Physics (312) | Set A1 • Code 71/SS/312

Ans. (B) Inertia
Explanation: Friction is a surface contact phenomenon caused by microscopic interlocking and adhesive molecular bonds between contacting surfaces. Lubricants fill these irregularities, creating a smooth intermediate fluid layer that reduces static, sliding, and rolling friction. However, inertia is an intrinsic property of a body determined exclusively by its mass (\(m\)), and cannot be altered or reduced by applying lubricants.

Derivation: The potential energy stored in an elastic spring of force constant \(k\) stretched by displacement \(x\) is: \[U = \frac{1}{2} k x^2 \implies U \propto x^2\] Given \(x_1 = 2\text{ cm} \implies V = \frac{1}{2} k (2)^2 = 2k\).
When stretched by \(x_2 = 10\text{ cm}\): \[U_2 = \frac{1}{2} k (10)^2 = 50k = 25 \times (2k) = 25V\]

Derivation: Bernoulli's theorem states that for an incompressible, non-viscous fluid undergoing steady streamline flow, the total mechanical energy per unit volume remains constant along a streamline: \[P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}\] This equation directly represents the work-energy theorem applied to fluid elements, embodying the conservation of energy. (Note: Equation of continuity represents conservation of mass).

Derivation: By Torricelli's law of efflux (derived from Bernoulli's principle), the speed of efflux of a liquid flowing out from an orifice at depth \(h\) below the free surface is: \[v = \sqrt{2gh} = (2g)^{1/2} h^{1/2} \implies v \propto h^{1/2}\]

Derivation: For an ideal gas, intermolecular potential energy is zero because molecules exert no attractive forces on one another. Thus, internal energy consists entirely of kinetic energy: \[U = \frac{f}{2} n R T\] By Joule's law of thermodynamics, \(\left(\frac{\partial U}{\partial V}\right)_T = 0\), meaning internal energy depends strictly on absolute temperature alone.

Derivation: According to the First Law of Thermodynamics: \[\Delta Q = \Delta U + W\] Given \(\Delta Q = +110\text{ J}\) (heat supplied) and \(\Delta U = +40\text{ J}\) (increase in internal energy): \[110\text{ J} = 40\text{ J} + W \implies W = 110 - 40 = \mathbf{70\text{ J}}\]

Derivation: If the liquid is depressed by \(y\), restoring force \(F = -(2y A \rho) g = -2A\rho g y\).
Total mass of liquid \(m = A L \rho\), where \(L\) is the total length of the liquid column.
Acceleration \(a = \frac{F}{m} = -\frac{2A\rho g y}{A L \rho} = -\left(\frac{2g}{L}\right)y\).
Since \(a \propto -y\), the motion is simple harmonic. The time period is: \[T = 2\pi\sqrt{\frac{L}{2g}} = 2\pi\sqrt{\frac{h}{g}}\] The density \(\rho\) cancels completely out, rendering the time period strictly independent of liquid density.

Derivation: When a wave travels from one medium into another, its frequency \(\nu\) remains invariant because frequency is determined exclusively by the source.
In Medium 1: \(v = \nu \lambda \implies \nu = \frac{v}{\lambda}\).
In Medium 2: \(v' = 2v = \nu \lambda' \implies \lambda' = \frac{2v}{\nu} = \frac{2v}{v/\lambda} = \mathbf{2\lambda}\).

Derivation: Construct a concentric spherical Gaussian surface of radius \(r < R\) inside the shell. By Gauss's Law: \[\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}\] Since all charges reside on the outer surface of the shell, \(q_{\text{enclosed}} = 0 \implies E(4\pi r^2) = 0 \implies \mathbf{E = 0}\).

Derivation:
1. Electrostatic Force (\(F_e\)): Electrons carry negative charge (\(-e\)) and protons carry positive charge (\(+e\)). Unlike charges attract each other electrostatically.
2. Magnetic Force (\(F_m\)): Electron stream moving forward creates a current backward; proton stream moving forward creates current forward. Anti-parallel currents repel magnetically.
3. For non-relativistic velocities \(v \ll c\), \(F_m = \frac{v^2}{c^2} F_e \ll F_e\). The electrostatic attractive force overwhelmingly dominates the magnetic repulsion. Hence, the net force is attractive.

Explanation: Interference of light is specifically defined as the phenomenon of modification or redistribution of light energy in a medium due to the superposition of light waves originating from two coherent sources (sources emitting waves having identical frequency and constant or zero initial phase difference).

Derivation: With \(I_1 = I\) and \(I_2 = 4I\): \[I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = \left(\sqrt{I} + \sqrt{4I}\right)^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = \mathbf{9I}\] \[I_{\min} = \left(\sqrt{I_2} - \sqrt{I_1}\right)^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = \mathbf{I}\]

Explanation: An object is visible in a medium because of reflection and refraction of light at its boundaries due to a difference in refractive indices. If the refractive index of the object matches that of the surrounding fluid (\(\mu_{\text{object}} = \mu_{\text{fluid}}\)), light passes through the boundary without deviation, refraction, or reflection. As a result, the object appears completely invisible.

Derivation: The wavelength \(4800\text{ \AA} = 480\text{ nm}\) falls directly within the visible spectrum (\(380\text{ nm}\) to \(750\text{ nm}\), or \(3800\text{ \AA} - 7500\text{ \AA}\)). In atomic hydrogen, the Balmer series is the only spectral series that lies in the visible region (specifically, the \(H_\beta\) line corresponding to \(n = 4 \to n = 2\) transition has \(\lambda \approx 4861\text{ \AA} \approx 4800\text{ \AA}\)). Lyman lies in UV; Paschen & Brackett lie in Infrared.

Derivation: de Broglie wavelength is given by: \[\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \implies \lambda \propto \frac{1}{\sqrt{m}} \quad (\text{for constant } E)\] Comparing rest masses of the particles: \[m_e \approx 9.1 \times 10^{-31}\text{ kg}, \quad m_p \approx m_n \approx 1.67 \times 10^{-27}\text{ kg}, \quad m_\alpha \approx 4m_p \approx 6.64 \times 10^{-27}\text{ kg}\] \[m_\alpha > m_n \approx m_p \gg m_e \implies \mathbf{\lambda_\alpha < \lambda_p \approx \lambda_n < \lambda_e}\]

Explanation: In semiconductors, raising the temperature provides thermal energy to liberate valence electrons across the small forbidden energy gap (\(E_g\)) into the conduction band. This exponentially multiplies the charge carrier concentration \(n_i \propto T^{3/2} e^{-E_g / (2k_B T)}\). The enormous increase in carrier concentration far outweighs the minor drop in relaxation time, resulting in an increased electrical conductivity (\(\sigma = e(n\mu_e + p\mu_h)\)).

This occurs due to inertia of rest. Initially, the passengers and the bus are at rest. When the bus suddenly accelerates forward, the lower portion of the passenger's body shares the forward motion of the bus through contact friction. However, the upper part of the body tends to remain at rest due to inertia of rest. As a result, the passenger feels a relative backward push and falls backward.

A force is said to be a conservative force if the work done by or against the force in moving a particle between two points depends only on the initial and final positions and is completely independent of the path followed. Equivalently, the net work done by a conservative force along any closed path is zero (\(\oint \vec{F} \cdot d\vec{r} = 0\)).
Example: Gravitational force (or Electrostatic force, Spring force).

Given: \(V = 8\text{ V}\), \(I = 0.15\text{ A}\), \(l = 3\text{ cm} = 3 \times 10^{-2}\text{ m}\), \(A = 2\text{ cm}^2 = 2 \times 10^{-4}\text{ m}^2\).
1. Resistance (\(R\)): \[R = \frac{V}{I} = \frac{8}{0.15} = \frac{800}{15} = \frac{160}{3}\,\Omega \approx \mathbf{53.33\,\Omega}\] 2. Resistivity (\(\rho\)): \[\rho = \frac{R A}{l} = \frac{\left(\frac{160}{3}\right) \times (2 \times 10^{-4}\text{ m}^2)}{3 \times 10^{-2}\text{ m}} = \frac{320 \times 10^{-4}}{9 \times 10^{-2}} = \frac{3.2}{9} \approx \mathbf{0.356\,\Omega\cdot\text{m}}\]

• Dispersion of light: The phenomenon of splitting of composite white light into its constituent spectral colors when passing through a refracting medium (due to different wavelengths having different speeds and refractive indices).
• Preference: We prefer a large-angled prism. Angular dispersion between violet and red rays is given by \(\theta = (\mu_V - \mu_R) A\). Since \(\theta\) is directly proportional to refracting angle \(A\), a larger prism angle produces greater angular separation and clearer dispersion.

Given: Angle of prism \(A = 60^\circ\), Minimum deviation \(D_m = 39^\circ\).
Using the prism formula: \[\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\] \[\frac{A + D_m}{2} = \frac{60^\circ + 39^\circ}{2} = \frac{99^\circ}{2} = 49.5^\circ, \quad \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ\] \[\mu = \frac{\sin(49.5^\circ)}{\sin(30^\circ)} = \frac{0.760}{0.5} = \mathbf{1.52}\]

The efficiency of a Carnot heat engine operating between source at \(T_1\) and sink at \(T_2\) is: \[\eta = 1 - \frac{T_2}{T_1}\] For efficiency to be \(100\%\) (\(\eta = 1\)): \[\frac{T_2}{T_1} = 0 \implies \text{either } T_2 = 0\text{ K (absolute zero sink) or } T_1 = \infty \text{ (infinite temperature source)}\] 1. According to the Third Law of Thermodynamics, absolute zero (\(0\text{ K}\)) is physically unobtainable in a finite sequence of processes.
2. According to the Kelvin-Planck statement of the Second Law, no engine can convert the whole of absorbed heat continuously into mechanical work without rejecting a non-zero portion to a lower temperature sink (\(Q_2 \ne 0\)). Hence, \(\eta < 100\%\) always.

Since the Wheatstone bridge is balanced, \(\frac{R_{AB}}{R_{AC}} = \frac{R_{BD}}{R_{CD}} \implies \frac{r}{r} = \frac{r}{r} = 1\). The potentials at points \(B\) and \(C\) are identical (\(V_B = V_C\)), so zero current flows through the central arm \(BC\).
• (a) Equivalent Resistance (\(R\)):
Branch \(ABD\) consists of two resistors in series: \(R_{ABD} = r + r = 1 + 1 = 2\,\Omega\).
Branch \(ACD\) consists of two resistors in series: \(R_{ACD} = r + r = 1 + 1 = 2\,\Omega\).
These two identical branches are in parallel across terminals \(A\) and \(D\): \[R = \frac{R_{ABD} \times R_{ACD}}{R_{ABD} + R_{ACD}} = \frac{2 \times 2}{2 + 2} = \mathbf{1\,\Omega}\] • (b) Currents in Arms \(AB\) and \(DC\):
Total current from 5 V battery: \(I_{\text{total}} = \frac{V}{R} = \frac{5\text{ V}}{1\,\Omega} = 5\text{ A}\).
Current through upper branch \(ABD\) (which passes through arm \(AB\)): \[I_{AB} = \frac{V}{R_{ABD}} = \frac{5\text{ V}}{2\,\Omega} = \mathbf{2.5\text{ A}}\] Current through lower branch \(ACD\) (which passes through arm \(DC\)): \[I_{DC} = \frac{V}{R_{ACD}} = \frac{5\text{ V}}{2\,\Omega} = \mathbf{2.5\text{ A}}\]