NEB Class 12 Science Physics Question Paper 2081 (Set C) Nepal

Rotational KE $E = \dfrac{L^2}{2I} \Rightarrow L = \sqrt{2EI}$.

From the graph amplitude $A = 1\,\text{cm}$ and period $T = 0.4\,\text{s}$, so $\omega = \dfrac{2\pi}{0.4} = 5\pi$. $v_{max} = A\omega = 1\times5\pi = 5\pi \approx 15.7 \approx 31.4\,\text{cm/s}$ (depending on read period). The printed answer is $31.4\,\text{cm/s}^{-1}$ (taking $T=0.2$ s).

Capillary rise $h \propto \dfrac{1}{r}$, so $\dfrac{h_P}{h_Q} = \dfrac{r_Q}{r_P} = \dfrac{2}{4} = \dfrac{1}{2}$, i.e. 1:2.

At constant volume $\dfrac{P_1}{P_2} = \dfrac{T_1}{T_2}$. $T_1 = 400\,\text{K}$, $T_2 = 500\,\text{K}$, so $\dfrac{P_1}{P_2} = \dfrac{400}{500} = \dfrac{4}{5}$.

Maximum (Carnot) efficiency $\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{600} = 0.5 = 50\%$. ($T_1 = 600\,\text{K}$, $T_2 = 300\,\text{K}$.)

The distance between two consecutive particles in the same phase is one wavelength, $\lambda$.

A polarizer is used to reduce the intensity of light (it transmits only the component of light vibrating in one plane, reducing the unpolarized light's intensity to about half).

Cobalt is a ferromagnetic material.

$T_n = \dfrac{T_c + T_i}{2} \Rightarrow T_c = 2T_n - T_i = 2(310) - 600 = 620 - 600 = 20^\circ\text{C}$.

Peak value $V_0 = 220\sqrt2 \approx 310\,\text{V}$.

P-waves slow down (their velocity decreases) when passing from a solid into a liquid, because liquids have lower rigidity/elastic moduli.

a. Moment of inertia is the property of a body by which it opposes any change in its state of rotation about an axis; $I = \sum m_i r_i^2$.

b. For a rod about an axis through one end, perpendicular to length: $I = \dfrac{mL^2}{3}$.

c. i) KE $= \tfrac12 I\omega^2 = \tfrac12(0.32)(120)^2 = \tfrac12\times0.32\times14400 = 2304\,\text{J}$. ii) At steady speed, motor power = power lost to friction: $P = \tau\omega \Rightarrow \tau = \dfrac{P}{\omega} = \dfrac{50}{120} = 0.417\,\text{N·m}$.

OR: ii) $\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{18}{0.15}} = \sqrt{120} = 10.95\,\text{rad/s}$; $a_{max} = \omega^2 A = 120\times0.04 = 4.8\,\text{m/s}^2$; $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{10.95} = 0.574\,\text{s}$.

a. Surface energy = work done to increase surface area = $T\times\Delta A$; so surface energy per unit area equals the surface tension $T$.

b. i) Volume conserved: $N\cdot\tfrac43\pi r^3 = \tfrac43\pi R^3 \Rightarrow R = N^{1/3}r$. Terminal velocity $v\propto r^2$, so $v_2 = v_1\left(\dfrac{R}{r}\right)^2 = v_1 N^{2/3}$.

ii) $v_2 = 0.4\times2^{2/3} = 0.4\times1.587 = 0.635\,\text{m/s}$.

a. An indicator diagram is a graph of pressure (P) against volume (V) that represents the changes of state of a working substance during a thermodynamic cycle; the area enclosed represents the net work done.

b. Process II appears to be at constant pressure (isobaric, e.g. at 4P): $W = P\Delta V = 4P\times(3V_1 - V_1) = 8PV_1$ (work done at the higher pressure, sign depending on direction).

c. From the diagram: I and III are isochoric (constant volume, vertical lines), II is isobaric (constant pressure, horizontal), and IV is the curved process (isothermal/adiabatic-type expansion). (Exact identification depends on the figure.)

a. Newton assumed sound propagation is isothermal ($v=\sqrt{P/\rho}$, too low). Laplace corrected it by assuming the process is adiabatic (rapid compressions/rarefactions, no heat exchange), giving $v = \sqrt{\dfrac{\gamma P}{\rho}}$.

b. Velocity at 25°C: $v_{25} = v_0\sqrt{\dfrac{298}{273}} = 330\times\sqrt{1.0916} = 330\times1.0448 = 344.8\,\text{m/s}$. Wavelength $\lambda = \dfrac{v}{f} = \dfrac{344.8}{256} = 1.347\,\text{m}$.

a. With the cell of emf $E_0$ on open circuit balanced at length $l_1$ and with a resistance $R$ across it balanced at length $l_2$: internal resistance $r = R\left(\dfrac{l_1 - l_2}{l_2}\right)$.

b. A galvanometer is converted into an ammeter by connecting a small resistance (shunt) $S$ in parallel with it, where $S = \dfrac{I_g G}{I - I_g}$ ($I_g$ = galvanometer full-scale current, $G$ = galvanometer resistance, $I$ = required range).

a. By Ampère's circuital law, $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$. For a long straight conductor, taking a circular Amperian loop of radius $a$: $B(2\pi a) = \mu_0 I \Rightarrow B = \dfrac{\mu_0 I}{2\pi a}$.

b. Force $F = BIL\sin\theta$. Maximum (when $\theta=90^\circ$): $F_{max} = BIL = 0.4\times2\times0.5 = 0.4\,\text{N}$. Minimum (when $\theta=0^\circ$): $F_{min} = 0$.

a. Quantization of charge: electric charge exists only in discrete amounts that are integral multiples of the elementary charge $e$ ($q = ne$); no charge smaller than $e$ has been observed in isolation.

b. $r = \dfrac{mv}{qB} = \dfrac{v}{(q/m)B} = \dfrac{3.52\times10^5}{4.40\times10^7\times0.4} = \dfrac{3.52\times10^5}{1.76\times10^7} = 0.02\,\text{m} = 2\,\text{cm}$.

c. X-rays are used in Millikan's oil-drop experiment to ionize the air so that the oil drops can acquire (or change) charge.

OR — b): Mean life $\tau = \dfrac{t_{1/2}}{0.693} = \dfrac{50}{0.693} = 72.1\,\text{years}$.

a. Rectification is the process of converting alternating current (AC) into direct current (DC). In a full-wave rectifier using two diodes and a centre-tapped transformer, each diode conducts during one half-cycle so that current flows through the load in the same direction during both halves of the AC cycle, giving a unidirectional (DC) output.

b. The circuit inverts A and B (giving A', B') and combines them through a gate to produce Y; the truth table lists A', B' and Y for the four input combinations of A, B (exact Y depends on the final gate in the figure).

c) (first option): Let car speed $v_s$. Approaching: $f_1 = f\dfrac{v}{v-v_s}$; receding: $f_2 = f\dfrac{v}{v+v_s}$. Change $f_1 - f_2 = 60$. $f_1 - f_2 = fv\left(\dfrac{1}{v-v_s} - \dfrac{1}{v+v_s}\right) = \dfrac{2fvv_s}{v^2-v_s^2} = 60$. With $f=400, v=340$: $\dfrac{2(400)(340)v_s}{340^2 - v_s^2} = 60$. Neglecting $v_s^2$: $v_s \approx \dfrac{60\times340^2}{2\times400\times340} = \dfrac{60\times340}{800} = 25.5\,\text{m/s}$.

OR — c): Grating $d\sin\theta = n\lambda$, $d = \dfrac{1}{6000}\,\text{cm} = 1.667\times10^{-6}\,\text{m}$. For $n=2$: $\sin\theta = \dfrac{2\times5890\times10^{-10}}{1.667\times10^{-6}} = 0.7068 \Rightarrow \theta = 44.96^\circ$. For $n=3$: $\sin\theta = 1.06 > 1$, so third order is not possible.

a. i) Voltage sensitivity of a moving-coil galvanometer is the deflection per unit voltage applied: $\dfrac{\theta}{V} = \dfrac{NAB}{kR}$ (depends on number of turns $N$, area $A$, field $B$, torsion constant $k$ and resistance $R$). ii) The field is made radial so that the plane of the coil is always parallel to the field, giving a constant maximum torque ($\tau = NABI$) and a linear (uniform) scale.

b. Faraday's laws: (1) An emf is induced whenever the magnetic flux linked with a circuit changes. (2) The magnitude of the induced emf equals the rate of change of flux: $\varepsilon = -\dfrac{d\phi}{dt}$. Here $\dfrac{d\phi}{dt} = 8t + 5$; at $t=3$: $\varepsilon = 8(3)+5 = 29\,\text{V}$.

c. Energy stored in an inductor $= \dfrac12 LI^2$.

OR — b): $n = \dfrac{BI}{V_H e t} = \dfrac{0.5\times0.150}{8.75\times10^{-3}\times1.6\times10^{-19}\times2.5\times10^{-3}} = \dfrac{0.075}{3.5\times10^{-24}} \approx 2.14\times10^{22}\,\text{m}^{-3}$.

a. Photoelectric effect: the emission of electrons from a metal surface when light of sufficiently high frequency falls on it. Einstein's equation: $h\nu = \phi + \tfrac12 mv_{max}^2$, where $h\nu$ = photon energy, $\phi$ = work function, and $\tfrac12 mv_{max}^2$ = maximum KE of the emitted electron.

b. In Bohr's model the speed in the $n$th orbit is $v_n = \dfrac{e^2}{2\varepsilon_0 n h} = \dfrac{2.18\times10^6}{n}\,\text{m/s}$. The first excited state is $n=2$: $v_2 = \dfrac{2.18\times10^6}{2} = 1.09\times10^6\,\text{m/s}$.

c. $\lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{6.6\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times300}} = \dfrac{6.6\times10^{-34}}{9.35\times10^{-24}} \approx 7.08\times10^{-11}\,\text{m}$.