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Test: Displacement & Velocity - NEET MCQ


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20 Questions MCQ Test Physics Class 11 - Test: Displacement & Velocity

Test: Displacement & Velocity for NEET 2024 is part of Physics Class 11 preparation. The Test: Displacement & Velocity questions and answers have been prepared according to the NEET exam syllabus.The Test: Displacement & Velocity MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Displacement & Velocity below.
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Test: Displacement & Velocity - Question 1

The displacement of a particle is given by x = (t – 2)2 where x is in meters and t in seconds. The distance covered by the particle in the first 4 seconds is:

Detailed Solution for Test: Displacement & Velocity - Question 1

 

Test: Displacement & Velocity - Question 2

The ratio of displacement to distance is:

Detailed Solution for Test: Displacement & Velocity - Question 2
  • Displacement is the shortest length between the starting and the end pointing of a journey, whereas the distance is the actual length of the path travelled.
  • Hence, the distance will always be equal to or greater than the displacement.
  • So, the ratio of displacement to distance will be always less than or equal to 1, since the denominator(distance) is either greater than or equal to the numerator (displacement).
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Test: Displacement & Velocity - Question 3

Observe the given distance-time graph.

Identify the respective speeds of the body when it moves from O to A and B to C.

Detailed Solution for Test: Displacement & Velocity - Question 3

The speeds of the body when it moves from O to A  is

The speeds of the body when it moves from B to C  is

Test: Displacement & Velocity - Question 4

The location of a particle is changed. What can we say about the displacement and distance covered by the particle?

Detailed Solution for Test: Displacement & Velocity - Question 4

 Displacement is a vector quantity that depends on the initial and final positions of the particle. If a particle returns to its original position after moving, its displacement is zero, even though the distance covered is non-zero. Therefore, it's possible for the displacement to be zero if the particle returns to its starting point, while the distance traveled is non-zero. Hence, one of the two quantities, displacement or distance, may be zero, but not both.

Test: Displacement & Velocity - Question 5

A particle is moving along a circle of radius R such that it completes one revolution in 40 seconds. What will be the displacement after 2 minutes 20 seconds?

Detailed Solution for Test: Displacement & Velocity - Question 5

1 revolution in 40 seconds
In 1 second it covers = 1/40 revolution
In 140 sec = 1/40 * 140 =7/2 rotation = 3 and half rotation

Then the particle will be on the diametrically opposite end. 

Therefore, Displacement = R + R = 2R

Test: Displacement & Velocity - Question 6

The numerical ratio of average velocity to average speed is:

Detailed Solution for Test: Displacement & Velocity - Question 6
  • When an object is moving along a straight path, the magnitude of average velocity is equal to the average speed.
  • Therefore, the numerical ratio of average velocity to average speed is one in a straight line motion.
  • But, during curved motion, the displacement < distance covered, so the velocity < speed.

Therefore, the numerical ratio of average velocity to average speed is equal to or less than one but can never be more than one. 

Test: Displacement & Velocity - Question 7

Two escalators move people up and down between floors of a shopping mall at the same rate, either up or down. What of the following statements regarding the motion of the escalators is true?

Detailed Solution for Test: Displacement & Velocity - Question 7

As the particles are moving at the same rate, their speed is the same but velocities will be different beacuse their directions will be different.

Test: Displacement & Velocity - Question 8

The displacement of a particle starting from rest (at t = 0) is given by s = 6t2-t. The time in seconds at which the particle will attain zero velocity again, is

Detailed Solution for Test: Displacement & Velocity - Question 8

Apply differentiation
v = ds/dt
v = 12t - 3t2
0 = 3t (4 - t)
⇒ t = 0 or 4
Here t = 0 is not realistic so at t = 4 the velocity will be zero.
Thus, the particle will attain zero velocity at the time of 4 seconds.

Test: Displacement & Velocity - Question 9

If a body does not change its direction during the course of its motion, then ______.

Detailed Solution for Test: Displacement & Velocity - Question 9

Since the object does not change its direction it means that the object is traveling in a straight line.
⇒ The path length and displacement will be equal.

Test: Displacement & Velocity - Question 10

36 km/h is equal to:

Detailed Solution for Test: Displacement & Velocity - Question 10

1 km = 1000 m and 1 hr = 3600 sec.

Therefore,

Test: Displacement & Velocity - Question 11

A car moves with a speed of 60 km/h for the first half an hour and with a speed of 45 km/h in the next half an hour. What is the total distance covered by the car?  

Detailed Solution for Test: Displacement & Velocity - Question 11

Calculation: 

the first half an hour when the car speed = 60 km/h.

the next half an hour when the car speed = 45 km/h.

Distance covered in the first half an hour = Speed × Time

Distance covered in the first half an hour = 60 km/h × 0.5 h

Distance covered in the first half an hour D1 = 30 km

Next, let's calculate the distance covered in the next half an hour when the car is moving at a speed of 45 km/h.

Distance covered in the next half an hour = Speed × Time

Distance covered in the next half an hour = 45 km/h × 0.5 h

Distance covered in the next half an hour D2 = 22.5 km

Total distance  = Distance covered in the first half an hour + Distance covered in the next half an hour

Total distance covered = 30 km + 22.5 km

Total distance covered = 52.5

Test: Displacement & Velocity - Question 12

Suppose our school is 1 km away from our house, and we go to school in the morning, and in the afternoon we come back. Then, the total displacement for the entire trip is:

Detailed Solution for Test: Displacement & Velocity - Question 12
  • Displacement quantifies both direction and distance of an imaginary motion along a straight line from the initial position to the final position of the point.
  • Here, since the same amount of distance is covered in the opposite direction, the total displacement for the entire trip is zero. 
Test: Displacement & Velocity - Question 13

The slope of velocity-time graph for motion with uniform velocity is equal to:

Detailed Solution for Test: Displacement & Velocity - Question 13

When velocity is uniform, velocity-time graph is straight line parallel to time axis. So, the slope is zero.

Test: Displacement & Velocity - Question 14

A man starts from his home with a speed of 4 km/h on a straight road up to his office 5 km away and returns to home, then the path length covered is:

Detailed Solution for Test: Displacement & Velocity - Question 14

The path length covered by the man is the total distance traveled, not the displacement. Here’s the breakdown:

  1. Distance from home to office: 5 km
  2. Return distance from office to home: 5 km

So, the total path length covered is:

5 km (to office)+5 km (back home)=10 km

Therefore, the correct answer is:

4. 10 km

Test: Displacement & Velocity - Question 15

Jagadeesh, on driving his way to school, calculates the speed for the trip to be 20 km/hr. After reaching the school he found the school was closed. So he immediately started returning home. While on his return trip, due to less traffic, he calculates the speed to be 40 km/hr. Calculate Jagadeesh's average speed for the entire journey.

Detailed Solution for Test: Displacement & Velocity - Question 15

Let t1 and t2 be the time taken for Jagadeesh to go to school and then back home. If s is the displacement, then

Total time taken

Total displacement=2 s
Avg Speed = 

Test: Displacement & Velocity - Question 16

A car covers a distance of 5 km in 5 mins, its average speed is equal to:

Detailed Solution for Test: Displacement & Velocity - Question 16

5 km is the distance covered in 5 mins, which means 1 km is the distance covered in 1 min
Hence, it will travel 60 km in 60 mins, i.e. speed = 60 km/hr

Test: Displacement & Velocity - Question 17

A car travels first half of the displacement between two places with a velocity of 40 km/hr and the second half at 60 km/hr. Calculate the average velocity for the entire journey.

Detailed Solution for Test: Displacement & Velocity - Question 17

Given,

v1=40 km/hr and v2=60 km/hr.

The average velocity when the car covers equal displacements would be,

Test: Displacement & Velocity - Question 18

Which of the following is not an example of a rectilinear motion?

Detailed Solution for Test: Displacement & Velocity - Question 18

Rectilinear motion is another name for straight-line motion
A body is said to experience rectilinear motion if any two particles of the body travel the same distance along two parallel straight lines.

A car moving in a circular path changes its direction at every instant. Thus, it is not the rectilinear motion.

Test: Displacement & Velocity - Question 19

A girl is running around a circular track with a uniform speed of 10 ms-1. What is the average velocity for movement of the girl from A to B ( in ms-1)?

Detailed Solution for Test: Displacement & Velocity - Question 19


The displacement made by the girl is AB

Test: Displacement & Velocity - Question 20

An upward, straight sloping line (left to right) on a distance time graph means:-

Detailed Solution for Test: Displacement & Velocity - Question 20

An upward, straight sloping line on a distance-time graph indicates constant speed because the distance covered is increasing at a steady rate over time. If the line was sloping upwards but becoming steeper, it would indicate speeding up. If the line was sloping upwards but becoming less steep, it would indicate slowing down. However, since the line is straight and upward, it suggests that the object is moving at a constant speed.

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