Step 1: Understanding the Concept:

The kinetic energy (\(K\)) of an object is a scalar quantity derived from its mass (\(m\)) and velocity (\(v\)).

When calculating a value from measurements that have uncertainties, the errors propagate through the mathematical operations.


Step 2: Key Formula or Approach:

The formula for kinetic energy is:
\[ K = \frac{1}{2}mv^2 \]
For propagation of errors in products and powers, the relative error in kinetic energy is:
\[ \frac{\Delta K}{K} = \frac{\Delta m}{m} + 2 \frac{\Delta v}{v} \]

Step 3: Detailed Explanation:

Given parameters: \(m = 5 kg\), \(\Delta m = 0.5 kg\), \(v = 20 m/s\), and \(\Delta v = 0.4 m/s\).

First, calculate the mean value of the kinetic energy:
\[ K = \frac{1}{2} \times 5 \times (20)^2 = \frac{1}{2} \times 5 \times 400 = 1000 J \]
Now, calculate the fractional error in \(K\):
\[ \frac{\Delta K}{1000} = \frac{0.5}{5} + 2 \left( \frac{0.4}{20} \right) \] \[ \frac{\Delta K}{1000} = 0.1 + 2(0.02) = 0.1 + 0.04 = 0.14 \]
To find the absolute error \(\Delta K\):
\[ \Delta K = 0.14 \times 1000 = 140 J \]
Thus, the kinetic energy is expressed as \((1000 \pm 140)\) J.


Step 4: Final Answer:

The kinetic energy with its error limit is \((1000 \pm 140)\) J.
Quick Tip: Remember that constants like \(1/2\) or \(\pi\) do not contribute to error propagation.
Always double the relative error of a squared term (like velocity) in the final error sum.

Step 1: Understanding the Concept:

To completely cross a tunnel, the distance traveled by a train must be the sum of the tunnel's length and the train's own length.

This is because the front of the train enters the tunnel, but the train is only "crossed" when its rear end exits.


Step 2: Key Formula or Approach:

Convert speeds to SI units (m/s):
\(v_A = 108 \times \frac{5}{18} = 30 m/s \)
\(v_B = 72 \times \frac{5}{18} = 20 m/s \)

Distance traveled by A: \(d_A = L + l\)

Distance traveled by B: \(d_B = L + 4l\)


Step 3: Detailed Explanation:

Given \(L = 60l\).

So, \(d_A = 60l + l = 61l\) and \(d_B = 60l + 4l = 64l\).

Time taken by train A: \(t_A = \frac{61l}{30}\)

Time taken by train B: \(t_B = \frac{64l}{20} = \frac{32l}{10}\)

According to the question, \(t_B - t_A = 35 s\):
\[ \frac{32l}{10} - \frac{61l}{30} = 35 \]
Multiply the first term by 3 to get a common denominator:
\[ \frac{96l - 61l}{30} = 35 \] \[ \frac{35l}{30} = 35 \implies l = 30 m \]
The length of the tunnel is \(L = 60l\):
\[ L = 60 \times 30 = 1800 m \]

Step 4: Final Answer:

The length 'L' of the tunnel is 1800 m.
Quick Tip: When dealing with multiple objects like trains, always use a single variable (like \(l\)) to express all lengths as given in the relationship \(L=60l\) to simplify equations.

Step 1: Understanding the Concept:

In pure rolling, the center of the disc moves forward by a distance equal to the arc length rotated.

For half a rotation, the center moves forward by \(\pi R\).


Step 2: Key Formula or Approach:

Let the initial position of A be \((0, 2R)\) relative to the point of contact on the ground at \(t=0\).

After half a rotation:

1. Horizontal displacement of the center is \(\pi R\).

2. The point A rotates from the top to the bottom of the disc relative to the center.


Step 3: Detailed Explanation:

Initial coordinates of A: \( (x_1, y_1) = (0, 2R) \)

Final coordinates of A:

The horizontal coordinate \(x_2\) is the distance the center moved: \(x_2 = \pi R\).

The vertical coordinate \(y_2\) is now 0 (since it moved from top to bottom relative to the center).

Final coordinates: \( (x_2, y_2) = (\pi R, 0) \)

Displacement \(S = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
\[ S = \sqrt{(\pi R - 0)^2 + (0 - 2R)^2} \] \[ S = \sqrt{\pi^2 R^2 + 4R^2} \] \[ S = R \sqrt{\pi^2 + 4} \]

Step 4: Final Answer:

The displacement of point A is \(R \sqrt{\pi^2 + 4}\).
Quick Tip: For any point on the rim, horizontal displacement is \(R(\theta + \sin\theta)\) and vertical is \(R(1 + \cos\theta)\). For half rotation, \(\theta = \pi\). This gives displacement components \(\pi R\) and \(-2R\).

Step 1: Understanding the Concept:

Kinetic energy (\(K\)) and momentum (\(p\)) are related through mass (\(m\)).

Understanding this inverse relationship allows us to find the ratio of masses when momentum is held constant.


Step 2: Key Formula or Approach:

The relationship is given by:
\[ K = \frac{p^2}{2m} \implies m = \frac{p^2}{2K} \]

Step 3: Detailed Explanation:

Given that momenta are equal: \(p_1 = p_2 = p\).

Ratio of kinetic energies: \( \frac{K_1}{K_2} = \frac{16}{9} \).

From the formula \(m \propto \frac{1}{K}\) (since \(p\) is constant):
\[ \frac{m_1}{m_2} = \frac{K_2}{K_1} \]
Substituting the given ratio:
\[ \frac{m_1}{m_2} = \frac{9}{16} \]

Step 4: Final Answer:

The ratio of their masses is 9 : 16.
Quick Tip: If you see "same momentum" in a question, the lighter body always has higher kinetic energy.
The ratio of masses is simply the inverse of the ratio of kinetic energies.

Step 1: Understanding the Concept:

Impulse is defined as the change in momentum.

According to the law of conservation of momentum, the change in momentum of the bullet is equal and opposite to the change in momentum of the gun.


Step 2: Key Formula or Approach:

Impulse (\(J\)) = Change in momentum (\(\Delta p\))
\[ J = m \times \Delta v \]

Step 3: Detailed Explanation:

Mass of bullet \(m = 10 g = 0.01 kg\).

Initial velocity of bullet \(u = 0 \).

Final velocity of bullet \(v = 600 m/s\).

Impulse supplied to the bullet:
\[ J_{bullet} = m(v - u) = 0.01 \times (600 - 0) = 6 Ns \]
By Newton's third law, the impulse supplied to the gun is equal in magnitude:
\[ J_{gun} = J_{bullet} = 6 Ns \]

Step 4: Final Answer:

The impulse supplied to the gun is 6 Ns.
Quick Tip: Ignore extra data like barrel length and gun mass if you can directly calculate the change in momentum of the projectile. It saves time during the exam.

Step 1: Understanding the Concept:

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a massive body.

It depends on the mass and radius of that body.


Step 2: Key Formula or Approach:

The escape velocity is given by:
\[ V_{esc} = \sqrt{\frac{2GM}{R}} \]
Thus, \(V_{esc} \propto \sqrt{\frac{M}{R}}\).


Step 3: Detailed Explanation:

Let \(V_p\) be the escape velocity of the planet and \(V_e\) be that of Earth.
\[ \frac{V_p}{V_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}} \]
Substitute the given values \(M_p = 9M_e\) and \(R_p = 4R_e\):
\[ \frac{V_p}{V_e} = \sqrt{\frac{9M_e}{M_e} \times \frac{R_e}{4R_e}} \] \[ \frac{V_p}{V_e} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]
Given \(V_e = 11.2 km/s\):
\[ V_p = 1.5 \times 11.2 = 16.8 km/s \]

Step 4: Final Answer:

The escape velocity on the planet is 16.8 km/s.
Quick Tip: Always solve such questions by comparing ratios rather than putting in the actual value of \(G\). It is much faster and reduces the chance of calculation errors.

Step 1: Understanding the Concept:

For an incompressible liquid in steady flow through a horizontal pipe, we use the Equation of Continuity and Bernoulli's Principle.


Step 2: Key Formula or Approach:

Continuity: \(A_A v_A = A_B v_B\)

Bernoulli (Horizontal): \(P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2\)


Step 3: Detailed Explanation:

Convert areas to same units: \(A_A = 1.5 cm^2 = 150 mm^2\).

Using Continuity:
\[ 150 \times v_A = 25 \times 60 \] \[ v_A = \frac{1500}{150} = 10 cm/s = 0.1 m/s \]
Given \(v_B = 60 cm/s = 0.6 m/s\).

Using Bernoulli's equation for pressure difference:
\[ P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2) \] \[ P_A - P_B = \frac{1}{2} \times 1000 \times (0.6^2 - 0.1^2) \] \[ P_A - P_B = 500 \times (0.36 - 0.01) \] \[ P_A - P_B = 500 \times 0.35 = 175 Pa \]

Step 4: Final Answer:

The pressure difference \((P_A - P_B)\) is 175 Pa.
Quick Tip: Be very careful with unit conversions of area (\(cm^2\) vs \(mm^2\)) and speed (\(cm/s\) vs \(m/s\)). One factor of 10 error can change the entire result.

Step 1: Understanding the Concept:

Bulk Modulus (\(B\)) is a measure of a substance's resistance to compression.

It is defined as the ratio of change in pressure to the fractional change in volume.


Step 2: Key Formula or Approach:

Bulk Modulus formula:
\[ B = -V \frac{dP}{dV} \]

Step 3: Detailed Explanation:

Given pressure equation: \(P = \alpha V^{-3}\).

Differentiate \(P\) with respect to \(V\):
\[ \frac{dP}{dV} = \alpha (-3) V^{-4} = -3\alpha V^{-4} \]
Now substitute this derivative into the formula for Bulk Modulus:
\[ B = -V \left( -3\alpha V^{-4} \right) \] \[ B = 3\alpha V^{-3} \]
Since \(P = \alpha V^{-3}\), we can substitute \(P\) back into the expression:
\[ B = 3 P \]

Step 4: Final Answer:

The bulk modulus at constant temperature is 3 P.
Quick Tip: For any process \(PV^n = constant\), the Bulk Modulus is always \(nP\). Here \(PV^3 = \alpha\), so \(n = 3\).

Step 1: Understanding the Concept:

Gas molecules have different statistical speeds based on the Maxwell-Boltzmann distribution.

The RMS speed and Average speed have different constant factors in their formulas.


Step 2: Key Formula or Approach:

Root Mean Square speed: \(v_{rms} = \sqrt{\frac{3RT}{M}}\)

Average speed: \(v_{avg} = \sqrt{\frac{8RT}{\pi M}}\)


Step 3: Detailed Explanation:

From the given condition: \(v_{rms} = \sqrt{1 + \frac{5}{x}} v_{avg}\)

Squaring both sides:
\[ \frac{v_{rms}^2}{v_{avg}^2} = 1 + \frac{5}{x} \] \[ \frac{3RT/M}{8RT/\pi M} = 1 + \frac{5}{x} \implies \frac{3\pi}{8} = 1 + \frac{5}{x} \]
Substitute \(\pi = 22/7\):
\[ \frac{3 \times 22}{8 \times 7} = 1 + \frac{5}{x} \] \[ \frac{66}{56} = 1 + \frac{5}{x} \] \[ \frac{33}{28} = 1 + \frac{5}{x} \implies \frac{5}{x} = \frac{33}{28} - \frac{28}{28} \] \[ \frac{5}{x} = \frac{5}{28} \implies x = 28 \]

Step 4: Final Answer:

The value of x is 28.
Quick Tip: Mnemonic for molecular speeds: RMS (\(\sqrt{3}\)) > Average (\(\sqrt{8/\pi}\)) > Most Probable (\(\sqrt{2}\)). The numbers are approx 1.73 > 1.6 > 1.41.

Step 1: Understanding the Concept:

In Simple Harmonic Motion (SHM), the total mechanical energy is conserved and remains constant at all positions.

The difference between total energy and potential energy is kinetic energy (\(K\)).


Step 2: Key Formula or Approach:

Total Energy \(E = \frac{1}{2} k A^2\)

Potential Energy \(U = \frac{1}{2} k x^2\)

Kinetic Energy \(K = E - U = \frac{1}{2} k (A^2 - x^2)\)


Step 3: Detailed Explanation:

The quantity to be plotted is \(f(x) = \frac{1}{2} k (A^2 - x^2)\).

This is a downward-opening parabola because the coefficient of \(x^2\) is negative.

1. At mean position (\(x=0\)), Kinetic Energy is maximum: \(K = \frac{1}{2} k A^2 = E\).

2. At extreme positions (\(x = \pm A\)), Kinetic Energy is zero.

Looking at the provided options, Graph 1 shows a maximum at the origin and zeros at \(\pm A\) with a parabolic shape.


Step 4: Final Answer:

The correct representation is Graph 1.
Quick Tip: Potential Energy graph is a 'U' shape (standard parabola). Kinetic Energy graph is the 'inverted U' shape. Always check values at \(x=0\) and \(x=A\) to identify the correct curve.

Step 1: Understanding the Concept:

Power is defined as the rate of doing work.

When raising a mass \(m\) to a height \(h\), the work done is equal to the change in potential energy, which is \(mgh\).

Power \(P\) is then calculated as \(P = \frac{Work}{Time} = \frac{mgh}{t}\).


Step 2: Key Formula or Approach:

For the first motor: \(P_1 = \frac{m_1 g h}{t_1}\)

For the second motor: \(P_2 = \frac{m_2 g h}{t_2}\)

Ratio of powers: \(\frac{P_1}{P_2} = \frac{m_1/t_1}{m_2/t_2} = \frac{m_1}{t_1} \times \frac{t_2}{m_2}\)


Step 3: Detailed Explanation:

Given:
\(m_1 = 300 kg\), \(t_1 = 5 minutes = 300 s\).
\(m_2 = 50 kg\), \(t_2 = 2 minutes = 120 s\).

Height \(h = 100 m\) is the same for both.

Calculate the ratio of powers:
\[ \frac{P_1}{P_2} = \frac{300 / 300}{50 / 120} = \frac{1}{5/12} = \frac{12}{5} = 2.4 \]

We are given the ratio as \(\frac{3\sqrt{x}}{\sqrt{x}+1}\). So:
\[ \frac{3\sqrt{x}}{\sqrt{x}+1} = 2.4 \]
\[ 3\sqrt{x} = 2.4(\sqrt{x} + 1) \]
\[ 3\sqrt{x} = 2.4\sqrt{x} + 2.4 \]
\[ 0.6\sqrt{x} = 2.4 \]
\[ \sqrt{x} = \frac{2.4}{0.6} = 4 \]
\[ x = 4^2 = 16 \]


Step 4: Final Answer:

The value of \(x\) is 16.
Quick Tip: Since height \(h\) and gravity \(g\) are common to both motors, they cancel out in the ratio.
Always convert time to seconds to maintain SI consistency, although in ratios, same units for time will also work.

Step 1: Understanding the Concept:

Maxwell's equations describe how electric and magnetic fields are generated and altered by each other and by charges and currents.

In static conditions, fields do not change with time (\(\frac{\partial}{\partial t} = 0\)).


Step 2: Detailed Explanation:

(A) \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I\): This is Ampere's Law in its basic form. It is valid for steady (static) currents but requires a displacement current term for time-varying fields.

(B) \(\oint \vec{E} \cdot d\vec{l} = -\frac{\partial \Phi_B}{\partial t}\): This is Faraday's Law of Induction. It states that a time-varying magnetic flux induces an electric field. In static conditions, \(\frac{\partial \Phi_B}{\partial t} = 0\), and the equation reduces to the conservative field condition \(\oint \vec{E} \cdot d\vec{l} = 0\). Thus, the non-zero RHS exists only for time-varying conditions.

(C) \(\oint \vec{D} \cdot d\vec{A} = Q\): This is Gauss's Law for electricity, valid for both static and dynamic conditions.

(D) \(\oint \vec{E} \cdot d\vec{l} = 0\): This is the electrostatic condition, valid only for static fields (conservative fields).

The question likely refers to the equation that specifically introduces the coupling of time-varying magnetic fields to electric fields.


Step 3: Final Answer:

The equation \(\oint \vec{E} \cdot d\vec{l} = -\frac{\partial \Phi_B}{\partial t}\) is the one that characterizes time-varying EM phenomena.
Quick Tip: Induced electric fields are non-conservative because their line integral over a closed loop is non-zero (\(\neq 0\)). This is a hallmark of time-varying electromagnetism.

Step 1: Understanding the Concept:

A time-varying magnetic field is produced by any process where the current or its equivalent (like displacement current) changes over time.


Step 2: Detailed Explanation:

(A) Permanent magnet: Produces a static magnetic field unless it is moved.

(B) Electric field changing linearly: If \(E \propto t\), then \(\frac{dE}{dt} = constant\). The displacement current \(I_d \propto \frac{dE}{dt}\) is constant. A constant current produces a static magnetic field.

(C) Direct current: Produces a constant (static) magnetic field.

(D) Decelerating charge particle: An accelerated or decelerated charge has a changing velocity, which corresponds to a changing current. This produces a time-varying magnetic field and electromagnetic radiation.

(E) Antenna fed with digital signal: While this produces varying fields, the options provided constrain the selection.

Among the given choices, (D) is a fundamental source of changing fields in physics problems.


Step 3: Final Answer:

Based on the choices, (D) only is the best fit.
Quick Tip: Always check if the "rate of change" of the source is itself zero. For a linear change in \(E\), the derivative is constant, leading to a constant (non-varying) \(B\) field.

Step 1: Understanding the Concept:

When an object is at the bottom of multiple transparent layers, the total apparent depth is the sum of the apparent depths of each individual layer.

The apparent depth of a layer of thickness \(t\) and refractive index \(n\) is \(t/n\).


Step 2: Key Formula or Approach:
\[ d_{apparent} = \sum \frac{t_i}{n_i} \]


Step 3: Detailed Explanation:

The vessel has a total real depth \(d\).

The first half (oil) has thickness \(t_1 = d/2\) and refractive index \(n_1\).

The second half (water) has thickness \(t_2 = d/2\) and refractive index \(n_2\).

Total apparent depth:
\[ d_{app} = \frac{d/2}{n_1} + \frac{d/2}{n_2} \]
\[ d_{app} = \frac{d}{2n_1} + \frac{d}{2n_2} \]

Taking common factor \(\frac{d}{2}\):
\[ d_{app} = \frac{d}{2} \left( \frac{1}{n_1} + \frac{1}{n_2} \right) \]
\[ d_{app} = \frac{d}{2} \left( \frac{n_2 + n_1}{n_1n_2} \right) = \frac{d(n_1 + n_2)}{2n_1n_2} \]


Step 4: Final Answer:

The apparent depth is \(\frac{d(n_1+n_2)}{2n_1n_2}\).
Quick Tip: If the refractive indices were the same (\(n_1=n_2=n\)), the formula simplifies to \(d/n\), which is consistent with the single-medium case. Use such limits to check your derivation.

Step 1: Understanding the Concept:

The work function (\(\Phi\)) of a metal is the minimum energy required to eject an electron. It is related to the threshold wavelength (\(\lambda_0\)) by the relation \(\Phi = \frac{hc}{\lambda_0}\).


Step 2: Key Formula or Approach:

Threshold wavelength \(\lambda_0 = \frac{hc}{\Phi}\)


Step 3: Detailed Explanation:

For metal A:
\[ \lambda_{0A} = \frac{1242 eV nm}{9 eV} = 138 nm \]

For metal B:
\[ \lambda_{0B} = \frac{1242 eV nm}{4.5 eV} = 276 nm \]

The difference in threshold wavelengths:
\[ \Delta \lambda = \lambda_{0B} - \lambda_{0A} = 276 nm - 138 nm = 138 nm \]


Step 4: Final Answer:

The difference between the threshold wavelengths is 138 nm.
Quick Tip: Since \(\Phi_B = \frac{1}{2} \Phi_A\), and \(\lambda \propto 1/\Phi\), it follows that \(\lambda_{0B} = 2\lambda_{0A}\). The difference is then simply equal to \(\lambda_{0A}\).

Step 1: Understanding the Concept:

The energy released (\(Q\)-value) in a nuclear reaction is given by the mass defect (\(\Delta m\)) multiplied by \(c^2\).

Mass defect \(\Delta m = Mass of Reactants - Mass of Products\).


Step 2: Key Formula or Approach:
\[ Q = (m_A - m_B - m_D) \times c^2 \]

Using atomic mass units (amu), \(1 amu \approx 931.5 MeV\).


Step 3: Detailed Explanation:

Calculate the mass defect \(\Delta m\) in amu:
\[ \Delta m = 238.05079 - (234.04363 + 4.00260) \]
\[ \Delta m = 238.05079 - 238.04623 = 0.00456 amu \]

Energy released \(Q\):
\[ Q = 0.00456 \times 931.5 MeV \approx 4.2476 MeV \]

Rounding to the nearest option, we get 4.25 MeV.


Step 4: Final Answer:

The approximate amount of energy released is 4.25 MeV.
Quick Tip: Be careful with the decimal subtractions. Small errors in the third or fourth decimal place will significantly change the final energy value in MeV.

Step 1: Understanding the Concept:

The circuit consists of logic gates. A NOR gate with its inputs tied together acts as a NOT gate.

The circuit shown has NOT A and NOT B as inputs to a NAND gate.


Step 2: Key Formula or Approach:

The output \(Y\) is:
\[ Y = \overline{\overline{A} \cdot \overline{B}} \]

Using De Morgan's Law: \( \overline{P \cdot Q} = \overline{P} + \overline{Q} \).

Thus, \( Y = \overline{\overline{A}} + \overline{\overline{B}} = A + B \).

The circuit behaves like an OR gate.


Step 3: Detailed Explanation:

For an OR gate, the output is high (1) if at least one input is high. It is low (0) only if both inputs are low.

Analyzing the given input waveforms:

- \(t_1\): A=1, B=1 \(\implies\) Y=1

- \(t_2\): A=0, B=0 \(\implies\) Y=0

- \(t_3\): A=1, B=0 \(\implies\) Y=1

- \(t_4\): A=0, B=1 \(\implies\) Y=1

- \(t_5\): A=1, B=1 \(\implies\) Y=1

- \(t_6\): A=0, B=0 \(\implies\) Y=0

The resulting sequence 1, 0, 1, 1, 1, 0 matches Waveform 2.


Step 4: Final Answer:

The correct output is Waveform 2.
Quick Tip: Memorizing common gate combinations is very helpful:
NOR + NOR \(\to\) OR (if arranged correctly)
NOTs into NAND \(\to\) OR
NOTs into NOR \(\to\) AND

Step 1: Understanding the Concept:

The Earth's atmosphere and ionosphere are divided into layers based on height, temperature, and ionization density.


Step 2: Detailed Explanation:

- Troposphere: The lowest layer, extending from the surface up to about 10-12 km. \(\implies\) (C) matches (I).

- D-Layer: The lowest part of the ionosphere, found at heights of about 65-75 km. \(\implies\) (B) matches (IV).

- E-Layer: Also known as the Kennelly-Heaviside layer, found at about 100 km. \(\implies\) (D) matches (III).

- \(F_1\)-Layer: Part of the upper ionosphere, found at higher altitudes, roughly 170-190 km during the day. \(\implies\) (A) matches (II).

Matching: A-II, B-IV, C-I, D-III.


Step 3: Final Answer:

The correct matching is (C).
Quick Tip: Remember the order of ionosphere layers: D, E, F (\(F_1\) then \(F_2\)) starting from the bottom up. D is lowest, F is highest.

Step 1: Understanding the Concept:

A dipole in a uniform electric field experiences a torque that tends to align it with the field.


Step 2: Key Formula or Approach:

Dipole moment \(p = q \times d\)

Torque \(\tau = pE \sin\theta\)


Step 3: Detailed Explanation:

Given:
\(q = 0.01 C\)
\(d = 0.4 mm = 0.4 \times 10^{-3} m\)

Electric field \(E = 10 dyne/C\). Since \(1 dyne = 10^{-5} N\):
\(E = 10 \times 10^{-5} N/C = 10^{-4} N/C\).

Calculate dipole moment \(p\):
\[ p = 0.01 \times 0.4 \times 10^{-3} = 4 \times 10^{-6} Cm \]

Calculate torque \(\tau\):
\[ \tau = (4 \times 10^{-6}) \times (10^{-4}) \times \sin(30^{\circ}) \]
\[ \tau = 4 \times 10^{-10} \times 0.5 = 2.0 \times 10^{-10} Nm \]


Step 4: Final Answer:

The magnitude of the torque is \(2.0 \times 10^{-10} Nm\).
Quick Tip: Always ensure units are in SI before calculation. The term 'dyne' is CGS; converting it early to Newtons is crucial.

Step 1: Understanding the Concept:

Power dissipated in a circuit is given by \(P = I^2 R_{eq}\) if current \(I\) is constant, or \(P = V^2/R_{eq}\) if voltage \(V\) is constant.

Since the diagram shows the same current \(I\) entering each combination, we use \(P = I^2 R_{eq}\). Thus, \(P \propto R_{eq}\).


Step 2: Detailed Explanation:

Calculate equivalent resistance for each case:

(A) Two resistors in parallel connected in series with one: \(R_{eq} = (R || R) + R = \frac{R}{2} + R = 1.5R\).

(B) Two resistors in series connected in parallel with one: \(R_{eq} = (R + R) || R = \frac{2R \times R}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3}R \approx 0.67R\).

(C) Three resistors in parallel: \(R_{eq} = \frac{R}{3} \approx 0.33R\).

(D) Three resistors in series: \(R_{eq} = R + R + R = 3R\).

Comparing the resistances:
\(0.33R < 0.67R < 1.5R < 3R\)
\(R_C < R_B < R_A < R_D\)

Since \(P \propto R_{eq}\), the power dissipation order is:
\(P_C < P_B < P_A < P_D\).


Step 3: Final Answer:

The increasing order for power dissipation is \(P_C < P_B < P_A < P_D\).
Quick Tip: For a constant current, more resistance means more power (\(P \propto R\)).
For a constant voltage, less resistance means more power (\(P \propto 1/R\)).
Always check the source type (Current vs Voltage) before deciding the order.

Step 1: Understanding the Concept:

In Bohr's atomic model, the radius of the \(n^{th}\) orbit for a hydrogen-like atom or ion is directly proportional to the square of the principal quantum number (\(n\)) and inversely proportional to the atomic number (\(Z\)).


Step 2: Key Formula or Approach:

The formula for the radius of the \(n^{th}\) orbit is:
\[ r_n = \frac{n^2 a_0}{Z} \]
where \(a_0\) is the Bohr radius (radius of the \(1^{st}\) orbit of hydrogen).


Step 3: Detailed Explanation:

For \(He^+\) ion: \(Z = 2\).

The radius of the \(2^{nd}\) orbit (\(n=2\)) is:
\[ r_1 = \frac{2^2 \times a_0}{2} = 2a_0 \]
For \(Be^{3+}\) ion: \(Z = 4\).

The radius of the \(4^{th}\) orbit (\(n=4\)) is:
\[ r_2 = \frac{4^2 \times a_0}{4} = 4a_0 \]
Now, calculating the ratio \(\frac{r_2}{r_1}\):
\[ \frac{r_2}{r_1} = \frac{4a_0}{2a_0} = 2 \]
Given the ratio is \(x : 1\), comparing the values:
\[ x = 2 \]

Step 4: Final Answer:

The value of \(x\) is 2.
Quick Tip: For Bohr's radius, always remember the proportionality \(r \propto \frac{n^2}{Z}\). It helps in quickly comparing different orbits and species without calculating actual values.

Step 1: Understanding the Concept:

When an observer in a denser medium (water) looks at an object in a rarer medium (air), the object appears to be further away than it actually is.

The apparent height (\(h'\)) is given by \(\mu h\), where \(\mu\) is the refractive index of the denser medium and \(h\) is the actual height.

Consequently, the apparent velocity of the object as seen by the stationary observer in water is \(\mu\) times the actual velocity.


Step 2: Key Formula or Approach:

Apparent height \(h_{app} = \mu h_{actual}\).

Differentiating with respect to time: \(v_{app} = \mu v_{actual}\).

Relative velocity of bird with respect to fish: \(v_{B/F} = v_{fish} + v_{bird, app}\) (since they move towards each other).


Step 3: Detailed Explanation:

Let \(v_B\) be the actual downward velocity of the bird and \(v_F = 8 ms^{-1}\) be the upward velocity of the fish.

The apparent velocity of the bird relative to the water surface as seen from inside water is \(v'_{B} = \mu v_B = \frac{4}{3} v_B\).

The fish observes the bird approaching with a relative velocity of \(12 ms^{-1}\).

Since they are moving in opposite directions:
\[ v_{rel} = v_F + v'_B \] \[ 12 = 8 + \frac{4}{3} v_B \] \[ 12 - 8 = \frac{4}{3} v_B \] \[ 4 = \frac{4}{3} v_B \implies v_B = 3 ms^{-1} \]

Step 4: Final Answer:

The actual velocity of the diving bird is 3 \(ms^{-1}\).
Quick Tip: Remember: Observer in Water \(\to\) Object in Air appears FARTHER and FASTER (\(v_{app} = \mu v_{act}\)).
Observer in Air \(\to\) Object in Water appears CLOSER and SLOWER (\(v_{app} = v_{act}/\mu\)).

Step 1: Understanding the Concept:

Energy is stored in an inductor when current increases through it. The rate of energy storage is the power delivered to the inductor, given by the derivative of the magnetic energy with respect to time.


Step 2: Key Formula or Approach:

Stored energy \(U = \frac{1}{2} L I^2\).

Rate of energy storage \(P_L = \frac{dU}{dt} = L I \frac{dI}{dt}\).

In a growing LR circuit: \(I = \frac{E}{R} (1 - e^{-Rt/L})\) and \(\frac{dI}{dt} = \frac{E}{L} e^{-Rt/L}\).


Step 3: Detailed Explanation:

Substitute the expressions for \(I\) and \(\frac{dI}{dt}\) into the power formula:
\[ P_L = L \left[ \frac{E}{R} (1 - e^{-Rt/L}) \right] \left[ \frac{E}{L} e^{-Rt/L} \right] \] \[ P_L = \frac{E^2}{R} (e^{-Rt/L} - e^{-2Rt/L}) \]
To find the maximum rate, let \(y = e^{-Rt/L}\). Then \(P_L = \frac{E^2}{R} (y - y^2)\).

Maximizing \((y - y^2)\) by setting \(\frac{d}{dy}(y - y^2) = 0 \implies 1 - 2y = 0 \implies y = \frac{1}{2}\).

Maximum Power \(P_{max} = \frac{E^2}{R} \left( \frac{1}{2} - \left( \frac{1}{2} \right)^2 \right) = \frac{E^2}{R} \left( \frac{1}{4} \right) = \frac{E^2}{4R}\).

Given \(R = 25 \Omega\):
\[ P_{max} = \frac{E^2}{4 \times 25} = \frac{E^2}{100} \]
Comparing this with the given form \(\frac{E^a}{2b}\):
\[ \frac{E^2}{100} = \frac{E^a}{2b} \implies a = 2 and 2b = 100 \implies b = 50 \]
The value of \(\frac{b}{a} = \frac{50}{2} = 25\).


Step 4: Final Answer:

The numerical value of \(\frac{b}{a}\) is 25.
Quick Tip: The maximum rate of energy storage in an inductor occurs when the back emf is exactly half the applied emf, which happens when \(e^{-Rt/L} = 1/2\). The result \(P_{max} = \frac{E^2}{4R}\) is a standard result for LR circuits.

Step 1: Understanding the Concept:

The resultant electric field at any point is the vector sum of fields produced by individual charge distributions. For points between a line charge and a sheet charge, the fields point in opposite directions if both charges are positive.


Step 2: Key Formula or Approach:

Electric field due to an infinite sheet: \(E_s = \frac{\sigma}{2\epsilon_0}\).

Electric field due to an infinite line: \(E_l = \frac{\lambda}{2\pi\epsilon_0 r}\).

Given \(2\sigma = \lambda\), we can substitute \(\sigma = \frac{\lambda}{2}\).


Step 3: Detailed Explanation:

At points between them, the line charge field points toward the sheet, and the sheet charge field points toward the line.

Magnitude of resultant field \(E = |E_l - E_s|\).

Field due to sheet: \(E_s = \frac{\lambda/2}{2\epsilon_0} = \frac{\lambda}{4\epsilon_0}\).

At point P (\(r = \frac{3}{\pi}\)):
\[ E_P = \left| \frac{\lambda}{2\pi\epsilon_0 (3/\pi)} - \frac{\lambda}{4\epsilon_0} \right| = \left| \frac{\lambda}{6\epsilon_0} - \frac{\lambda}{4\epsilon_0} \right| = \left| \frac{2\lambda - 3\lambda}{12\epsilon_0} \right| = \frac{\lambda}{12\epsilon_0} \]
At point Q (\(r = \frac{4}{\pi}\)):
\[ E_Q = \left| \frac{\lambda}{2\pi\epsilon_0 (4/\pi)} - \frac{\lambda}{4\epsilon_0} \right| = \left| \frac{\lambda}{8\epsilon_0} - \frac{\lambda}{4\epsilon_0} \right| = \left| \frac{\lambda - 2\lambda}{8\epsilon_0} \right| = \frac{\lambda}{8\epsilon_0} \]
Ratio of fields:
\[ \frac{E_P}{E_Q} = \frac{\lambda/12\epsilon_0}{\lambda/8\epsilon_0} = \frac{8}{12} = \frac{2}{3} \]
Given \(\frac{E_P}{E_Q} = \frac{4}{a}\):
\[ \frac{2}{3} = \frac{4}{a} \implies 2a = 12 \implies a = 6 \]

Step 4: Final Answer:

The value of \(a\) is 6.
Quick Tip: Always check the direction of fields before adding or subtracting. Since both charges are positive and the points are between them, the fields must be subtracted.

Step 1: Understanding the Concept:

Elastic potential energy is the work done in stretching a wire within its elastic limit. It is stored in the wire as potential energy per unit volume.


Step 2: Key Formula or Approach:

Potential Energy \(U = \frac{1}{2} \times Stress \times Strain \times Volume\).
\[ U = \frac{1}{2} \times \left( Y \frac{\Delta L}{L} \right) \times \left( \frac{\Delta L}{L} \right) \times (A L) = \frac{Y A (\Delta L)^2}{2 L} \]

Step 3: Detailed Explanation:

Given: \(L = 20 m\), \(\Delta L = 2 cm = 0.02 m\), \(U = 80 J\), \(Y = 2.0 \times 10^{11} Nm^{-2}\).

Substitute values into the energy formula:
\[ 80 = \frac{2.0 \times 10^{11} \times A \times (0.02)^2}{2 \times 20} \] \[ 80 = \frac{2.0 \times 10^{11} \times A \times 4 \times 10^{-4}}{40} \] \[ 80 = \frac{8.0 \times 10^7 \times A}{40} \] \[ 80 = 2.0 \times 10^6 \times A \] \[ A = \frac{80}{2 \times 10^6} = 40 \times 10^{-6} m^2 \]
Convert to \(mm^2\):
\[ A = 40 \times 10^{-6} \times (10^3)^2 mm^2 = 40 \times 10^{-6} \times 10^6 mm^2 = 40 mm^2 \]

Step 4: Final Answer:

The cross sectional area of the wire is 40 \(mm^2\).
Quick Tip: Be extremely careful with units. Convert everything to SI (meters, Joules, Newtons) before calculating, and convert to the requested unit (\(mm^2\)) only at the final step.

Step 1: Understanding the Concept:

In rolling without slipping, the body has both translational and rotational kinetic energy. The angular momentum about the center of mass is related to its rotation.


Step 2: Key Formula or Approach:

Angular momentum \(L = I_{cm} \omega\).

Total Energy \(K_{total} = \frac{1}{2} M v^2 + \frac{1}{2} I_{cm} \omega^2\).

Condition for no slipping: \(v = \omega R\).

For a solid sphere, \(I_{cm} = \frac{2}{5} M R^2\).


Step 3: Detailed Explanation:

Calculate Angular Momentum:
\[ L = \frac{2}{5} M R^2 \omega \]
Calculate Total Kinetic Energy:
\[ K_{total} = \frac{1}{2} M (\omega R)^2 + \frac{1}{2} \left( \frac{2}{5} M R^2 \right) \omega^2 = \frac{1}{2} M R^2 \omega^2 + \frac{1}{5} M R^2 \omega^2 = \frac{7}{10} M R^2 \omega^2 \]
Take the ratio \(\frac{L}{K_{total}}\):
\[ \frac{L}{K_{total}} = \frac{\frac{2}{5} M R^2 \omega}{\frac{7}{10} M R^2 \omega^2} = \frac{2}{5} \times \frac{10}{7 \omega} = \frac{4}{7 \omega} \]
Given ratio is \(\frac{\pi}{22}\). Using \(\pi = \frac{22}{7}\):
\[ \frac{\pi}{22} = \frac{22/7}{22} = \frac{1}{7} \]
Equating the ratios:
\[ \frac{4}{7 \omega} = \frac{1}{7} \implies \omega = 4 rad/s \]

Step 4: Final Answer:

The angular speed is 4 rad/s.
Quick Tip: For pure rolling, total energy is always \(\frac{1}{2} M v^2 \left( 1 + \frac{K^2}{R^2} \right)\). For a solid sphere, \(1 + K^2/R^2 = 7/5\). Using this helps calculate energy ratios very quickly.

Step 1: Understanding the Concept:

The kinetic energy (\(K\)) of a simple harmonic oscillator at any displacement \(y\) is given by the difference between the total energy and the potential energy.


Step 2: Key Formula or Approach:

Kinetic energy \(K = \frac{1}{2} k (A^2 - y^2)\).


Step 3: Detailed Explanation:

Given: \(A = 40 cm = 0.4 m\), \(y = A \cos(30^\circ) = 0.4 \times \frac{\sqrt{3}}{2} = 0.2\sqrt{3} m\).

Given Kinetic Energy \(K = 200 J\).

Calculate \((A^2 - y^2)\):
\[ A^2 - y^2 = A^2 - (A \cos 30^\circ)^2 = A^2 (1 - \cos^2 30^\circ) = A^2 \sin^2 30^\circ \] \[ A^2 - y^2 = (0.4)^2 \times \left( \frac{1}{2} \right)^2 = 0.16 \times 0.25 = 0.04 m^2 \]
Substitute into the kinetic energy formula:
\[ 200 = \frac{1}{2} \times k \times 0.04 \] \[ 200 = 0.02 k \] \[ k = \frac{200}{0.02} = 10000 = 1.0 \times 10^4 Nm^{-1} \]
Comparing with \(1.0 \times 10^x\), we get \(x = 4\).


Step 4: Final Answer:

The value of \(x\) is 4.
Quick Tip: Notice that \(\sin^2\theta + \cos^2\theta = 1\). If displacement is given as \(A \cos\theta\), kinetic energy is proportional to \(A^2 \sin^2\theta\) and potential energy is proportional to \(A^2 \cos^2\theta\).

Step 1: Understanding the Concept:

A galvanometer can be converted into an ammeter by connecting a small resistance (shunt) in parallel, and into a voltmeter by connecting a large resistance in series. In both cases, the current through the galvanometer remains its full-scale deflection current \(I_g\).


Step 2: Key Formula or Approach:

For Ammeter conversion: \(S = \frac{I_g G}{I - I_g}\).

For Voltmeter conversion: \(V = I_g (G + R)\).


Step 3: Detailed Explanation:

Let \(G\) be the galvanometer resistance and \(I_g\) be its full-scale deflection current.

Case 1 (Ammeter): Shunt \(S = 5 \Omega\), Total current \(I = 250 mA = 0.25 A\).
\[ 5 = \frac{I_g G}{0.25 - I_g} \implies 1.25 - 5I_g = I_g G \implies I_g(G+5) = 1.25 \quad --- (i) \]
Case 2 (Voltmeter): Series resistance \(R = 1050 \Omega\), Voltage \(V = 25 V\).
\[ 25 = I_g (G + 1050) \quad --- (ii) \]
Divide equation (ii) by equation (i):
\[ \frac{25}{1.25} = \frac{I_g (G + 1050)}{I_g (G + 5)} \] \[ 20 = \frac{G + 1050}{G + 5} \] \[ 20G + 100 = G + 1050 \] \[ 19G = 950 \implies G = 50 \Omega \]

Step 4: Final Answer:

The resistance of the galvanometer is 50 \(\Omega\).
Quick Tip: Whenever a problem involves both ammeter and voltmeter conversions of the same galvanometer, the key is to eliminate \(I_g\) by equating it or taking a ratio of equations.

Step 1: Understanding the Concept:

In a Common Emitter (CE) configuration, the power gain is determined by the square of the current gain (\(\beta\)) and the ratio of the output resistance (load) to the input resistance.


Step 2: Key Formula or Approach:

Current gain \(\beta = \frac{\Delta I_C}{\Delta I_B}\).

Power gain \(P.G. = \beta^2 \times \frac{R_C}{R_B}\).


Step 3: Detailed Explanation:

From the transfer characteristic graph, for \(\Delta I_B = 100 \muA = 0.1 mA\), the corresponding \(\Delta I_C = 10 mA\).

Calculate current gain \(\beta\):
\[ \beta = \frac{10 mA}{0.1 mA} = 100 \]
Given resistances: \(R_B = 10 k\Omega = 10000 \Omega\) and \(R_C = 1 k\Omega = 1000 \Omega\).

Calculate Power Gain:
\[ P.G. = (100)^2 \times \frac{1000}{10000} \] \[ P.G. = 10000 \times 0.1 = 1000 = 10^3 \]
Comparing with \(10^x\), we get \(x = 3\).


Step 4: Final Answer:

The value of \(x\) is 3.
Quick Tip: Power gain is also equal to Voltage Gain \(\times\) Current Gain. If you find voltage gain \(A_v = \beta \frac{R_c}{R_b}\), then \(P.G. = A_v \cdot \beta\).

Step 1: Understanding the Concept:

Power dissipated in a conductor depends on the applied potential and the resistance. When a wire is cut, its resistance changes proportionally to its length.


Step 2: Key Formula or Approach:

Power \(P = \frac{V^2}{R}\).

Resistance \(R = \rho \frac{l}{A}\). If length is halved, resistance is halved.


Step 3: Detailed Explanation:

Initially: Potential = \(V_0\), Resistance = \(R\).
\[ P_1 = \frac{V_0^2}{R} \]
Finally: Wire is cut into two halves. Resistance of each half = \(R/2\).

Potential \(V_0\) is applied to each half separately.

Power in the first half: \(P_{half1} = \frac{V_0^2}{R/2} = \frac{2V_0^2}{R} = 2P_1\).

Power in the second half: \(P_{half2} = \frac{V_0^2}{R/2} = \frac{2V_0^2}{R} = 2P_1\).

Total power dissipation \(P_2 = P_{half1} + P_{half2} = 4P_1\).

Ratio \(\frac{P_2}{P_1} = 4\).

Given the ratio is \(\sqrt{x} : 1\), comparing the values:
\[ \sqrt{x} = 4 \implies x = 16 \]

Step 4: Final Answer:

The value of \(x\) is 16.
Quick Tip: Cutting a wire into \(n\) pieces and applying the same voltage to each piece increases the total power by a factor of \(n^2\). Here \(n=2\), so power increases by \(2^2 = 4\).