MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 3

A body of mass m is rotated at uniform speed along a vertical circle with the help of light string. If

$T_{1} and \ T_{2}$ are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively, then which of the following expressions is correct?

0%
$T_{2}-T_{1}=6mg$
0%
$T_{2}-T_{1}=4mg$
0%
$T_{2}-T_{1}=2mg$
0%
$T_{2}-T_{1}=mg$

Explanation

Centripetal Force is:

$\dfrac { m{ v }^{ 2 } }{ r }$
At the highest point:

${ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } -mg$
At the lowest point:

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

$\therefore \quad { T }_{ 2 }-{ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } +mg-(\dfrac { m{ v }^{ 2 } }{ r } -mg)=2mg$

A body is moving in a vertical circle such that the velocities of body at different points are critical. The ratio of velocities of body at angular displacements

$60^{0}$ and

$120^{0}$ from the lowest point is:

0%
$\sqrt{5}:\sqrt{2}$
0%
$\sqrt{3}:\sqrt{2}$
0%
$\sqrt{3}:1$
0%
$\sqrt{2}:1$

Explanation

Given that velocities at different points are critical

so velocity at top point (

$T$ )

$\Rightarrow$

${ V }_{ T }$

$=\sqrt { gR } ...(1)$

From the diagram

in right triangle

$\triangle OMA$

$\Rightarrow$

$MA=OP=OP\cos { \theta }$

At point

$A$ ,

$\theta =60°$

${ \Rightarrow H }_{ A }=OG-OP\\ { \Rightarrow H }_{ A }=R-R\cos { 60° } =\frac { R }{ 2 } ...(2)$

in right triangle

$\triangle OMB$

$\Rightarrow$

$MB=QP=QP\cos { (180-\theta) }$

At point

$B$ ,

$\theta =120°$

${ \Rightarrow H }_{ B }=OG+OQ\\ { \Rightarrow H }_{ B }=R+R\cos { 60° } =\frac { 3R }{ 2 } ...(4)$

Using energy conservation equation at point

$T$ and

$A$

$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg{ H }_{ 1 }...(2)$

putting the value of eqn (1), eqn (2) in eqn (4)

$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg\frac { R }{ 2 }$

$\Rightarrow V_{ A }=2\sqrt { gR }$

Using energy conservation equation at point

$T$ and

$B$

$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B } }^{ 2 }+mg{ H }_{2}...(2)$

putting the value of eqn (1), eqn (3) in eqn (4)

$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B} }^{ 2 }+mg\frac { 3R }{ 2 }$

$\Rightarrow V_{ B }=\sqrt { 2gR }$

Ratio of

${ V }_{ A }$ and

${ V }_{ B }$

$=\sqrt { 2 } :1$

So the correct option is (

$D$ )

A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius 0.75m such that it has speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is:

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

0%
$30N$
0%
$26N$
0%
$20N$
0%
$6N$

Explanation

Given:

$m=0.6 kg; R=0.75 m$

Force equilibrium

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

$=\dfrac{(0.6)(5)^{2}}{(0.75)}$

$=20 N$

A simple pendulum is oscillating with an angular amplitude

$60^{0}$ . If mass of bob is

$50\ g$ , the tension in the string at mean position is:

Consider:

$g=10ms^{-2}$ , length of the string,

$L = 1\ m$ .

0%
$0.5\ N$
0%
$1\ N$
0%
$1.5\ N$
0%
$2\ N$

Explanation

Apply work energy theorem between A and B

$mg L(1-cos\theta)=\dfrac{1}{2}mv^{2}$

$10\times 1\times (1-\dfrac{1}{2})=\dfrac{1}{2}mv^{2}$

$v^{2}=10$

Then, the tension in,

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

$=\dfrac{50\times 10}{1000}+\dfrac{50}{1000}\times \dfrac{10}{1}$

$=1 N$

Regarding linear momentum of a body
a. It is a measure of quantity of motion contained by the body
b. Change in momentum is the measure of impulse
c. Impulse and acceleration act in opposite direction to the change in momentum
d. In the case of uniform circular motion the linear momentum is conserved.

0%
a & b are true
0%
b & c are true
0%
c & d are true
0%
a, b & c are ture

Explanation

Linear momentum (

$\vec{P}$ ) of a body of mass m and velocity

$\vec{v}$ is expressed as:

$\vec{P}=m \times \vec{v}$ .

SO the linear momentum is the measure of the quantity of motion contained by the body.

And impulse of a force is the change in momentum occurring due to the force.
So, Impulse acts in the same direction to the change in momentum.
Acceleration,

$\vec{a}= \dfrac{\vec{F}}{m}=\dfrac{1}{m} \dfrac{d \vec{P}}{dt}$ . So, acceleration also acts in the same direction as impulse.
Uniform circular motion corresponds to zero angular acceleration leading to constant angular momentum. Linear momentum, in this case, is not conserved when an external force is present.

A ball of mass 'm' is projected from the ground with a speed 'u' at an angle '

$\alpha$ ' with the horizontal. The magnitude of the change in momentum of the ball over a time interval from begining till it strikes the ground again is

0%
$\dfrac {mu\sin{ \alpha } }{ 2 }$
0%
$2mu\cos { \alpha }$
0%
$\dfrac { mu{ \cos { \alpha } } }{ 2 }$
0%
$2mu \sin { \alpha }$

Explanation

Over the course of the journey, the velocity of the ball in the x-direction is constant. So the momentum in that direction remains the same.
Initial momentum in y-direction

$p_{yi}=mu sin\alpha$
Final momentum in y-direction

$p_{yf}=-musin\alpha$

$\therefore$ change in momentum

$= -2musin\alpha$
Magnitude of change

$= 2musin\alpha$

A water bucket of mass '

$m$ ' is revolved in a vertical circle with the help of a rope of length '

$r$ '. If the velocity of the bucket at the lowest point is

$\sqrt{7gr}$ . Then the velocity and tension in the rope at the highest point are:

0%
$\sqrt{3gr},2mg$
0%
$\sqrt{2gr},mg$
0%
$\sqrt{gr},mg$
0%
$Zero$ , $Zero$

Explanation

$V_{b} = \sqrt{7gr}$

Work Energy Theorem applied between top and bottom.

$mg(2R) = \dfrac{1}{2}m (V_{b} )^{2} - \dfrac{1}{2}m (V_{T} )^{2}$

$2mgr = \dfrac{7}{2}mgr - \dfrac{1}{2}m (V_{T} )^{2}$

$V_{T} = \sqrt{3gr}$

At top point:

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

$=m \dfrac{3gr}{r} -mg$

$= 2mg$

A person carries a hammer on his shoulder and holds the other end of its light handle in his hand. Let

$y$ be the distance between his hand and the point of support. If the person changes

$y$ , the pressure on his hand will be proportional to:

0%
$y$
0%
$y^{2}$
0%
$\dfrac{1}{y}$
0%
$\dfrac{1}{y^{2}}$

Explanation

As it is clear from the Free body diagram, for the hammer to be in equilibrium condition following should be followed :

$F \times y = w \times (l-y)$
which gives;

$F \times y = w \times l - w \times y$

$y = \dfrac {w \times l}{F+w}$

$\dfrac {F+w}{w \times l} = \dfrac {1}{y}$
Let

$w \times l =d$ and

$\dfrac {1}{l} = c$
Hence the new equation becomes;

$\dfrac {F}{d} + c = \dfrac {1}{y}$
which gives

$F = \dfrac {d}{y} - c\times d$
Let

$c \times d =a$ , which gives

$F+ a = \dfrac {d}{y}$
As "

$a$ " being a constant does not effect proportionality we have

$F$ inversely proportional to

$y$ .

An impulse is supplied to a moving object with the force at an angle of

$120^{0}$ with the velocity vector. The angle between the impulse vector and the change in momentum vector is

0%
$120^{0}$
0%
$0^{0}$
0%
$60^{0}$
0%
$240^{0}$

A pilot of mass

$m$ can bear a maximum apparent weight

$7$ times of

$mg$ . The aeroplane is moving in a vertical circle. If the velocity of aeroplane is

$210\ m/s$ while driving up from the lowest point of vertical circle, the minimum radius of vertical circle should be:

0%
$375\ m$
0%
$420\ m$
0%
$750\ m$
0%
$840\ m$

Explanation

Given maximum apparent weight is

$7mg$
Apparent weight at the lowest point is (

$mg + m \dfrac{v^2}{r}$ )
So,

$7mg = mg + m \dfrac{v^2}{r}$
or,

$6mg = m \dfrac{v^2}{r}$
or,

$r = \dfrac{v^2}{6g}$
or,

$r = \dfrac{210^2}{6g}$
or,

$r = 750\ m$
So , minimum radius of circle is

$750\ m$

A body of mass

$2\ kg$ attached at one end of light string is rotated along a vertical circle of radius

$2\ m$ . If the string can withstand a maximum tension of

$140.6\ N$ , the maximum speed with which the stone can be rotated is:

0%
$22\ m/s$
0%
$44\ m/s$
0%
$33\ m/s$
0%
$11\ m/s$

Explanation

$m= 2 Kg \quad R = 2m$

$T_{max} = 1406\ N$
The tension is max at bottom point.
At bottom:

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

$\dfrac{2(140.6- 2X9.81)}{2} = V^{2}$

${V = 10.99\ m/s}$

A body is revolving in a vertical circle of radius r such that the sum of its

$K.E.$ and

$P.E.$ is constant. If the speed of the body at the highest point is

$\sqrt{2gr}$ then the speed of the body at the lowest point will be:

0%
$\sqrt{7gr}$
0%
$\sqrt{6gr}$
0%
$\sqrt{8gr}$
0%
$\sqrt{9gr}$

Explanation

Applying WET between A and B.

$\dfrac{m(\sqrt{2gR})^{2}}{2} + mg(2R) = \dfrac{m}{2}v_{0}^{2}$

$mgr + 2 mgR = \dfrac{m}{2}v_{0}^{2}$

$v_{0} = \sqrt{6gr}$

A vehicle is travelling with uniform speed along concave road and then along convex road of same radius of curvature. If the normal reaction on the vehicle is half of its weight as it reaches highest point of convex road, then normal reaction on the vehicle as it reaches lowest point of concave road is:

0%
twice weight of vehicle
0%
thrice weight of vehicle
0%
$1.5$ times weight of vehicle
0%
$\sqrt{2}$ times weight of vehicle

Explanation

In case of convex

$mg - N_2 = \dfrac {mV^2}{R} - (1)$

In case of concave

$N_1 - mg = \dfrac {mV^2}{R} - (2)$
Given

$N_2 = \dfrac {mg}{2}$

$\therefore In - (1), \dfrac {mg}{2} = \dfrac {mV^2}{R}$
put it in - (2)

$N_1 - mg = \dfrac {mg}{2}$

$N_1 = \dfrac {3}{2}mg$

A vehicle is travelling at uniform speed

$36\ kmph$ along a fly-over bridge and a person of mass

$50\ kg$ in the vehicle experiences a normal reaction

$390\ N$ as the vehicle crosses the highest point. The radius of curvature of the fly-over bridge is:

0%
$40\ m$
0%
$50\ m$
0%
$80\ m$
0%
$100\ m$

Explanation

$V = 36\ km/h$

$= 36 \times \dfrac {5}{18}$

$= 10\ m/s$

$mg - 390 = \dfrac {mV^2}{R}$

$490 - 390 = \dfrac {50 \times 10}{R}^2$

$100 = \dfrac {5000}{R}$

$R = 50\ m$

The bob of a simple pendulum at rest position is given a velocity

$V$ in horizontal direction so that the bob describes vertical circle of radius equal to length of pendulum

$l$ . If the tension in string is

$4$ times weight of bob when the string is horizontal, the velocity of bob when it is crossing highest point of vertical circle is:

0%
$\sqrt{\dfrac{gl}{2}}$
0%
$\sqrt{gl}$
0%
$\sqrt{\dfrac{3gl}{2}}$
0%
$\sqrt{2gl}$

Explanation

At pt B.

$T = \dfrac{mv^{2}} {L} and T = 4mg.$

$4\dfrac{mg}{m}L = v^{2}$

$V= \sqrt{4gL}$

Now WET between B and C.

$mgL = \dfrac{1}{2}mv^{2} - \dfrac{m}{2}v_{0}^{2}$

$mgL = 2mgL - \dfrac{m}{2}v_{0}^{2}$

$v_{0}= \sqrt{2gL}$

A simple pendulum is oscillating with an angular amplitude

$60^{0}$ . If mass of bob is

$50\ gram$ , the tension in the string at mean position is:

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

0%
$0.5\ N$
0%
$1\ N$
0%
$1.5\ N$
0%
$2\ N$

Explanation

$H = R(1 - cos\ 60^0)$

$H = \dfrac {R}{2}$

Apply Work Energy Theorem,

$mg\dfrac {R}{2} = \dfrac {1}{2}mV^2$

$V = \sqrt {Rg}$

At mean position,

$T - mg = \dfrac {mV^2}{R}$

$T = 2\ mg$

$= 2 \times 0.05 \times 10$

$= 1\ N$

An army vehicle is travelling at uniform speed along convex road then along concave road of same radius of curvature and finally along horizontal road.

$R_{1},R_{2}$ and

$R_{3}$ are normal reactions on the vehicle while crossing highest point of convex road, lowest point of concave road and horizontal road respectively. Which of the following relation is true?

0%
$R_{1}< R_{2}< R_{3}$
0%
$R_{1} > R_{2} > R_{3}$
0%
$R_{2} > R_{3} > R_{1}$
0%
$R_{2}=\dfrac{R_{1}+R_{3}}{2}$

Explanation

At A:

$R_1 = mg-\dfrac{mV^2}{R}$

At B:

$R_2 = mg+\dfrac{mV^2}{R}$

At C:

$R_3 = mg$

$\therefore R_2 > R_3 > R_1$

A particle starts sliding down from the top of the smooth convex hemisphere of radius

$30$

$cm$ , placed horizontally with convex side up. The vertical distance at which particle leaves the surface from it's highest point is:

0%
$5$ $cm$
0%
$10$ $cm$
0%
$15$ $cm$
0%
$20$ $cm$

Explanation

$H = R(1-cos\theta)$
Using Work, Energy Theorem between A and B

$mgR (1 - cos\theta) = \frac {1}{2} mV^2$

$V = \sqrt {2gR(1-cos\theta)}$

at B, normal becoms zero

$mgcos\theta = m\frac {V^2}{R}$

$gcos\theta = 2g - 2gcos\theta$

$cos\theta = \frac {2}{3}$

$H = R(1 - cos\theta)$

$=\dfrac {R}{3}$

$=10\ cm$

The bob of a simple pendulum of mass

$'m'$ is performing oscillations such that the tension in the string is equal to twice the weight of the bob while it is crossing the mean position. The tension in the string when the bob reaches extreme position is :

0%
$\dfrac{mg}{2}$
0%
$mg$
0%
$\dfrac{3mg}{2}$
0%
zero

Explanation

At bottom

$T = 2mg$

$T - mg = \dfrac{mV^2}{R}$

$2mg - mg = \dfrac{mV^2}{R}$

$\sqrt {Rg} = V$
Applying Work Energy Theorem between pt. (1) and (2)

$mg R(1-cos\theta) = \dfrac{1}{2}mV^2$

$mg R(1-cos\theta) = \dfrac{mRg}{2}$

here,

$\theta$ is the angle made by the particle with the axis at its extreme point,

$1-cos\theta = \dfrac{1}{2}$

$\theta = 60^0$
at,

$\theta = 60^0$

$T = mgcos\theta$

$\therefore T = mg\dfrac{1}{2}$

A simple pendulum consists of a light string from which a spherical bob of mass,

$M$ , is suspended. The distance between the point of suspension and the center of bob is

$L$ . At the lowest position, the bob is given tangential velocity of

$\sqrt{5gL}$ . The K.E of the bob when the string becomes horizontal is:

0%
$0$
0%
$\dfrac{MgL}{2}$
0%
$\dfrac{3MgL}{2}$
0%
$\dfrac{5MgL}{2}$

Explanation

From Work Energy Theorem,

$(WD)_G = K_f - K_i$ , where

$(WD)_G$ is work done by gravity

$-MgL = \dfrac{-1}{2}M\times 5gL + K_i$

$K_i = \dfrac {3}{2}MgL$

The bob of a simple pendulum is given a velocity in horizontal direction when the bob is at lowest position ,so that the bob describes vertical circle of radius equal to length of pendulum and tension in the string is

$10 \ N$ when the bob is at an angle

$60^0$ from lowest position of vertical circle. The tension in the string when the bob reaches highest position is (The mass of bob is

$100$ gram.

$g = 10\ ms^{–2}$ ):

0%
$9\ N$
0%
$7 \ N$
0%
$5.5\ N$
0%
$3.5 \ N$

Explanation

For a body of mass

$'m'$ moving in a vertical circle, Tension in the string at highest point and Tension at any angle

$\theta$ with vertical is related as:-

$T-{ T }_{ H }=3mg(1+cos\theta )$

Substituting the values

$10-{ T }_{ H }=3(0.1)(10)(1+cos60^0)=4.5\\ \Rightarrow { T }_{ H }=5.5\ N$

A small block is freely sliding down from the top of a rough inclined plane whose angle of inclination is

$45^{0}$ . The block reaches bottom then it completes a smooth vertical circle. If the coefficient of friction is

$0.5$ , the ratio of minimum vertical height of inclined plane to radius of vertical circle is:

0%
$3:1$
0%
$5:1$
0%
$5:2$
0%
$7:3$

Explanation

$\mu= 0.5$ .

$W_o\ by\ friction = \mu N a$

$=(0.5)(\dfrac{mg}{\sqrt{2}})(\sqrt{2H})$

$=\dfrac{mgH}{2}$

minimum velocity required for completely velocity lively

$=\sqrt{5gR}$ work energy therefore on work

$mgH =\dfrac{mV_o^{2}}{2}+\dfrac{mgH}{2}.$

$\dfrac{mgH}{2}=\dfrac{m}{2}\left ( \sqrt{5gR} \right )^{2}$

$gH = 5gR.$

$\dfrac{H}{R}=\dfrac{5}{1}$

A simple pendulum is oscillating with an angular amplitude

$60^o$ . If

$m$ is mass of bob and

$T_1$ ,

$T_2$ are tensions in the string, when the bob is at extreme position, mean position respectively then is:

$T_1 = \dfrac{mg}{2}$
B)

$T_2 = 2\ mg$
C)

$T_1 = 0$
D)

$T_2 = 3\ mg$

0%
A and B are true.
0%
A and D are true.
0%
B and C are true.
0%
C and D are true.

Explanation

Angular Magnitude is given

$\theta=60^o$

As we see in figure, At extreme position:

$T_A=mg\cos\theta$

$T_A=mg\cos 60^o$

$T_A=T_1=\dfrac{mg}{2}$

As we see in figure, At middle position:

$T_B=mg+F_c$ where

$F_c$ is centrifugal force

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

$v$ is velocity of pendulum.

Using energy conservation for extreme position and middle position

$v=\sqrt{gR}$

putting values :

$T_B=T_2=mg+\dfrac{mgR}{R}=2mg$

Hence option A is correct.

The length of a simple pendulum is '

$L$ '. Its bob from rest position is projected horizontally with a velocity

$\sqrt{\dfrac{7gL}{2}}$ . The maximum angular displacement of bob, such that the string does not slack, is:

0%
$30^o$
0%
$60^o$
0%
$120^o$
0%
$150^o$

Explanation

At point where tension become just zero, the centripetal force must be balanced by gravitational force

$mg cos\theta = \dfrac{mv_{1}^{2}}{L}$ ------ (1)
BY conservation of energy

$\dfrac{1}{2}m \dfrac{7gL}{2} = \dfrac{mv_{1}^{2}}{2}+mgL(1+cos\theta)$

$\sqrt{2gL} \sqrt{\dfrac{3}{4}-cos\theta} = v_{1}$
Put

$v_{1}$ in (1)

$mg$

$cos\theta =$

$\dfrac{m\times2gL}{L}\left ( \dfrac{3}{4}-cos\theta \right ).$

$cos\theta =\dfrac{3}{2}- 2cos\theta$

$cos\theta=\dfrac{1}{2}$

$\theta=60^{0}$

$\therefore$ Total maximum angular displacement

$= 180^0-60^0$

$=120^{0}$

The length of a ballistic pendulum is

$1 m$ and mass of its block is

$1.9 kg$ . A bullet of mass

$0.1 kg$ strikes the block of ballistic pendulum in horizontal direction with a velocity

$100ms^{–1}$ and got embedded in the block. After collision the combined mass (block & bullet) swings away from lowest point. The tension in the string when it makes an angle

$60°$ with vertical is

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

0%
$20\ N$
0%
$30\ N$
0%
$40\ N$
0%
$50\ N$

Explanation

Momentum Conservation

$0.1\times 100= \left ( 1.9+0.1 \right )V_{o}$

$V_o = 5m s^{-1}.$

$H= L(1-cos60^{\circ}). =\dfrac{1}{2}= \dfrac{1}{2}m.$

Energy Conservation

$\dfrac{1}{2}(2)(5)^{2} = \dfrac{1}{2} \times 2 \times V^{2}+2\times10 \times \dfrac{1}{2}$

$25-10= V^{2}.$

$V=\sqrt{15}mvs$

By Force Balance

$T-mg\ cos\ 60^{\circ}= m\dfrac{V^{2}}{L}.$

$T= 2\times10\times\dfrac{1}{2}+2\times\dfrac{15}{1}$

$= 40N.$

A small mass lying at the top of a smooth convex hemisphere is just pushed horizontally. The angle with the vertical where it looses contact with surface is:

0%
$tan^{-1}(\dfrac{2}{3})$
0%
$sin^{-1}(\dfrac{2}{3})$
0%
$cos^{-1}(\dfrac{2}{3})$
0%
$cot^{-1}(\dfrac{2}{3})$

Explanation

At point of loosing contact, mass will still be in circular motion. Gravity is responsible for it.
Conservation of energy
0+0

$=\dfrac{1}{2}mv^{2}-mgR(1-cos\theta)$

$v=\sqrt{2gR(1-cos\theta)}$

At any instant partical looses contact ,the normal reaction on it becomes zero,

$mg cos\theta = \dfrac{mv^{2}}{R.}$

$g cos\theta =2g(1-cos\theta)$ ;

$cos\theta = 2-2cos\theta$

$\theta =cos^{-1}\left ( \dfrac{2}{3} \right ).$

A car is travelling with uniform speed on a flyover bridge which is a part of vertical circle whose radius of curvature is

$60\ m$ . A person in that car feels

$75\%$ decrease in his weight when the car is crossing highest position. The speed of car at that position is:

0%
$28\ m/s$
0%
$21\ m/s$
0%
$14\ m/s$
0%
$7\ m/s$

Explanation

Since the weight decreased by

$75$ %, the centripetal force acting upwards must be equal to

$\dfrac{3mg}{4}$

Hence

$\dfrac{mv^2}{R}=\dfrac{3mg}{4}$

$\implies v=\sqrt{\dfrac{3gR}{4}}$

$\implies v=21\ m/s$

A simple pendulum of length

$50 cm$ is suspended from a fixed point

$O$ and a nail is fixed at a point

$P$ which is vertically below

$O$ at some distance. The bob is released when string is horizontal. The bob reaches lowest position then it describes vertical circle whose centre coincides with point

$P$ . The minimum distance between

$O$ and

$P$ is:

0%
20 cm
0%
25 cm
0%
30 cm
0%
40 cm

Explanation

When the ball just reaches the bottom, from conservation of energy,

$v^2=2gL$

But when it reaches the bottom, it describes a vertical circle. hence

$v^2=5gr$

where

$r$ is the radius of the circle.

Hence

$\dfrac{r}{L}=\dfrac{2}{5}$

$\implies r=20cm$

Hence minimum distance between O and P=

$50cm-20cm=30cm$

Mass of the bob of a simple pendulum of length

$L$ is

$m$ . If the bob is projected horizontally from its mean position with velocity

$\sqrt{4gL}$ , then the tension in the string becomes zero after a vertical displacement of :

0%
$L/3$
0%
$3L/4$
0%
$4L/3$
0%
$5L/3$

Explanation

Let the angle

$\theta$ be when tension becomes zero.

Under this condition

$mg sin\theta = \dfrac{mv^{2}}{L}$ ....... eq(1)

Also, by conservation of energy:

$\dfrac { m (\sqrt {4gL})^{2}}{2} = mg(L+L sin \theta) + \dfrac{mv^{2}}{2}$

$\dfrac{mv^{2}}{L} = 2gmL - 2gmLsin \theta$ ------- eq(2)

From eq (1) and eq(2):

$mg sin\theta = 2mg - 2mg sin\theta$

$sin \theta = 2/3$

So, the tension in the string becomes zero after a vertical displacement of:

$L+Lsin\theta = 5L/3$

Two bodies A, B of masses

$m_1, m_2$ are knotted to a mass-less string at different points rotated along concentric circles in horizontal plane. The distances of A, B from common centre are 50cm, 1m. If the tensions in the string between centre to A and A to B are in the ratio 5:4, then the ratio of

$m_{1}$ to

$m_{2}$ is:

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

Explanation

The tension will be more between center to A as it has to bear the centripetal force of 2 bodies.

applying force balance on the masses,

$T_{1}-T_2=m_1 \omega^{2}r_1$ .........(1)

$T_2=m_2 \omega^{2}r_2$ .........(1)

$\Rightarrow \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\dfrac { 4 }{ 5 } \quad$

Now:

${ r }_{ 1 } = 0.5$ ,

${ r }_{ 2 } =1$

$\dfrac { 5 }{ 4 } = \dfrac { { m }_{ 1 }(0.5) }{ { m }_{ 2 }+1 }$

$\dfrac{m_1}{m_2} = \dfrac{2\times 1}{4} = \dfrac{1}{2}$

A curved road is

$7.5\ m$ wide and its outer edge is raised by

$1.5\ m$ over the inner edge. The radius of curvature is

$50\ m$ . For what speed of the car is this road suited?

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

0%
$5\ m/s$
0%
$10\ m/s$
0%
$15\ m/s$
0%
$20\ m/s$

Explanation

$N sin \theta =\dfrac{mv^2}{R}$

$N cos \theta =mg$

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

$tan \theta =\dfrac{1.5}{7.5}$

$\dfrac{1}{5}=\dfrac{v^2}{50\times 10}$

$U=10 \ m/s$

A block of mass m is in contact with the cart as shown in figure

The coefficient of static friction between the block and the cart is . The acceleration of the cart that will prevent the block from falling satisfies

0%
$\alpha > \frac{mg}{\mu }$
0%
$\alpha > \frac{g}{\mu m}$
0%
$\alpha \geq \frac{g}{\mu }$
0%
$\alpha < \frac{g}{\mu }$

Explanation

$\textbf{Step}$ :

Given a block of mass

$m$ is in contact with the cart.

Here the coefficient of static friction between the block and the cart is

$\mu$ .

Here the figure shows all the forces acting on the block of mass

$m$

The pseudo force

$=m \alpha$

Frictional force

$f_s \geq mg$

Her

$N=m \alpha$

So, frictional force

$\mu_N \geq mg$

$\implies \mu m \alpha \geq mg$

$\implies \alpha \geq \dfrac{g}{\mu}$

STATEMENT-1
It is easier to pull a heavy object than to push it on a level ground.
STATEMENT-2
The magnitude of frictional force depends on the nature of the two surfaces in contact.

0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
0%
STATEMENT -1 is True, STATEMENT-2 is False
0%
STATEMENT -1 is False, STATEMENT-2 is True

Instantaneous power of constant force acting on a particle moving in a straight line under the action of this force

0%
is constant
0%
increases linearly with time
0%
decreases linearly with time
0%
either increases or decreases linearly with time

Explanation

Power is defined as rate of work done

$\implies$ Power,

$P = F.v$

Even though the force is constant, due to the force the velocity increases because of which its power increases.

Also from equations of motion we know that

$v = u + at$

$\implies$ instantaneous power increases linearly with time

A block is freely starts sliding down on a rough inclined plane from a vertical height

$5\ m$ . The block reaches the bottom of inclined plane then it just describes vertical circle of radius

$1.2\ m$ along smooth track. The percentage of energy wasted due to friction is:

0%
$20$ %
0%
$30$ %
0%
$40$ %
0%
$60$ %

Explanation

Minimum height from which the body has to be dropped on a plane to just complete the vertical circular motion is

$\dfrac {5r}{2}=\dfrac {5\times 1.2}{2}=3\ m$
So, a minimum height of

$3\ m$ is enough for the body to complete circular motion of radius

$1.2\ m$ .
But here the body is released from a height of

$5\ m$ , still is just completes the loop, implies that the loss in energy is because of the friction.
The percentage energy wasted due to friction

$=\dfrac {h-h_{min}}{h}\times 100=\dfrac {5-3}{5}\times 100=40$ %

In the following figure all surfaces are assumed to be frictionless and pulley is assumed to be ideal. The block 'A' is projected towards the pulley 'P' with an initial velocity

$u_0$ then select incorrect option

0%
the string would become tight at $t=\dfrac{2u_0}{g}$
0%
the distance travelled by 'A' before the string is taut is $\dfrac{u_0^2}{g}$
0%
the distance travelled by 'B' before string is taut is $\dfrac{2u_0^2}{g}$
0%
the common speed of the blocks just after the string is taut is $\left [ \dfrac{n+2}{n+1} \right ]u_0$

Explanation

When the string is taut again both the block would travel the same distance
hence

$u_0t=\dfrac{1}{2}gt^2$ [as 'B' is falling freely]

hence

$t=\dfrac{2u_0}{g}$ in this time 'A' and 'B' would travel a distance of

$\dfrac{2u_0^2}{g}$

also when same impulse acts along the wire the change in momentum would be same for both blocks

$J = nm\upsilon - nmu_0$ (for block A)

$-J = m\upsilon - m(2u_0)$ (for block B)

$\therefore nm\upsilon - nmu_0= 2mu_0-m\upsilon$

$(n+1)\upsilon=(n+2)u_0$

$\nu=\left [ \dfrac{n+2}{n+1} \right ]u_0$

A body is moving in a vertical circle of radius '

$r$ ' by a string. If the ratio of maximum to minimum speeds is

$\sqrt{3}:1$ , the ratio of maximum to minimum tensions in the string is:

0%
3 : 1
0%
5 : 1
0%
7 : 1
0%
9 : 1

Explanation

Let

$v_1$ and

$v_2$ be the velocities at the lowest and the highest points.

$\therefore \dfrac {v_1}{v_2}=\dfrac {\sqrt 3}{1}$
Conserving energy between the lowest and highest points gives

${v_1}^2={v_2}^2+2gh$

$\Rightarrow 3{v_2}^2={v_2}^2+2gh$

$\Rightarrow {v_2}^2=gh$

$\Rightarrow {v_1}^2=3gh$

$h$ is distance between the lowest and the highest points which is

$2r$ .

$h=2r$

Tensions at the lowest and highest point

$T_1=\dfrac {m{v_1}^2}{r}+mg$

$T_2=\dfrac {m{v_2}^2}{r}-mg$

$\Rightarrow \dfrac {T_1}{T_2}=\dfrac {{v_1}^2/r+g}{{v_2}^2/r-g}$

$\Rightarrow \dfrac {T_1}{T_2}=\dfrac {6gr/r+g}{2gr/r-g}$

$\Rightarrow \dfrac {T_1}{T_2}=7/1$

A block of mass

$m$ is freely sliding down on a smooth inclined plane from its top. The block reaches the bottom of inclined plane and then describes a vertical circle of radius

$r$ along smooth track. If vertical height of inclined plane is

$3r$ , the normal reaction on the block when it is crossing highest point of vertical circle:

0%
$2mg$
0%
$mg$
0%
$\dfrac{mg}{2}$
0%
$Zero$

Explanation

$By\quad conservation\quad of\quad Energy\\ \dfrac { 1 }{ 2 } m{ v }^{ 2 }+mg(2r)=mg(3r)\\ \Rightarrow { v }^{ 2 }=2gr\\ N+mg=\dfrac { m{ v }^{ 2 } }{ r } \Rightarrow N=2mg-mg=mg$

A horse continues to apply a force in order to move a cart with a constant
speed. This is because:

0%
the force applied by the horse balances the gravitational force.
0%
the force applied by the horse balances the force of friction.
0%
Horse does not apply any force.
0%
None of these

Explanation

A horse is harnessed to a cart. If it exerts a force on ( or pulls ) the cart, then by the

$\text{Newton's Third Law of Motion}$ the cart must then exert an equal and opposite force on the horse.

For the horse to move with constant velocity, hence zero net acceleration, the

$\text{frictional force}$ acting on him from the surface and the mechanical force it applies on the art should and indeed balance each other.

So, B. The force applied by the horse balances the force of friction.

A mass

$0.1\ kg$ is rotated in a vertical circle using the cord of length

$1\ m$ , when the cord makes an angle

$60^0$ with the vertical, the speed of the mass is

$3\ m/s$ the resultant radial acceleration of mass in that position is:

0%
$9m/s^{2}$
0%
$4.9m/s^{2}$
0%
$4.1m/s^{2}$
0%
zero

Explanation

Radial acceleration at that position ,

$a_r = \dfrac{v^2}{r} - g cos \theta$

$a_r = 9 - 4.9 = 4.1\ m/s^2$

A tennis ball and an iron ball is thrown towards us with the same velocity. Then, the tennis ball is easy to catch as compared to an iron ball.

0%
True
0%
False

A system of

$n$ particles is free from any external force. Which of the following is true for the magnitude of the total momentum of the system?

0%
It must be zero
0%
It could be non-zero, but it must be constant
0%
It could be non-zero, and it might not be constant
0%
The answer depends on the nature of the internal forces in the system

S.I. unit of momentum is :

0%
$kg\ ms^{-1}$
0%
$kg\ mg^{-2}$
0%
$kg\ mg^2$
0%
$kg\ m^{-1}s^{-1}$

Explanation

Hint: Momentum $p=m\times v$

Correct opton is Option (A) $\bf{\mathrm{kg} \mathrm{m} / \mathrm{s}^{-1}}$

Explanation:

$\bullet$ Momentum depends upon the mass and velocity.

$\bullet$ In terms of equation, the momentum of an object is equal to the mass of the object times velocity of object .

$\bullet$ Momentum = $\bf\text { mass } \times \text { velocity }$

$\bullet$ The SI unit of momentum is

$\mathrm{kg} \mathrm{m} / \mathrm{s}^{-1}$

A rubber ball of mass 250 g hits a wall normally with a velocity

$10 m s^{-1}$ and bounces back with a velocity of

$8 m s^{-1}$ . The impluse is _______N s.

0%
$-0.5$
0%
$+0.5$
0%
$-4.5$
0%
$+4.5$

Explanation

The ball initially has 10 m/s in the positive x - direction and 8 m/s in the negative x-direction
v

$= -8 m/s$

$u = 10 m/s$
Impulse is the change in momentum

$= mv - mu$

$I = 0.25 \times (-8 - 10) N s$

$I = -4.5 Ns$

A force of 10 N acts on a body for 3 microsecond

$(\mu s)$ . Calculate the impulse. If mass of the body is 5 g, calculate the change of velocity.

0%
$6 \times 10^{-3} m s^{-1}$
0%
$3 \times 10^{-3} m s^{-1}$
0%
$8 \times 10^{-3} m s^{-1}$
0%
$16 \times 10^{-3} m s^{-1}$

Explanation

Here,

$F=10 N, t=3 \mu s$

$=3 \times 10^{-6}s$ ;

$M =5 g = 5 \times 10^{-3} kg$
Now, impulse

$= Ft = 10 \times 3 \times 10^{-6}$

$= 3 \times 10^{-5} Ns$
Also, impulse

$=$ change in momentum

$=Mv_2 - Mv_1$

$= M(v_2-v_1)$

$\therefore (v_2 - v_1) = \displaystyle \frac{impulse}{M}=\frac{3 \times 10^{-5}}{5 \times 10^{-3}}$

$=6 \times 10^{-3} m s^{-1}$

The increase in base area leads to decrease in stability of an object.

0%
True
0%
False

..... of Roads is done by raising the outer edges of the road slightly above the level of inner edge, to avoid accidents.

0%
Finishing
0%
Banking
0%
Inclination
0%
None of these

A stone of mass

$1 kg$ tied to a light inextensible string of length

$L=\dfrac{10}{3}m$ is whirling in a circular path of radius

$L$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum is

$4$ and if

$g$ is taken to be

$10 \ m/s^2$ , then speed of stone at the highest point of the circle is

0%
$20 m/sec$
0%
$10\sqrt{3} \ m/sec$
0%
$5\sqrt{2} \ m/sec$
0%
$10 m/sec$

Explanation

Let '

$v$ ' be the velocity at the top point
Velocity at the bottom point is

$= v^2 + 2gd$

$= v^2 + 4gr$
Maximum tension in string appears when the stone is at the bottom of the circle.

$T_{max} = \dfrac{m(v^2 + 4gr)}{r} + mg$

Minimum tension appears in a string when the stone is at the top of the circle.

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

$\dfrac{T_{max}}{T_{min}} = \dfrac{\dfrac{v^2 + 4gr}{r} + g}{\dfrac{v^2}{r} - g}$

$= 4$

On solving the equation we get,

$v$

$= 10\ m/s$

China wares are wrapped in straw or paper before packing. This is the application of concept of :

0%
Impulse
0%
Momentum
0%
Acceleration
0%
Force

A particle of mass

$m$ is projected with a velocity

$6\hat { i } +8\hat { j }$ . Then the magnitude of change in momentum when it just touches ground will be

0%
$0$
0%
$12m$
0%
$16m$
0%
$20m$

Explanation

Initial velocity,

$\overrightarrow { v }_{ i }=6\hat { i } +8\hat { j }$ and final velocity,

$\overrightarrow { v }_{ f }=6\hat { i } -8\hat { j }$
change in momentum

$=\overrightarrow\Delta p=m\left( \overrightarrow{ v }_{ f }-\overrightarrow{ v }_{ i } \right) =-16m\hat { j }$

$\left| \Delta p \right| =16m$

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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