| 1. | \(8\) N in \(-z\text-\)direction. |
| 2. | \(4\) N in the \(z\text-\)direction. |
| 3. | \(8\) N in the \(y\text-\)direction. |
| 4. | \(8\) N in the \(z\text-\)direction. |
| 1. | Curves \(a\) and \(b\) represent incident radiations of different frequencies and different intensities. |
| 2. | Curves \(a\) and \(b\) represent incident radiation of the same frequency but of different intensities. |
| 3. | Curves \(b\) and \(c\) represent incident radiation of different frequencies and different intensities. |
| 4. | Curves \(b\) and \(c\) represent incident radiations of the same frequency having the same intensity. |
The power dissipated in an L-C-R series circuit connected to an AC source of emf E is:
| 1. | Putting in parallel, a resistance of \(24~ \Omega\) |
| 2. | Putting in series, a resistance of \(15~ \Omega\) |
| 3. | Putting in series, a resistance of \(240~ \Omega\) |
| 4. | Putting in parallel, a resistance of \(15~ \Omega\) |
The two ends of a rod of length \(L\) and a uniform cross-sectional area \(A\) are kept at two temperatures \(T_1\text{ and }T_2~ (T_1> T_2).\) The rate of heat transfer \(\dfrac{dQ}{dt}\) through the rod in a steady state is given by:
1. \(\dfrac{dQ}{dt} = \dfrac{KL \left(\right. T_{1} - T_{2} \left.\right)}{A}\)
2. \(\dfrac{dQ}{dt} = \dfrac{K \left(\right. T_{1} - T_{2} \left.\right)}{LA}\)
3. \(\dfrac{dQ}{dt} = KLA \left(\right. T_{1} - T_{2} \left.\right)\)
4. \(\dfrac{dQ}{dt} = \dfrac{KA \left(\right. T_{1} - T_{2} \left.\right)}{L}\)
See the electrical circuit shown in this figure. Which of the following is a correct equation for it?

| 1. | \(\varepsilon_1-(i_1+i_2)R-i_1r_1=0\) |
| 2. | \(\varepsilon_2-i_2r_2-\varepsilon_1-i_1r_1=0\) |
| 3. | \(-\varepsilon_2-(i_1+i_2)R+i_2r_2=0\) |
| 4. | \(\varepsilon_1-(i_1+i_2)R+i_1r_1=0\) |