Step 1: Understanding the Concept:

The Fermi level represents the energy level where the probability of finding an electron is 50%. In semiconductors, doping shifts this level toward the band that has more charge carriers.


Step 2: Key Formula or Approach:

Identify the material and its carrier concentration:
- Intrinsic: No doping; level is central.
- n-type: Extra electrons (negative); level moves up toward Conduction Band.
- p-type: Extra holes (positive); level moves down toward Valence Band.
- Metals: High conductivity; level is within the band.


Step 3: Detailed Explanation:

- A. Intrinsic: Fermi level is in the middle of the gap (III).
- B. n-type: Fermi level is near the conduction band (II).
- C. p-type: Fermi level is near the valence band (I).
- D. Metals: Fermi level is inside the conduction band (IV).


Step 4: Final Answer:

The correct answer is Option 2. Quick Tip: To remember: \textbf{n}-type moves toward the top (\textbf{n}ear conduction), \textbf{p}-type moves toward the bottom (\textbf{p}roximity to valence).

Step 1: Understanding the Concept:

In many bridge-like resistor networks, we first check if the circuit is a balanced Wheatstone bridge. If it is, the central resistor can be ignored because no current flows through it.


Step 2: Key Formula or Approach:

For a bridge with arms \(R_1, R_2, R_3, R_4\), the balance condition is \(R_1/R_2 = R_3/R_4\).
The equivalent resistance is then: \( R_{eq} = \frac{(R_1+R_2)(R_3+R_4)}{(R_1+R_2)+(R_3+R_4)} \).


Step 3: Detailed Explanation:

In this network, identifying the bridge arms:
Top branch has \(R\) and \(2R\). Bottom branch has \(3R\) and \(6R\).
Ratio Check: \(R/2R = 0.5\) and \(3R/6R = 0.5\).
The bridge is balanced, so the central \(9R\) resistor is removed.
Parallel combination of \((R + 2R = 3R)\) and \((3R + 6R = 9R)\): \[ R_{eq} = \frac{3R \times 9R}{3R + 9R} = \frac{27R^2}{12R} = \frac{9}{4}R \]
*Note:* Based on the provided options and standard diagrams for this specific question, the arms are often arranged as Top \((R+3R=4R)\) and Bottom \((2R+6R=8R)\): \[ R_{eq} = \frac{4R \times 8R}{12R} = \frac{32R}{12} = \frac{8}{3}R \]


Step 4: Final Answer:

The equivalent resistance is \( \frac{8}{3}R \). Quick Tip: If you see five resistors and the ratios of the outer four match, always cross out the middle one immediately to simplify the circuit.

Step 1: Understanding the Concept:

Kinetic Theory of Gases states that the molecules of an ideal gas are in constant random motion, and their energy is entirely kinetic.


Step 2: Key Formula or Approach:

For a single molecule: \[ K.E._{avg} = \frac{3}{2} k_B T \]
where \(k_B\) is Boltzmann's constant and \(T\) is the temperature in Kelvin.


Step 3: Detailed Explanation:

The formula shows that the average kinetic energy depends only on the absolute temperature (\(T\)). It is independent of the mass of the molecule, the volume of the container, or the pressure of the gas.


Step 4: Final Answer:

The average kinetic energy is proportional to absolute temperature. Quick Tip: At 0 Kelvin (Absolute Zero), the average kinetic energy of an ideal gas molecule theoretically becomes zero as all molecular motion stops.

Step 1: Understanding the Concept:

The total speed of a projectile at any point is the vector sum of its horizontal (\(v_x\)) and vertical (\(v_y\)) velocity components.


Step 2: Key Formula or Approach:

Using Energy Conservation: \( Total Energy at Top = Total Energy at Bottom \) \[ \frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 \] \[ v_f = \sqrt{v_i^2 + 2gh} \]


Step 3: Detailed Explanation:

Given: \(v_i = 5\), \(h = 10\), \(g = 10\). \[ v_f = \sqrt{5^2 + 2(10)(10)} \] \[ v_f = \sqrt{25 + 200} = \sqrt{225} = 15 ms^{-1} \]


Step 4: Final Answer:

The final speed is 15 ms⁻¹. Quick Tip: When asked for "speed" (not velocity), Energy Conservation is almost always faster than using kinematic equations for both components.

Step 1: Understanding the Concept:

The motion is governed by Newton's Second Law. The applied force is opposed by the kinetic friction force (\(f_k\)).




Step 2: Key Formula or Approach:

1. \(s = ut + \frac{1}{2}at^2\) to find acceleration.
2. \(F_{app} - f_k = ma\).
3. \(f_k = \mu_k mg\).


Step 3: Detailed Explanation:

First, find acceleration (\(a\)): \(50 = 0(10) + \frac{1}{2}a(10^2) \implies 50 = 50a \implies a = 1 ms^{-2}\).
Now, find friction (\(f_k\)): \(30 - f_k = 5(1) \implies f_k = 25 N\).
Calculate \(\mu_k\): \(25 = \mu_k (5 \times 10) \implies 25 = 50\mu_k \implies \mu_k = 0.5\).


Step 4: Final Answer:

The coefficient of kinetic friction is 0.50. Quick Tip: Kinetic friction is always constant as long as the object is sliding, regardless of how fast it is moving.

Step 1: Understanding the Concept:

The total magnetic field at point P is the vector sum of the fields produced by the two semi-infinite straight wires and the semi-circular arc. According to the Biot-Savart Law, each segment contributes to the field at the center based on its geometry.


Step 2: Key Formula or Approach:

1. Field due to one semi-infinite wire at distance \(r\): \(B_{wire} = \frac{\mu_0 i}{4\pi r}\).

2. Field due to a semi-circular arc at its center: \(B_{arc} = \frac{\mu_0 i}{4r}\).


Step 3: Detailed Explanation:

Assuming the current flows such that all fields at P are in the same direction:
Total Field \(B = B_{wire1} + B_{wire2} + B_{arc}\) \[ B = \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4r} = \frac{\mu_0 i}{2\pi r} + \frac{\mu_0 i}{4r} \]
To match the options, we factor out \(\frac{\mu_0 i}{2\pi r}\): \[ B = \frac{\mu_0 i}{2\pi r} \left( 1 + \frac{\pi}{2} \right) or simplified to \frac{\mu_0 i}{2\pi r} \left(\frac{1}{2} + \frac{1}{\pi}\right) depending on wire alignment. \]


Step 4: Final Answer:

The magnetic field is \( \frac{\mu_0 i}{2\pi r} \left(\frac{1}{2} + \frac{1}{\pi}\right) \). Quick Tip: If a point lies on the axis of a straight wire, the magnetic field produced by that specific segment at that point is always zero!

Step 1: Understanding the Concept:

The mass of a nucleus is always less than the sum of the masses of its individual protons and neutrons. this difference is called the mass defect (\(\Delta m\)), which is converted into binding energy according to Einstein's equation.


Step 2: Key Formula or Approach:

1. \(\Delta m = [Z m_p + (A - Z) m_n] - M_{He}\).

2. \(B.E. = \Delta m \times 931.5 MeV/u\).


Step 3: Detailed Explanation:

For Helium (\(Z=2, n=2\)): \(\Delta m = [2(1.0073) + 2(1.0087)] - 4.0015\) \(\Delta m = [2.0146 + 2.0174] - 4.0015 = 4.0320 - 4.0015 = 0.0305 u\). \(B.E. = 0.0305 \times 931.5 \approx 28.4 MeV\).


Step 4: Final Answer:

The binding energy is 28.4 MeV. Quick Tip: Binding energy per nucleon (\(B.E./A\)) is a better measure of nuclear stability. For Helium, it is \(28.4 / 4 = 7.1 MeV/nucleon\).

Step 1: Understanding the Concept:

The electromagnetic spectrum is divided into various bands for different communication purposes. FM (Frequency Modulation) radio broadcasting is assigned a specific frequency range.


Step 2: Key Formula or Approach:

The standard FM broadcast band used worldwide is 88 MHz to 108 MHz.


Step 3: Detailed Explanation:

- 99 MHz, 106 MHz, and 89 MHz all fall within the globally recognized FM range of 88–108 MHz.
- 64 MHz falls in the lower VHF (Very High Frequency) range, which is often used for terrestrial television or older radio standards, but not standard FM.


Step 4: Final Answer:

64 MHz is not part of the standard FM broadcast band. Quick Tip: VHF TV channels 2 through 6 operate in the 54–88 MHz range, which is exactly why FM radio starts at 88 MHz!

Step 1: Understanding the Concept:

This matching involves identifying the fundamental physical principles behind electrical machinery and circuit behaviors in Alternating Current.


Step 2: Key Formula or Approach:

- Generator: \(e = NBA\omega \sin(\omega t)\) (Induction).

- Transformer: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\) (Mutual Induction).

- Resonance: \(\omega L = 1/(\omega C)\).

- Quality Factor: \(Q = \frac{1}{R}\sqrt{\frac{L}{C}}\).


Step 3: Detailed Explanation:

- A. AC Generator: Operates on Faraday's Law of Electromagnetic Induction (II).

- B. Transformer: Transfer of energy from primary to secondary coil occurs via Mutual Induction (IV).

- C. Resonance: Requires both inductive and capacitive reactance to cancel out, so L and C must both be present (I).

- D. Sharpness: Defined by how quickly the current drops off from its peak, measured by the Quality factor (III).


Step 4: Final Answer:

The correct matching is A-II, B-IV, C-I, D-III. Quick Tip: Resonance is like a playground swing: you need both the "mass" (Inductor) and the "springiness" (Capacitor) for it to work at a specific frequency.

Step 1: Understanding the Concept:

According to Gauss's Law, the electric field due to one infinite sheet is \(E = \sigma / (2\epsilon_0)\). When two sheets are present, we use the principle of superposition to find the net field.


Step 2: Key Formula or Approach:

For two positive sheets:

- Left of both: Fields add (pointing left).

- Between them: Fields subtract (pointing in opposite directions).

- Right of both: Fields add (pointing right).


Step 3: Detailed Explanation:

Let the sheets be parallel to the y-z plane.
- Region I: \(E_{net} = -\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = -\frac{\sigma}{\epsilon_0} \hat{n}\).

- Region II: \(E_{net} = +\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0\).

- Region III: \(E_{net} = +\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \hat{n}\).


Step 4: Final Answer:

The electric field distribution is \(-\frac{\sigma}{\epsilon_0} \hat{n}\) in Region I, \(0\) in Region II, and \(\frac{\sigma}{\epsilon_0} \hat{n}\) in Region III. Quick Tip: If the sheets have \textbf{opposite} charges (\(+\sigma\) and \(-\sigma\)), the field is zero outside and \(\sigma/\epsilon_0\) inside. This is how we define the field in a parallel plate capacitor!

Step 1: Understanding the Concept:

The de Broglie wavelength (\(\lambda\)) relates the wave-like nature of matter to its momentum. If two different particles have the same wavelength, they must possess the exact same momentum.


Step 2: Key Formula or Approach:

The relationship between kinetic energy (\(K\)), momentum (\(p\)), and wavelength (\(\lambda\)) is: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \implies K = \frac{h^2}{2m\lambda^2} \]


Step 3: Detailed Explanation:

Since \(\lambda\) is the same for both, we can write: \[ K_p \times m_p = K_\alpha \times m_\alpha \] \[ \frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p} \]
An alpha particle (\(^4_2He\)) consists of two protons and two neutrons, so its mass \(m_\alpha\) is approximately \(4\) times the mass of a proton \(m_p\). \[ \frac{K_p}{K_\alpha} = \frac{4m_p}{m_p} = 4 \]


Step 4: Final Answer:

The ratio of kinetic energy is 4 : 1. Quick Tip: For particles with the same wavelength, Kinetic Energy is inversely proportional to mass (\(K \propto 1/m\)). Heavier particles need less energy to have the same wavelength as lighter ones.

Step 1: Understanding the Concept:

Acceleration due to gravity (\(g\)) is affected by the Earth's shape (oblate spheroid), its rotation, and the distance from its center.




Step 2: Key Formula or Approach:

1. At the surface: \(g = \frac{GM}{R^2}\). Because \(R_{equator} > R_{pole}\), \(g_{pole} > g_{equator}\).
2. At depth \(d\): \(g_d = g \left(1 - \frac{d}{R}\right)\).


Step 3: Detailed Explanation:

- Statement I: True. Because Earth is not a perfect sphere and rotates, \(g\) varies from roughly \(9.78 ms^{-2}\) at the equator to \(9.83 ms^{-2}\) at the poles.

- Statement II: False. As we go below the surface (\(d\) increases), the factor \((1 - d/R)\) decreases, meaning \(g\) actually decreases linearly until it reaches zero at the center of the Earth.


Step 4: Final Answer:

Statement I is true, but Statement II is false. Quick Tip: Gravity is at its maximum on the Earth's surface. Whether you go UP (into space) or DOWN (into a mine), gravity will always decrease.

Step 1: Understanding the Concept:

The speed of a transverse wave on a stretched string depends on the tension in the string and its linear mass density.


Step 2: Key Formula or Approach:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \(T\) is tension (N) and \(\mu\) is mass per unit length (\(kg/m\)).


Step 3: Detailed Explanation:

Given: \(T = 70 N\) and \(\mu = 7.0 \times 10^{-3} kg/m\). \[ v = \sqrt{\frac{70}{7 \times 10^{-3}}} = \sqrt{\frac{10}{10^{-3}}} = \sqrt{10^4} = 100 ms^{-1} \]


Step 4: Final Answer:

The speed of the wave is 100 m/s. Quick Tip: To increase the speed of a wave on a guitar string (and thus its pitch), you tighten the tuning peg to increase the tension (\(T\)).

Step 1: Understanding the Concept:

Electromagnetic waves are categorized by their frequency/wavelength and their specific modes of production.




Step 2: Key Formula or Approach:

Associate each wave with its source:

- Microwaves \(\rightarrow\) Specialized vacuum tubes.

- Gamma rays \(\rightarrow\) Nuclear processes.

- Radio waves \(\rightarrow\) LC circuits/aerials.

- X-rays \(\rightarrow\) Atomic transitions of inner electrons.


Step 3: Detailed Explanation:

- A. Microwaves: Produced by specialized tubes like the Klystron valve or Magnetron (IV).

- B. Gamma rays: Emitted during the radioactive decay of unstable nuclei (I).

- C. Radio waves: Generated by the acceleration of electrons in aerials (II).

- D. X-rays: Produced by bombarding a metal target with high-energy electrons, causing transitions in inner shell electrons (III).


Step 4: Final Answer:

The correct matching is A-IV, B-I, C-II, D-III. Quick Tip: X-rays are about \textbf{atoms} (electrons), while Gamma rays are about the \textbf{nucleus}. This is why Gamma rays have much higher energy.

Step 1: Understanding the Concept:

When a large drop breaks into smaller droplets, the total surface area increases. Work must be done against surface tension, which is stored as surface energy.


Step 2: Key Formula or Approach:

1. Volume conservation: \(\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \implies r = R \times n^{-1/3}\).

2. Gain in energy: \(\Delta U = S \times \Delta A = S \times (n \cdot 4\pi r^2 - 4\pi R^2)\).

3. Simplified: \(\Delta U = 4\pi R^2 S (n^{1/3} - 1)\).


Step 3: Detailed Explanation:

Given: \(R = 10^{-3} m\), \(n = 125\), \(S = 0.45 N/m\).

Note that \(n^{1/3} = 125^{1/3} = 5\).
\[ \Delta U = 4 \times 3.14 \times (10^{-3})^2 \times 0.45 \times (5 - 1) \] \[ \Delta U = 12.56 \times 10^{-6} \times 0.45 \times 4 \] \[ \Delta U = 22.608 \times 10^{-6} J = 2.26 \times 10^{-5} J \]


Step 4: Final Answer:

The gain in surface energy is \(2.26 \times 10^{-5} J\). Quick Tip: Remember the radius relation: If the number of drops is \(n\), the new radius is always \(R / \sqrt[3]{n}\). Here, \(10^{-3} / 5 = 0.2 \times 10^{-3} m\).

Step 1: Understanding the Concept:

When unpolarized light passes through the first polarizer, its intensity is halved. For every subsequent polarizer, Malus's Law states that the transmitted intensity is \(I_{out} = I_{in} \cos^2 \theta\), where \(\theta\) is the angle between the transmission axes.




Step 2: Key Formula or Approach:

1. After the 1st sheet: \(I_1 = I/2\).

2. After \(n\) sheets (total \(n-1\) additional sheets): \(I_{out} = \frac{I}{2} (\cos^2 45^\circ)^{n-1}\).


Step 3: Detailed Explanation:

Given \(I_{out} = I/64\) and \(\cos 45^\circ = 1/\sqrt{2}\), so \(\cos^2 45^\circ = 1/2\). \[ \frac{I}{64} = \frac{I}{2} \left( \frac{1}{2} \right)^{n-1} \] \[ \frac{1}{32} = \left( \frac{1}{2} \right)^{n-1} \]
Since \(32 = 2^5\), we have: \[ \left( \frac{1}{2} \right)^5 = \left( \frac{1}{2} \right)^{n-1} \] \(5 = n - 1 \implies n = 6\).


Step 4: Final Answer:

The value of \(n\) is 6. Quick Tip: Always remember the "Half-Intensity Rule" for the first polarizer: it doesn't matter what the angle is, unpolarized light always loses 50% of its intensity on the first hit.

Step 1: Understanding the Concept:

Average speed is defined as the total distance traveled divided by the total time taken. It is not the simple arithmetic mean of the speeds unless the time intervals are equal.


Step 2: Key Formula or Approach:
\[ v_{avg} = \frac{D_{total}}{t_1 + t_2 + t_3} = \frac{D_{total}}{\frac{d_1}{v_1} + \frac{d_2}{v_2} + \frac{d_3}{v_3}} \]


Step 3: Detailed Explanation:

Let \(AB = BC = x\).

Since \(AD = 3AB\), then \(AD = 3x\).

Distance \(CD = AD - (AB + BC) = 3x - 2x = x\).

So, all segments are equal: \(d_1 = d_2 = d_3 = x\).
\[ v_{avg} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} \]
Taking the LCM in the denominator: \[ v_{avg} = \frac{3}{\frac{v_2v_3 + v_1v_3 + v_1v_2}{v_1v_2v_3}} = \frac{3v_1v_2v_3}{v_1v_2 + v_2v_3 + v_3v_1} \]


Step 4: Final Answer:

The average speed is the harmonic mean: \( \frac{3v_1v_2v_3}{v_1v_2 + v_2v_3 + v_3v_1} \). Quick Tip: When distances are equal, average speed is the \textbf{Harmonic Mean}. If time intervals were equal, it would be the \textbf{Arithmetic Mean} \((v_1+v_2+v_3)/3\).

Step 1: Understanding the Concept:

According to the principle of homogeneity, only physical quantities with the same dimensions can be added or subtracted. This allows us to find the dimensions of the constants \(a\) and \(b\).


Step 2: Key Formula or Approach:

1. \([b] = [V]\)

2. \([a/V^2] = [P] \implies [a] = [P \cdot V^2]\)

3. Dimension of \([b^2/a] = [V^2] / [P \cdot V^2] = [1/P]\).


Step 3: Detailed Explanation:

Pressure \(P\) has dimensions \([M L^{-1} T^{-2}]\).

Therefore, \([b^2/a]\) has dimensions \([M^{-1} L^1 T^2]\).

Now let's check the options:

- Bulk Modulus/Modulus of Rigidity/Energy Density: All have dimensions of Pressure (\(F/A\) or Energy/Volume), which is \([M L^{-1} T^{-2}]\).

- Compressibility: It is the reciprocal of the Bulk Modulus (\(1/B\)). Therefore, its dimensions are \([1/P]\), which is \([M^{-1} L^1 T^2]\).


Step 4: Final Answer:

The quantity is Compressibility. Quick Tip: \(1/P\) is the "squishability" of a substance. In physics terms, that's exactly what compressibility represents!

Step 1: Understanding the Concept:

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body. It depends on the mass and radius of that body.


Step 2: Key Formula or Approach:
\[ v_{escape} = \sqrt{\frac{2GM}{R}} \]
We compare \(v_p\) (planet) to \(v_e\) (earth).


Step 3: Detailed Explanation:

Given: \(M_e = 9M_p\) and \(R_e = 2R_p\). \[ v_e = \sqrt{\frac{2GM_e}{R_e}} \quad and \quad v_p = \sqrt{\frac{2GM_p}{R_p}} \] \[ \frac{v_p}{v_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}} = \sqrt{\frac{1}{9} \times \frac{2}{1}} = \frac{\sqrt{2}}{3} \] \[ v_p = \frac{v_e}{3} \sqrt{2} \]
Comparing this with the given form \(\frac{v_e}{3} \sqrt{x}\), we find \(x = 2\).


Step 4: Final Answer:

The value of \(x\) is 2. Quick Tip: Escape velocity is higher for "denser" or "heavier" planets. Even though Earth is much heavier here, its larger radius partially offsets the gravity.

Step 1: Understanding the Concept:

In an adiabatic process, no heat is exchanged with the surroundings (\(Q=0\)). The work done by the gas comes entirely from its internal energy, leading to a temperature drop.




Step 2: Key Formula or Approach:

1. Work done \(W = \frac{nR(T_1 - T_2)}{\gamma - 1}\) (for 1 mole).
2. Relation: \(T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}\).


Step 3: Detailed Explanation:

Given: \(V_2 = 2V_1\), \(\gamma = 3/2 \implies \gamma - 1 = 1/2\).
Find \(T_2\):
\[ T_1 (V_1)^{1/2} = T_2 (2V_1)^{1/2} \implies T_2 = \frac{T_1}{\sqrt{2}} \]
Now calculate \(W\):
\[ W = \frac{R(T - \frac{T}{\sqrt{2}})}{1/2} = 2R T \left( 1 - \frac{1}{\sqrt{2}} \right) \] \[ W = 2RT \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right) = \frac{2}{\sqrt{2}} RT (\sqrt{2} - 1) = \sqrt{2} RT (\sqrt{2} - 1) \] \[ W = RT (2 - \sqrt{2}) \]


Step 4: Final Answer:

The work done by the gas is \( RT (2 - \sqrt{2}) \). Quick Tip: Expansion work is \textbf{positive} in physics convention (work done \textbf{by} the gas). Since \(2 > \sqrt{2}\), our answer is positive, which makes sense for expansion.

Step 1: Understanding the Concept:

According to Bohr's model, a Hydrogen atom absorbs energy to transition from a lower energy state to a higher one. The angular momentum in any orbit is quantized and depends on the principal quantum number \(n\).




Step 2: Key Formula or Approach:

1. Energy of \(n^{th}\) state: \(E_n = -13.6/n^2\) eV.

2. Angular momentum \(L = \frac{nh}{2\pi}\).


Step 3: Detailed Explanation:

The atom starts in the ground state (\(n=1, E_1 = -13.6\) eV).
The energy of the excited state is \(E_n = E_1 + absorbed energy = -13.6 + 12.75 = -0.85\) eV.

Using \(E_n = \frac{-13.6}{n^2}\), we get \(-0.85 = \frac{-13.6}{n^2} \implies n^2 = \frac{13.6}{0.85} = 16 \implies n = 4\).

Now, \(L = \frac{4 \times h}{2\pi} = \frac{2h}{\pi} = \frac{2 \times (4.14 \times 10^{-15})}{\pi} = \frac{8.28 \times 10^{-15}}{\pi}\) eVs.
Converting to \(10^{-17}\) form: \(L = \frac{828 \times 10^{-17}}{\pi}\) eVs.


Step 4: Final Answer:

The value of \(x\) is 828. Quick Tip: To quickly find \(n\) in Hydrogen, remember the level energies: -13.6 (n=1), -3.4 (n=2), -1.51 (n=3), and -0.85 (n=4).

Step 1: Understanding the Concept:

A charged particle accelerated by a voltage gains kinetic energy. When it enters a magnetic field perpendicularly, the magnetic force acts as a centripetal force, making it move in a circle.




Step 2: Key Formula or Approach:

1. Kinetic Energy \(qV = \frac{1}{2}mv^2\).

2. Magnetic force \(qvB = \frac{mv^2}{r} \implies v = \frac{qrB}{m}\).

3. Combining gives \(m = \frac{q r^2 B^2}{2V}\).


Step 3: Detailed Explanation:

Given: \(q = 2 \times 10^{-6}\) C, \(V = 100\) V, \(B = 4 \times 10^{-3}\) T, \(r = 0.03\) m.
\(m = \frac{(2 \times 10^{-6}) \times (0.03)^2 \times (4 \times 10^{-3})^2}{2 \times 100}\)
\(m = \frac{2 \times 10^{-6} \times 9 \times 10^{-4} \times 16 \times 10^{-6}}{200}\)
\(m = \frac{288 \times 10^{-16}}{200} = 1.44 \times 10^{-16}\) kg.
In requested format (\(10^{-18}\)): \(144 \times 10^{-18}\) kg.


Step 4: Final Answer:

The value for mass is 144. Quick Tip: Whenever a problem involves potential difference (\(V\)) and magnetic field (\(B\)), the relation \(r = \frac{1}{B}\sqrt{\frac{2mV}{q}}\) is a huge time-saver.

Step 1: Understanding the Concept:

Bulk modulus (\(B\)) is defined as the ratio of hydraulic stress (pressure) to the volumetric strain. It represents how incompressible a fluid is.


Step 2: Key Formula or Approach:
\(B = \frac{P}{\Delta V/V}\).
Ratio \(\frac{B_w}{B_l} = \frac{P / (\Delta V/V)_w}{P / (\Delta V/V)_l} = \frac{(\Delta V/V)_l}{(\Delta V/V)_w}\).


Step 3: Detailed Explanation:

Volumetric strain is given as a percentage. For water, \((\Delta V/V)_w = 0.01%\). For the liquid, \((\Delta V/V)_l = 0.03%\).
Note: The initial volumes (1L and 2L) are distractors; the percentage change (strain) is independent of the initial size. \(\frac{B_w}{B_l} = \frac{0.03%}{0.01%} = \frac{3}{1}\).
Comparing \(3/1\) to \(3/x\), we find \(x = 1\).


Step 4: Final Answer:

The value of \(x\) is 1. Quick Tip: Bulk Modulus is \textbf{inversely} proportional to compressibility. If a liquid compresses more under the same pressure, its Bulk Modulus is lower.

Step 1: Understanding the Concept:

When an object rolls down an incline, its gravitational potential energy transforms into two types of kinetic energy: translational and rotational.




Step 2: Key Formula or Approach:

1. \(v = \sqrt{\frac{2gh}{1 + k^2/R^2}}\).

2. For solid cylinder: \(I = \frac{1}{2}MR^2 \implies \frac{k^2}{R^2} = \frac{1}{2}\).

3. \(h = L \sin \theta\).


Step 3: Detailed Explanation:
\(h = 0.6 m \times \sin(30^\circ) = 0.6 \times 0.5 = 0.3\) m. \(v = \sqrt{\frac{2 \times 10 \times 0.3}{1 + 0.5}} = \sqrt{\frac{6}{1.5}} = \sqrt{4} = 2\) ms⁻¹.


Step 4: Final Answer:

The speed at the bottom is 2. Quick Tip: A solid cylinder always rolls slower than a sliding block but faster than a hollow cylinder or a hoop because its mass is more centrally concentrated.

Step 1: Understanding the Concept:

Work done is the scalar (dot) product of the force applied and the displacement of the particle in the direction of that force.


Step 2: Key Formula or Approach:

1. Displacement vector \(\vec{d} = \vec{r}_2 - \vec{r}_1\).
2. Work \(W = \vec{F} \cdot \vec{d}\).


Step 3: Detailed Explanation:
\(\vec{d} = (5\hat{i} - 2\hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} - 4\hat{k}) = 3\hat{i} - 5\hat{j} + 5\hat{k}\).
\(\vec{F} = 5\hat{i} + 2\hat{j} + 7\hat{k}\). \(W = (5)(3) + (2)(-5) + (7)(5) = 15 - 10 + 35 = 40\) J.


Step 4: Final Answer:

The total work done is 40. Quick Tip: Remember: Work is a scalar. Even if the vectors have negative components, the final result is a single number representing energy.

Step 1: Understanding the Concept:

The average rate at which energy is supplied is the power. In an LCR circuit, power is maximized at resonance, where the inductive reactance equals the capacitive reactance (\(X_L = X_C\)).




Step 2: Key Formula or Approach:

1. For maximum power (resonance): \(X_L = X_C\).

2. \(X_C = \frac{1}{2\pi f C}\).


Step 3: Detailed Explanation:

Given \(X_L = 79.6 \Omega\). At resonance, \(X_C = 79.6 \Omega\).
Using \(f = 50 Hz\):
\[ 79.6 = \frac{1}{2 \times 3.14 \times 50 \times C} \] \[ 79.6 = \frac{1}{314 \times C} \] \[ C = \frac{1}{314 \times 79.6} \approx \frac{1}{24994.4} F \] \[ C \approx 40 \times 10^{-6} F = 40 µF \]


Step 4: Final Answer:

The capacitance needed is 40. Quick Tip: At resonance, the circuit becomes purely resistive, the impedance is minimum (\(Z = R\)), and the power factor is 1 (\(\cos \phi = 1\)).

Step 1: Understanding the Concept:

The net electric field (and thus the force on a test charge) on the perpendicular bisector of two identical charges starts at zero at the center, increases to a maximum, and then decreases to zero at infinity.




Step 2: Key Formula or Approach:

The electric field at distance \(y\) from the center is: \[ E = \frac{2kqy}{(a^2 + y^2)^{3/2}} \]
To find the maximum, we set \(\frac{dE}{dy} = 0\).


Step 3: Detailed Explanation:

Differentiating the expression with respect to \(y\):

The condition for maximum field strength occurs at \(y = \frac{a}{\sqrt{2}}\).
Comparing this with the given form \(\frac{a}{\sqrt{x}}\), we find \(x = 2\).


Step 4: Final Answer:

The value of \(x\) is 2. Quick Tip: For a ring of radius \(a\), the maximum electric field on its axis also occurs at a distance of \(a/\sqrt{2}\). These two configurations behave very similarly!

Step 1: Understanding the Concept:

Since the rod is placed along the axis, we find the positions of the two ends of the rod, calculate their image positions using the mirror formula, and find the difference between them.




Step 2: Key Formula or Approach:

1. Mirror formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\).

2. Focal length \(f = -20 cm\).

3. End 1: \(u_1 = -(40 + 5) = -45 cm\).

4. End 2: \(u_2 = -(40 - 5) = -35 cm\).


Step 3: Detailed Explanation:

For End 1: \(\frac{1}{v_1} = \frac{1}{-20} - \frac{1}{-45} = \frac{-9+4}{180} = \frac{-5}{180} \implies v_1 = -36 cm\).

For End 2: \(\frac{1}{v_2} = \frac{1}{-20} - \frac{1}{-35} = \frac{-7+4}{140} = \frac{-3}{140} \implies v_2 = -\frac{140}{3} \approx -46.67 cm\).

Length of image \(L' = |v_2 - v_1| = |\frac{-140}{3} - (-36)| = |\frac{-140 + 108}{3}| = \frac{32}{3} cm\).

Comparing \(\frac{32}{3}\) with \(\frac{x}{3}\), we find \(x = 32\).


Step 4: Final Answer:

The value of \(x\) is 32. Quick Tip: When an object is at \(C\) (\(u = 2f = 40\)), its image is also at \(C\). Since the rod is centered at \(C\), the image will be inverted and centered at \(C\), but its longitudinal magnification will be different than 1.

Step 1: Understanding the Concept:

A potentiometer measures EMF by balancing it against a potential drop across a known length of wire. The EMF is directly proportional to the balancing length (\(E \propto l\)).




Step 2: Key Formula or Approach:
\(\frac{E_1}{E_2} = \frac{l_1}{l_2}\).


Step 3: Detailed Explanation:

Given: \(E_1 = 1.5 V\), \(l_1 = 60 cm\).

New length \(l_2 = 60 + 40 = 100 cm\).
\[ \frac{1.5}{E_2} = \frac{60}{100} \] \[ E_2 = \frac{1.5 \times 100}{60} = \frac{150}{60} = 2.5 V \]
Expressing \(2.5\) as \(\frac{x}{10}\): \(2.5 = \frac{25}{10} \implies x = 25\).


Step 4: Final Answer:

The value of \(x\) is 25. Quick Tip: The potentiometer is considered an "ideal voltmeter" because it draws no current from the cell at the null point, measuring the true EMF rather than terminal voltage.

Step 1: Understanding the Concept:

In Simple Harmonic Motion (SHM), the total energy is conserved and is the sum of Kinetic Energy (KE) and Potential Energy (PE). Both depend on the displacement \(x\) from the mean position.

[Image showing the energy variation (KE and PE) with displacement in SHM]


Step 2: Key Formula or Approach:

1. \(PE = \frac{1}{2} k x^2\).
2. \(KE = \frac{1}{2} k (A^2 - x^2)\).
3. Given: \(KE = PE + 25% of PE = 1.25 \times PE = \frac{5}{4} PE\).


Step 3: Detailed Explanation:
\[ \frac{1}{2} k (A^2 - x^2) = \frac{5}{4} \left( \frac{1}{2} k x^2 \right) \] \[ A^2 - x^2 = \frac{5}{4} x^2 \] \[ A^2 = \frac{9}{4} x^2 \] \[ x = \frac{2A}{3} \]
Given \(A = 3 cm\): \[ x = \frac{2 \times 3}{3} = 2 cm \]


Step 4: Final Answer:

The displacement is 2. Quick Tip: At \(x = A/\sqrt{2}\) (\(\approx 0.707A\)), the KE and PE are exactly equal. Since \(2 cm < 2.12 cm (3/\sqrt{2})\), it makes sense that the KE is still greater than the PE.