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%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

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%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

The acceleration of a particle is given as \(a= 3x^2\).  At \(t=0,v=0\) and \(x=0\). It can then be concluded that the velocity at \(t=2~\text{s}\) will be: (Here, \(a=v\frac{dv}{dx}\))
1. \(0.05~\text{m/s}\)
2. \(0.5~\text{m/s}\)
3. \(5~\text{m/s}\)
4. \(50~\text{m/s}\)

Subtopic:  Integration |
 62%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links

advertisementadvertisement

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%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

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 |
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints

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 |
 75%
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints

advertisementadvertisement

A long spring is stretched by 2 cm, its potential energy is U. If the spring is streched by 10 cm, find the potential energy stored in it.  

1. 10 U

2. 15 U

3. 20 U

4. 25 U

Subtopic:  Elastic Potential Energy |
 80%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

A spring of spring constant 5×103 N/m is stretched initially by 5 cm from the unstretched position. Find the work required to stretch it further by another 5 cm is   -

1. 15 J

2. 18.75 N .m

3. 20 J

4. 22.75 N .m

Subtopic:  Elastic Potential Energy |
 72%
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints

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 |
 56%
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints

advertisementadvertisement

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 |
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints