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Test: Motion in a Plane- 2 - NEET MCQ


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5 Questions MCQ Test Physics Class 11 - Test: Motion in a Plane- 2

Test: Motion in a Plane- 2 for NEET 2024 is part of Physics Class 11 preparation. The Test: Motion in a Plane- 2 questions and answers have been prepared according to the NEET exam syllabus.The Test: Motion in a Plane- 2 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Motion in a Plane- 2 below.
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Test: Motion in a Plane- 2 - Question 1

A stone is thrown upwards with initial velocity of 20 m s-1, the height that stone will reach would be:

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

Initial velocity = u = 20 m/s 
Final velocity = v = 0     ( at maximum height , v = 0 )
Acceleration due to gravity(g)  in this case, is taken as negative.
This is because when the direction of motion of object is opposite to "g" , then value of g is taken as -ve.
Hence, g = -9.8 m/s²
Using v2 - u2 = 2gh
02 - 202 = 2 * (- 9.8) * h
- 400 = -19.6 h
h = - 400 /(-19.6) = 20.408 m (approximately)

Test: Motion in a Plane- 2 - Question 2

A particle has an initial velocity of 3i + 4j and an acceleration of 0.4i + 0.3j. Its speed after 10s is:

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

Use equation of kinematics: v = u + at
Initial velocity, u = 3i + 4j
Acceleration, a = 0.4i + 0.3j
Time, t = 10s
v = (3i + 4j) + 10(0.4i + 0.3j)
v = (3i + 4j) + (4i + 3j)
v = 7i + 7j
|v| = √(7+ 72) = 7√2

Test: Motion in a Plane- 2 - Question 3

A body sliding on a smooth inclined plane requires 4 seconds to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth distance starting from the rest at the top?

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

A body start from rest so, u = 0 
Let total distance covered = s
Let a body moves with accerlation = a 
s = 1/2 × a× t2
s = 1/2× a × 42
s = 8a
The one-fourth of total distance, s'  = 8a/4
= 2a
s' = 1/2 × a × t2
2a / a = 1/2 × t2
4 = t2
t = 2 seconds

Test: Motion in a Plane- 2 - Question 4

A car travelling at 36km/h-1 due North turns West in 5 seconds and maintains the same speed. What is the acceleration of the car?

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

Initial velocity of car = 36 km/hr due north
Final velocity of car = 36 km/hr due west

The magnitude of change in velocity:

(since velocity is a vector, so direction has to be taken into account)

Acceleration = (Change in Velocity) / Time 

= [36(√2) * 5/18] / 5
= 2(√2) m/s2 in South West

Test: Motion in a Plane- 2 - Question 5

Water drops fall at regular intervals from a tap that is 5m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?

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

Time taken by first drop to cover 5 cm (u = 0), is: 
h = 1/2 gt2
5 = 1/2 x 10 x t2
t = 1 sec
Hence, the interval is 0.5 sec for each drop.
Now distance fallen by second drop in 0.5 sec
h1 = 1/2 gt2
= 1/2 x 10 x (0.5)2         
= 5 x 0.25 
= 1.25 m
Height above the ground ( of 2nd drop) = 5 - 1.25 = 3.75 m

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