Two small spherical metal balls, having equal masses, are made from materials of densities \(\rho_1\) and \(\rho_2\) such that \(\rho_1=8\rho_2\)
| 1. | \(\dfrac{79}{72}\) | 2. | \(\dfrac{19}{36}\) |
| 3. | \(\dfrac{39}{72}\) | 4. | \(\dfrac{79}{36}\) |
The distance covered by a particle undergoing SHM in one time period is: (amplitude \(= A\))
1. zero
2. \(A\)
3. \(2 A\)
4. \(4 A\)
Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass \(m\) when taken to a height \(h\) from the surface of the earth (of radius \(R\) and mass \(M\)), is given by:
| 1. | \(-\frac{GMm}{R+h}\) | 2. | \(\frac{GMmh}{R(R+h)}\) |
| 3. | \(mgh\) | 4. | \(\frac{GMm}{R+h}\) |
A deep rectangular pond of surface area \(A\), containing water (density = \(\rho,\) specific heat capacity = \(s\)), is located in a region where the outside air temperature is at a steady value of \(-26^{\circ}\text{C}\). The thickness of the ice layer in this pond at a certain instant is \(x\). Taking the thermal conductivity of ice as \(k\), and its specific latent heat of fusion as \(L\), the rate of increase of the thickness of the ice layer, at this instant, would be given by:
| 1. | \(\dfrac{26k}{x\rho L-4s}\) | 2. | \(\dfrac{26k}{x^2\rho L}\) |
| 3. | \(\dfrac{26k}{x\rho L}\) | 4. | \(\dfrac{26k}{x\rho L+4s}\) |
The variation of EMF with time for four types of generators is shown in the figures. Which amongst them can be called AC voltage?
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| (a) | (b) |
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| (c) | (d) |
| 1. | (a) and (d) |
| 2. | (a), (b), (c), and (d) |
| 3. | (a) and (b) |
| 4. | only (a) |
The metre bridge shown is in a balanced position with \(\frac{P}{Q} = \frac{l_1}{l_2}\). If we now interchange the position of the galvanometer and the cell, will the bridge work? If yes, what will be the balanced condition?
| 1. | Yes, \(\frac{P}{Q}=\frac{l_1-l_2}{l_1+l_2}\) | 2. | No, no null point |
| 3. | Yes, \(\frac{P}{Q}= \frac{l_2}{l_1}\) | 4. | Yes, \(\frac{P}{Q}= \frac{l_1}{l_2}\) |
An object kept in a large room having an air temperature of \(25^\circ \text{C}\) takes \(12 ~\text{min}\) to cool from \(80^\circ \text{C}\) to \(70^\circ \text{C}.\) The time taken to cool for the same object from \(70^\circ \text{C}\) to \(60^\circ \text{C}\) would be nearly:
| 1. | \(10 ~\text{min}\) | 2. | \(12 ~\text{min}\) |
| 3. | \(20 ~\text{min}\) | 4. | \(15 ~\text{min}\) |
| 1. | \(\dfrac{1}{v} = \dfrac{1}{v_1}+\dfrac{1}{v_2}\) | 2. | \(\dfrac{2}{v} = \dfrac{1}{v_1}+\dfrac{1}{v_2}\) |
| 3. | \(\dfrac{v}{2} = \dfrac{v_1+v_2}{2}\) | 4. | \(v = \sqrt{v_1v_2}\) |
An object of mass \(500~\text g\) initially at rest is acted upon by a variable force whose \(x\)-component varies with \(x\) in the manner shown. The velocities of the object at the points \(x=8~\text m\) and \(x=12~\text m\) would have the respective values of nearly:

| 1. | \(18~\text {m/s}\) and \(22.4~\text {m/s}\) | 2. | \(23~\text {m/s}\) and \(22.4~\text {m/s}\) |
| 3. | \(23~\text {m/s}\) and \(20.6~\text {m/s}\) | 4. | \(18~\text {m/s}\) and \(20.6~\text {m/s}\) |
A mass falls from a height \(h\) and its time of fall \(t\) is recorded in terms of time period \(T\) of a simple pendulum. On the surface of the earth, it is found that \(t=2T\). The entire setup is taken on the surface of another planet whose mass is half of that of the Earth and whose radius is the same. The same experiment is repeated and corresponding times are noted as \(t'\) and \(T'\). Then we can say:
| 1. | \(t' = \sqrt{2}T\) | 2. | \(t'>2T'\) |
| 3. | \(t'<2T'\) | 4. | \(t' = 2T'\) |