Physics Test 200

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Physics Test 200

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Weekly Quiz Competition

Question 1
4 / -1

A large mass M moving with velocity v makes an elastic head-on collision with a small mass m at rest. What will be the expression for energy lost by mass M?

A

B

C

D

Solution
Question 2
4 / -1

A body’s displacement in metres fluctuates as y = t²– t – 2 with time t in seconds. The displacement is zero for a value of t equal to

A
1 s

B
2 s

C
3 s

D
4 s

Solution

To find when the displacement (y) is zero, we set the displacement equation equal to zero and solve for t:

y = t² − t − 2

Setting y to zero:

0 = t² − t − 2

Now, we can factor the quadratic equation or use the quadratic formula to find the values of t when y is zero.

Factoring:

(t − 2)(t + 1) = 0

Setting each factor to zero:

t − 2= 0 or t + 1 = 0

Solving for t:

t = 2 or t = −1

Since time (t) cannot be negative in this context, we discard t = −1.

Therefore, the correct answer is (b) 2 s. The displacement is zero when t = 2 seconds.
Question 3
4 / -1

A mass 1 kg is suspended by a thread. It is

(i) lifted up with an acceleration 4.9 ms^–2.

(ii) lowered down with an acceleration 4.9 ms^–2. The ratio of tensions in the thread is

A
3 : 1

B
1 : 3

C
1 : 2

D
2 : 1

Solution
Question 4
4 / -1

Four girls skating on circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in figure. For which girl displacement is equal to the actual length of path?

A
A

B
B

C
C

D
D

Solution
Question 5
4 / -1

A
53^o with x-axis

B
37^o with x-axis

C
30^o with y-axis

D
60^o with y-axis

Solution
Question 6
4 / -1

A block of mass 'm' is slipping down a rough inclined plane with constant speed. The force on block by plane is -

A
mg

B
mg /2

C
Depends upon coefficient of friction

D
Depends upon angle of inclination

Solution

Since the block is moving down with constant speed, the net force acting on it along the plane is zero. This means the component of gravitational force down the incline is balanced by the frictional force. The force exerted by the plane on the block is the normal reaction, which depends on the angle of inclination. The normal force is given by:

N = mg cosθ

Thus, the force on the block by the plane is determined by the angle of inclination.
Question 7
4 / -1

What is the mass of a body that accelerates at a rate of 2.6 m/s² with a force of 90 N?

A
44.6 kg

B
34.6 kg

C
54.6 kg

D
None of the mentioned

Solution

Newton's second law of motion relates force (F), mass (m), and acceleration (a) with the equation:

F = m ⋅ a

We can rearrange this equation to solve for mass (m):

m = F/a

In this case, the force (F) is given as 90 N, and the acceleration (a) is given as 2.6 m/s².

m = 90N / 2.6m/s²

Calculating this gives:

m ≈ 34.615kg

So, the closest answer is (b) 34.6 kg.
Question 8
4 / -1

The speed of boat is 5 km/h in still water. It crosses a river of width 1 km along shortest possible path in 15 min. The velocity of river water is

A
1 km/h

B
3 km/h

C
4 km/h

D
5 km/h

Solution

Step 1: Understanding the problem

Speed of boat in still water = 5 km/h

Width of river = 1 km

Time taken to cross the river (shortest path, straight across) = 15 minutes
(15 min = 15/60 = 0.25 hours)

We need to find: Speed of river water

Step 2: Finding effective speed across the river

To cross the river via shortest possible path means going straight across the river without drifting downstream.

Effective speed across the river = distance ÷ time

Effective speed = 1 km ÷ 0.25 hr = 4 km/h

So, the boat moves straight across at a speed of 4 km/h.

Step 3: Using Pythagoras theorem

Since the boat's speed of 5 km/h is at an angle to counteract the river's current, we have a right triangle:

Boat’s speed in still water (hypotenuse) = 5 km/h

Effective speed across the river (vertical component) = 4 km/h

Speed of river current (horizontal component) = ?

According to Pythagoras theorem:

Boat speed² = (effective speed)² + (river speed)²

Thus:

5² = 4² + (river speed)²
25 = 16 + (river speed)²

Step 4: Solving for the river speed

(river speed)² = 25 – 16 = 9

river speed = √9 = 3 km/h
Question 9
4 / -1

A body starts off at rest and moves in a straight line with a constant acceleration. If its velocity is 8 m/s with a displacement of 32 m, its acceleration is

A
1 m/s²

B
2 m/s²

C
3 m/s²

D
4 m/s²

Solution

We can use the following kinematic equation to relate initial velocity (u), final velocity (v), acceleration (a), and displacement (s):

v² = u²+ 2as

In this scenario, the body starts at rest, so the initial velocity u is 0 m/s. The final velocity v is given as 8 m/s, and the displacement s is 32 m. We want to find the acceleration a.

8²= 0²+ 2a(32)

64 = 64a

a = 1 m/s²

So, the correct answer is (a) 1 m/s².
Question 10
4 / -1

A bomb of mass 4 kg explodes in air into two pieces of masses 3 kg and 1 kg. The smaller mass goes at a speed of 90 m/s. The total energy imparted to two fragments is.

A
2.4 kJ

B
5.4 kJ

C
5.9 kJ

D
3.8 kJ

Solution

By conservation of momentum we get the speed of the bigger part let say, v = 1 x90 / 3

Hence we get v = 30

Thus the total KE of the system after collision is ½ (3 X 900 + 1 X 8100)

Thus KE = ½ (10800) = 5400

Now if we apply WET to the system, as no external force has acted upon it, we get

W = ΔKE

= 5400 - 0

= 5.4 kJ