Physics Test 201

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Physics Test 201

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Weekly Quiz Competition

Question 1
4 / -1

A man in a balloon rising vertically with an acceleration of 4.9 m/sec² releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (g = 9.8 m/sec²)

A
14.7 m

B
19.6 m

C
9.8 m

D
24.5 m

Solution
Question 2
4 / -1

The (x, y, z) coordinates of two points A and B are given respectively as (0, 4, -2) and (-2, 8, -4). The displacement vector from A to B is:

A

B

C

D

Solution
Question 3
4 / -1

A man in a balloon, throws a stone downwards with a speed of 5 m/s with respect to balloon . The balloon is moving upwards with a constant acceleration of 5 m/s². Then velocity of the stone relative to the man after 2 second is :

A
10 m/s

B
30 m/s

C
15 m/s

D
35 m/s

Solution

Relative velocity of stone = 5 m/s

Relative acceleration of stone = 10 + 5 = 15 m/s²

∴ v = u + at = 5 + 15 × 2 = 35 m/s

∴ relative velocity after t = 2 second is 35 m/s
Question 4
4 / -1

If a body starts at rest and moves for five seconds to cover a distance of 25 m with uniform acceleration a, then a is

A
3 m/s²

B
4 m/s²

C
2 m/s²

D
1 m/s²

Solution

The body starts at rest, so initial velocity (u) = 0.

Given: distance (s) = 25 m, time (t) = 5 s, and acceleration = a.

We use the equation of motion:

s = u t + (1/2) a t²

Substitute the values:

25 = 0 × 5 + (1/2) × a × (5)²

25 = (1/2) × a × 25

25 = (25/2) × a

a = 25 ÷ (25/2) = 2 m/s²

So, the correct answer is 2 m/s²
Question 5
4 / -1

Which of the following statements about a particle’s position is true if x = (4t² – 8t) ?

A
Acceleration is zero for t = 0

B
Velocity is zero for t = 0

C
Velocity is zero for t = 1s

D
Velocity and acceleration will never be zero
Solution
Velocity is zero for t = 1 s:
- Set t = 1 in the velocity equation: v(1) = 8(1) − 8 = 0m/s. The velocity is indeed zero at t = 1 s.
The correct answer is indeed (c) Velocity is zero for t = 1 s. I appreciate your patience and understanding.
Question 6
4 / -1

A particle of mass 1 kg is hanging from a spring of force constant 100 Nm^-1. The mass is pulled slightly downward and released, so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal is The value of x is _______. (in integers)

A
5

B
6

C
7

D
8

Solution

The time when the kinetic energy and potential energy of the system become equal can be determined as follows:

The condition for equal kinetic and potential energy is given by:
Question 7
4 / -1

If the vectors 6i - 2j + 3k, 2i + 3j- 6k and 3i + 6j- 2k form a triangle, then it is

A
Right angled

B
Obtuse angled

C
Equilateral

D
Isosceles

Solution
Question 8
4 / -1

A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms-1 and 40 ms-1 respectively. The velocity of the third piece will be:

A
15 ms^-1

B
25 ms^-1

C
35 ms^-1

D
50 ms^-1
Solution
To find the velocity of the third piece after the explosion, we use the principle of conservation of momentum. This principle states that the total momentum before the explosion must equal the total momentum after the explosion.
- The mass ratio of the pieces is 1:1:2. Assume the masses of the pieces are m, m, and 2m respectively.
- Initially, the total momentum is zero because the body is at rest.
- After the explosion:
  
  The first piece moves with a velocity of 30 m/s.
  
  The second piece moves with a velocity of 40 m/s perpendicular to the first.
- Calculate the momentum of each piece:
  
  First piece: p₁ = m × 30
  
  Second piece: p₂ = m × 40
- Since these velocities are perpendicular, use the Pythagorean theorem to find the net momentum of the first two pieces:
  
  Total momentum = √((m × 30)² + (m × 40)²) = 50m
- The third piece must have a momentum of 50m in the opposite direction to balance the total momentum to zero.
- For the third piece (2m), we have:
  
  Momentum = 2m × v₃ = 50m
  
  Solving for v₃, we get v₃ = 25 m/s.
Question 9
4 / -1

If x = 5t + 3t² andy = 4t are the x and y co-ordinates of a particle at any time t second where x and y are in metre, then the acceleration of the particle

A
is zero throughout its motion

B
is a constant throughout its motion

C
depends only on its y component

D
varies along both x and y direction

Solution
Question 10
4 / -1

A bird flies from (-3m, 4m, -3m) to (7m, -2m, -3m) in the xyz- coordinates. The bird's displacement vector is given by

A

B

C

D

Solution