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Q. No.A small block of mass \(m\) lies on a frictionless wedge of mass \(M,\) which is pushed horizontally to the right by means of a constant force \(F\). There is no relative motion between the block and the wedge. Let the work done by \(F\) on \(M\) be \(W_F\). The work done by the normal force (between \(M\) & \(m\)) on \(m\) be \(W_m\). Both are measured for the same time interval.
Then:
1. \(\dfrac{W_F}{M}=\dfrac{W_m}{m}\)
2. \(W_F\cdot M=W_m\cdot m\)
3. \(\dfrac{W_F}{M+m}=\dfrac{W_m}{m}\)
4. \(\dfrac{W_F}{M}=\dfrac{W_m}{m+M}\)
Subtopic: Work done by constant force |
63%
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A ball of mass \(m\) is projected with a speed \(u,\) at an angle of \(\theta\) with the horizontal. At its highest point, it moves on a smooth horizontal platform with a spring of spring constant \(k\) attached, and the ball compresses the spring. The maximum compression in the spring is \(x.\) Then:
| 1. | \(\frac12mu^2=\frac12kx^2\) | 2. | \(\frac12mu^2cos^2\theta=\frac12kx^2\) |
| 3. | \(\frac12mu^2=\frac12kx^2cos^2\theta\) | 4. | \(\frac12mu^2sin^2\theta=\frac12kx^2\) |
Subtopic: Elastic Potential Energy |
79%
From NCERT
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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. \(34500\)
(Take \(g=10\) ms–2)
1. \(23500\)
2. \(23000\)
3. \(20000\)
4. \(34500\)
Subtopic: Power |
66%
From NCERT
NEET - 2022
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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%
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Given below are two statements:
| Assertion (A): | Frictional forces are conservative forces. |
| Reason (R): | Potential energy 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%
From NCERT
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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%
From NCERT
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\(300\) 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 |
83%
From NCERT
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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%
From NCERT
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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%
From NCERT
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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%
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