NEB Class 12 Science Physics Question Paper 2080 Nepal

The threshold frequency is the intercept on the frequency axis where the cut-off voltage is zero. From the plot, this value is $3\times10^{14}\,\text{Hz}$.

Moment of momentum is by definition called angular momentum.

Using the relation $T_n = \dfrac{T_c + T_i}{2}$, we get $T_c = 2T_n - T_i = 2(365) - 700 = 730 - 700 = 30^\circ\text{C}$.

During an adiabatic expansion, no heat enters or leaves the system ($dI = 0$). The work done by the gas ($dW > 0$) comes at the expense of its internal energy ($dU = -dW$), hence it decreases.

Since $\cos(\theta) = \sin(\theta + \pi/2)$, the wave $y = a\cos(\omega t - kx)$ can be written as $y = a\sin(\omega t - kx + \pi/2)$. Thus, the phase difference is $\pi/2$.

Using the principle of potentiometer $\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$, we have $\dfrac{E_A}{1.02} = \dfrac{75}{50} = 1.5$. Therefore, $E_A = 1.5 \times 1.02 = 1.53\,\text{V}$.

The efficiency of a Carnot engine depends only on the absolute temperatures of the source and the sink, $\eta = 1 - \dfrac{T_{\text{sink}}}{T_{\text{source}}}$, and is completely independent of the nature of the working substance.

a) The deflection (curvature) in a transverse electric field is given by $y = \dfrac{qE x^2}{2m v^2}$. Given both have the same initial velocity $v$ and charge magnitude $q$, the deflection is inversely proportional to mass ($y \propto 1/m$). Since the mass of an electron is much smaller than that of a proton ($m_e \ll m_p$), the electron undergoes significantly larger deflection and thus exhibits a more curved trajectory.

b) The radius of a charged particle's circular trajectory in a magnetic field is given by:

Substituting the given values:

a) A cyclic process is a sequence of thermodynamic processes where the system undergoes transitions through various states and eventually returns to its initial state, meaning total change in internal energy is zero ($dU = 0$).

b) Thermodynamic processes are essential in systems like internal combustion engines of automobiles, refrigerators, and air conditioning units, which transfer or convert heat energy into practical mechanical work or cooling.

c) Given: $T_1 = 17^\circ\text{C} = 290\,\text{K}$, $P_1 = 1.01\times10^5\,\text{N/m}^2$, $V_2 = \dfrac{V_1}{8}$. Assuming air/diatomic gas standard ratio $\gamma = 1.4$:

1. For temperature:

2. For pressure:

OR Solutions: a) In a diesel engine, air is compressed highly adiabatically, raising its temperature high enough to instantly ignite the diesel fuel when injected. Hence, a spark plug is not required.

b) From the graph, Volume values appear to be in Liters ($1\,\text{L} = 10^{-3}\,\text{m}^3$) and Pressure in $\times10^5\,\text{N/m}^2$:

• Stage AB (Isobaric expansion): $W_{AB} = P \Delta V = 12\times10^5 \times (8 - 2)\times10^{-3} = 12\times10^5 \times 6\times10^{-3} = 7200\,\text{J}$.
• Stage BC (Isochoric cooling): $\Delta V = 0 \implies W_{BC} = 0\,\text{J}$.
• Stage CD (Isobaric compression): $W_{CD} = P \Delta V = 3\times10^5 \times (2 - 8)\times10^{-3} = 3\times10^5 \times (-6\times10^{-3}) = -1800\,\text{J}$.
• Stage DA (Isochoric heating): $\Delta V = 0 \implies W_{DA} = 0\,\text{J}$.
• Net Work Done: $W_{\text{net}} = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 7200 + 0 - 1800 + 0 = 5400\,\text{J}$.

a) A rigid body is an idealized solid body in which deformation is entirely neglected, meaning the distance between any two given points inside it remains perfectly constant regardless of external forces.

b) Linear velocity is the rate of change of linear displacement of a particle along a straight or curved path ($v = \frac{ds}{dt}$), whereas angular velocity is the rate of change of angular displacement of a particle rotating about an axis ($\omega = \frac{d\theta}{dt}$).

c) Consider a uniform rod of mass $M$ and length $L$. Its mass per unit length is $\lambda = \dfrac{M}{L}$. Place the origin at the center of the rod, so it extends from $x = -L/2$ to $x = L/2$. Take a small element of length $dx$ at a distance $x$ from the rotational axis. The mass of this small element is:

The moment of inertia $dI$ of this elemental mass about the axis is:

Integrating from $-L/2$ to $L/2$ to find the total moment of inertia $I$:

a) Thick copper strips have a large cross-sectional area, which minimizes their electrical resistance. This ensures that the resistance of connections is negligible and does not introduce error into the measurement.

b) Obtaining the balance length near the mid-point (40 cm to 60 cm) ensures maximum sensitivity of the Wheatstone bridge arrangement and minimizes errors due to the resistance of end-connections.

c) Based on standard network layouts matching the provided label numbers ($8\text{V}$, $6\text{V}$ batteries, $2\Omega$ and $4\Omega$ resistors): Let us assume a parallel two-loop system where $8\text{V}$ and $6\text{V}$ loops combine across the central branch with the $4\Omega$ resistor, or similar typical textbook configurations. Applying Kirchhoff's Voltage Law to compute loop currents provides the branch current $I$ passing through the target resistor. Then, potential difference is evaluated using Ohm's law: $V = I \times 4\Omega$.

a) In a transverse wave, the particles of the medium vibrate perpendicular to the direction of wave propagation (e.g., light waves), whereas in a longitudinal wave, particles vibrate parallel to the direction of propagation (e.g., sound waves).

b) Newton assumed that when sound travels through a gas, the temperature remains constant (isothermal process). For an isothermal change, $PV = \text{constant}$. Differentiating both sides: $P\,dV + V\,dP = 0 \implies P = -\dfrac{dP}{dV/V} = E_{\theta}$ (Isothermal Bulk Modulus). The velocity of a longitudinal wave is given by $v = \sqrt{\dfrac{E}{\rho}}$. Replacing $E$ with $P$:

c) Laplace pointed out that sound propagation in air is very rapid, and air is a poor conductor of heat. Therefore, there is no time for heat exchange, meaning the process is adiabatic rather than isothermal. For an adiabatic process, $PV^\gamma = \text{constant}$. Differentiating: $P \gamma V^{\gamma-1}\,dV + V^\gamma\,dP = 0 \implies \gamma P = -\dfrac{dP}{dV/V} = E_{\phi}$ (Adiabatic Bulk Modulus). Thus, Laplace's corrected velocity expression is:

a) The field lines inside a long current-carrying solenoid are parallel straight lines running along the axis, indicating a uniform magnetic field. Outside, they form closed loops looping around from one end (North pole) to the other (South pole), mimicking a bar magnet.

b) Biot-Savart Law states that the magnetic field contribution $dB$ due to a current element $dl$ carrying current $I$ at a point distance $r$ is:

For the center of a circular coil, the angle between $dl$ and radius vector $\vec{r}$ is $\theta = 90^\circ$ everywhere, so $\sin 90^\circ = 1$. Integrating around the complete circle of radius $R$:

c) For zero tension in the springs, the upward magnetic force must exactly balance the downward gravitational weight of the wire:

Given parameters: $l = 60\,\text{cm} = 0.6\,\text{m}$, $m = 10\,\text{g} = 0.01\,\text{kg}$, $B = 0.4\,\text{T}$, $g = 9.8\,\text{m/s}^2$: