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Rest and Motions Kinematics Obj-1 HC Verma Solutions Vol-1 Chapter-3

Rest and Motions Kinematics Obj-1 (MCQ-1) HC Verma Solutions Ch-3 Vol-1 Concept of Physics for Class-11. Step by Step Solution of Objective -1 (MCQ-1) Questions of Chapter-3 Rest and Motions Kinematics (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Rest and Motions Kinematics Obj-1 (MCQ-1) HC Verma Solutions Ch-3 Vol-1 Concept of Physics for Class-11

Board ISC and other board
Publications Bharti Bhawan Publishers
Chapter-3 Rest and Motions Kinematics
Class 11
Vol  1st
writer HC Verma
Book Name Concept of Physics
Topics Solution of Objective-1 (MCQ-1) Questions
Page-Number 49 , 50

-: Select Topics :-

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Exercise


Rest and Motions Kinematics Obj-1 (MCQ-1)

HC Verma Solutions of Ch-3  Vol-1 Concept of Physics for Class-11

(Page-49)

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Question-1

A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about

(a) 50 km/h towards west

(b) 70 km/h towards south-west

(c) 70 km/h towards north-west

(d) zero

Answer-1

The option (b) 70 km/h towards south-west. is correct

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Explanation:

HC Verma Solutions Rest and Motion Obj-1 Ans-1

HC Verma Solutions Rest and Motion Obj-1 Ans-1.1

Question-2

In figure shows the displacement-time graph of a particle moving on the X-axis.

HC Verma Solutions Rest and Motion MCQ-1 Q -2

 

(a)the particle is continuously going in positive x direction

(b) the particle is at rest

(c) the velocity increases up to a time t0, and them become constant

(d) the particle moves at a constant velocity up to a time t0, and then stops.

Answer-2

The option (d) the particle moves at a constant velocity up to a time t0, and then stops. is correct

Explanation:

The particle moves at a constant velocity up to a time t0 and then stops.

The slope of the x–t graph gives the velocity. In the graph, the slope is constant from t = 0 to t = t0, so the velocity is constant. After t = t0, the displacement is zero; i.e., the particle stops.

Question-3

A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant. Let xA and xB be the magnitude of displacements in the first 10 seconds and the next 10 seconds

(a) xA < xB

(b) xA = xB

(c) xA > xB

(d) the information is insufficient to decide the relation of xA with xB.

Answer-3

The option (d) the information is insufficient to decide the relation of xA with xB. is correct

Explanation:

As velocity and acceleration are in opposite directions, velocity will become zero after some time (t) and the particle will return.

∴0 = u− a t
⇒  t=  u/a

Because the value of acceleration is not given, we cannot say that the particle will return after/before 10 seconds.

Question-4

A person travelling on a straight line moves with a uniform velocity v1 for some time and with uniform velocity v2 for the next equal time. The average velocity v is given by
HC Verma Solutions Rest and Motion MCQ-1 Q -4

Answer-4

The option (a) v = (v1+ v2) / 2  is correct

Explanation:

HC Verma Solutions Rest and Motion Obj-1 Ans-4

Question-5

A person travelling on a straight line moves with a uniform velocity v1 for a distance x and with a uniform velocity v2 for the next equal distance. The average velocity v is given by
HC Verma Solutions Rest and Motion MCQ-1 Q -4

Answer-5

The option (c)  is correct

Explanation:

HC Verma Solutions Rest and Motion Obj-1 Ans-5

Question-6

A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is

(a)  a upward

(b) (g − a) upward

(c) (g − a) downward

(d) g downward

Answer-6

The option (d)  g downward is correct

Explanation:

Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction

Question-7

A person standing near the edge of the top of a building throws two balls A and B. the ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with a speed vA and the ball B this the ground with a speed vB. We have

(a) vA > vB

(b)  vA < vB

(c) vA = vB

(d) the relation between vA and vB depends on height of the building above the ground.

Answer-7

The option (c) vA = vB is correct

Explanation:

Total energy of any particle =

HC Verma Solutions Rest and Motion Obj-1 Ans-7

Both the particles were at the same height and thrown with equal initial velocities, so their initial total energies are equal. By the law of conservation of energy, their final energies are equal.

At the ground, they are at the same height. So, their P.E. are also equal; this implies that their K.E. should also be equal. In other words, their final velocities are equal

Question-8

In a projectile motion the velocity

(a) is always perpendicular to the acceleration

(b)  is never perpendicular to the acceleration

(c) is perpendicular to the acceleration for one instant only

(d)  is perpendicular to the acceleration for two instants.

Answer-8

The option (c) is perpendicular to the acceleration for one instant only is correct

Explanation:

 In projectile motion, velocity is perpendicular to acceleration only at the highest point. Here, velocity is along the horizontal direction and acceleration is along the vertically downward direction.

Question-9

Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit he ground first?

(a) the faster one

(b) the slower one

(c) both will reach simultaneously

(d) depends on the masses.

Answer-9

The option (c) both will reach simultaneously is correct

Explanation:

Because the downward acceleration and the initial velocity in downward direction of the two bullets are the same, they will take the same time to hit the ground and for a half projectile.

HC Verma Solutions Rest and Motion Obj-1 Ans-10

Question-10

The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be

(a) 25 m

(b)  37 m

(c) 50 m

(d) 100 m

Answer-10

The option (d) 100 m is correct

Explanation:

HC Verma Solutions Rest and Motion Obj-1 Ans-10.1

Question-11

Two projectiles A and B are projected with angle of projection 15° for the projectile A and 45° for the projectile B. If RA and RB be the horizontal range for the two projectiles, then

(a) R< RB

(b)  RA > RB

(c)  RA = RB

(d) the information is insufficient to decide the relation of RA with RB.

Answer-11

The option (d) the information is insufficient to decide the relation of RA with RB. is correct

Explanation:

Horizontal range for the projectile,

HC Verma Solutions Rest and Motion Obj-1 Ans-11

Information of the initial velocity is not given in the question

Question-12

A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction.

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