ is correct answer.
1380481_1132935_ans_19e2235363804e628b47ae162028617f.jpg

The moment P (in

$kg\, ms^{-1}$ ) of a particle is varying with time t (in second) as

$p=2+3t^2$ . The force acting on the particle at

$t=3s$ will be

0%
18 N
0%
54 N
0%
9 N
0%
15 N

Explanation

Given,

$p=2+3t^2$

From Newton's second law,

$F=\dfrac{dp}{dt}$

$F=\dfrac{d(2+3t^2)}{dt}$

$F=0+6t=6t$

$t=3s$

$F=6\times 3=18N$

The correct option is A.

What is corrct statement about impulse?

0%
change in velocity
0%
change in momentum due to force in small time
0%
change in acceleration
0%
change in speed

Explanation

Impulse:- It is defined as a change in momentum due to force in small time.

Mathematically,

From Newton's law,

$F=\dfrac{\Delta p}{\Delta t}$

Impulse,

$I=F\times \Delta t=\Delta P$

The correct option is B.

A thin uniform rod of mass

$M$ and length

$L$ is hinged at its upper end released from rest in horizontal position. The tension at a point located at a point located at a point, when the rod becomes vertical, will be:

0%
$\dfrac {22Mg}{27}$
0%
$\dfrac {11Mg}{13}$
0%
$\dfrac {6Mg}{11}$
0%
$2Mg$

Explanation

$2mg$

given,

$0+mg \times 0=\dfrac{1}{2}Iw^2+mg\left ( -\dfrac{L}{2} \right )$

$\dfrac{1}{2}\left ( \dfrac{mL^2}{3} \right )w^2=\dfrac{mgL}{2}$

$w=\sqrt{\dfrac{3g}{L}}$

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

$=m[g+rw^2]$

$=m\left [ g+\dfrac{2L}{3} \times \dfrac{3g}{L} \right ]$

$=\dfrac{2m}{3} \times [g+2g]=2mg$

The coefficient of friction between the tyres and the road is

$0.25$ . The maximum speed with which a car can be driven around a curve of radius

$40m$ , without skidding is ? (assume

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

0%
$40\ ms^{-1}$
0%
$20\ ms^{-1}$
0%
$15\ ms^{-1}$
0%
$10\ ms^{-1}$

$10$ kkg object attached to a nylon cord outside a space vehicle is rotating at a speed of

$5 m/s$ . If the force acting on the cord is

$125 N$ its radius of path is

0%
2 m
0%
4 m
0%
6 m
0%
1 m

A car moves around a curved road of radius

$R_1$ at constant speed v without sliding. If we double the car's speed, what is the least radius that would now keep the car safe from sliding?

0%
$2R_1$
0%
$4R_1$
0%
$6R_1$
0%
None of these

The length of simple pendulum is 1 m and mass of its bob is 50 g. The bob is given sufficient velocity so that the bob describe vertical circle whose radius equal to length of pendulum. The maximum difference in the kinetic energy of bob during one revolution is.

0%
0.98 J
0%
1.96 J
0%
4.9 J
0%
9.8 J

Explanation

A bob of mass

$m = 0.05\ Kg$ is performing vertical circle of radius

$(r) = 1m$ .

In the boundary case.

a minimum velocity of

$\mu = \sqrt{5gr}$ has to provided to it

so it completes a vertical circle.

So, kinetic energy

$(K_1)$ at the lowest point,

$K_1 = \dfrac{1}{2}m\mu^2 = \dfrac{5}{2}mgr$

and

$PE = 0$ , raking it as reference at highest point

$PE : \mu_2 =mgh = 2mgr$

$(as \, h = 2r)$

From conservation of energy.

$K_1 + \mu_1= K_2+\mu_2 \Rightarrow K_2 = K_1 - \mu_2 = \dfrac{1}{2}mgr$

Thus, loss in

$KE$ :

$\Delta K = K_1-K_2 = 2mgr = 2\times 0.05 \times 9.8 \times 1 = 0.98\ J$ (Ans)

1373012_1162993_ans_f8a6ff3e4dc94002a20245803142ea11.png

A player kicks a football of mass

$0.5$ kg and the football begins to move with a velocity of

$10 { m }/{ s }$ . If the contact between the leg and the football lasts for

$\dfrac { 1 }{ 50}$ sec, then the force acted on the football should be

0%
$2500 N$
0%
$1250 N$
0%
$250 N$
0%
$625 N$

Explanation

Momentum transferred to the ball is

$\Delta P=m\Delta v=0.5kg\times (10-0)=5Kgm/s$

so the force will be

$F=\dfrac{\Delta p}{\Delta t}=\dfrac{5}{1/50}=250Kgm/s^2=250N$

Safe maximum seed of the vehicle along an unbanked curved rough road of radius

$500m$ and coefficient of static friction

$0.1$ is

0%
$3.141m/s$
0%
$11.2m/s$
0%
$19m/s$
0%
$22m/s$

A homogeneous disc of mass 2 kg and radius 15 cm is rotating about its axis (which is fixed) with an angular velocity of 4 rad/s. The linear momentum of the disc is :

0%
1.2 kg-m/s
0%
1.0 kg-m/s
0%
0.6 kg-m/s
0%
None of the above

Explanation

The formula for angular momentum is:

$L=Iw$ , where is the angular velocity and

$I$ is the moment of inertia.

For a disk

$,$

$I = \dfrac{1}{2}M{R^2},$ where

$M$ is the mass of the disk and

$R$ is the radius of the disk.

For this disk,

$I = \frac{1}{2}M{R^2} = \dfrac{1}{2}\left( 2 \right) \times {\left( {0.15} \right)^2} = 0.0225\,kg\,{m^2}$

$L = Iw = 0.0225\,kg\,{m^2} \times \left( {4\,rad/s} \right) = 0.09\,kg\,{m^2}\,rad/s$

The linear momentum is related to the the angular momentum by the formula;

$L=pr.$

We can solve for p, the linear momentum in this equation to get,

$p = \dfrac{L}{r} = \dfrac{{0.09}}{{0.15}} = 0.6kgs$

Hence,

option

$(C)$ is correct answer.

A body of mass tied at the end of a strong of length

$l$ is projected with velocity

$\sqrt{4lg}$ , at what height will it leave the circular path:

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$\dfrac{5}{3}l$
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$\dfrac{3}{5}l$
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$\dfrac{1}{3}l$
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$\dfrac{2}{3}l$

Explanation

REF.Image

In a vertical circle the tension at the string is given by :-

$T= \dfrac{mu^{2}}{l}-mg(2-3 cos \theta )$

Here

$u^{2}= 4gl$ , and particle stop circular motion

when

$T=0$

$\Rightarrow T=0\Rightarrow \dfrac{mu^{2}}{L}= mg(2-3 cos\theta )$

$\Rightarrow \dfrac{m.4gL}{L}= mg(2-3 cos \theta )\Rightarrow cos\theta = -2/3$

$\therefore$ Height,

$h=l\left | cos\theta \right |=\frac{2}{3}l$

Hence it attains maximum height,

$H= h+L= \dfrac{5}{3}L$ (Ans)

1102738_1174252_ans_2513c8f5ddca48d3b02195a944deccc4.JPG

What is the impulse of force shown in the following fingre?

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205 Ns
0%
450 Ns
0%
800 Ns
0%
1000 Ns

Explanation

Impulse, $I=\int{F(t).dt}=$ Area of F and t graph.

Where,

$F=\,force$

$t=\,time$

$I=\dfrac{1}{2}Ft=\dfrac{1}{2}\times 100\times \left( 10-1 \right)=450\,Ns$

Hence, impulse is $450\,Ns$

A thread tightly wound over wound over a uniform solid cylinder of mass

$M$ , which lies on a smooth horizontal surface as shown. The free end of the thread is pulled by horizontal force

$F$ as shown. Then the acceleration of centre of cylinder at the shown instant is:-

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$\dfrac{F}{2M}$
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$\dfrac{F}{M}$
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$\dfrac{3F}{2M}$
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$\dfrac{2F}{M}$

Choose the correct statement(s) about the frictional force between two solid surfaces in contact

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Static friction is a variable force
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$\overset { - }{ { { f }_{ lim } } } ={ \mu }_{ s }\overset { - }{ N }$ (where symbols have their usual meanings)
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Kinetic friction is self adjusting
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None of these

Explanation

Important points about the frictional force:-

A) Static friction is a variable force

$f_{lim}=\mu_sN$

C) Static friction is considered as a self- adjusting.

From the given option,

Option A and B are correct option.

A force of

$25$ N acts on a body at rest for

$0.2$ s and a force of

$70$ N acts for the next

$0.1$ s in opposite direction. If the final velocity of the body is

$5ms^{-1}$ , the mass of the body is:

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$1$ kg
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$2$ kg
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$0.8$ kg
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$0.4$ kg

Explanation

Resulatant impule = Change in momentum

$7N\cdot s-5N\cdot s=2N\cdot s$

$mv=2N\cdot s$

$m=\dfrac { 2 }{ 5 } kg$

In which of the following cases , the net force is zero?
I) A ball freely falling from a certain height
II) A cork floating on the surface of water
III) An object floating in air

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I and II only
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II and III only
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III and I only
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I, II and III

Explanation

$\begin{array}{l} \text { In first case gravity is there so net fore is } \\ \text { not zero? while in others it is balanced by } \\ \text { buoyancy of air and liquid. } \\ \text { so net force is zero in II, } III \text { case } \\ \text { so Ans is (B). } \end{array}$

Block

$A$ of mass

$4kg$ is to be kept at rest against a smooth vertical wall by applying a force

$F$ as shown in fig. The force required is

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

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$40 \sqrt { 2 } N$
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$20 \sqrt { 2 } N$
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$10 \sqrt { 2 } N$
0%
$15 \sqrt { 2 } N$

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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