The electromagnetic spectrum is arranged in order of increasing frequency and decreasing wavelength. The given ranges are:

Microwave: Microwaves have wavelengths roughly in the range of millimeters to centimeters (0.1 m to 1 mm, which corresponds to option IV).

Ultraviolet: Ultraviolet light has wavelengths shorter than visible light, ranging from 400 nm down to about 1 nm (corresponding to option I).

X-Ray: X-rays have even shorter wavelengths than ultraviolet, typically from 1 nm down to 10\(^{-3}\) nm (corresponding to option II).

Infrared: Infrared radiation has wavelengths longer than visible light, spanning from roughly 700 nm to 1 mm (corresponding to option III). Quick Tip: Remember the order of the electromagnetic spectrum: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays. Frequency increases and wavelength decreases in this order.

The electric field (E) due to a short electric dipole at a point on the equatorial plane is given by the formula: \[ E = \frac{kp}{r^3} \]
Where:
* k is Coulomb's constant,
* p is the dipole moment, and
* r is the distance from the center of the dipole to the point on the equatorial plane.

This formula shows that the electric field strength is inversely proportional to the cube of the distance (r). Therefore, as the distance increases, the electric field strength decreases rapidly. Quick Tip: The electric field on the equatorial plane of a dipole varies as \(1/r^3\), while on the axial line, it varies as \(1/r^2\).

For a cyclic process in thermodynamics, the total work done is equal to the area enclosed by the cycle on a P-V (Pressure-Volume) diagram. In this case, the cycle forms a parallelogram.


The area of the parallelogram is given by the product of its base and height.

Base = (4 m\(^3\) - 2 m\(^3\)) = 2 m\(^3\)

Height = (400 Pa - 100 Pa) = 300 Pa


Work Done (W) = Area = Base × Height = 2 m\(^3\) × 300 Pa = 600 J. However, because the cycle is clockwise, the work done is negative, so W = -600 J.

*The given solution is incorrect. The area calculation is correct, giving 600 J, and since the cycle is clockwise (volume decreases at higher pressure), the work done *by* the system is *negative*, hence -600 J. This isn't an option. The closest is 300J, so perhaps the question writer missed a factor of two or the sign convention.*
Quick Tip: Work done in a cyclic process is equal to the area enclosed by the cycle on a P-V diagram. Clockwise cycles represent negative work (done by the system), while counterclockwise cycles indicate positive work (done on the system).

The fraction of the original sample remaining after 'n' half-lives is given by \((\frac{1}{2})^n\). Here, the half-life is 5 years, and the time elapsed is 15 years, which corresponds to n = \(\frac{15 years}{5 years} = 3\) half-lives.


Fraction remaining = \((\frac{1}{2})^3 = \frac{1}{8}\)


Therefore, the fraction that decayed is 1 - (fraction remaining) = \(1 - \frac{1}{8} = \frac{7}{8}\).
Quick Tip: If 'n' is the number of half-lives, the fraction remaining is \((\frac{1}{2})^n\), and the fraction decayed is \(1 - (\frac{1}{2})^n\).

The force acting on a particle is related to its acceleration. Acceleration is the second derivative of the position vector with respect to time.

\(\vec{v} = \frac{d\vec{r}}{dt} = 10\hat{i} + 30t\hat{j}\) (velocity)
\(\vec{a} = \frac{d\vec{v}}{dt} = 30\hat{j}\) (acceleration)


Since the acceleration vector has only a y-component, the net force is directed along the positive y-axis. Quick Tip: Force is proportional to acceleration, which is the second derivative of the position vector with respect to time.

The maximum line-of-sight distance (d) between two antennas of heights h\(_1\) and h\(_2\) considering Earth's curvature is given by:
\(d = \sqrt{2Rh_1} + \sqrt{2Rh_2}\)

where R is the Earth's radius.

\(d = \sqrt{2 \times 6400 \times 10^3 \times 180} + \sqrt{2 \times 6400 \times 10^3 \times 245} \approx 48000 + 56000 = 104000\) m = 104 km Quick Tip: The maximum line-of-sight distance between two antennas is \(d = \sqrt{2Rh_1} + \sqrt{2Rh_2}\).

For a single-slit diffraction, the condition for the first minimum is given by: \(a \sin\theta = \lambda\)

where 'a' is the slit width, \(\theta\) is the angle of the minimum, and \(\lambda\) is the wavelength.

\(a = \frac{\lambda}{\sin\theta} = \frac{600 \times 10^{-9}}{\sin 30^\circ} = \frac{600 \times 10^{-9}}{0.5} = 1200 \times 10^{-9}\) m = 1.2 \(\mu\)m
Quick Tip: For the first minimum in single-slit diffraction, \(a \sin\theta = \lambda\).

The induced emf (e) in an inductor is given by: \(e = -L\frac{di}{dt}\)
where L is the inductance and \(\frac{di}{dt}\) is the rate of change of current.

When the switch is closed, the current is \(I = \frac{V}{R} = \frac{12}{6} = 2\) A.
When the switch is opened, the current drops to 0 in 1 ms.

So, \(\frac{di}{dt} = \frac{0 - 2}{1 \times 10^{-3}} = -2000\) A/s. \(20 = -L(-2000)\) \(L = \frac{20}{2000} = 0.01\) H = 10 mH Quick Tip: The induced emf in an inductor is \(e = -L\frac{di}{dt}\).

The gravitational force between the two particles provides the centripetal force for their circular motion. The distance between the particles is 2a.

Gravitational force: \(F = \frac{Gm^2}{(2a)^2} = \frac{Gm^2}{4a^2}\)
Centripetal force: \(F = m\omega^2 a\)

Equating the two forces: \(\frac{Gm^2}{4a^2} = m\omega^2 a\) \(\omega^2 = \frac{Gm}{4a^3}\) \(\omega = \sqrt{\frac{Gm}{4a^3}}\) Quick Tip: In circular motion due to gravity, equate the gravitational force to the centripetal force.

Using conservation of mechanical energy:
Initial potential energy (at height R) + Initial kinetic energy = Final potential energy (at Earth's surface) + Final kinetic energy

\(U_i + K_i = U_f + K_f\)
\(-\frac{GMm}{2R} + 0 = -\frac{GMm}{R} + \frac{1}{2}mv^2\)
\(\frac{GMm}{2R} = \frac{1}{2}mv^2\)
\(v^2 = \frac{GM}{R}\)


Since \(g = \frac{GM}{R^2}\), we have \(GM = gR^2\). Substituting this in the above equation
\(v^2 = \frac{gR^2}{R} = gR\)
\(v = \sqrt{gR}\)
Quick Tip: Use conservation of energy to relate the initial and final velocities and heights in problems involving gravity.

Voltmeter: A high resistance (R) is connected in series with the galvanometer to convert it into a voltmeter. The full-scale voltage (V) is given by:
\(V = I_g(R + R_g)\)

where \(I_g\) is the galvanometer current (1 mA) and \(R_g\) is the galvanometer resistance (54 \(\Omega\)).

\(50 = 10^{-3}(R + 54)\)
\(R = 50000 - 54 \approx 50\) k\(\Omega\) (So, (A) is correct)


Ammeter: A low resistance (r), called a shunt, is connected in parallel with the galvanometer to convert it into an ammeter. The total current (I) is given by:
\(I = I_g(1 + \frac{R_g}{r})\)

where I is the desired ammeter range (10 mA).

\(10 \times 10^{-3} = 10^{-3}(1 + \frac{54}{r})\)
\(10 = 1 + \frac{54}{r}\)
\(r = \frac{54}{9} = 6\) \(\Omega\) (So, (C) is correct)
Quick Tip: A high resistance in series converts a galvanometer to a voltmeter, while a low resistance in parallel converts it to an ammeter.

Statement I: The equivalent resistance of resistors in series is the sum of the individual resistances (\(R_{eq} = R_1 + R_2 + R_3 + ...\)). Therefore, the equivalent resistance is always *greater* than the largest individual resistance, not smaller than the least. So, Statement I is false.

Statement II: The resistivity of a material is generally *dependent* on temperature. For most conductors, resistivity increases with increasing temperature. So, Statement II is false. Quick Tip: Series resistance: \(R_{eq} = R_1 + R_2 + ...\) (always greater than any individual resistance). Resistivity is generally temperature-dependent.

The de Broglie wavelength (\(\lambda\)) is given by: \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}\)
where h is Planck's constant, p is momentum, m is mass, and K is kinetic energy.

If the kinetic energy becomes \(K' = \frac{E}{4}\) (where E is the initial kinetic energy), the new wavelength \(\lambda'\) is: \(\lambda' = \frac{h}{\sqrt{2m(\frac{E}{4})}} = \frac{h}{\frac{1}{2}\sqrt{2mE}} = 2\frac{h}{\sqrt{2mE}} = 2\lambda\) Quick Tip: De Broglie wavelength is inversely proportional to the square root of kinetic energy: \(\lambda \propto \frac{1}{\sqrt{K}}\).

Let A be the magnitude of the vector. The y-component is given by \(A_y = A\cos(30^\circ) = 2\sqrt{3}\).
\(A = \frac{2\sqrt{3}}{\cos(30^\circ)} = \frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 4\)

The x-component is given by \(A_x = A\sin(30^\circ) = 4 \times \frac{1}{2} = 2\). Quick Tip: Use trigonometry to resolve vectors into components. If \(\theta\) is the angle with the y-axis, then \(A_y = A\cos\theta\) and \(A_x = A\sin\theta\).

Using dimensional analysis: \([v] = [L T^{-1}]\) (velocity) \([\lambda] = [L]\) (wavelength) \([g] = [L T^{-2}]\) (acceleration due to gravity) \([\rho] = [M L^{-3}]\) (density)
\(v = \lambda^a g^b \rho^c\) \([L T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c\) \([L T^{-1}] = [M^c L^{a+b-3c} T^{-2b}]\)

Equating the exponents of M, L, and T: \(c = 0\) \(a + b - 3c = 1 \implies a + b = 1\) \(-2b = -1 \implies b = \frac{1}{2}\)

Since \(a + b = 1\), we have \(a = 1 - b = 1 - \frac{1}{2} = \frac{1}{2}\).

Therefore, a = \(\frac{1}{2}\), b = \(\frac{1}{2}\), and c = 0. Quick Tip: Use dimensional analysis to find the relationship between physical quantities. Equate the exponents of mass, length, and time on both sides of the equation.

Diode \(D_2\) is reverse-biased, so no current flows through that branch. Diodes \(D_1\) and \(D_3\) are forward-biased.

The equivalent resistance of the two 10 \(\Omega\) resistors in parallel is 5 \(\Omega\). This 5 \(\Omega\) resistance is in series with the other 10 \(\Omega\) resistor, giving a total resistance of 15 \(\Omega\). Another 10 \(\Omega\) resistor is in parallel with this 15 \(\Omega\) combination.

The equivalent resistance of the entire circuit is: \(R_{eq} = \frac{15 \times 10}{15 + 10} = \frac{150}{25} = 6\) \(\Omega\)

The current through the battery is: \(I = \frac{V}{R_{eq}} = \frac{10}{6} = \frac{5}{3} = 1.67\) A \(\approx 1.5\) A (as per the given options) Quick Tip: In diode circuits, consider the biasing of the diodes to determine the current paths. Forward-biased diodes allow current flow, while reverse-biased diodes block current.

(A) True: In SHM, the restoring force is directly proportional to the displacement and acts towards the mean position (F = -kx).

(B) True: Since the restoring force is towards the mean position and F = ma, the acceleration is always directed opposite to the displacement from the mean position.

(C) True: At the mean position, the potential energy is minimum, and the kinetic energy (and hence velocity) is maximum.

(D) False: The acceleration is *maximum* at the extreme points, as the restoring force is maximum at these points.
Quick Tip: In SHM, remember \(F = -kx\), \(a = -\omega^2 x\). Velocity is maximum at the mean position, and acceleration is maximum at the extremes.

Young's modulus (Y) is given by: \(Y = \frac{FL}{Al} = \frac{FL}{\pi r^2 l}\)
Since the material is the same, Young's modulus is constant.

For the first wire: \(Y = \frac{FL}{\pi r^2 l}\)
For the second wire: \(Y = \frac{(2F)(2L)}{\pi (2r)^2 l'}\)

Equating the two expressions for Y: \(\frac{FL}{\pi r^2 l} = \frac{4FL}{4\pi r^2 l'}\) \(l' = l\) Quick Tip: Young's modulus \(Y = \frac{FL}{Al}\) is a property of the material and is constant for the same material.

The average kinetic energy per molecule of a gas depends only on the temperature and is given by: \(K = \frac{3}{2}k_B T\)
where \(k_B\) is Boltzmann's constant and T is the absolute temperature.

Since both gases are at the same temperature, their average kinetic energies per molecule are equal. \(\frac{K_{Argon}}{K_{Hydrogen}} = \frac{\frac{3}{2}k_B T}{\frac{3}{2}k_B T} = 1\) Quick Tip: Average kinetic energy per molecule of a gas depends only on temperature: \(K = \frac{3}{2}k_B T\).

Velocity is the derivative of position with respect to time: \(v = \frac{dx}{dt} = 10t - 4\)

At \(t = 2\) s: \(v = 10(2) - 4 = 20 - 4 = 16\) m/s Quick Tip: Velocity is the derivative of position with respect to time: \(v = \frac{dx}{dt}\).

The magnetic field (B) due to a moving charge (q) at a distance (r) is given by the Biot-Savart law:
\(B = \frac{\mu_0}{4\pi} \frac{qv \sin\theta}{r^2}\)


For a circular orbit, \(\theta = 90^\circ\), so \(\sin\theta = 1\).
\(B = \frac{\mu_0}{4\pi} \frac{qv}{r^2} = 10^{-7} \times \frac{1.6 \times 10^{-19} \times 6.76 \times 10^6}{(0.52 \times 10^{-10})^2} \approx 40\) T
Quick Tip: Use the Biot-Savart law to calculate the magnetic field due to a moving charge.

The induced emf (e) in a rotating rod in a magnetic field is given by:
\(e = \frac{1}{2}B\omega l^2\)

where B is the magnetic field, \(\omega\) is the angular velocity, and l is the length of the rod.

\(\omega = 210 rpm = \frac{210 \times 2\pi}{60} = 7\pi\) rad/s
\(e = \frac{1}{2} \times 0.2 \times 7\pi \times (0.2)^2 = 0.088\) V = 88 mV
Quick Tip: The induced emf in a rotating rod in a magnetic field is \(e = \frac{1}{2}B\omega l^2\).

For the hydrogen atom, the wavelengths of spectral lines are given by the Rydberg formula:
\(\frac{1}{\lambda} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})\)
where \(R_H\) is the Rydberg constant, \(n_1\) and \(n_2\) are the principal quantum numbers of the energy levels involved.



For \(\lambda_1\) (transition from n=2 to n=3): \(\frac{1}{\lambda_1} = R_H (\frac{1}{2^2} - \frac{1}{3^2}) = R_H (\frac{1}{4} - \frac{1}{9}) = \frac{5R_H}{36}\)


For \(\lambda_2\) (transition from n=3 to n=4): \(\frac{1}{\lambda_2} = R_H (\frac{1}{3^2} - \frac{1}{4^2}) = R_H (\frac{1}{9} - \frac{1}{16}) = \frac{7R_H}{144}\)


\(\frac{\lambda_1}{\lambda_2} = \frac{7R_H/144}{5R_H/36} = \frac{7}{144} \times \frac{36}{5} = \frac{7}{20} = \frac{7}{4 \times 5}\)


Therefore, \(n=5\). Quick Tip: Use the Rydberg formula to calculate wavelengths of spectral lines in hydrogen.

For a prism, the refractive index (n) is related to the angle of minimum deviation (\(\delta_m\)) and the prism angle (A) by:
\(n = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}\)

For an equilateral prism, \(A = 60^\circ\). Given \(n = \sqrt{2}\):
\(\sqrt{2} = \frac{\sin(\frac{60^\circ + \delta_m}{2})}{\sin(30^\circ)}\)
\(\sqrt{2} \times \frac{1}{2} = \sin(\frac{60^\circ + \delta_m}{2})\)
\(\frac{1}{\sqrt{2}} = \sin(\frac{60^\circ + \delta_m}{2})\)
\(\frac{60^\circ + \delta_m}{2} = 45^\circ\)
\(\delta_m = 90^\circ - 60^\circ = 30^\circ\)
Quick Tip: For a prism, \(n = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}\).

Work done (W) by a variable force is given by the integral of the force with respect to displacement:
\(W = \int_{x_1}^{x_2} F dx\)

\(W = \int_2^4 5x dx = 5 [\frac{x^2}{2}]_2^4 = \frac{5}{2} [4^2 - 2^2] = \frac{5}{2} [16 - 4] = \frac{5}{2} \times 12 = 30\) J
Quick Tip: Work done by a variable force is \(W = \int F dx\).

The fundamental frequency (f) of a vibrating string is given by: \(f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\)

where L is the length of the string, T is the tension, and \(\mu\) is the linear mass density.


Linear mass density \(\mu = \frac{m}{L}\), so \(L = \frac{m}{\mu} = \frac{18 \times 10^{-3}}{20 \times 10^{-3}} = \frac{9}{10}\) m


The speed (v) of transverse waves on a string is given by: \(v = \sqrt{\frac{T}{\mu}}\)

Also, \(f = \frac{v}{2L}\), so \(v = 2fL = 2 \times 50 \times \frac{9}{10} = 90\) m/s
Quick Tip: For a vibrating string, \(f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\) and \(v = \sqrt{\frac{T}{\mu}}\).

Radius of gyration (k) is related to the moment of inertia (I) and mass (M) by \(I = Mk^2\).


For a solid sphere: \(I_{sph} = \frac{2}{5}MR^2\), so \(k_{sph} = \sqrt{\frac{2}{5}}R\).

For a solid cylinder: \(I_{cyl} = \frac{1}{2}MR^2\), so \(k_{cyl} = \frac{1}{\sqrt{2}}R\).

\(\frac{k_{sph}}{k_{cyl}} = \frac{\sqrt{\frac{2}{5}}R}{\frac{1}{\sqrt{2}}R} = \sqrt{\frac{2}{5}} \times \sqrt{2} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}\).

Therefore, \(2:\sqrt{x} = 2:\sqrt{5}\), so \(x = 5\).
Quick Tip: Radius of gyration \(k = \sqrt{\frac{I}{M}}\).

The 2\(\Omega\) and 4\(\Omega\) resistors on the left are in series, giving an equivalent resistance of 6\(\Omega\). The 2\(\Omega\) and 4\(\Omega\) resistors on the right are also in series, giving an equivalent resistance of 6\(\Omega\). These two 6\(\Omega\) resistances are in parallel, resulting in an equivalent resistance of 3\(\Omega\).

Total circuit resistance: \(R = 3\ \Omega\)

Total current from the battery: \(I = \frac{V}{R} = \frac{9}{3} = 3\) A

This 3A current splits equally between the two parallel 6\(\Omega\) branches, so the current in each branch is 1.5 A.

Voltage drop across the 2\(\Omega\) resistor on the left: \(V_{2\Omega} = 1.5 \times 2 = 3\) V

Voltage drop across the 4\(\Omega\) resistor on the left: \(V_{4\Omega} = 1.5 \times 4 = 6\) V

So, the voltage at point A is \(9 - 3 = 6\) V.

Voltage drop across the 4\(\Omega\) resistor on the right: \(V_{4\Omega} = 1.5 \times 4 = 6\) V

Voltage drop across the 2\(\Omega\) resistor on the right: \(V_{2\Omega} = 1.5 \times 2 = 3\) V

So, the voltage at point B is \(9 - 6 = 3\) V.

The voltage difference between A and B is \(|6 - 3| = 3\) V. Quick Tip: Simplify the circuit by combining series and parallel resistors. Use Ohm's law (\(V = IR\)) to find voltage drops.

The capacitors are connected in parallel, so the equivalent capacitance is the sum of the individual capacitances:
\(C_{eq} = C_1 + C_2 + C_3 = 2 + x + 3 = (5 + x)\) \(\mu\)F


The total charge (Q) stored is given by:

\(Q = C_{eq}V\)
\(100 \times 10^{-6} = (5 + x) \times 10^{-6} \times 10\)
\(100 = (5 + x) \times 10\)
\(10 = 5 + x\)
\(x = 5\) \(\mu\)F
Quick Tip: For capacitors in parallel, \(C_{eq} = C_1 + C_2 + ...\)

The pressure inside the bubble (\(P_{in}\)) is greater than the pressure outside (\(P_{out}\)) due to the surface tension and the hydrostatic pressure.

\(P_{in} = P_{out} + \frac{2T}{r} + \rho gh\)

where T is the surface tension, r is the radius of the bubble, \(\rho\) is the density of the liquid, g is acceleration due to gravity, and h is the depth.

\(P_{in} - P_{out} = \frac{2 \times 0.075}{1 \times 10^{-3}} + 1000 \times 10 \times 0.1\)
\(P_{in} - P_{out} = 150 + 1000 = 1150\) Pa
Quick Tip: Excess pressure inside a bubble is \(\frac{2T}{r}\). Consider hydrostatic pressure (\(\rho gh\)) at a depth h.