Q. No.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. \(5~\text{N}\)
\(U(x,y)= -3xy+2y^2\) (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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A particle of mass '\(m\)' is released from the origin, and it moves under the action of a force: \(F(x)= F_0-kx\)
The maximum speed of the particle is, \(v= \)
| 1. | \(\sqrt{\dfrac{F_0^2}{mk}}\) | 2. | \(\sqrt{\dfrac{2F_0^2}{mk}}\) |
| 3. | \(\sqrt{\dfrac{F_0^2}{2mk}}\) | 4. | \(2\sqrt{\dfrac{F_0^2}{mk}}\) |
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A projectile is launched from a cliff of height \(h,\) with an initial speed \(u,\) at an angle \(\theta.\) The speed with which it hits the ground:
| 1. | depends on the vertical component, \(u \text{sin}\theta\) |
| 2. | depends on the horizontal component, \(u \text{cos}\theta\) |
| 3. | depends on \(u,\) but not on \(\theta\) |
| 4. | depends on the quantity \(u \text{tan}\theta\) |
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A small block of mass '\(m\)' is placed against a compressed spring, of spring constant \(k\). The initial compression in the spring is '\(d\)'. The block is released and the spring relaxes, while the block is projected up to a height \(H\) relative to its initial position. Then, \(H\) =
| 1. | \(\dfrac{kd^2}{2mg}\) | 2. | \(\dfrac{kd^2}{2mg}+d\) |
| 3. | \(\dfrac{kd^2}{2mg}-d\) | 4. | \(\dfrac{kd^2}{mg}+d\) |
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Given below are two statements:
| Assertion (A): | The net work done by gravity is equal to the loss in the vertical component of the kinetic energy for a projectile. |
| Reason (R): | The work-energy theorem applies to all systems including projectiles. |
| 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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The minimum speed that a simple pendulum's bob should be given so that it rises to a level where its string makes \(60^\circ\) with the vertical is:
2. \(\sqrt{2gL}\) \(\text{cos} 60^\circ\)
3. \(\sqrt{2g(L-L~\text{cos}60^\circ)}\)
4. \(\sqrt{2gL}\) \(\text{sin} 60^\circ\)
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The net work done by all forces (internal and external) equals the change in:
| 1. | potential energy |
| 2. | kinetic energy |
| 3. | total energy |
| 4. | kinetic energy and conservative potential energy |
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