The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle θ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

(1) 2gl(1-sinθ)

(2) 2gl(1+cosθ)

(3) 2gl(1-cosθ)

(4) 2gl(1+sinθ)

Subtopic:  Gravitational Potential Energy |
 73%
From NCERT
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A spring 40 mm long is stretched by the application of force. If 10 N force is required to stretch the spring through 1 mm, then work done to stretch the spring 40 mm is equal to:

1. 84 J 2. 68 J
3. 23 J 4. 8 J
Subtopic:  Elastic Potential Energy |
 71%
From NCERT
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Two springs with spring constants k1 = 1500 N/m and k2 = 3000 N/m are stretched by the same force. The ratio of potential energy stored in the springs will be 

1. 2:1

2. 1:2

3. 4:1

4. 1:4

Subtopic:  Concept of Work | Work Done by Variable Force | Work Energy Theorem | Potential Energy: Relation with Force | Gravitational Potential Energy | Elastic Potential Energy | Power |
 67%
From NCERT
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A block of mass \(2\) kg moving with a velocity of \(10\) m/s on a smooth surface hits a spring of force constant \(80\times10^3\) N/m as shown in the figure. The maximum compression in the spring will be:
               
1. \(5\) cm
2. \(10\) cm
3. \(15\) cm
4. \(20\) cm

Subtopic:  Elastic Potential Energy |
 81%
From NCERT
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A particle of mass 10 kg is moving with velocity of 10x m/s, where x is displacement . The work done by net force during the displacement of particle form x = 4 to x = 9 m is 

1. 1250 J

2. 1000 J

3. 3500 J

4. 2500 J

Subtopic:  Concept of Work |
 70%
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The relation between velocity (v) and time (t) is t, then which one of the following quantity is constant:

1.  Force

2.  Power

3.  Momentum

4.  Kinetic Energy

Subtopic:  Power |
 62%
From NCERT
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A particle is moving on the circular path of the radius (R) with centripetal acceleration ac=k2Rt2. Then the correct relation showing power (P) delivered by net force versus time (t) is 

1. 1

2. 2

3. 3

4. 4

Subtopic:  Power |
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A body is displaced from \((0,0)\) to \((1~\text{m}, 1~\text{m})\) along the path \(x=y\) by a force \(F = (x^2\hat j+y\hat i)~\text{N}.\) The work done by this force will be:
1. \(\frac{4}{3}~\text{J}\)
2. \(\frac{5}{6}~\text{J}\)
3. \(\frac{3}{2}~\text{J}\)
4. \(\frac{7}{5}~\text{J}\)

Subtopic:  Work Done by Variable Force |
 78%
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A force F is applied on a body which moves with a velocity v in the direction of the force, then the power will be

1.  Fv2

2.  Fv

3.  F/v2

4.  F/v

 

Subtopic:  Power |
 93%
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A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is [ MP PMT 1993]

(1) Rg

(2) 2Rg

(3) 2πRg

(4) πRg

Subtopic:  Gravitational Potential Energy |
 83%
From NCERT
PMT - 1993
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