Select Question Set:
Q. No.An electric lift with a maximum load of \(2000\) kg (lift+passengers) is moving up with a constant speed of \(1.5\) ms–1. The frictional force opposing the motion is \(3000\) N. The minimum power delivered by the motor to the lift in watts is:
(Take \(g=10\) ms–2)
1. \(23500\)
2. \(23000\)
3. \(20000\)
4. \(34500\)
Subtopic: Power |
66%
Level 2: 60%+
NEET - 2022
Hints
Given below are two statements:
| Assertion (A): | Work done by friction on a body sliding down an inclined plane is negative. |
| Reason (R): | Work done is less than zero if the angle between force and displacement is acute or both are in the same direction. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | Both (A) and (R) are False. |
Subtopic: Concept of Work |
77%
Level 2: 60%+
Hints
Given below are two statements:
| Assertion (A): | Frictional forces are conservative in nature. |
| Reason (R): | A potential energy function can be associated with frictional forces. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | Both (A) and (R) are False. |
Subtopic: Potential Energy: Relation with Force |
86%
Level 1: 80%+
Hints
It is given that the block never loses contact (in the figure given below) with the smooth horizontal surface, and the force always acts at an angle \(\theta\) with the horizontal. The speed of the block when it covers a horizontal distance \(l\) will be:
| 1. | \(\sqrt{\dfrac{l F_{} \cos \theta}{m}}\) | 2. | \(\dfrac{2 l \mathrm{~F}_{} \cos \theta}{\mathrm{m}}\) |
| 3. | \(\sqrt{\dfrac{2 l}{m} F_{} \cos \theta}\) | 4. | \(\dfrac{l \mathrm{~F}_{} \cos \theta}{\mathrm{m}}\) |
Subtopic: Work Energy Theorem |
86%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
\(300\) J of work is done in sliding a \(2\) kg block up an inclined plane of height \(10\) m. Taking \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), work done against friction is:
1. \(1000\) J
2. \(200\) J
3. \(100\) J
4. zero
1. \(1000\) J
2. \(200\) J
3. \(100\) J
4. zero
Subtopic: Work Done by Variable Force |
81%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
A body of mass \(8\) kg and another of mass \(2\) kg are moving with equal kinetic energy. The ratio of their respective momenta will be:
1. \(1:1\)
2. \(2:1\)
3. \(1:4\)
4. \(4:1\)
1. \(1:1\)
2. \(2:1\)
3. \(1:4\)
4. \(4:1\)
Subtopic: Work Energy Theorem |
84%
Level 1: 80%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
A boat of mass \(200\) kg is accelerated by an engine of power \(16\) kW. If the boat covers a distance of \(72\) km in \(2\) hr, the acceleration of the boat is:
1. \(8\times10^{-2}\) m/s2
2. \(8\times10^{-1}\) m/s2
3. \(8\) m/s2
4. \(80\) m/s2
1. \(8\times10^{-2}\) m/s2
2. \(8\times10^{-1}\) m/s2
3. \(8\) m/s2
4. \(80\) m/s2
Subtopic: Power |
80%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
A particle experiences a variable force \(\vec F = (4x \hat i + 3y^2 \hat j)\) in a horizontal x-y plane. Assume distance in meters and force is in newton. If the particle moves from point \((1,2)\) to point \((2,3)\) in the x-y plane, the kinetic energy changes by:
1. \(50.0\) J
2. \(12.5\) J
3. \(25.0\) J
4. \(0\)
1. \(50.0\) J
2. \(12.5\) J
3. \(25.0\) J
4. \(0\)
Subtopic: Work Energy Theorem |
81%
Level 1: 80%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
A ball with a mass of \(100\) g is dropped from a height of \(h=10\) cm onto a platform fixed at the top of a vertical spring (as shown in the figure). The ball remains on the platform, and the platform is depressed by a distance of \(\dfrac {h} {2}.\) The spring constant is: (use \(g=10\) ms-2)
| 1. | \(100\) Nm–1 | 2. | \(110\) Nm–1 |
| 3. | \(120\) Nm–1 | 4. | \(130\) Nm–1 |
Subtopic: Elastic Potential Energy |
76%
Level 2: 60%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
Potential energy as a function of \({r}\) is given by; \({U}=\dfrac{{A}}{{r}^{10}}-\dfrac{{B}}{{r}^5},\) where \({r}\) is the interatomic distance, and \({A}\) and \({B}\) are positive constants. The equilibrium distance between the two atoms will be:
| 1. | \(\left ( \dfrac{{A}}{{B}}\right )^{1/5}\) | 2. | \(\left ( \dfrac{{B}}{{A}}\right )^{1/5}\) |
| 3. | \(\left ( \dfrac{{2A}}{{B}}\right )^{1/5}\) | 4. | \(\left ( \dfrac{{B}}{{2A}}\right )^{1/5}\) |
Subtopic: Potential Energy: Relation with Force |
81%
Level 1: 80%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
Select Question Set:
© 2025 GoodEd Technologies Pvt. Ltd.