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# Test: Motion In A Plane 1 - From Past 28 Years Questions

## 31 Questions MCQ Test Physics 28 Years Past year papers for NEET/AIPMT Class 11 | Test: Motion In A Plane 1 - From Past 28 Years Questions

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This mock test of Test: Motion In A Plane 1 - From Past 28 Years Questions for NEET helps you for every NEET entrance exam. This contains 31 Multiple Choice Questions for NEET Test: Motion In A Plane 1 - From Past 28 Years Questions (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Motion In A Plane 1 - From Past 28 Years Questions quiz give you a good mix of easy questions and tough questions. NEET students definitely take this Test: Motion In A Plane 1 - From Past 28 Years Questions exercise for a better result in the exam. You can find other Test: Motion In A Plane 1 - From Past 28 Years Questions extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

### A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km/hr is [1998, 2000]

Solution:

Speed along the shortest path

4 km/hr

Speed of water

QUESTION: 2

Solution:

QUESTION: 3

### Two particles of mass M and m are moving in a circle of radii R and r. If their time-periods are same, what will be the ratio of their linear velocities? [2001]

Solution:

Linear velocity v = rω

[ω is same in both cases because time period is same]

QUESTION: 4

The angle between the two vectors will be

Solution:

QUESTION: 5

From a 10 m high building a stone 'A' is dropped, and simultaneously another identical stone 'B' is thrown horizontally with an initial speed of 5 ms–1. Which one of the following statements is true?[2002]

Solution:

h and g are same for both the balls, so time of fall ‘t’ will also be the same for both of them (h is vertical height]

QUESTION: 6

A body of 3 kg moves in the XY plane under the action of a force given by . Assuming that the body is at rest at time t = 0, the velocity of the body at t = 3s is [2002]

Solution:

QUESTION: 7

The vector sum of two forces is perpendicular to their vector differences. In that case, the forces[2003]

Solution:

QUESTION: 8

A particle moves along a circle of radius  with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is [2003]

Solution:

Circumference

Distance travelled in 2 revolutions  = 2 × 40 = 80 m
Initial velocity = u = 0
Final velocity v = 80m/sec
Applying the formula, v2 = u2 + 2as
(80)2 = 02 + 2 × a × 80 ⇒ a = 40 m/sec2

QUESTION: 9

If then the value of is

Solution:

QUESTION: 10

A stone is tied to a string of length ℓ and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed u.
The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is [2004]

Solution:

QUESTION: 11

If a vector is perpendicular to the vector then the value of α is  [2005]

Solution:

For two vectors to be perpendicular to each other

QUESTION: 12

If the angle between the vectors  the value of the product is equal to [2005]

Solution:

QUESTION: 13

Two boys are standing at the ends A and B of a ground where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and catches the oth er boy in a time t, where t is [2005]

Solution:

Velocity of A relative to B is given by

By taking x-components of equation (1), we get

By taking Y-components of equation (1), we get

.....(3)

Time taken by boy at A to catch the boy at B is given by

[From equation (1)]

QUESTION: 14

A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone? [2005]

Solution:

and direction along the radius towards the centre.

QUESTION: 15

The cir cular motion of a particle with constant speed is [2005]

Solution:

In circular motion of a particle with constant speed,  particle repeats its motion after a regular interval of time but does not oscillate about a fixed point. So, motion of particle is periodic but not simple harmonic.

QUESTION: 16

A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 seconds in every circular loop. The average velocity and average speed for each circular loop respectively, is [2006]

Solution:

Distance covered in one circular loop = 2πr = 2 × 3.14 × 100 = 628 m

Speed

Displacement in one circular loop = 0

Velocity

QUESTION: 17

For angles of projection of a projectile (45° – θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio of [2006]

Solution:

(45º – θ) & (45º + θ) are complementary angles as 45º – θ + 45º + θ = 90º. We know that if angle of projection of two projectiles make complementary angles, their ranges are equal.
In this case also, the range will be same. So the ratio is 1 : 1.

QUESTION: 18

The vectors are such that  The angle between the two vectors is [1991, 1996, 2001, 2006]

Solution:

So, angle between A & B is 90º.

QUESTION: 19

A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are . The path of the particle makes with the x-axis an angle of [2007]

Solution:

Let θ be the angle which the particle makes with x axis.
From figure,

QUESTION: 20

are two vectors and θ is the angle between them, if , the valueof θ is[2007]

Solution:

QUESTION: 21

Three forces acting on a body are shown in the figure. To have the resultant force only along the y- direction, the magnitude of the minimum additional force needed is: [2008]

Solution:

Th e componen ts of 1N and 2N forces along + x axis = 1 cos60° + 2 sin30°

The component of 4 N force along –x-axis

Therefore, if a force of 0.5N is applied along + x-axis, the resultant force along x-axis will become zero and the resultant force will be obtained only along y-axis.

QUESTION: 22

A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be: [2008]

Solution:

The magnitude of the resultant velocity at the point of projection and the landing point is same.

Clearly, change in momentum along horizontal (i.e along x-axis) = mvcosθ – mv cosθ = 0
Change in momentum along vertical (i.e. along y–axis) = mv sinθ – (–mv sinθ) = 2 mvsinθ = 2mv × sin 45°

Hence, resultant change in momentum

QUESTION: 23

A block of mass m is in contact with the cart C as shown in the Figure. [2010]

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

Solution:

Forces acting on the block are as shown in the fig. Normal reaction N is  provided by the force mα due to acceleration α

∴ N = mα For the block not to fall, frictional force,

QUESTION: 24

Six vectors, have the magnitudes and directions indicated in the figure. Which of the following statements is true? [2010]

Solution:

Using the law of vector addition, is as shown in the fig.

QUESTION: 25

A particle moves in a circle of radius 5 cm with constant speed and time period 0.2πs. The acceleration of the particle is [2011]

Solution:

Centripetal acceleration ac = ω2r

QUESTION: 26

A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s2, the range of the missile is

Solution:

For maximum range, the angle of projection, θ = 45°.

QUESTION: 27

A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is [2011M]

Solution:

QUESTION: 28

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectiles is : [2012]

Solution:

Horizontal range

....(1)

Maximum height

....(2)

According to the problem   R = H

QUESTION: 29

A particle ha s initial velocity and acceleration . The magnitude of velocity after 10 seconds will be : [2012]

Solution:

QUESTION: 30

The velocity of a projectile at the initial point A is m/s. It’s velocity (in m/s) at point B is [NEET 2013]

Solution:

At point B the direction of velocity component of the projectile along Y - axis reverses.

QUESTION: 31

Vectors are such that  and Then the vector parallel to is [NEET Kar. 2013]

Solution:

Vector triple product