MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 14

The variation of force F acting on a body moving along x-axis varies with its position (x) shown in figure.
The body is in stable equilibrium state at

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P
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Q
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R
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both P & Q

A toy car mass 2 kg moving with constant speed 2 m/s in vertical plane as shown in figure.Tangent at point A is horizontal and radius of curvature at A is 1 m then normal reaction at point A is

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16 N
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20 N
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24 N
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28 N

A car is circulating on a circular path of radius r. At some instant its velocity is v and rate of increase of speed is a. The resultant acceleration of the car will be

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$\sqrt { \dfrac { { v }^{ 2 } }{ { a }^{ 2 } } +{ r }^{ 2 } }$
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$\sqrt { \dfrac { { v }^{ 2 } }{ r } +a }$
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$\sqrt { \dfrac { { v }^{ 4 } }{ { r }^{ 2 } } +{ a }^{ 2 } }$
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$\left( { \dfrac { { v }^{ 2 } }{ { r }^{ } } +{ a }^{ } } \right)$

A particle located in the one-dimensional potential field has potential energy function $U(x) = \dfrac{a}{{{x^2}}} - \dfrac{b}{{{x^3}}}$ where a and b are positive constants. The position of equilibrium corresponds to x?

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$\dfrac{{3a}}{{2b}}$
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$\dfrac{{2b}}{{3a}}$
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$\dfrac{{2a}}{{3b}}$
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$\dfrac{{3b}}{{2a}}$

An electron moves along the line AB, which lies in the same plane as a circular loop of conducting wires as shown in the diagram will be the direction of current induced if any, in the loop

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no current will be induced
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the current will be clockwise
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the current will be anticlockwise
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the current will change as the electron passes by

$F = 2 {x^2} - 3x-2$ . Choose correct option

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$x= -\dfrac{1}{2}$ is position of stable equilibrium
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$x= 2$ is position of stable equilibrium
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$x= -\dfrac{1}{2}$ is position of unstable equilibrium
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$x= 2$ is position of neutral equilibrium

A cyclist is riding at a speed of

$14 \sqrt { 3 } m/s$ and takes a turn around a circular road of radius

$20\sqrt { 3 } m$ . Its inclination to the vertical is

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$30^0$
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$40^0$
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$60^0$
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$70^0$

If the action and reaction forces are always equal in magnitude, then these forces

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will produce accelerations of equal magnitudes.
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may not produce accelerations of equal magnitudes.
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produce velocities of equal magnitudes.
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will not produce accelerations of equal magnitudes

Explanation

According to newton's 3rd law of motion, Every action has equal and opposite reaction. Action and reaction always act in pairs, and on different bodies.

Let two masses

$m_1$ and

$m_2$ are pairs of masses applying force on each other ie

$F$ .

Acceleration of masses will be

$a_1=\dfrac F{m_1}$ and

$a_2=\dfrac F{m_2}$ Hence acceleration of masses also depends on the mass.

So, these forces may not produce accelerations of equal magnitudes.

A bullet of mass

$m$ and speed

$v$ hits a pendulum bob of mass

$M$ at time

$t _ { 1 }$ and passes completely through the bob. The bullet emerges at time

$t _ { 2 }$ with a speed of

$v / 2$ . The pendulum bob is suspended by a stiff rod of length

$l$ and negligible mass. After the collision, the bob can barely swing through a complete vertical circle.At time

$t _ { 3 }$ , the bob reaches the highest position.What quantities are conserved in this process?

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Total kinetic energy of the bob and the bullet during the time interval $\Delta t = t _ { 2 } - t _ { 1 }$
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Total momentum of the bob and the bullet during the time interval $\Delta t = t _ { 2 } - t _ { 1 }$
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Total mechanical energy of the bob and the bullet during the time interval $t _ { 3 } - t _ { 1 }$
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Momentum of the bob after $t _ { 2 }$

A uniform rod of mass

$M$ has an impulse applied at right angles to one end begins to move with speed

$V$ ,the magnitude of the impulse is

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$MV$
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$\dfrac {MV}{2}$
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$3MV$
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$\dfrac {2MV}{3}$

A 10 kg block is pulled a along a frictional surface in the form of an are of a circle of radius 10 m. The applied force F is 200 N as shown in figure If a block started from rest at P

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the velocity at Q would be 20 m/s
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the velocity at Q would be 17 m/s
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the velocity at Q would be 15 .5 m.s
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the velocity at Q would be 18 m/s

Explanation

We know,

change in KE

$=$ work done

$(KE)_f-(KE)_i=F$ (length of arc)

$(KE)_f-O=F\times 60^o\times \dfrac{\pi}{80^o}\times$ (radius) [Initially at rest]

$(KE)_f=\dfrac{\pi F}{3}\times 10$

$(KE)_f=\dfrac{\pi\times 200}{3}\times 10$

$\dfrac{1}{2}mv^2_f=\dfrac{2000\pi}{3}$

$v^2_f=\dfrac{2}{m}\times \dfrac{2000\pi}{3}$

$v_f=\sqrt{\dfrac{2}{10}\times \dfrac{2000\pi}{3}}$

$v_f\approx 20m/s$ ,

1211892_1400901_ans_bcd8aef53bb64fac9202ad3dd50c0e51.jpg

A bucket tied at the end of a

$1.6m$ long string is whirled in a vertical circle with a constant speed. What should be the minimum speed so that the water from the bucket does not spill when the bucket is at the highest position, (Take g=10m/s^2$$)

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$16m/s$
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$6.25m/s$
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$4m/s$
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$2m/s$

A heavy particle hanging from a fixed point by a light inextensible string of length l h projected horizontally with a speed of

$\sqrt { { g }^{ l } }$ . Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion, when the tension in the string is equal to the weight of the particle.

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$\sqrt { gl }$
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$\sqrt { 2gl }$
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$\sqrt { 3gl }$
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$\sqrt { \frac { gl }{ 3 } }$

Which of the following sets of concurrent forces may be in equilibrium?

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$F_1=3N, F_2=5N, F_3=1N$
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$F_1=3N, F_2=5N, F_3=9N$
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$F_1=3N, F_2=5N, F_3=6N$
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$F_1=3N, F_2=5N, F_3=15N$

A stone tied to a piece of string is shrilled in a vertical cirlce with uniform speed. In what position of the stone is the tension in the string greatest

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in the highest position of stone
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in the lowest position of stone
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in the position when string is horizontal
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is same for all positions of stone

A strign of length

$1=1m$ is fixed at one end and carries a mass of

$100gm$ at other end. The string makes

$\sqrt { 5 } \pi$ revolutions per second about a vertical axis passing through its second end. What is the angle of inclination of the string with the vertical?

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$30^0$
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$45^0$
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$60^0$
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$75^0$

The vertical section of the road over a bridge is a circle of radius 8.8 m. A car has its center of mass one meter above the ground. If g is 9.8

$m/{ s }^{ 2 }$ , then the greatest speed at which the car crosses the bridge without losing contact with road at the highest point is :

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7.8 m/s
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10.8 m/s
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9.8 m/s
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9.286m/s

A body of mass 20 Kg moving with a velocity of 3 m/s, rebounds on a wall with same velocity. The impulse on the body is-

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60 Ns
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120 Ns
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30 Ns
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180 Ns

A motorcyclist is moving in circular path of radius

$40\ m$ with

$72\ km/h$ . The angle with vertical at which motorcyclist will bend is

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

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$60^ {o}$
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$30^ {o}$
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$45^ {o}$
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$75^ {o}$

The moment of inertia of a disc about its axis is

$I = 0.1 \ kg-m^2$ . A tangential force of 2 kg wt is applied round the circumference of flywheel with the help of string and mass m. The acceleration of the mass is

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$18.6 \ rad/sec^2$
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$17.6 \ rad/sec^2$
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$14.6 \ rad/sec^2$
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$16.7 \ rad/sec^2$

It is easier to pull a body than to push, because-

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the coefficient of friction is more in pushing
than that in pulling
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the friction force is more in pushing than
that in pulling
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the body does not move forward when
pushed
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None of these

The maximum and minimum tensions in a string of length 0.1 m with a stone 5 g tied to it and whirled in a vertical circle are in the ratio 13 :The velocity of the body at highest point is

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$0.7\ ms^{-1}$
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$0.7\sqrt 3\ ms^{-1}$
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$\sqrt 3\ ms^{-1}$
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$7\sqrt 3 ms^{-1}$

The figure shows an $11.7$ ft wide ditch with the approach roads at an angle of $15^\circ$ with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch? Assume that the length of the bike is $5 ft$ , and it leaves the road when the front part runs out of the approach road.

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$42.7 \ ft \ s^{-1}$
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$36.7 \ ft \ s^{-1}$
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$34.7 \ ft \ s^{-1}$
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$32.7 \ ft \ s^{-1}$

We can say that a body is acted upon by balanced forces

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if it is in rest only
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if it is moving with constant speed
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if even number of forces are acting on it
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if it is not accelerating

Explanation

If a body is acted upon by balanced forces then the net force acting on the body is

$zero$ .

We know from Newton's law of motion

$F=ma$

$F=0$ then

$a=0$

Thus, the body will not accelerate. It will either be at rest or move with a constant velocity.

A stone of mass m tied to a string of length L is whirled in a vertical circle. If the string remains just stretched when the stone is at the top of the circle, the tension in the string when the stone is at bottom of the circle is

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3 mg
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6 mg
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4 mg
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2 mg

A water jet, whose cross sectional are is 'a' strikes a wall making an angle

$'\theta '$ with the normal and rebounds elastically. The velocity of water of density 'd' is v. Force exerted on wall is

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2 $av^2d$ $cos \theta$
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2 $av^2d$ $sin \theta$
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2 $avd$ $cos \theta$
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$avd$ $cos \theta$

A circular road of radius 1000 m has banking angle

$45^{\circ}$ The maximum safe speed

$\left ( in ms^{-1} \right )$ of a car having a mass 2000 kg will be (if the coefficient of the friction between tyre and road is 0.5)

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172
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124
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99
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86

Total impulse imposed by ball of mass 2 kg to the ball of mass 4 kg has the magnitude equal to

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10 N-s
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5.3 N-s
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40 N-s
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16 N-s

A car of mass

$1500kg$ takes a turn of a curved road of radius

$300m$ with a speed of

$9m/s$ . The centripetal force acting on the car is:

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$45N$
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$225N$
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$810N$
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$405N$

The radius of curvature of a road at a certain turn is 50m. The width of the road is 10m and its outer edge is 1.5m higher than the inner edge. The safe speed for such an inclination will be

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$6.5 m/s$
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$8.6 m/s$
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$8 m/s$
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$10 m/s$

In a conical pendulum, the centripetal force

$\left[ \dfrac { { mv }^{ 2 } }{ r } \right]$ acting on the bob is given by

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$\dfrac { mgr }{ \sqrt { { L }^{ 2 }-{ r }^{ 2 } } }$
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$\dfrac { mgr }{ { L }^{ 2 }-{ r }^{ 2 } }$
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$\dfrac { { L }^{ 2 }-{ r }^{ 2 } }{ mgt }$
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$\dfrac { mgL }{ \left[ { L }^{ 2 }-{ r }^{ 2 } \right] ^{ 1/2 } }$

A curved road of 50 m in radius is banked to correct angle for a given speed. If the speed is to be doubled keeping the same banking angle, the radius of curvature of the road should be changed to

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250 m
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100 m
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150 m
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200 m

Rocket propulsion theory is analogues to-

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Newton`s third law
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Relative theory
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First law of thermodynamics
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none

Race car A follows path abcde while race car

$\mathrm{B}$ follows path12

$\mathrm{c} 34$ track. If each car has a constant speed corresponding to a normal acceleration of 8

$\mathrm{m} / \mathrm{s}^{2}$ . The tracks abcde and 2

$\mathrm{c} 3$ are semicircular track while tracks

$1-2$ and

$3-4$ are straight track. Point a and point 1 are the starting points of race and point 4 and point e are finishing points of the race choose the correct statements.

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Car A wins the race with time difference $\dfrac{14+3}{3}$ $s$
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Car A wins the race with time difference $\dfrac{14-3}{3}$ $s$
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Car B wins the race with time difference $\dfrac{14+3}{3}$ $s$
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Car B wins the race with time difference $\dfrac{14-3}{3}$ $s$

A vessel filled with water is placed on the edge of a board as shown in figure. A small board carrying some weight is placed on the water surface. Then choose the correct statement -

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The equilibrium would be violated
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The equilibrium would not be violance
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The vessel would toppic down
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Nothing could be said

Metal wire in the form of a ring of radius

$R$ and mass m placed on a smooth horizontal table is set rotating about its own axis in such a way that each part of the ring moves with speed

$V$ .The tension in the ring Will be;-

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$\dfrac {2\pi m v^2 }{R}$
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$\dfrac {mv^2}{4\pi R}$
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$\dfrac {mv^2}{R}$
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$2 \pi m v^2$

The angle of banking

$\theta$ for a cyclist taking curve is given by

$tan\theta=\dfrac { { v }^{ n } }{ rg }$ , where symbols( v= speed of the cyclist, r= radius of the curved path, g= acceleration due to gravity) have their usual meanings.Then the value of

$n$ is equal to

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$1$
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$3$
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$2$
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$4$

Explanation

Mass of cyclist + cycle

$= M$

radius of curvature of the curved path

$= R$

ANgle of inclination from vertical

$= \theta$

acceleration due to gravity

$= g$

velocity of cyclist

$= v$

Let the normal force : N from the ground on to cyclist. Balance forces in horizontal and vertical directions.

$N sin \theta = \dfrac{M v^2}{ R}$ = centripetal force for the cyclist to move in a curve of radius R

$N cos \theta = M g$ = weight balanced by normal reaction from ground

SO divide equ 1 by equ 2

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

Compairing with the equation

$tan \theta= \dfrac{ v^n}{ Rg}$ as given in question

Required answer is

$n= 2$

A 20 g bullet pierces through a plate of mass

${ m }_{ 1 }=1$ kg and then comes to rest inside a second plate of mass

${ m }_{ 2 }=2.98$ kg. It is found that the two plates, initially at rest, now move with equal velocities. The percentage loss in the initial velocity of bullet when it is in between the plates is (neglect any loss of material due to action of bullet)

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20%
0%
25%
0%
30%
0%
45%

A stone of mass

$1$ kg is tied to a string

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

$40$ m/s in a vertical circle. The ratio of the tension at the top and the bottom is:

$(g=10m/s^2)$

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$11:12$
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$39:41$
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$41:39$
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$12:11$

A stone of mass

$0.3$ kg attached to a

$1.5$ m long string is whirled around in a horizontal circle at a speed of

$6$ m/s. The tension in the string is?

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$10$ N
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$20$ N
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$7.2$ N
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None of these

Two particles A and B having mass m each and charge

$q_1$ and

$-q_2$ respectively, are connected at the ends of a non conducting flexible and inextensible string of the length l. The particle A is fixed and B is whirled along a vertical circle with centre at A. If a vertically upward electric field of strength E exists in the space, then for minimum velocity of particle B:

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The tension force in string at lowest point is zero
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The tension force at the lowest point is non-zero
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The tension force at the highest point is zero
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The work done by interaction force between particles A and B is non-zero

In which example is it not possible for the underlined body to be in equilibrium?

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An aeroplane climbs at a steady rate.
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An aeroplane tows a glider at a constant altitude.
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A speedboat changes direction at a constant speed.
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Two boats tow a ship into harbour.

A uniform bar of mass

$m$ is supported by a pivot at its top about which the bar can swing like a pendulum. If a force

$F$ is applied perpendicular to the lower end of the bar as shown in figure, what is the value of

$F$ in order to hold the bar in equilibrium at an angle

$(\theta )$ from the vertical

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$2mg\sin\theta$
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$mg\sin\theta$
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$mg\cos\theta$
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$\dfrac{mg}{2}\sin \theta$
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$\dfrac{mg}{2}\cos \theta$

Explanation

Let the length of the bar is, then according to the problem,

Refer image,

For equilibrium of the bar

$F^{net}_{ext}=0$ and

$\tau^{net}_{ext}=0$

taking torque about pivot

$0$ ,

$OB\times Mg+OA\times F=0$

$\Rightarrow -OB\,Mg\sin\theta+OA.F\sin 90^0=0$

(clockwise) (anti-clockwise)

$\therefore 1\times F=\dfrac{1}{2}Mg\sin\theta$

$\Rightarrow F=\dfrac{Mg\sin\theta}{2}$

1490828_1698511_ans_ddb5892d941f4c499e5928c4916f7229.png

A heavy stone hanging from a massless string of length

$15$ m is projected horizontally with speed

$147$ m/s. The speed of the particle at the point where the tension in the string equals the weight of the particle is?

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$10$ m/s
0%
$7$ m/s
0%
$12$ m/s
0%
None of these

Explanation

Let at angle

$\theta$ from vertical

speed be

$u$ m/s and tension equal weight of particle i.e

$\boxed{T=mg}$ .........(1) [given]

at this location,

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

$mg-mg\cos\theta=\dfrac{mu^2}{l}$ [From (1)]

$\boxed{lg(1-\cos\theta)=u^2}$

by work energy theorem

$W_{gravity}=\Delta K.E$

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

$-2gl(1-\cos\theta)=u^2-v^2$

$-2u^2=u^2-v^2$ [From (2)]

$3u^2=v^2$

$u=\dfrac{v}{\sqrt{3}}=\dfrac{147}{\sqrt{3}}=85m/s$

So answer is

$85m/s$

1536487_1702179_ans_df34943380ce467584dcab13c3bc550e.png

A particle originally at rest at the hoghest point of a smooth vertical circle is slightly displaced. it will leave the circle at a vertical distance h below the highest point, such that h is equal to

0%
R
0%
R/4
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R/2
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R/3

A block of mass 10 kg is suspended through two light spring balances as shown below

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Both the scales will read 5 kg
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The upper scale will read 10 kg & the lower zero
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Both the scale will read 10 kg
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The readings may be anything but their sum will be 10 kg

The tension in the string at the highest position is

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$166\ N$
0%
$67.56\ N$
0%
$68\ N$
0%
$152\ N$

In the following figure, the pulley

$P_1$ is fixed and the pulley

$P_2$ is movable. If

$W_1=W_2=100N$ , what is the angle

$AP_2P_1$ ?
The pulley are fritionless and assume equilibrium.

0%
$30^o$
0%
$60^o$
0%
$90^o$
0%
$120^o$

Explanation

Let the angle

$Ap_2p_1 = \theta$

By glancing force:

$T = W_1 = 100N \\ T\cos\theta/2+T\cos\theta/2 = w_2 \\ 2T\cos\theta/2 = 100N \\ 2 *100 * \cos\theta/2 = 100 \\ \cos\theta/2 = \dfrac{1}{2} \\ \dfrac{\theta}{2} = 60 \\ \theta = 120$

Hence option D is correct.

In the diagram shown the ground is smooth and

$F_1$ &

$F_2$ are both horizontal forces .The mass of the upper block is 10 kg while that of lower block is 15 kg. The correct statement is :

0%
$m_1$ experiences frictional force towards west only if $F_1 > F_2$
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If $F_1 \neq F_2$ then it is possible to keep the system in equilibrium for certain suitable value of $F_1$ & $F_2$
0%
If the system is to remain in equilibrium then $F_1$ must be equal to $F_2$ & $F_2 \le 10 N$
0%
If $\frac {F_1}{m_1} = \frac {F_2}{m_2}$ then frictional force between the blocks is zero blocks is zero.

Find impulse acted on the ball during the time interval

$\triangle t$

0%
$\frac { 2MJ}{ ( 3m + M) }$
0%
$\frac { 2MJ}{ ( m + 3M) }$
0%
$\frac { 2mJ}{ ( 3m + M) }$
0%
$\frac { 2mJ}{ ( m + M) }$

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 14 - 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 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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