Step 1: Understanding the Concept:

The maximum distance for line-of-sight (LOS) communication between a transmitter and a receiver depends on their respective heights above the earth's surface.

The total line-of-sight distance \(d\) is the sum of the horizons of both antennas.

If the transmitting antenna is on the surface (\(h_t = 0\)), the entire distance \(d\) must be covered by the horizon of the receiving antenna.


Step 2: Key Formula or Approach:

The distance to the horizon for an antenna of height \(h\) is given by \(\sqrt{2Rh}\).

Given \(d = \sqrt{2Rh_r}\), where \(R\) is the radius of the earth and \(h_r\) is the height of the receiving antenna.


Step 3: Detailed Explanation:

First, convert all given values to standard SI units.

Radius of earth \(R = 6400 km = 6.4 \times 10^6 m\).

Distance \(d = 4 km = 4000 m\).

Using the formula \(d^2 = 2Rh_r\):
\[ (4000)^2 = 2 \times 6.4 \times 10^6 \times h_r \]
\[ 16 \times 10^6 = 12.8 \times 10^6 \times h_r \]
\[ h_r = \frac{16}{12.8} = 1.25 m \]

The question asks for the value in the form \(x \times 10^{-2}\) m.
\[ 1.25 m = 125 \times 10^{-2} m \]

Therefore, \(x = 125\).


Step 4: Final Answer:

The value of \(x\) is 125.
Quick Tip: For rapid calculation, remember that for \(R = 6400\) km, the horizon distance \(d\) in km is approximately \(3.57\sqrt{h}\) where \(h\) is in meters. Squaring this gives \(h \approx d^2 / 12.8\).

Step 1: Understanding the Concept:

To find the equivalent gate, we need to trace the logic states from the inputs \(a\) and \(b\) to the output \(Y\).

The circuit consists of two NOT gates at the input stage and one NOR gate at the output stage.


Step 2: Key Formula or Approach:

Output of NOT gate: \(Output = \overline{Input}\).

Output of NOR gate: \(Y = \overline{A + B}\).

De Morgan's Theorem: \(\overline{\bar{A} + \bar{B}} = A \cdot B\).


Step 3: Detailed Explanation:

1. Input \(a\) passes through a NOT gate, resulting in \(\bar{a}\).

2. Input \(b\) passes through a NOT gate, resulting in \(\bar{b}\).

3. These two signals \(\bar{a}\) and \(\bar{b}\) act as inputs to the NOR gate.

4. The output \(Y\) of the NOR gate is:
\[ Y = \overline{\bar{a} + \bar{b}} \]

5. According to De Morgan's first law, the complement of a sum is equal to the product of the complements:
\[ Y = \bar{\bar{a}} \cdot \bar{\bar{b}} \]
\[ Y = a \cdot b \]

The expression \(Y = a \cdot b\) corresponds to the AND logic operation.


Step 4: Final Answer:

The circuit is equivalent to an AND gate.
Quick Tip: This specific configuration (Bubbled NOR) is logically equivalent to an AND gate. Conversely, a Bubbled NAND gate is equivalent to an OR gate.

Step 1: Understanding the Concept:

Radioactive decay is characterized by the decay constant \(\lambda\).

The half-life (\(T_{1/2}\)) is the time required for half the atoms to decay.

The average life (\(\tau\)) is the mean lifetime of the radioactive atoms.


Step 2: Key Formula or Approach:

Formula for half-life: \(T_{1/2} = \frac{\ln 2}{\lambda}\).

Formula for average life: \(\tau = \frac{1}{\lambda}\).


Step 3: Detailed Explanation:

According to the problem statement:
\[ T_{1/2, A} = \tau_B \]

Substituting the standard formulas for each element:
\[ \frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B} \]

To find the relationship for \(\lambda_A\), we cross-multiply:
\[ \lambda_A = \lambda_B \ln 2 \]


Step 4: Final Answer:

The correct relationship is \(\lambda_A = \lambda_B \ln 2\).
Quick Tip: Always remember: \(T_{1/2} = 0.693 \tau\). If the half-life of A equals the average life of B, then A must decay slower than B to reach its half-point at the same time B reaches its average life point. Thus \(\lambda_A < \lambda_B\).

Step 1: Understanding the Concept:

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons (expressed as \(eV_s\)) is equal to the energy of incident photons minus the work function of the metal.


Step 2: Key Formula or Approach:

Photoelectric equation: \(eV_s = \frac{hc}{\lambda} - \phi = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}\).

Where \(\lambda_0\) is the threshold wavelength.


Step 3: Detailed Explanation:

For the first case:
\[ eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad ---(1) \]

For the second case:
\[ e\left(\frac{V_0}{4}\right) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \]

Multiply both sides of the second equation by 4 to facilitate elimination:
\[ eV_0 = \frac{4hc}{2\lambda} - \frac{4hc}{\lambda_0} = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_0} \quad ---(2) \]

Equating (1) and (2) as both represent \(eV_0\):
\[ \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_0} \]

Divide the entire equation by \(hc\):
\[ \frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{2}{\lambda} - \frac{4}{\lambda_0} \]

Rearrange the terms to group \(\lambda_0\) on one side:
\[ \frac{4}{\lambda_0} - \frac{1}{\lambda_0} = \frac{2}{\lambda} - \frac{1}{\lambda} \]
\[ \frac{3}{\lambda_0} = \frac{1}{\lambda} \]
\[ \lambda_0 = 3\lambda \]


Step 4: Final Answer:

The threshold wavelength is \(3\lambda\).
Quick Tip: When wavelength is doubled, photon energy is halved. If stopping potential drops by more than half, it indicates that the incident photon energy was close to the threshold.

Step 1: Understanding the Concept:

Critical angle (\(C\)) is the angle of incidence in a denser medium that results in refraction along the interface (\(90^\circ\)).

The refractive index of the denser medium relative to the rarer medium is \(n = \frac{1}{\sin C}\).


Step 2: Key Formula or Approach:

Refractive index in terms of speeds: \(n = \frac{v_{rarer}}{v_{denser}}\).

Therefore, \(\frac{v_{rarer}}{v_{denser}} = \frac{1}{\sin C} \implies v_{denser} = v_{rarer} \cdot \sin C\).


Step 3: Detailed Explanation:

Given:
\(C = 45^\circ\)
\(v_{rarer} = 3 \times 10^8 m/s\)

Calculating the speed in the denser medium:
\[ v_{denser} = (3 \times 10^8) \times \sin(45^\circ) \]
\[ v_{denser} = (3 \times 10^8) \times \frac{1}{\sqrt{2}} \]
\[ v_{denser} = \frac{3 \times 10^8}{1.414} \]
\[ v_{denser} \approx 2.1213 \times 10^8 m/s \]


Step 4: Final Answer:

The speed of light in the denser medium is \(2.12 \times 10^8 m/s\).
Quick Tip: Always remember \(\sin 45^\circ = 0.707\). Multiplying \(3 \times 0.707 = 2.121\). This shortcut avoids heavy division during exams.

Step 1: Understanding the Concept:

The graph depicts the variation of reactance/impedance against frequency (\(f\)).

Different components in an AC circuit respond differently to changes in signal frequency.


Step 2: Key Formula or Approach:

- Inductive Reactance: \(X_L = \omega L = 2\pi f L\). (Linear variation with \(f\))

- Capacitive Reactance: \(X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}\). (Inversely proportional to \(f\))

- Resistance: \(R\) is constant regardless of \(f\).


Step 3: Detailed Explanation:

1. Analysis of Curve B: Curve B is a straight line starting from the origin. This indicates that the impedance increases linearly with frequency (\(B \propto f\)). This matches the behavior of an inductor, where \(X_L = 2\pi f L\). Thus, \(B = X_L\).

2. Analysis of Curve A: Curve A is a rectangular hyperbola that decreases as frequency increases. This indicates that the impedance is inversely proportional to frequency (\(A \propto 1/f\)). This matches the behavior of a capacitor, where \(X_C = \frac{1}{2\pi f C}\). Thus, \(A = X_C\).


Step 4: Final Answer:

Curve A represents \(X_C\) and Curve B represents \(X_L\).
Quick Tip: An inductor "chokes" high frequencies (high impedance), while a capacitor "blocks" DC but passes high-frequency AC (low impedance). Use these physical properties to identify the graphs instantly.

Step 1: Understanding the Concept:

The energy in an electromagnetic wave is stored in its electric and magnetic fields. The average energy density \(u_{avg}\) represents the total energy (electric + magnetic) per unit volume.


Step 2: Key Formula or Approach:

The average energy density for an EM wave in vacuum is:
\[ u_{avg} = \frac{1}{2} \epsilon_0 E_0^2 \]

Total energy \(U = u_{avg} \times Volume\).


Step 3: Detailed Explanation:

From the wave equation, the peak electric field amplitude is \(E_0 = 20 N/C\).

Volume \(V = 5 \times 10^{-4} m^3\).

Substituting values into the energy formula:
\[ U = \left( \frac{1}{2} \epsilon_0 E_0^2 \right) \times V \]
\[ U = \frac{1}{2} \times 8.85 \times 10^{-12} \times (20)^2 \times 5 \times 10^{-4} \]
\[ U = \frac{1}{2} \times 8.85 \times 10^{-12} \times 400 \times 5 \times 10^{-4} \]
\[ U = 8.85 \times 10^{-12} \times 200 \times 5 \times 10^{-4} \]
\[ U = 8.85 \times 10^{-12} \times 1000 \times 10^{-4} \]
\[ U = 8.85 \times 10^{-12} \times 10^{-1} \]
\[ U = 8.85 \times 10^{-13} J \]


Step 4: Final Answer:

The total energy contained in the volume is \(8.85 \times 10^{-13} J\).
Quick Tip: Note that \(u_{total} = \epsilon_0 E_{rms}^2\). Since \(E_{rms} = E_0/\sqrt{2}\), squaring it leads back to the \(\frac{1}{2} \epsilon_0 E_0^2\) factor. Always check if you are given peak or RMS field.

Step 1: Understanding the Concept:

The magnetic field inside a toroid (or solenoid) depends on the permeability of the core material. Introducing a material with susceptibility \(\chi\) changes the magnetic field.


Step 2: Key Formula or Approach:

Original field: \(B_0 = \mu_0 n I\).

Field with material: \(B = \mu n I = \mu_r \mu_0 n I\).

Relationship between relative permeability and susceptibility: \(\mu_r = 1 + \chi\).


Step 3: Detailed Explanation:

Given susceptibility \(\chi = 2 \times 10^{-2} = 0.02\).

The new magnetic field is:
\[ B = B_0 (1 + \chi) \]

The fractional increase in the magnetic field is:
\[ \frac{B - B_0}{B_0} = \frac{B_0(1 + \chi) - B_0}{B_0} = \chi \]

To find the percentage increase:
\[ Percentage Increase = \chi \times 100 \]
\[ Percentage Increase = 0.02 \times 100 = 2% \]


Step 4: Final Answer:

The magnetic field increases by \(2%\).
Quick Tip: Percentage increase in \(B\) is numerically equal to \(\chi \times 100\). This is a direct consequence of the linear relationship \(B \propto \mu_r\).

Step 1: Understanding the Concept:

Current sensitivity (\(S_i\)) is the deflection per unit current. Voltage sensitivity (\(S_v\)) is the deflection per unit voltage. They are linked via Ohm's law.


Step 2: Key Formula or Approach:

Relationship: \(S_v = \frac{\theta}{V} = \frac{\theta}{IR} = \frac{S_i}{R}\).


Step 3: Detailed Explanation:

We are given that \(S_i\) increases by \(25%\). Let the initial current sensitivity be \(S_{i1}\) and the final be \(S_{i2}\).
\[ S_{i2} = S_{i1} + 0.25 S_{i1} = 1.25 S_{i1} \]

The problem explicitly states that the resistance \(R\) of the galvanometer is kept constant.

Initial voltage sensitivity \(S_{v1} = \frac{S_{i1}}{R}\).

Final voltage sensitivity \(S_{v2} = \frac{S_{i2}}{R} = \frac{1.25 S_{i1}}{R}\).

Substituting \(S_{v1}\):
\[ S_{v2} = 1.25 S_{v1} \]

The change is an increase of \(25%\).


Step 4: Final Answer:

The percentage change in voltage sensitivity is \(+25%\).
Quick Tip: In general, increasing turns \(N\) increases \(S_i\) but often increases \(R\) as well, sometimes leaving \(S_v\) unchanged. However, since this question says \(R\) is constant, \(S_v\) must change exactly as \(S_i\) does.

Step 1: Understanding the Concept:

Heat production in a resistor connected to a voltage source \(V\) is inversely proportional to the equivalent resistance of the circuit.


Step 2: Key Formula or Approach:

Formula for heat: \(H = \frac{V^2}{R_{eq}} t\).

For identical resistors \(R\):

In series: \(R_s = R + R = 2R\).

In parallel: \(R_p = \frac{R \times R}{R + R} = \frac{R}{2}\).


Step 3: Detailed Explanation:

Calculate heat produced in parallel:
\[ H_p = \frac{V^2}{R_p} t = \frac{V^2}{(R/2)} t = \frac{2V^2}{R} t \]

Calculate heat produced in series:
\[ H_s = \frac{V^2}{R_s} t = \frac{V^2}{(2R)} t = \frac{V^2}{2R} t \]

Now, take the ratio of parallel to series:
\[ \frac{H_p}{H_s} = \frac{(2V^2/R) t}{(V^2/2R) t} \]

The constants \(V, t,\) and \(R\) cancel out:
\[ \frac{H_p}{H_s} = \frac{2}{1/2} = 4 \]

The ratio is \(4 : 1\).


Step 4: Final Answer:

The ratio of heat produced in parallel to series is \(4 : 1\).
Quick Tip: For \(n\) identical resistors, the ratio of power (or heat) in parallel to series for a constant voltage is always \(n^2 : 1\). Since \(n = 2\), the ratio is \(2^2 : 1 = 4 : 1\).

Step 1: Understanding the Concept:

When a charged capacitor is connected to another uncharged capacitor, the total charge is conserved but redistributed. Due to the change in equivalent capacitance, the total energy stored in the system changes.


Step 2: Key Formula or Approach:

- Energy stored in a capacitor: \( E = \frac{1}{2}CV^2 \) or \( E = \frac{Q^2}{2C} \).

- Charge on a capacitor: \( Q = CV \).

- In parallel combination: \( C_{eq} = C_1 + C_2 \).


Step 3: Detailed Explanation:

Initially, for the first capacitor:

Capacitance \( C = 2 F \), Potential = \( V \).

Initial energy \( E_1 = \frac{1}{2}CV^2 = \frac{1}{2}(2)V^2 = V^2 \).

Initial charge \( Q = CV = 2V \).


Now, it is connected to an uncharged identical capacitor (\( C = 2 F \)) in parallel.

Equivalent capacitance \( C_{eq} = 2 + 2 = 4 F \).

Since the battery is disconnected (as not mentioned otherwise), total charge \( Q \) remains conserved.

Final energy \( E_2 = \frac{Q^2}{2C_{eq}} = \frac{(2V)^2}{2(4)} = \frac{4V^2}{8} = \frac{V^2}{2} \).


The ratio \( \frac{E_2}{E_1} \) is:
\[ \frac{E_2}{E_1} = \frac{V^2/2}{V^2} = \frac{1}{2} \]


Step 4: Final Answer:

The ratio of the energy stored in the combination to the initial energy is 1 : 2.
Quick Tip: When a charged capacitor is connected to \( n-1 \) identical uncharged capacitors in parallel, the total energy decreases by a factor of \( n \). Here \( n=2 \), so the final energy is half the initial energy.

Step 1: Understanding the Concept:

In Simple Harmonic Motion (SHM), the kinetic energy is a function of displacement. At the mean position (\(x=0\)), velocity is maximum, so KE is maximum. At the extreme position (\(x=A\)), the particle momentarily stops, so KE is zero.


Step 2: Key Formula or Approach:

The formula for kinetic energy in SHM is:
\[ KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2) \]


Step 3: Detailed Explanation:

- At the mean position (\(x = 0\)): \( KE = \frac{1}{2}m\omega^2A^2 \). (Maximum value)

- At the extreme position (\(x = A\)): \( KE = \frac{1}{2}m\omega^2(A^2 - A^2) = 0 \).

- The relationship \( KE \propto (A^2 - x^2) \) represents a downward-opening parabola.

- Between \(x=0\) and \(x=A\), the graph must show a parabolic curve starting from a high value on the y-axis and decreasing to zero at \(x=A\).


Step 4: Final Answer:

The correct graph is the inverted parabolic curve shown in option (D).
Quick Tip: Remember that Kinetic Energy is maximum at the mean position and zero at extremes, while Potential Energy (\(U = \frac{1}{2}kx^2\)) is zero at the mean and maximum at extremes. The graphs are always parabolas.

Step 1: Understanding the Concept:

The root mean square speed of gas molecules depends on the absolute temperature of the gas and its molar mass.


Step 2: Key Formula or Approach:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

For the same temperature \( T \), \( v_{rms} \propto \frac{1}{\sqrt{M}} \).


Step 3: Detailed Explanation:

Let's look at the molar masses (\(M\)) of the given gases:

1. Neon (monoatomic): \( M_{Ne} \approx 20 g/mol \).

2. Chlorine (diatomic): \( M_{Cl_2} \approx 2 \times 35.5 = 71 g/mol \).

3. Uranium hexafluoride (polyatomic): \( M_{UF_6} \approx 238 + (6 \times 19) = 352 g/mol \).


Comparing the masses: \( M_{mono} < M_{dia} < M_{poly} \).

Since \( v_{rms} \) is inversely proportional to the square root of molar mass:
\( v_{rms}(mono) > v_{rms}(dia) > v_{rms}(poly) \).


Step 4: Final Answer:

The monoatomic gas (lightest) has the highest \(v_{rms}\), and the polyatomic gas (heaviest) has the lowest.
Quick Tip: At a constant temperature, "the heavier the gas molecule, the slower it moves." Always check the molar mass for \(v_{rms}\) comparisons.

Step 1: Understanding the Concept:

According to the First Law of Thermodynamics, the heat supplied to a system is used to increase its internal energy and to do work against external pressure.


Step 2: Key Formula or Approach:

- First Law of Thermodynamics: \( Q = \Delta U + W \).

- Heat supplied during phase change: \( Q = mL \).

- Work done at constant pressure: \( W = P \Delta V = P(V_2 - V_1) \).


Step 3: Detailed Explanation:

1. Calculate Heat Supplied (\(Q\)):

Mass \( m = 1 kg \), \( L = 2257 kJ/kg \).
\( Q = 1 \times 2257 = 2257 kJ \).


2. Calculate Work Done (\(W\)):
\( P = 1 \times 10^5 Pa \).
\( \Delta V = V_{steam} - V_{water} = 1.671 - 1.00 \times 10^{-3} = 1.671 - 0.001 = 1.670 m^3 \).
\( W = 10^5 \times 1.670 = 167,000 J = 167 kJ \).


3. Calculate Change in Internal Energy (\(\Delta U\)):
\( \Delta U = Q - W \)
\( \Delta U = 2257 kJ - 167 kJ = 2090 kJ \).

Since heat is absorbed and volume increases, internal energy increases (\(\Delta U\) is positive).


Step 4: Final Answer:

The change in internal energy is +2090 kJ.
Quick Tip: Be very careful with units! Atmospheric pressure is in Pa (\(N/m^2\)), which yields work in Joules. Convert it to kJ to match the latent heat unit before subtracting.

Step 1: Understanding the Concept:

For any linear temperature scale, the ratio of (Temperature - Lower Fixed Point) to (Upper Fixed Point - Lower Fixed Point) remains constant.


Step 2: Key Formula or Approach:
\[ \frac{X - LFP_x}{UFP_x - LFP_x} = \frac{F - LFP_f}{UFP_f - LFP_f} \]

Where LFP is the Freezing Point and UFP is the Boiling Point.


Step 3: Detailed Explanation:
For scale 'X':

LFP\(_x = -15^{\circ} \), UFP\(_x = 65^{\circ} \).

For Fahrenheit scale:

LFP\(_f = 32^{\circ} \), UFP\(_f = 212^{\circ} \).

Given temperature \( X = -95^{\circ} \).


Substituting the values:
\[ \frac{-95 - (-15)}{65 - (-15)} = \frac{F - 32}{212 - 32} \]
\[ \frac{-80}{80} = \frac{F - 32}{180} \]
\[ -1 = \frac{F - 32}{180} \]
\[ F - 32 = -180 \]
\[ F = -180 + 32 = -148^{\circ} F \]


Step 4: Final Answer:

The equivalent temperature is \(-148^{\circ} F\).
Quick Tip: This formula \( \frac{T - LFP}{Range} = Constant \) works for any thermometer calibration problem. Memorize the fixed points for Celsius (\(0, 100\)), Fahrenheit (\(32, 212\)), and Kelvin (\(273, 373\)).

Step 1: Understanding the Concept:

The acceleration due to gravity on the surface of a planet depends on its mass and radius. If we know the density and radius, we can express \(g\) in terms of these two variables.


Step 2: Key Formula or Approach:
\[ g = \frac{GM}{R^2} \]

Since \( Mass = Volume \times Density \), \( M = \frac{4}{3}\pi R^3 \rho \).

Substituting this into the \(g\) formula:
\[ g = \frac{G (\frac{4}{3}\pi R^3 \rho)}{R^2} = \frac{4}{3} \pi G R \rho \]

Thus, \( g \propto R \rho \).


Step 3: Detailed Explanation:

Let \( g_A \) and \( g_B \) be the gravity on planets A and B.
\[ \frac{g_A}{g_B} = \frac{R_A \rho_A}{R_B \rho_B} \]

Given: \( R_A = R \), \( R_B = 4R \), \( \rho_A = \rho \), \( \rho_B = \rho/3 \).

Substituting these:
\[ \frac{g_A}{g_B} = \frac{R \cdot \rho}{4R \cdot (\rho/3)} \]
\[ \frac{g_A}{g_B} = \frac{1}{4/3} = \frac{3}{4} \]


Step 4: Final Answer:

The ratio \( g_A : g_B \) is 3 : 4.
Quick Tip: When comparing gravity on surfaces, if radius and density are given, use \( g \propto R \rho \). If mass and radius are given, use \( g \propto M/R^2 \). Choosing the right dependency saves time.

Step 1: Understanding the Concept:

The area under a velocity-time (\(v-t\)) graph represents the position change.

- Displacement: Sum of areas (considering sign).

- Distance: Sum of the absolute values of areas (ignoring sign).


Step 2: Key Formula or Approach:

Area of Triangle = \( \frac{1}{2} \times base \times height \).

Area of Trapezium = \( \frac{1}{2} \times (sum of parallel sides) \times height \).


Step 3: Detailed Explanation:

Let's divide the graph into parts:

1. \(t=0\) to \(5 s\) (Triangle): Area \( A_1 = \frac{1}{2} \times 5 \times 10 = 25 m \).

2. \(t=5\) to \(10 s\) (Rectangle): Area \( A_2 = 5 \times 10 = 50 m \).

3. \(t=10\) to \(15 s\) (Trapezium): Area \( A_3 = \frac{1}{2} \times (10 + 20) \times 5 = 75 m \).

4. \(t=15\) to \(20 s\) (Triangle): Area \( A_4 = \frac{1}{2} \times 5 \times 20 = 50 m \).

5. \(t=20\) to \(25 s\) (Triangle below axis): Area \( A_5 = \frac{1}{2} \times 5 \times (-20) = -50 m \).


Displacement = \( 25 + 50 + 75 + 50 - 50 = 150 m \).

Distance = \( 25 + 50 + 75 + 50 + |-50| = 250 m \).


Ratio \(\frac{Distance}{Displacement} = \frac{250}{150} = \frac{25}{15} = \frac{5}{3} \).


Step 4: Final Answer:

The ratio of distance to displacement is 5/3.
Quick Tip: In a \(v-t\) graph, if the entire graph is above the x-axis, distance equals displacement. If it goes below, they will differ. Distance is always \(\ge\) Displacement.

Step 1: Understanding the Concept:

A coin on a rotating table stays in place due to static friction acting as the centripetal force. It "just slips" when the required centripetal force equals the maximum static friction (\( \mu mg \)).


Step 2: Key Formula or Approach:

Centripetal Force required = \( m \omega^2 r \).

At the point of slipping: \( \mu mg = m \omega^2 r \).

Since \( \mu \) and \( g \) are constant: \( \omega^2 r = constant \implies r \propto \frac{1}{\omega^2} \).


Step 3: Detailed Explanation:

Let initial values be \( r_1 = 1 cm \) and \( \omega_1 = \omega \).

Let new angular velocity be \( \omega_2 = \omega/2 \). We need to find \( r_2 \).

Using the relation \( r_1 \omega_1^2 = r_2 \omega_2^2 \):
\[ 1 \times \omega^2 = r_2 \times \left(\frac{\omega}{2}\right)^2 \]
\[ \omega^2 = r_2 \times \frac{\omega^2}{4} \]
\[ r_2 = 4 cm \]


Step 4: Final Answer:

The coin will slip at a distance of 4 cm from the center.
Quick Tip: If the speed is reduced, you can place the object further out to experience the same limiting friction. Halving the speed allows for 4 times the radius because of the squared relationship (\( 1/0.5^2 = 4 \)).

Step 1: Understanding the Concept:

The force required to hold the gun is equal to the rate of change of momentum of the bullets being fired. According to Newton's Second Law, \( F = \frac{dp}{dt} \).


Step 2: Key Formula or Approach:

For \( n \) bullets fired per second, each with mass \( m \) and velocity \( v \):

Total momentum change per second = \( n \times m \times v \).

Thus, \( F = n \cdot m \cdot v \).


Step 3: Detailed Explanation:

Given:
\( F = 125 N \).
\( m = 10 g = 0.01 kg \).
\( v = 250 m/s \).


Substituting the values:
\[ 125 = n \times 0.01 \times 250 \]
\[ 125 = n \times 2.5 \]
\[ n = \frac{125}{2.5} = \frac{1250}{25} = 50 \]


Step 4: Final Answer:

The number of bullets fired per second is 50.
Quick Tip: Always ensure mass is in kg and speed in m/s before calculating force in Newtons. For recoil problems, Force = (bullets/sec) \(\times\) (mass per bullet) \(\times\) (muzzle velocity).

Step 1: Understanding the Concept:

The question tests the knowledge of units of astronomical distance and their relative magnitudes.


Step 2: Key Values:

- 1 Astronomical Unit (AU) \(\approx 1.5 \times 10^{11} m \) (Distance from Earth to Sun).

- 1 Light Year (ly) \(\approx 9.46 \times 10^{15} m \) (Distance light travels in 1 year).

- 1 Parsec (pc) \(\approx 3.08 \times 10^{16} m \approx 3.26 ly \).


Step 3: Detailed Explanation:

- Statement I: AU, pc, and ly are indeed standard units for measuring vast distances in space. So, Statement I is correct.

- Statement II: Based on the values above, the correct order is \( AU < ly < pc \).

- Statement II claims \( pc < ly \), which is false as 1 Parsec is about 3.26 Light years. Thus, Statement II is incorrect.


Step 4: Final Answer:

Statement I is correct, but Statement II is incorrect.
Quick Tip: Remember: "Parsec" is the largest of these three common astronomical units. 1 pc \(\approx\) 3.26 ly. 1 ly \(\approx\) 63,241 AU.

Step 1: Understanding the Concept:

When a hydrogen atom in the ground state absorbs energy, it is excited to a higher energy level \(n\). Upon de-excitation, it emits photons. The number of unique spectral lines (wavelengths) emitted when returning from state \(n\) to the ground state is given by \(N = \frac{n(n-1)}{2}\).


Step 2: Key Formula or Approach:

- Number of emission lines: \(N = \frac{n(n-1)}{2}\).

- Energy of incident photon: \(E = E_n - E_1 = hf\).

- Energy levels of Hydrogen: \(E_n = -\frac{13.6}{n^2}\) eV.


Step 3: Detailed Explanation:

1. Determine the excited state \(n\):

Given \(N = 6\), we solve for \(n\):
\[ 6 = \frac{n(n-1)}{2} \implies n(n-1) = 12 \implies n^2 - n - 12 = 0 \]

Solving the quadratic equation: \((n-4)(n+3) = 0\). Since \(n > 0\), we have \(n = 4\).


2. Calculate the energy absorbed:

The atom was excited from \(n=1\) to \(n=4\). The energy of the incident photon must be:
\[ E = E_4 - E_1 = -0.85 eV - (-13.6 eV) = 12.75 eV \]


3. Calculate the frequency (\(f\)):

Using \(E = hf\):
\[ f = \frac{E}{h} = \frac{12.75 eV}{4.25 \times 10^{-15} eVs} \]
\[ f = 3 \times 10^{15} Hz \]

Comparing with the given form \(x \times 10^{15}\) Hz, we find \(x = 3\).


Step 4: Final Answer:

The value of \(x\) is 3.
Quick Tip: Memorize the number of spectral lines for the first few levels: \(n=2 \to 1\), \(n=3 \to 3\), \(n=4 \to 6\), \(n=5 \to 10\). This saves time in solving the quadratic equation.

Step 1: Understanding the Concept:

The power of a lens depends on its refractive index relative to the surrounding medium and its geometry. This is described by the Lens Maker's Formula.


Step 2: Key Formula or Approach:

Lens Maker's Formula: \(P = \frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).


Step 3: Detailed Explanation:

For a double convex lens with radii of curvature \(R = 20 cm\):
\(R_1 = +20 cm\), \(R_2 = -20 cm\).

Geometric factor \(G = \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{1}{20} - \frac{1}{-20} \right) = \frac{2}{20} = \frac{1}{10} cm^{-1} \).


1. Power in Air (\(P_a\)):

Here, \(\mu_{rel} = \mu_g = 1.8 \).
\[ P_a = (1.8 - 1) \cdot G = 0.8 \cdot G \]


2. Power in Liquid (\(P_l\)):

Here, \(\mu_{rel} = \frac{\mu_g}{\mu_l} = \frac{1.8}{1.5} = 1.2 \).
\[ P_l = (1.2 - 1) \cdot G = 0.2 \cdot G \]


3. Calculate the ratio \(x\):
\[ \frac{P_a}{P_l} = \frac{0.8 \cdot G}{0.2 \cdot G} = \frac{0.8}{0.2} = 4 \]

Thus, the ratio is \(4 : 1\), so \(x = 4\).


Step 4: Final Answer:

The value of \(x\) is 4.
Quick Tip: Notice that the actual values of radii of curvature often cancel out in ratio problems. Focusing only on the term \((\mu_{rel} - 1)\) makes the calculation much faster.

Step 1: Understanding the Concept:

Faraday's Law of Induction states that the magnitude of induced emf is equal to the rate of change of magnetic flux through the circuit. Since the magnetic field is normal to the plate, flux \(\Phi = BA\).


Step 2: Key Formula or Approach:

Induced emf \(|e| = \left| \frac{d\Phi}{dt} \right| = A \left| \frac{dB}{dt} \right| \).

The term \(\frac{dB}{dt}\) is the slope of the B-t graph.


Step 3: Detailed Explanation:

1. Determine the slope of the B-t graph:

The graph is a straight line passing through the origin. From the figure:

At \(t = 0 s, B = 0 mT\).

At \(t = 5 s, B = 10 mT\).

Slope \(\frac{dB}{dt} = \frac{10 - 0}{5 - 0} = 2 mT/s = 2 \times 10^{-3} T/s\).


2. Calculate the induced emf:

Area \(A = 4 m^2\).
\[ |e| = A \cdot \frac{dB}{dt} = 4 m^2 \times 2 \times 10^{-3} T/s \]
\[ |e| = 8 \times 10^{-3} V = 8 mV \]


Step 4: Final Answer:

The magnitude of the induced emf is 8 mV.
Quick Tip: Induced emf is constant when the rate of change of magnetic field is linear. Since the slope is constant from \(t=0\) to \(t=5\), the answer remains the same for any sub-interval in this range.

Step 1: Understanding the Concept:

This is a standard DC circuit problem involving cells and resistors. We need to find the equivalent resistance and the total current, then apply the current division rule.


Step 2: Key Formula or Approach:

- Net EMF for series-aiding cells: \(E_{net} = E_1 + E_2\).

- Equivalent parallel resistance: \(R_p = \frac{R_1 R_2}{R_1 + R_2}\).

- Ohm's Law: \(I = \frac{E}{R}\).


Step 3: Detailed Explanation:

1. Identify the circuit structure:

The two batteries (4V and 8V) are connected in series aiding (positive to negative).

Net EMF \(E_{net} = 4 + 8 = 12 V\).


2. Calculate equivalent resistance:

- Internal resistances: \(r_1 = 0.5 \Omega, r_2 = 1 \Omega\).

- External resistor: \(R_3 = 4.5 \Omega\).

- Parallel combination of \(3 \Omega\) and \(6 \Omega\):
\[ R_p = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \Omega \]

Total Resistance \(R_{total} = 0.5 + 1 + 4.5 + 2 = 8 \Omega\).


3. Find total current and specific branch current:

Main current \(I = \frac{12 V}{8 \Omega} = 1.5 A\).

Using the current divider for the \(3 \Omega\) resistor:
\[ I_{3\Omega} = I \times \left( \frac{6}{3 + 6} \right) = 1.5 \times \frac{2}{3} = 1 A \]

Given \(I_{3\Omega} = \frac{x}{3}\), then \(1 = \frac{x}{3} \implies x = 3\).


Step 4: Final Answer:

The value of \(x\) is 3.
Quick Tip: In parallel branches, current divides inversely as the resistance. Since \(3 \Omega\) is half of \(6 \Omega\), it carries twice the current. Thus, \(I_{3\Omega} = \frac{2}{3}I_{total}\).

Step 1: Understanding the Concept:

For the upper charge to be in equilibrium on a frictionless incline, the component of gravity pulling it down the slope must be balanced by the electrostatic repulsion from the lower charge.


Step 2: Key Formula or Approach:

- Component of gravity down incline: \(F_g = mg \sin\theta\).

- Coulomb's Law: \(F_e = \frac{k q^2}{r^2}\).

- Relation between distance \(r\) and height \(h\): \(r = \frac{h}{\sin\theta}\).


Step 3: Detailed Explanation:

Let \(r\) be the separation between the charges along the incline.

The height \(h\) is related to \(r\) by \(h = r \sin(30^\circ) \implies r = 2h\).

Equilibrium condition:
\[ \frac{k q_0^2}{r^2} = mg \sin(30^\circ) \]

Substituting the values:
\[ \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{(2h)^2} = (0.02 kg) \cdot (10 m/s^2) \cdot (0.5) \]
\[ \frac{9 \times 10^9 \times 4 \times 10^{-12}}{4h^2} = 0.1 \]
\[ \frac{9 \times 10^{-3}}{h^2} = 0.1 \]
\[ h^2 = 0.09 \implies h = 0.3 m \]

Converting to the required form: \(0.3 m = 300 \times 10^{-3} m\).

Therefore, \(x = 300\).


Step 4: Final Answer:

The value of \(x\) is 300.
Quick Tip: Ensure the mass is in kilograms (20 g = 0.02 kg) and charge is in Coulombs (\(2 \muC = 2 \times 10^{-6} C\)) to maintain consistency with SI units.

Step 1: Understanding the Concept:

The standard equation of a travelling wave is \(y = A \sin(\omega t - kx + \phi)\). The wave speed \(v\) is given by the ratio of angular frequency \(\omega\) to wave number \(k\).


Step 2: Key Formula or Approach:

Wave speed \(v = \frac{\omega}{k}\).


Step 3: Detailed Explanation:

Rewrite the given equation by distributing \(2\pi\):
\[ Y = 10^{-2} \sin (320\pi t - \pi x + \pi/2) \]

Comparing with the standard form:

Angular frequency \(\omega = 320\pi rad/s\).

Wave number \(k = \pi rad/m\).

Speed in m/s:
\[ v = \frac{320\pi}{\pi} = 320 m/s \]

Convert the speed to km/h:
\[ v (km/h) = 320 \times \frac{18}{5} = 64 \times 18 = 1152 km/h \]


Step 4: Final Answer:

The speed of the wave is 1152 km/h.
Quick Tip: Speed of wave \(v\) is simply the coefficient of \(t\) divided by the coefficient of \(x\). Here, \(v = \frac{160}{0.5} = 320 m/s\). This saves expansion steps!

Step 1: Understanding the Concept:

Hooke's Law states that within the elastic limit, extension is directly proportional to the applied force (tension). The total length is the natural length plus the extension.


Step 2: Key Formula or Approach:
\(F = k(l - L)\), where \(L\) is the natural length.

Thus, \(l = L + \frac{F}{k}\).


Step 3: Detailed Explanation:

Let the natural length be \(L\) and spring constant be \(k\).

For \(F = 100 N\): \(l_1 = L + \frac{100}{k} \implies \frac{100}{k} = l_1 - L \quad ---(1)\)

For \(F = 120 N\): \(l_2 = L + \frac{120}{k} \implies \frac{120}{k} = l_2 - L \quad ---(2)\)

Dividing (1) by (2):
\[ \frac{100}{120} = \frac{l_1 - L}{l_2 - L} \implies \frac{5}{6} = \frac{l_1 - L}{l_2 - L} \]
\[ 5l_2 - 5L = 6l_1 - 6L \implies L = 6l_1 - 5l_2 \]

Given \(10l_2 = 11l_1 \implies l_2 = 1.1 l_1\).

Substitute \(l_2\) in the expression for \(L\):
\[ L = 6l_1 - 5(1.1 l_1) = 6l_1 - 5.5 l_1 = 0.5 l_1 \]
\[ L = \frac{1}{2} l_1 \]

Comparing with \(L = \frac{1}{x} l_1\), we find \(x = 2\).


Step 4: Final Answer:

The value of \(x\) is 2.
Quick Tip: Linear proportionality problems like this can be solved using the ratio of force changes: \(\frac{\Delta F_1}{\Delta F_2} = \frac{\Delta l_1}{\Delta l_2}\). It avoids complex substitution.

Step 1: Understanding the Concept:

We need to calculate the Moment of Inertia (MI) about a tangent using the parallel axis theorem and the angular momentum (\(L\)) about the diameter.


Step 2: Key Formula or Approach:

- MI about diameter: \(I_d = \frac{2}{5} mR^2\).

- MI about tangent: \(I_t = I_d + mR^2 = \frac{7}{5} mR^2\).

- Angular momentum: \(L = I_d \omega\).


Step 3: Detailed Explanation:

Given:
\(m = 0.5 kg\), \(R = 0.05 m\), \(\omega = 10 rad/s\).

Ratio required: \(I_t = (x \times 10^{-2}) \cdot L\).
\[ \frac{7}{5} mR^2 = (x \times 10^{-2}) \cdot \left( \frac{2}{5} mR^2 \cdot \omega \right) \]

The terms \(m\) and \(R^2\) cancel out:
\[ \frac{7}{5} = (x \times 10^{-2}) \cdot \frac{2}{5} \cdot 10 \]
\[ 7 = (x \times 10^{-2}) \cdot 20 \]
\[ 7 = x \cdot 0.2 \]
\[ x = \frac{7}{0.2} = 35 \]


Step 4: Final Answer:

The value of \(x\) is 35.
Quick Tip: Notice that the mass and radius were not needed for the calculation because they appeared on both sides of the equation. Always look for cancellations to avoid unnecessary arithmetic.

Step 1: Understanding the Concept:

Work done by a variable force is calculated by integrating the force over the displacement interval.


Step 2: Key Formula or Approach:
\(W = \int_{x_1}^{x_2} F dx \).


Step 3: Detailed Explanation:

Substitute the expression for \(F\):
\[ W = \int_{0}^{4} (2 + 3x) dx \]

Perform the integration:
\[ W = [2x + \frac{3x^2}{2}]_{0}^{4} \]

Substitute the upper and lower limits:
\[ W = \left( 2(4) + \frac{3(4)^2}{2} \right) - (0) \]
\[ W = 8 + \frac{3 \times 16}{2} = 8 + 3 \times 8 = 8 + 24 \]
\[ W = 32 J \]


Step 4: Final Answer:

The work done is 32 J.
Quick Tip: Since the force is linear in \(x\), you can also calculate work as: \(Average Force \times Displacement\).
\(F_{avg} = \frac{F(0) + F(4)}{2} = \frac{2 + 14}{2} = 8 N\).
\(W = 8 \times 4 = 32 J\).

Step 1: Understanding the Concept:

In projectile motion, for a given height \(h\) above the ground (where \(h < H_{max}\)), the projectile reaches this height at two distinct times: \(t_1\) during its ascent and \(t_2\) during its descent.

Due to the symmetry of the parabolic trajectory, the sum of these two times is equal to the total time of flight \(T\) of the projectile.


Step 2: Key Formula or Approach:

The total time of flight \(T\) for a projectile fired with velocity \(u\) at an angle \(\theta\) is given by:
\[ T = \frac{2u \sin\theta}{g} \]

Given that the projectile is at the same height at \(t_1\) and \(t_2\), we use the property:
\[ t_1 + t_2 = T \]


Step 3: Detailed Explanation:

From the problem, we have the following data:

Time 1, \(t_1 = 3 s\).

Time 2, \(t_2 = 5 s\).

Angle of projection, \(\theta = 30^\circ\).

Acceleration due to gravity, \(g = 10 m/s^2\).


First, calculate the total time of flight \(T\):
\[ T = t_1 + t_2 = 3 s + 5 s = 8 s \]


Next, substitute the values into the time of flight formula to solve for the initial speed \(u\):
\[ 8 = \frac{2 \times u \times \sin(30^\circ)}{10} \]

Since \(\sin(30^\circ) = 0.5\):
\[ 8 = \frac{2 \times u \times 0.5}{10} \]
\[ 8 = \frac{u}{10} \]
\[ u = 8 \times 10 = 80 m/s \]


Step 4: Final Answer:

The speed of projection of the projectile is 80 m s\(^{-1}\).
Quick Tip: Whenever a question mentions two different times for the same height in projectile motion, always use the shortcut \(t_1 + t_2 = T\). This avoids the need to set up and solve quadratic equations for vertical displacement.