Q. No.When a 'rectangular' tank is filled upto a level \(h\) with water of mass \(m\), the work done is:
(assuming that the lower part of the tank is at ground level from where water is pumped into it)
1.
\(mgh\)
2.
\(\dfrac{mgh}{2}\)
3.
\(\dfrac{mgh}{4}\)
4.
\(2~mgh\)
(assuming that the lower part of the tank is at ground level from where water is pumped into it)
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If a tank is placed as shown in the figure and it is filled with water of mass '\(m\)' pumped from the ground below, then the work done is:
1. \(mg(H+h)\)
2. \(mg\left(\dfrac{H+h}{2}\right)\)
3. \(mg\left(H+\dfrac{h}{2}\right)\)
4. \(mg\left(\dfrac{H-h}{2}\right)\)
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Two block \(A\) and \(B\) of equal mass are connected by means of a spring, and the system is placed on a smooth horizontal table. A force \(F=\dfrac{mg}{2}\) acts on the block \(B\) and the spring gets extended, while the block \(B\) moves to the right. At the given moment, both blocks are moving and \(v_B>v_A\). The acceleration of \(B\) is also towards right.
Then,
| 1. | The kinetic energy of block \(A\) is decreasing. |
| 2. | The rate of work done by \(F\) on \(B\) is negative. |
| 3. | The rate of work done by the spring on \(B\) is negative. |
| 4. | The rate of work done by \(F\) on \(A\) is positive. |
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| Assertion (A): | Work done by friction is always negative. |
| Reason (R): | Kinetic friction is a non-conservative force. |
| 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. | (A) is False but (R) is True. |
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A block of mass \(m\) is being lowered by means of a string attached to it. The system moves down with a constant velocity. Then:
| 1. | the work done by gravity on the block is positive. |
| 2. | the work done by force, \(F \) (the force of the string) on the block is negative. |
| 3. | the work done by gravity is equal in magnitude to that done by the string. |
| 4. | All of the above are true. |
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A block of mass \(m\) is placed on a spring and then it is pressed down by means of an external force, and then released. The block moves upwards. Let the work done by the spring be \(W_s\) and that done by gravity be \(W_g\).
Let the P.E. of the spring be \(P_s\),
the gravitational P.E. of the block be \(P_{mg}\)
and the K.E. of the block be \(K_m\).
Then,
| 1. | loss in \(P_s\) = gain in \(K_m\) |
| 2. | loss in \(P_s\) = gain in \(K_m\) + gain in \(P_{mg}\) |
| 3. | \(W_g = W_s\) |
| 4. | \(W_s= \) gain in \(K_m\) |
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The potential energy due to a force is given by:
\(U(x,y)= -3xy+2y^2\) (in joule)
where \(x,y\) are in metres.
The force acting when \(x=0,y=1\) (m) is: (in magnitude)
1. \(2~\text{N}\)
2. \(1~\text{N}\)
3. \(3~\text{N}\)
4. \(5~\text{N}\)
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The force acting on a particle is shown in the diagram as a function of \(x\). Work done by this force when the particle moves from \(x=0~\text{to}~x=2~\text{m}\) equals:
| 1. | \(5~\text{J}\) | 2. | \(10~\text{J}\) |
| 3. | \(7.5~\text{J}\) | 4. | \(2.5~\text{J}\) |
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A simple pendulum consisting of a bob of mass \(m\), and a string of length \(L\) is given a horizontal speed \(u\), at its lowest point as shown in the figure. As a result, it rises to \(B\), where it just comes to rest momentarily with \(OB\) horizontal.
During the motion \(AB,\)
| 1. | Work done by the string is zero |
| 2. | Work done by gravity is \(-mgL\) |
| 3. | Change in K.E. of the bob is \(-\dfrac{1}{2}mu^2\) |
| 4. | All the above are true |
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A person of mass \(m\) ascends the stairs and goes up slowly through a height \(h\). Then,
| 1. | Work done by gravity is \(mgh\) |
| 2. | Work done by normal reaction is \(mgh\) |
| 3. | Work done by normal reaction is zero |
| 4. | Work done by gravity is stored as gravitational \(P.E\). |
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