If a curve is governed by the equation y = sinx, then the area enclosed by the curve and x-axis between x = 0 and x = π is (shaded region):

              
1. \(1\) unit
2. \(2\) units
3. \(3\) units
4. \(4\) units

Subtopic:  Integration |
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The acceleration of a particle starting from rest varies with time according to relation, a=α t+β. The velocity of the particle at time instant \(t\) is: \(\left(\text{Here,}~ a=\frac{dv}{dt}\right)\)

1. αt2+βt

2. αt2+βt2

3. αt22+βt

4. 2αt2+βt

Subtopic:  Integration |
 85%
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The displacement of the particle is zero at \(t=0\) and at \(t=t\) it is \(x\). It starts moving in the \(x\)-direction with a velocity that varies as \(v = k \sqrt{x}\), where \(k\) is constant. The velocity will: (Here, \(v=\frac{dx}{dt}\))

1. vary with time.
2. be independent of time.
3. be inversely proportional to time.
4. be inversely proportional to acceleration.
Subtopic:  Integration |
 52%
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The acceleration of a particle is given as a=3x2.  At t = 0, v = 0 and x = 0. It can then be concluded that the velocity at t = 2 sec will be: (Here, \(a=v\frac{dv}{dx}\))

1.  0.05 m/s

2. 0.5 m/s

3. 5 m/s

4. 50 m/s 

Subtopic:  Integration |
 62%
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The acceleration of a particle is given by \(a=3t\) at \(t=0\), \(v=0\), \(x=0\). The velocity and displacement at \(t = 2~\text{sec}\) will be:
\(\left(\text{Here,} ~a=\frac{dv}{dt}~ \text{and}~v=\frac{dx}{dt}\right)\)
1. \(6~\text{m/s}, 4~\text{m}\)
2. \(4~\text{m/s}, 6~\text{m}\)
3. \(3~\text{m/s}, 2~\text{m}\)
4. \(2~\text{m/s}, 3~\text{m}\)

Subtopic:  Integration |
 85%
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The 9 kg block is moving to the right with a velocity of 0.6 m/s on a horizontal surface when a force F, whose time variation is shown in the graph, is applied to it at time t = 0. Calculate the velocity v of the block when t= 0.4s. The coefficient of kinetic fricton is μk=0.3[This question includes concepts from Work, Energy & Power chapter]

1. 0.6 m/s

2. 1.2 m/s

3. 1.8 m/s

4. 2.4 m/s

Subtopic:  Friction |
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The relationship between force and position is shown in the figure given (in one dimensional case). Find the work done by the force in displaying a body from x= 1 cm to x= 5cm is [This question includes concepts from Work, Energy and Power chapter]

1. 10 erg

2. 20 erg

3. 30 erg

4. 40 erg

Subtopic:  Concept of Work |
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The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. The mass of the body is 25 kg and initial velocity is 2 m/s. When the distance covered by the body is 4 m, its kinetic energy would be  [This question includes concepts from Work, Energy & Power chapter]

1. 50 J

2. 40 J

3. 20 J

4. 10 J

Subtopic:  Work Energy Theorem |
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A constant force F is applied on a body. The power (P) generated is related to the time elapsed (t) as [This question includes concepts from Work, Energy and Power chapter]

1. P  t2

2. P  t

3. P  t

4. P  t3/2

Subtopic:  Power |
 55%
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The gravitational field due to a mass distribution is given by I= kx2i^, where k is a constant. Assuming the potential to be zero at infinity, find the potential at a point x = a.[This question includes concepts from Gravitation chapter]

1. ka2

2. -ka2

3. ka

4. -ka

Subtopic:  Gravitational Potential |
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