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Test: Motion in a Plane - 2 - NEET MCQ


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Test: Motion in a Plane - 2 - Question 1

A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is   

Detailed Solution for Test: Motion in a Plane - 2 - Question 1

Test: Motion in a Plane - 2 - Question 2

A particle is fired with velocity v making an angle q with the horizontal. What is the magnitude of change in velocity when it is at the highest point?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 2

velocity at the highest point 

velocity at the starting point 

 

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Test: Motion in a Plane - 2 - Question 3

A person can throw a stone to a maximum distance of 100 m. The greatest height to which he can throw the stone is  

Detailed Solution for Test: Motion in a Plane - 2 - Question 3

 

This is a projectile motion problem. Given that, the range is 100m, we have to find the maximum height that a person can throw a stone.
Range is given as R = u2sin2θ / g where u is the initial velocity, θ is the launch angle and g is the acceleration due to gravity.
Since, R=100m, thus u2sin2θ / g=100 and the acceleration due to gravity g = 9.8.
Thus, it can be written that, u2sin2θ =100g where g=9.8.
It is known to us that, the greatest range is achieved using a 45launch angle:
u2sin(2×45)=980
⇒ u2sin90= 980

⇒ u2sin⁡90=980
We know that, sin 90=1,
∴ u= 980
The height is achieved when thrown straight up at 45 angle.
Thus, height H=u2sin2θ / 2g.
Since, u2=980, θ=45 and g=9.8
H=980 / 39.2
⇒ H = 25m

Correct Answer is C.

Test: Motion in a Plane - 2 - Question 4

The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 m/s without hitting the ceiling of the hall is  

Detailed Solution for Test: Motion in a Plane - 2 - Question 4

 

Test: Motion in a Plane - 2 - Question 5

A bomb is dropped from an aeroplane when it is directly above a target at a height of 1254.4 m. The aeroplane is going horizontally with a speed of 150 m/s. The distance by which it will miss the target is  

Detailed Solution for Test: Motion in a Plane - 2 - Question 5

using the formula:
D = u√2H/g
Where D is distance, u is initial velocity, H is height and g is acceleration due to gravity.
Putting the values,
D= 150√2×1254.4/9.8
D= 150√2508.8/9.8
D= 150√256
D= 150×16
D= 2400m
D= 2.4 km

Test: Motion in a Plane - 2 - Question 6

The  range R of projectile is same when its maximum height are h1 and h2. What is the relation between R, h1 and h2?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 6

Range is same for angles of projection θ and 900 - θ

Test: Motion in a Plane - 2 - Question 7

A projectile is fired from level ground at an angle θ above the horizontal. The elevation angle of the highest point as seen from the launch point is related to θ by the relation   

Detailed Solution for Test: Motion in a Plane - 2 - Question 7

It is clear from the adjoining figure, 

 

Test: Motion in a Plane - 2 - Question 8

A projectile has a time of flight T and range R. For a given angle of projection if the time of flight is doubled, then what happens to the range?   

Detailed Solution for Test: Motion in a Plane - 2 - Question 8

 

If T is doubled, then R becomes 4 times. 

Test: Motion in a Plane - 2 - Question 9

hree particles A, B and C are projected from the same point with the same initial speeds making angles 30o, 45o and 60o respectively with the horizontal. Which of the following statements is correct?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 9

when a projectile is projected at an angle θ or at angle ( 90o - θ ) with the horizontal, the horizontal range remains the same. The horizontal range is maximum  when the angle of projection is 45o

Test: Motion in a Plane - 2 - Question 10

A particle is projected from the ground with an initial speed of v at an angle q with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

Detailed Solution for Test: Motion in a Plane - 2 - Question 10

Average velocity = Displacement/Time

Here,  H = maximum height =

R = range = 

nd T = time of flight = 

Test: Motion in a Plane - 2 - Question 11

With what minimum speed a particle be projected from origin so that it is able to pass through a given point (30 m, 40 m)?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 11

Test: Motion in a Plane - 2 - Question 12

A particle is moving along a circular path. The angular velocity, linear velocity, angular acceleration and centripetal acceleration of the particle at any instant respectively are   . Which of the following relations is not correct?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 12
Linear velocity is the tangential velocity angular velocity acts in the same direction. So they can never be perpendicular.
Test: Motion in a Plane - 2 - Question 13

A particle is moving along a circular path with uniform speed. Through what angle does its angular velocity change when it completes half of the circular path?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 13

Angular velocity is always directed perpendicular to the plane of the circular path. Hence, required change in angle = 0o.

Test: Motion in a Plane - 2 - Question 14

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle q1; in the next 2 sec it rotates through an additional angle θ2, the ratio θ21 is   

Detailed Solution for Test: Motion in a Plane - 2 - Question 14

Test: Motion in a Plane - 2 - Question 15

If the equation for the displacement of a particle moving on a circular path is given by  θ  = 2t3 + 0.5 where θ is in radian and t in second, then the angular velocity of the particle at  t = 2 s is  

Detailed Solution for Test: Motion in a Plane - 2 - Question 15

Test: Motion in a Plane - 2 - Question 16

Which of the following statements is false for a particle moving in a circle with a constant angular speed?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 16

For a particle moving in a circle with constant angular speed, the velocity vector is always tangent to the circle and the acceleration vector always points towards the centre of the circle or is always along the radius of the circle.
Since the tangential vector is perpendicular to radial vector, therefore, velocity vector will be perpendicular to the acceleration vector. Acceleration only causes change in direction of velocity at all instants of time because in no case, acceleration vector is tangent to the circle.

Hence option B is the only false statement.

Test: Motion in a Plane - 2 - Question 17

Which motion does not require force to maintain it?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 17

Uniform straight line motion is a type of linear motion. It works according to Newton's First law of motion i.e. objects that do not experience any net force will continue to move in a straight line with a constant velocity until they are subjected to a net force.

Test: Motion in a Plane - 2 - Question 18

Two trains are each 50 m long moving parallel to each other at speeds 10 m/s and 15 m/s respectively, at what time will they pass each other?  

Detailed Solution for Test: Motion in a Plane - 2 - Question 18

 As both the trains are moving in the opposite direction, the relative speed of each train with respect to each other be, v=10+15=25m/s
Here distance covered by each train = sum of their lengths =50+50=100m
∴ Required time=100/25​=4sec

Test: Motion in a Plane - 2 - Question 19

Two cars are moving in same direction with speed of 30 kmph. They are separated by a distance of 5 km. What is the speed of a car moving in opposite direction if it meets the two cars at an interval of 4 min?   

Detailed Solution for Test: Motion in a Plane - 2 - Question 19

Lets say the speed of second car be v. Now with respect to car 1, speed of car 2 is v - 30. Now if the cross each other at just after 4 min then the distance traveled by car 2 relative to car 1 is (v -30) x 4/60 and we know the distance travelled relatively is 5km
Thus we get 
(v -30) x 4/60 = 5
V = 300/4 - 30
= 45 km/h

Test: Motion in a Plane - 2 - Question 20

A man can swim in still water at a speed of 6 kmph and he has to cross the river and reach just opposite point on the other bank. If the river is flowing at a speed of 3 kmph, and the width of the river is 2 km, the time taken to cross the river is (in hours)   

Detailed Solution for Test: Motion in a Plane - 2 - Question 20

To reach the exact opposite end the parallel component of velocity would nullify the velocity of the river, hence we get 6.sin a = 3 , where a is the angle with perpendicular to river flow.
Thus we get sin a = ½ and cos a =√3/2
Thus the perpendicular component of velocity, 6.cos a =  3√3
Thus time taken is 2 / 3√3

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