Question 51:
easy37. The current in self inductance $L = 40 mH$ is to be increased uniformly from 1 amp to 11 amp in 4 milliseconds. The e.m.f. induced in inductor during process is (1990)
Induced e.m.f. is given by $e = L \frac{\Delta I}{\Delta t}$.
Given $L = 40 mH = 40 \times 10^{-3} H$, $\Delta I = 11 - 1 = 10 A$, and $\Delta t = 4 ms = 4 \times 10^{-3} s$.
$e = 40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}} = 100 volt$.
Question 52:
easy38. The magnetic potential energy stored in a certain inductor is $25 mJ$, when the current in the inductor is $60 mA$. This inductor is of inductance: (2018)
The magnetic energy stored is $U = \frac{1}{2} L I^2$.
Given $U = 25 mJ = 25 \times 10^{-3} J$ and $I = 60 mA = 60 \times 10^{-3} A$.
$L = \frac{2U}{I^2} = \frac{2 \times 25 \times 10^{-3}}{(60 \times 10^{-3})^2} = \frac{50 \times 10^{-3}}{3600 \times 10^{-6}} = \frac{50000}{3600} \approx 13.89 H$.
Question 53:
easy39. For a inductor coil $L = 0.04 H$, then work done by source to establish a current of $5 A$ in it is: (1999)
Work done to establish a current in an inductor is stored as magnetic energy: $W = \frac{1}{2} L I^2$.
Given $L = 0.04 H$ and $I = 5 A$.
$W = \frac{1}{2} \times 0.04 \times 5^2 = 0.02 \times 25 = 0.5 J$.
Question 54:
easy11. For photoelectric emission from certain metal the cut-off frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be: (m is the electron mass) (2013)
From Einstein's photoelectric equation, $K_{max} = h\nu_{incident} - h\nu_{threshold}$. Here, incident frequency is $2\nu$ and threshold frequency is $\nu$. So, $K_{max} = h(2\nu) - h\nu = h\nu$. Since $K_{max} = \frac{1}{2}mv^2$, we have $\frac{1}{2}mv^2 = h\nu \implies v^2 = \frac{2h\nu}{m} \implies v = \sqrt{\frac{2h\nu}{m}}$.
Question 55:
easy12. Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is: (2012 Mains)
Maximum kinetic energy is given by $K = E - W$. For the first radiation, $K_1 = 1.0 - 0.5 = 0.5 eV$. For the second radiation, $K_2 = 2.5 - 0.5 = 2.0 eV$. The ratio of their kinetic energies is $K_1/K_2 = 0.5/2.0 = 1/4$. Since $K \propto v^2$, the ratio of speeds is $v_1/v_2 = \sqrt{K_1/K_2} = \sqrt{1/4} = 1/2$.
Question 56:
easy13. A 200 W sodium street lamp emits yellow light of wavelength $0.6 \mu m$. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is. (2012 Pre)
Useful power for light emission is $P = 25\% \text{ of } 200 W = 50 W$. The energy of one photon is $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{0.6 \times 10^{-6}} = 3.3 \times 10^{-19} J$. The number of photons emitted per second is $n = \frac{P}{E} = \frac{50}{3.3 \times 10^{-19}} \approx 1.5 \times 10^{20}$.
Question 57:
easy14. Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the materials is: (2012 Pre)
Energy of incident radiation $E = 13.6 (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times \frac{3}{4} = 10.2 eV$. The stopping potential is $3.57 V$, so $K_{max} = 3.57 eV$. Work function $W = E - K_{max} = 10.2 - 3.57 = 6.63 eV$. Using $W = h\nu_0$, we get threshold frequency $\nu_0 = \frac{6.63 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} Hz$.
Question 58:
easy15. The threshold frequency for a photosensitive metal is $3.3 \times 10^{14} Hz$. If light of frequency $8.2 \times 10^{14} Hz$ is incident on this metal, the cut-off voltage for the photoelectric emission is nearly: (2011 Mains)
From photoelectric equation, $eV_0 = h(\nu - \nu_0)$. Thus, $V_0 = \frac{h(\nu - \nu_0)}{e} = \frac{6.6 \times 10^{-34} \times (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{-19}} = \frac{6.6 \times 4.9 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 2 V$.
Question 59:
easy16. In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is: (2011 Pre)
The stopping potential $V_0$ is numerically equal to the maximum kinetic energy of the emitted photoelectrons expressed in electron-volts (eV). Since $K_{max} = 0.5 eV$, the stopping potential is $0.5 V$.
Question 60:
easy17. Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be: (2011 Pre)
Maximum kinetic energy $K_{max} = E - W$. For the first light, $K_1 = 1 - 0.5 = 0.5 eV$. For the second light, $K_2 = 2.5 - 0.5 = 2.0 eV$. The ratio of their kinetic energies is $K_1/K_2 = 1/4$. The ratio of maximum speeds is $v_1/v_2 = \sqrt{K_1/K_2} = \sqrt{1/4} = 1/2$.