Uncategorized - NEET Physics Chapterwise MCQs & PYQs

NEET Uncategorized MCQs & PYQs

Practice NEET Uncategorized Questions

Question 121:

easy

43. A long solenoid of diameter $0.1 m$ has $2 \times 10^4$ turns per metre. At the centre of solenoid, a coil of 100 turns and radius $0.01 m$ is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to $0 A$ from $4 A$ in $0.05 s$. If the resistance of the coil is $10 \pi^2 \Omega$, the total charge flowing through the coil during this time is: (2017-Delhi)

Total charge flowing is $q = \frac{\Delta \Phi}{R} = \frac{N_{coil} (\Delta B) A_{coil}}{R}$.
Change in magnetic field is $\Delta B = \mu_0 n \Delta I = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 4 = 32\pi \times 10^{-3} T$.
$q = \frac{100 \times (32\pi \times 10^{-3}) \times [\pi (0.01)^2]}{10 \pi^2} = \frac{100 \times 32\pi^2 \times 10^{-7}}{10 \pi^2} = 32 \times 10^{-6} C = 32 \mu C$.

💬 Discuss this Question

Question 122:

easy

44. Two coils of self-inductances $2 mH$ and $8 mH$ are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is: (2006)

For complete coupling between two coils, the coupling coefficient is $k = 1$.
The mutual inductance is given by $M = k \sqrt{L_1 L_2}$.
$M = 1 \times \sqrt{2 mH \times 8 mH} = \sqrt{16} = 4 mH$.

💬 Discuss this Question

Question 123:

easy

45. Two coil have a mutual inductance $0.005 H$. The current changes in first coil according to equation $I = I_0 \sin \omega t$ where $I_0 = 2 A$ and $\omega = 100pi rad/sec$. The maximum value of emf in second coil is: (1998)

Induced emf in the second coil is $e = -M \frac{dI}{dt} = -M \frac{d}{dt}(I_0 \sin \omega t) = -M I_0 \omega \cos \omega t$.
The maximum (peak) value of induced emf is $e_{max} = M I_0 \omega$.
$e_{max} = 0.005 \times 2 \times 100\pi = 0.01 \times 100\pi = \pi V$.

💬 Discuss this Question

Question 124:

easy

32. A long solenoid has 1000 turns. When a current of $4 A$ flows through it, the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} Wb$. The self inductance of the solenoid is: (2016 – I)

Total magnetic flux linked with the solenoid is $N \Phi = L I$.
Given $N = 1000$, $\Phi = 4 \times 10^{-3} Wb$, and $I = 4 A$.
$L = \frac{N \Phi}{I} = \frac{1000 \times 4 \times 10^{-3}}{4} = 1 H$.

💬 Discuss this Question

Question 125:

easy

14. 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 $\nu_0 = \frac{6.63 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} Hz$.

💬 Discuss this Question

Question 126:

easy

15. 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, $V_0 = \frac{h(\nu - \nu_0)}{e} = \frac{6.6 \times 10^{-34} \times (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{-19}} = \frac{6.6 \times 4.9 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 2 V$.

💬 Discuss this Question

Question 127:

easy

16. 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 $0.5 V$.

💬 Discuss this Question

Question 128:

easy

17. 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 $v_1/v_2 = \sqrt{K_1/K_2} = \sqrt{1/4} = 1/2$.

💬 Discuss this Question

Question 129:

easy

18. Photoelectric emission occurs only when the incident light has more than a certain minimum: (2011 Pre)

Photoelectric emission takes place only when the frequency of the incident light is greater than a characteristic minimum frequency for the given metal, known as the threshold frequency.

💬 Discuss this Question

Question 130:

easy

11. 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 $\frac{1}{2}mv^2 = h\nu \implies v^2 = \frac{2h\nu}{m} \implies v = \sqrt{\frac{2h\nu}{m}}$.

💬 Discuss this Question