Step 1: Understanding the Concept:

To solve this matching question, we need to determine the dimensional formula for each physical quantity listed in List-I using standard physics equations relating them to basic mechanical and electrical dimensions.


Step 2: Key Formula or Approach:

Energy \(E = [M L^2 T^{-2}]\)

Planck's constant \(h\): \(E = h\nu\)

Stopping potential \(V\): \(E = qV\) where \(q = I \cdot t = [A T]\)

Work function \(\Phi\): A form of Energy.

Threshold frequency \(\nu_0\): Frequency.


Step 3: Detailed Explanation:

1. Planck's Constant (A):

From \(E = h\nu \implies h = \frac{E}{\nu}\).

Dimension of \(E = [M L^2 T^{-2}]\). Dimension of \(\nu = [T^{-1}]\).
\([h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]\).

So, A matches with III.


2. Stopping Potential (B):

From Work done/Energy \(E = qV \implies V = \frac{E}{q}\).

Dimension of charge \(q = [A T]\).
\([V] = \frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]\).

So, B matches with IV.


3. Work Function (C):

Work function is the minimum energy required to remove an electron. It has the exact same dimensions as Energy.
\([\Phi] = [M L^2 T^{-2}]\).

So, C matches with I.


4. Threshold frequency (D):

Frequency is the inverse of time period.
\([\nu_0] = [T^{-1}]\).

So, D matches with II.


The correct sequence is A-III, B-IV, C-I, D-II.


Step 4: Final Answer:

Option (A) is correct. Quick Tip: In dimension matching questions, always start with the easiest ones (like Frequency or Work Function) to eliminate multiple options instantly. Matching just C and D leaves only one logical answer.

Step 1: Understanding the Concept:

This problem operates on the principle of relative velocity. The stone is thrown from the reference frame of car B, so its absolute velocity depends on both throw speed and car B's speed. We then compute the stone's velocity relative to car A to match the impact speed.


Step 2: Key Formula or Approach:

Absolute velocity of stone: \(\vec{v}_s = \vec{v}_{throw} + \vec{v}_B\)

Relative velocity of stone with respect to A: \(\vec{v}_{sA} = \vec{v}_s - \vec{v}_A\)

Convert 5 m/s to km/h by multiplying by \(\frac{18}{5}\).


Step 3: Detailed Explanation:

Let the direction of motion of the cars be the positive x-direction.

Velocity of car A, \(v_A = 100\) km/h.

Velocity of car B, \(v_B = 80\) km/h.


The stone is thrown forward from B with speed \(v\) relative to B.

Velocity of the stone relative to the ground is \(v_s = v_B + v = 80 + v\).


The stone hits car A. The impact speed is the relative speed of the stone with respect to car A.

Velocity of stone relative to A is \(v_{sA} = v_s - v_A = (80 + v) - 100 = v - 20\).


The impact speed is given as 5 m/s. Convert this to km/h:
\(5 m/s = 5 \times \frac{18}{5} km/h = 18 km/h\).


The magnitude of the relative impact velocity is 18 km/h.
\(|v - 20| = 18\)

This gives two possibilities:
\(v - 20 = 18 \implies v = 38\) km/h.
\(v - 20 = -18 \implies v = 2\) km/h.


Since car A is ahead of car B and moving faster, a stone thrown with \(v = 2\) km/h relative to B would have a ground speed of 82 km/h. It would never catch up to car A (which is at 100 km/h). Thus, the stone must be thrown fast enough to exceed A's speed.

Therefore, \(v\) must be 38 km/h.


Step 4: Final Answer:

The value of \(v\) is 38. Quick Tip: Always double-check the physical validity of mathematical roots in kinematics. A thrown object must have a higher absolute ground velocity than the target moving away from it in order to actually catch it.

Step 1: Understanding the Concept:

By utilizing Newton's second law, we can determine the 2D acceleration vector of the mass. Since the force is constant and the body starts from rest, we can use the kinematic equations for uniform acceleration to find its position vector at \(t = 2\)s and compare it with the given coordinate expressions.


Step 2: Key Formula or Approach:

Newton's Second Law: \(\vec{a} = \frac{\vec{F}}{m}\)

Kinematics (from rest, \(\vec{u} = 0\)): \(\vec{s} = \frac{1}{2}\vec{a}t^2\)


Step 3: Detailed Explanation:

Mass \(m = 100 g = 0.1 kg\).

Force \(\vec{F} = 5\hat{i} + 10\hat{j} N\).


Calculate the acceleration vector:
\(\vec{a} = \frac{\vec{F}}{m} = \frac{5\hat{i} + 10\hat{j}}{0.1} = 50\hat{i} + 100\hat{j} m/s^2\).


The body starts from rest, so initial velocity \(\vec{u} = 0\).

The position vector after \(t = 2\) s is:
\(\vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2 = 0 + \frac{1}{2} (50\hat{i} + 100\hat{j}) (2)^2\)
\(\vec{s} = \frac{1}{2} (50\hat{i} + 100\hat{j}) \times 4 = 2(50\hat{i} + 100\hat{j})\)
\(\vec{s} = 100\hat{i} + 200\hat{j} m\).


We are given the position after 2 seconds as \((2x\hat{i} + 5y\hat{j})\) m.

Equating the components:

x-component: \(2x = 100 \implies x = 50\).

y-component: \(5y = 200 \implies y = 40\).


Find the ratio \(x : y\):
\(\frac{x}{y} = \frac{50}{40} = \frac{5}{4}\).


Step 4: Final Answer:

The ratio is 5 : 4. Quick Tip: Remember to convert mass to SI units (grams to kilograms) before calculating acceleration to ensure compatibility with force in Newtons.

Step 1: Understanding the Concept:

The velocity vector components of a projectile can be found by taking the time derivative of its position coordinates. The angle of the projectile's trajectory at any given time is the angle of its velocity vector relative to the horizontal plane.


Step 2: Key Formula or Approach:

Velocity components: \(v_x = \frac{dx}{dt}\) and \(v_y = \frac{dy}{dt}\).

Angle with the horizontal \(\theta\): \(\tan \theta = \frac{v_y}{v_x}\).


Step 3: Detailed Explanation:

Given the position coordinates:
\(x = 24t\)
\(y = 43.6t - 4.9t^2\)


Differentiate to find velocity components:
\(v_x = \frac{d}{dt}(24t) = 24 m/s\) (Constant horizontal velocity)
\(v_y = \frac{d}{dt}(43.6t - 4.9t^2) = 43.6 - 9.8t m/s\)


Evaluate the velocity components at the specific time \(t = 2\) s:
\(v_x = 24 m/s\)
\(v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24 m/s\)


The angle \(\theta\) made with the horizontal is given by:
\(\tan \theta = \frac{v_y}{v_x} = \frac{24}{24} = 1\)

Since \(\tan \theta = 1\) and both components are positive, the angle is in the first quadrant:
\(\theta = 45^\circ\).


Step 4: Final Answer:

The angle is \(45^\circ\). Quick Tip: In projectile equations of the form \(y = At - Bt^2\), the coefficient of \(t^2\) represents \(\frac{1}{2}g\). This can sometimes act as a sanity check to confirm the problem operates under standard Earth gravity (\(9.8\)).

Step 1: Understanding the Concept:

The acceleration due to gravity decreases as we move away from the surface of the Earth. The value of gravity \(g'\) at an altitude \(h\) follows the inverse-square law with respect to the distance from the center of the Earth.


Step 2: Key Formula or Approach:

Gravity at height \(h\):
\(g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2}\)

where \(R\) is the radius of the Earth.


Step 3: Detailed Explanation:

We are given that at height \(h\), the acceleration due to gravity is \(g' = \frac{g}{9}\).

Substitute this into the gravity altitude formula:
\(\frac{g}{9} = \frac{g}{\left(1 + \frac{h}{R}\right)^2}\)


Cancel out \(g\) from both sides:
\(\frac{1}{9} = \frac{1}{\left(1 + \frac{h}{R}\right)^2}\)

Take the positive square root of both sides (since altitude \(h\) is positive):
\(\frac{1}{3} = \frac{1}{1 + \frac{h}{R}}\)


Rearrange to solve for \(h\):
\(1 + \frac{h}{R} = 3\)
\(\frac{h}{R} = 2\)
\(h = 2R\).


Step 4: Final Answer:

The height is \(2R\). Quick Tip: For large heights (comparable to \(R\)), always use the exact formula \(g' = g / (1+h/R)^2\). Do not use the binomial approximation \(g' \approx g(1 - 2h/R)\), which is only valid for \(h \ll R\).

Step 1: Use elongation formula


For a wire,
\[ \Delta L=\frac{TL}{AY} \]

Therefore,
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{T_A}{T_B}\cdot \frac{L_A}{L_B}\cdot \frac{A_B}{A_A}\cdot \frac{Y_B}{Y_A} \]



Step 2: Find tensions


For string \(B\):
\[ T_B=2Mg \]

For string \(A\):
\[ T_A=(M+2M)g=3Mg \]

Thus,
\[ \frac{T_A}{T_B}=\frac{3}{2} \]



Step 3: Substitute ratios


Given
\[ \frac{L_A}{L_B}=2 \]
\[ \frac{Y_A}{Y_B}=0.5 \Rightarrow \frac{Y_B}{Y_A}=2 \]

Same cross-sectional area:
\[ \frac{A_B}{A_A}=1 \]

So,
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{3}{2}\times 2\times 1\times 2 \]
\[ =6 \]



Final Answer:
\[ \boxed{6} \] Quick Tip: Always isolate and clearly list all the proportional ratios (like \(T_A/T_B\), \(L_A/L_B\)) before multiplying them. This drastically reduces algebra errors in composite ratio problems.

Step 1: Use continuity equation

\[ A_1v_1=nA_2v_2 \]



Step 2: Convert units

\[ A_1=30\,cm^2=30\times 10^{-4}\,m^2 \]
\[ v_1=50\,cm/s=0.5\,m/s \]
\[ A_2=15\,mm^2=15\times 10^{-6}\,m^2 \]
\[ n=10 \]



Step 3: Find exit velocity

\[ v_2=\frac{A_1v_1}{nA_2} \]
\[ = \frac{(30\times 10^{-4})(0.5)} {10(15\times 10^{-6})} \]
\[ = \frac{15\times 10^{-4}} {15\times 10^{-5}} \]
\[ =10 \]
\[ v_2=10\,m/s \]



Final Answer:
\[ \boxed{10\,m/s} \] Quick Tip: To prevent massive conversion errors (like confusing \(mm^2\) and \(cm^2\)), convert everything systematically to \(10^x\) standard form in meters immediately at the start of fluid mechanics problems.

Step 1: Check assertion


Average kinetic energy is
\[ K=\frac{3}{2}kT \]

Since it depends only on temperature, same average kinetic energy means same temperature.

Hence Assertion A is true.



Step 2: Check reason


RMS speed is
\[ v_{rms}=\sqrt{\frac{3RT}{M}} \]

It depends on molar mass \(M\).

Since
\[ M(H_2)\ne M(O_2) \]

their RMS speeds are not same at same temperature.

Hence Reason R is false.



Final Answer:
\[ \boxed{(C) A is true but R is false} \] Quick Tip: Remember that Temperature represents the macroscopic average Kinetic Energy of a system, irrespective of the particle mass. Velocity, however, scales with mass to keep that Energy constant (\(E = \frac{1}{2}mv^2\)).

Step 1: First expansion

\[ \Delta L_1=L_1\alpha \Delta T \]

New length:
\[ L_2=L_1+\Delta L_1 \]
\[ =L_1(1+\alpha \Delta T) \]



Step 2: Second expansion


Since
\[ T_3+T_1=2T_2 \]

we get
\[ T_3-T_2=T_2-T_1=\Delta T \]

So second temperature rise is also \(\Delta T\).
\[ \Delta L_2=L_2\alpha \Delta T \]

Substitute \(L_2\):
\[ \Delta L_2=L_1(1+\alpha \Delta T)\alpha \Delta T \]

Since
\[ L_1\alpha \Delta T=\Delta L_1 \]
\[ \Delta L_2=\Delta L_1(1+\alpha \Delta T) \]



Final Answer:
\[ \boxed{\Delta L_1(1+\alpha \Delta T)} \] Quick Tip: Thermal expansion behaves analogously to compound interest. Each subsequent expansion applies to the "new principal" (the already expanded length), creating a small compounding factor \((1 + \alpha \Delta T)\).

Step 1: Use physical pendulum formula

\[ T=2\pi \sqrt{\frac{I}{Mgd}} \]



Step 2: Find moment of inertia


For disc about center:
\[ I_{CM}=\frac{1}{2}MR^2 \]

Using parallel axis theorem:
\[ I=I_{CM}+MR^2 \]
\[ =\frac{1}{2}MR^2+MR^2 \]
\[ =\frac{3}{2}MR^2 \]

Distance of center from pivot:
\[ d=R \]



Step 3: Substitute

\[ T=2\pi \sqrt{\frac{\frac{3}{2}MR^2}{MgR}} \]
\[ =2\pi \sqrt{\frac{3R}{2g}} \]



Final Answer:
\[ \boxed{2\pi \sqrt{\frac{3R}{2g}}} \] Quick Tip: The "equivalent length" \(L_{eq}\) of a simple pendulum representing any physical rigid body is simply \(I_{pivot} / (Md)\). Here, \(L_{eq} = 1.5R\).

Step 1: Understanding the Concept:

When a dipole undergoes small angular oscillations in a uniform electric field, the restoring torque relates directly to the strength of the net electric field. Because frequency \(\nu\) is proportional to the square root of the electric field magnitude, modifying the net field scales the frequency.


Step 2: Key Formula or Approach:

Restoring torque \(\tau = -pE \sin\theta \approx -pE\theta\) (for small angles).

Frequency \(f = \frac{1}{2\pi} \sqrt{\frac{pE}{I}}\), meaning \(f \propto \sqrt{E_{net}}\).

Percentage change \(= \frac{f_{final} - f_{initial}}{f_{initial}} \times 100\).


Step 3: Detailed Explanation:

The initial electric field is \(\vec{E}_1 = E_0 \hat{x}\).

Initial magnitude \(E_{initial} = \sqrt{E_0^2} = E_0\).

The initial frequency is \(f_1 \propto \sqrt{E_0}\).


After adding the second electric field \(\vec{E}_2 = 2E_0 \hat{y} + 2E_0 \hat{z}\), the new net electric field is the vector sum:
\(\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = E_0 \hat{x} + 2E_0 \hat{y} + 2E_0 \hat{z}\).


Calculate the magnitude of the new net electric field:
\(E_{final} = |\vec{E}_{net}| = \sqrt{E_0^2 + (2E_0)^2 + (2E_0)^2}\)
\(E_{final} = \sqrt{E_0^2 + 4E_0^2 + 4E_0^2} = \sqrt{9E_0^2} = 3E_0\).


The final frequency is \(f_2 \propto \sqrt{E_{final}} = \sqrt{3E_0}\).

Taking the ratio of the frequencies:
\(f_2 = \sqrt{3} f_1\).


Calculate the percentage change in frequency:
\(% change = \left( \frac{f_2 - f_1}{f_1} \right) \times 100%\)
\(= \left( \frac{\sqrt{3} f_1 - f_1}{f_1} \right) \times 100%\)
\(= (\sqrt{3} - 1) \times 100%\)

Using the approximation \(\sqrt{3} \approx 1.732\):
\(% change = (1.732 - 1) \times 100% = 0.732 \times 100% = 73.2%\).

The approximate percentage change is 73%.


Step 4: Final Answer:

The percentage change is roughly 73%. Quick Tip: Since frequency relates to the restoring force constant via a square root relationship (\(f \propto \sqrt{k}\)), scaling the net acting field by a factor of \(N\) changes the frequency by a factor of \(\sqrt{N}\).

Step 1: Understanding the Concept:

By carefully analyzing the schematic nodes, we can redraw the circuit into a standard parallel/series representation. The key is tracing the continuous wires (nodes) to see which components are effectively bridged across the same potential differences.


Step 2: Key Formula or Approach:

Series capacitors: \(\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \dots\)

Parallel capacitors: \(C_p = C_1 + C_2 + \dots\)


Step 3: Detailed Explanation:

Let's analyze the circuit diagram connections carefully.

The main top branch consists of three capacitors in series: \(C_1\), \(C_2\), and \(C_3\).

There are two vertical wires connecting a parallel bottom branch containing \(C_4\).

- The first vertical wire drops down from the terminal A line, strictly before the plate of \(C_1\). This means the left plate of \(C_4\) is directly connected to Node A.

- The second vertical wire drops down from the terminal B line, strictly after the plate of \(C_3\). This means the right plate of \(C_4\) is directly connected to Node B.


Because \(C_4\) spans the entire length from Node A to Node B, it is wired perfectly in parallel with the entire top series branch.


First, calculate the equivalent capacitance of the top series branch (\(C_s\)):

Since \(C_1 = C_2 = C_3 = 1 \muF\),
\(\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \muF^{-1}\).
\(C_s = \frac{1}{3} \muF\).


Now, add the parallel capacitor \(C_4 = 2 \muF\):

The total equivalent capacitance \(C_{eq} = C_s + C_4\)
\(C_{eq} = \frac{1}{3} + 2 = \frac{1 + 6}{3} = \frac{7}{3} \muF\).


Step 4: Final Answer:

The equivalent capacitance is \(7/3\) \(\muF\). Quick Tip: When deciphering schematic diagrams, completely trace a continuous wire with your pencil and label the entire wire as a single "Node" letter. If two capacitors connect to the exact same pair of Node letters, they are strictly in parallel.

Step 1: Understanding the Concept:

Evaluate the physical correctness of both the Assertion and the Reason independently. If both are factually correct physics statements, then determine if the mechanism described in the Reason is the fundamental cause producing the phenomenon described in the Assertion.


Step 2: Key Formula or Approach:

Gauss's Law inside a conductor: \(E_{in} = 0 \implies Q_{in} = 0\).

Electric force on a charge: \(\vec{F} = q\vec{E}\).


Step 3: Detailed Explanation:

Let's analyze Assertion (A):

In electrostatics, the electric field strictly vanishes inside the bulk material of a conductor. By Gauss's Law (\(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}\)), since \(\vec{E} = 0\) everywhere inside the Gaussian surface drawn within the bulk, the net enclosed charge must be zero. Any excess charge resides entirely on the outer surface. Thus, Assertion A is true.


Let's analyze Reason (R):

Between the plates of a charged capacitor (in the vacuum gap), a uniform electric field \(\vec{E}\) exists. If any free charge carrier (like an electron) is placed in this region, it will experience an electric force \(\vec{F} = q\vec{E}\) and accelerate/drift towards the oppositely charged plate. Thus, Reason R is an entirely factual and true statement about electric fields acting on charges in a vacuum gap.


Relationship Check:

Does R explain A? No. Assertion A discusses the macroscopic equilibrium property of a conductive lattice (the shielding effect forcing charge to the surface so that internal fields neutralize). Reason R simply describes the trivial definition of electric force operating in an empty space between two plates. Although both involve charges moving due to forces until equilibrium, R does not explicitly explain the mechanism of surface-charge accumulation inherent to A.


Step 4: Final Answer:

Both statements are true, but R does not correctly explain A. Quick Tip: In Assertion-Reason questions, inject "because" between the statements. Read: "A conductor has no charge inside BECAUSE free charges drift inside a capacitor gap." The disconnected context immediately reveals R isn't the correct explanation.

Step 1: Understanding the Concept:

The magnetic intensity \(H\) (also called the magnetizing field) is related directly to the total induced magnetic field \(B\) and the properties of the core material (its permeability \(\mu\)).


Step 2: Key Formula or Approach:

The relationship between Magnetic Field (\(B\)) and Magnetic Intensity (\(H\)):
\(B = \mu H\)

The permeability of the material is \(\mu = \mu_0 \mu_r\), where \(\mu_r\) is relative permeability.
\(H = \frac{B}{\mu_0 \mu_r}\)


Step 3: Detailed Explanation:

Given values:

Relative permeability, \(\mu_r = 400\).

Magnetic field, \(B = 1.0 T\).

Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} T\cdotm/A\).


Calculate the magnetic intensity \(H\):
\(H = \frac{B}{\mu_0 \mu_r}\)
\(H = \frac{1.0}{(4\pi \times 10^{-7}) \times 400}\)
\(H = \frac{1}{1600\pi \times 10^{-7}}\)
\(H = \frac{1}{16\pi \times 10^2 \times 10^{-7}}\)
\(H = \frac{1}{16\pi \times 10^{-5}}\)

Bringing the power of 10 to the numerator:
\(H = \frac{10^5}{16\pi} A/m\).


The problem states that the magnetic intensity is \(\alpha \times 10^5\).

Equating our result to this format:
\(\alpha \times 10^5 = \left(\frac{1}{16\pi}\right) \times 10^5\).


Thus, the value of \(\alpha\) is \(\frac{1}{16\pi}\).


Step 4: Final Answer:

The value of \(\alpha\) is \(1/16\pi\). Quick Tip: Be extremely careful distinguishing between Magnetic Field \(B\) (Tesla) and Magnetic Intensity \(H\) (A/m). The intensity \(H\) describes the "effort" of the external coil, while \(B\) describes the final "result" amplified by the core.

Step 1: Understanding the Concept:

In an electromagnetic wave, the electric field \(\vec{E}\), magnetic field \(\vec{B}\), and direction of wave propagation \(\vec{v}\) are mutually perpendicular. Their directions follow the right-hand cross product rule. Also, the amplitudes of the fields are firmly related by the wave velocity.


Step 2: Key Formula or Approach:

Direction of propagation: \(\hat{c} = \hat{E} \times \hat{B}\)

Amplitude relation: \(E_0 = c \cdot B_0\)

Wave speed: \(c = \nu \lambda\) (In the options, the letter \(v\) is used to denote the wave phase velocity \(c\)).


Step 3: Detailed Explanation:

1. Determine the direction of propagation:

The argument of the sine function is \((2\pi\nu t - \frac{2\pi x}{\lambda}) = (\omega t - kx)\).

The negative sign between the time and space components indicates the wave is propagating in the positive x-direction.

So, the propagation direction unit vector is \(\hat{c} = \hat{i}\).


2. Determine the direction of the Electric Field:

The magnetic field acts in the \(\hat{j}\) direction.

We must satisfy the relation \(\hat{E} \times \hat{B} = \hat{c}\).

Let \(\hat{E} = \hat{u}\). Then, \(\hat{u} \times \hat{j} = \hat{i}\).

From standard cross products, we know \(\hat{k} \times \hat{j} = -\hat{i}\).

Therefore, \((-\hat{k}) \times \hat{j} = \hat{i}\).

This means the electric field must oscillate in the \(-\hat{k}\) direction.


3. Determine the Amplitude:

The amplitude of the electric field is \(E_0 = v B_0\), where \(v\) is the wave speed.

From wave properties, speed \(v = \nu \lambda\).

So, \(E_0 = (v \lambda) B_0\) (Wait, looking at the exact text in options, they used \(v\) where usually \(\nu\) goes, or it's simply defining velocity \(v = \nu\lambda\). Since the option states \(v \lambda B_0\), it directly maps to the standard formulation \(\nu \lambda B_0\) if \(v\) represents frequency, or \(v B_0\) if \(v\) is speed. Given the explicit structure of the options, it represents the exact constant coefficient.)

Following the pattern, \(E_0 = \nu \lambda B_0\). The OCR prints \(v \lambda B_0\).


Combining magnitude and direction:
\(\vec{E} = -v \lambda B_0 \sin \left( 2\pi\nu t - \frac{2\pi x}{\lambda} \right) \hat{k}\).


This matches option (A).


Step 4: Final Answer:

The associated electric field vector is given by \(\vec{E} = -v \lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{k}\). Quick Tip: The mnemonic \(\vec{E} \times \vec{B} = \vec{v}_{prop}\) is non-negotiable for EM waves. Write the standard \(i,j,k\) cross product circle in the margin to avoid silly sign errors under pressure.

Step 1: Understanding the Concept:

The focal length of a thin lens is determined by the refractive index of its material and the radii of curvature of its two surfaces. For a biconvex lens with symmetric surfaces, we can use the Lens Maker's Formula to find a direct relation between the focal length \(f\) and the radius \(R\).


Step 2: Key Formula or Approach:

Lens Maker's Formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For an equiconvex lens, by sign convention: \(R_1 = +R\) and \(R_2 = -R\).


Step 3: Detailed Explanation:

Given the refractive index \(\mu = 1.4\).

Substitute the radii into the Lens Maker's Formula:
\[ \frac{1}{f} = (1.4 - 1) \left( \frac{1}{R} - \left( -\frac{1}{R} \right) \right) \] \[ \frac{1}{f} = (0.4) \left( \frac{1}{R} + \frac{1}{R} \right) \] \[ \frac{1}{f} = 0.4 \times \frac{2}{R} \] \[ \frac{1}{f} = \frac{0.8}{R} \]
We need the ratio of focal length to radius of curvature, which is \(\frac{f}{R}\).

Rearranging the equation:
\[ \frac{f}{R} = \frac{1}{0.8} = \frac{10}{8} = 1.25 \]

Step 4: Final Answer:

The ratio is 1.25. Quick Tip: For any equiconvex lens, the formula simplifies instantly to \(f = \frac{R}{2(\mu - 1)}\). Memorizing this reduced form saves crucial time during exams.

Step 1: Understanding the Concept:

When unpolarized light passes through the first polarizer, its intensity is halved, and it becomes polarized parallel to the polarizer's axis. As it passes through subsequent polarizers, the transmitted intensity is governed by Malus's Law, which depends on the angle between the transmission axes of adjacent polarizers.


Step 2: Key Formula or Approach:

Unpolarized light through first polarizer: \(I_1 = \frac{I_0}{2}\).

Malus's Law for subsequent polarizers: \(I_{out} = I_{in} \cos^2(\theta)\), where \(\theta\) is the relative angle between the two consecutive transmission axes.


Step 3: Detailed Explanation:

Let the initial intensity of the unpolarized light be \(I_0\).


Case 1: Without the third polarizer (Only \(P_1\) and \(P_2\))

Light passes through \(P_1\) (at \(30^\circ\)):
\(I_1 = \frac{I_0}{2}\).

Light then passes through \(P_2\) (at \(90^\circ\)). The relative angle is \(\theta_1 = 90^\circ - 30^\circ = 60^\circ\).
\(I_{out1} = I_1 \cos^2(60^\circ) = \frac{I_0}{2} \left( \frac{1}{2} \right)^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}\).


Case 2: With the third polarizer \(P_3\) (at \(60^\circ\)) inserted between \(P_1\) and \(P_2\)

Light passes through \(P_1\) (at \(30^\circ\)):
\(I_1 = \frac{I_0}{2}\).

Light passes through \(P_3\) (at \(60^\circ\)). The relative angle is \(\theta_2 = 60^\circ - 30^\circ = 30^\circ\).
\(I_2 = I_1 \cos^2(30^\circ) = \frac{I_0}{2} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8}\).

Light passes through \(P_2\) (at \(90^\circ\)). The relative angle is \(\theta_3 = 90^\circ - 60^\circ = 30^\circ\).
\(I_{out2} = I_2 \cos^2(30^\circ) = \frac{3I_0}{8} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3I_0}{8} \times \frac{3}{4} = \frac{9I_0}{32}\).


Ratio:

We need the ratio of intensity with the third polarizer to that without it:
\(Ratio = \frac{I_{out2}}{I_{out1}} = \frac{\frac{9I_0}{32}}{\frac{I_0}{8}} = \frac{9}{32} \times \frac{8}{1} = \frac{9}{4}\).


Step 4: Final Answer:

The ratio is 9/4. Quick Tip: Always calculate the \textbf{relative} angle between consecutive polarizers for Malus's Law. Do not simply plug the absolute axis angles into the cosine formula.

Step 1: Understanding the Concept:

Rutherford's scattering experiment revealed the structure of the atom. Rebounding (scattering at \(180^\circ\)) requires a massive repulsive force, which only happens if the alpha particle scores a direct, head-on hit with the dense, positively charged nucleus. Because the nucleus is astronomically small compared to the atom, these specific head-on paths are extremely rare.


Step 2: Key Formula or Approach:

Scattering angle \(\theta\) depends on the impact parameter \(b\):
\(b = \frac{1}{4\pi\epsilon_0} \frac{Z e^2 \cot(\theta/2)}{K}\)

For rebounding (\(\theta = 180^\circ\)), \(\cot(90^\circ) = 0 \implies b = 0\).


Step 3: Detailed Explanation:

Let's evaluate each statement:

A. True. The nucleus is incredibly small (\(\sim 10^{-15}\) m) compared to the atom (\(\sim 10^{-10}\) m). Most of the atom is empty space, explaining why so few particles encounter the nucleus at all.

B. False. The alpha particle has a charge of \(+2e\), whereas the gold nucleus has a charge of \(+79e\). They are not equal.

C. True. The impact parameter \(b\) is the perpendicular distance between the particle's initial trajectory and the parallel line running through the nucleus's center. For an alpha particle to rebound (\(\theta \approx 180^\circ\)), it must be aimed almost exactly at the center of the nucleus, meaning \(b \approx 0\) (minimum).

D. False. A higher kinetic energy would actually make it harder to rebound completely and would allow the particle to penetrate closer to the nucleus before stopping. The number of rebounding particles is dictated by geometry (the tiny target area), not variance in particle kinetic energy (the emitted alpha particles had relatively uniform energy).

E. True. Rebounding is precisely defined as a head-on collision where the repulsive Coulomb force completely stops and reverses the particle. Because the nucleus is so small, only a few particles happen to be on a direct collision course.


Thus, statements A, C, and E are the correct physical explanations.


Step 4: Final Answer:

Options A, C, and E only are correct. Quick Tip: Remember that the "impact parameter" dictates the scattering angle. An impact parameter of zero means a direct hit (head-on collision) resulting in a \(180^\circ\) backscatter.

Step 1: Understanding the Concept:

When a charged particle accelerates through a potential difference \(V\), it gains kinetic energy equal to \(qV\). The de Broglie wavelength connects this kinetic energy to the particle's momentum and mass.


Step 2: Key Formula or Approach:

Kinetic Energy gained: \(K = qV\).

Momentum: \(p = \sqrt{2mK} = \sqrt{2mqV}\).

de Broglie wavelength: \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}\).


Step 3: Detailed Explanation:

Both the electron and the proton carry the same magnitude of elementary charge, so \(q_e = q_p = e\).

Both are accelerated through the exact same potential difference \(V\).


For the electron:
\[ \lambda_e = \frac{h}{\sqrt{2m_eeV}} \]

For the proton:
\[ \lambda_p = \frac{h}{\sqrt{2m_peV}} \]

Take the ratio of the two wavelengths:
\[ \frac{\lambda_e}{\lambda_p} = \frac{ \frac{h}{\sqrt{2m_eeV}} }{ \frac{h}{\sqrt{2m_peV}} } \]
Cancel out the common terms \(h\), \(\sqrt{2}\), \(\sqrt{e}\), and \(\sqrt{V}\):
\[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_p}}{\sqrt{m_e}} = \sqrt{\frac{m_p}{m_e}} \]

Step 4: Final Answer:

The ratio is \(\sqrt{\frac{m_p}{m_e}}\). Quick Tip: For particles with the same charge accelerating through the same voltage, \(\lambda\) is strictly inversely proportional to the square root of their masses. \(\lambda \propto \frac{1}{\sqrt{m}}\).

Step 1: Understanding the Concept:

Evaluate the physical mechanism of a p-n junction diode under reverse bias. The reverse saturation current is created by minority carriers, not majority carriers. The breakdown at high voltages happens due to avalanche or Zener breakdown mechanisms.


Step 2: Key Formula or Approach:

In reverse bias, the depletion region widens, acting as a massive barrier to majority carriers. The small current that trickles through is purely due to minority carriers thermally generated near the junction.


Step 3: Detailed Explanation:

Let's analyze Assertion (A):

Under reverse bias, a p-n junction diode blocks the main current, yielding only a tiny leakage current (in \(\muA\) or \(nA\)). This current remains relatively constant and independent of the applied voltage up to a certain point. When the voltage hits a critical threshold (Zener or Avalanche breakdown voltage), the current suddenly skyrockets. This correctly describes the macroscopic V-I characteristic of a diode. Thus, Assertion A is true.


Let's analyze Reason (R):

Reason R claims that this leakage current is due to the flow of majority charge carriers. This is fundamentally incorrect. The reverse applied voltage pushes majority carriers (holes in p-side, electrons in n-side) away from the junction, increasing the depletion width. The tiny leakage current is actually sustained entirely by the flow of minority charge carriers (electrons in p-side, holes in n-side) sweeping across the junction due to the strong built-in electric field.

Since the fundamental mechanism stated in R is wrong, Reason R is false.


Step 4: Final Answer:

Assertion A is true, but Reason R is false. Quick Tip: Reverse bias = Reverse carriers. Always link reverse-bias leakage current exclusively to minority charge carriers.

Step 1: Understanding the Concept:

A Zener diode is used as a voltage regulator. To prevent it from overheating and being destroyed, the current flowing through it must not exceed a maximum threshold determined by its power rating. A series resistor is used to drop the excess voltage and limit this current.


Step 2: Key Formula or Approach:

Maximum allowed Zener current: \(I_{Z(max)} = \frac{P_{max}}{V_Z}\)

Voltage across the series resistor: \(V_s = V_{in} - V_Z\)

Minimum safe series resistance: \(R_{min} = \frac{V_s}{I_{Z(max)}}\)


Step 3: Detailed Explanation:

Given parameters:

Zener voltage \(V_Z = 10 V\)

Maximum power dissipation \(P_{max} = 0.5 W\)

Source voltage \(V_{in} = 25 V\)


First, calculate the maximum safe current that the Zener diode can handle.
\[ I_{Z(max)} = \frac{P_{max}}{V_Z} = \frac{0.5 W}{10 V} = 0.05 A \]

When connected in the circuit, the Zener diode will lock the voltage across its terminals at \(10 V\). The remaining voltage from the supply must be dropped across the series resistor \(R\).
\[ V_s = V_{in} - V_Z = 25 V - 10 V = 15 V \]

To ensure the Zener diode does not blow out under the worst-case scenario (when no load is attached, and all current flows through the Zener), the series resistor must restrict the total circuit current to exactly \(I_{Z(max)}\).
\[ R_{min} = \frac{V_s}{I_{Z(max)}} = \frac{15 V}{0.05 A} \] \[ R_{min} = \frac{15}{0.05} = 300\ \Omega \]

Step 4: Final Answer:

The minimum resistance required is \(300\ \Omega\). Quick Tip: The "minimum safe resistance" is evaluated under the "no-load" condition because attaching any parallel load resistor would siphon current away from the Zener, making it safer.

Step 1: Understanding the Concept:

When a projectile is fired with a fixed speed, its horizontal range depends on the angle of projection. The maximum possible range is achieved when the angle of projection is \(45^\circ\).


Step 2: Key Formula or Approach:

Horizontal range formula: \(R = \frac{u^2 \sin(2\theta)}{g}\)

For maximum range, \(\theta = 45^\circ \implies \sin(90^\circ) = 1\).

Maximum range: \(R_{max} = \frac{u^2}{g}\)


Step 3: Detailed Explanation:

We are given that the farthest distance (maximum range) the bullets can reach is \(R_{max} = 6.4 m\).

The acceleration due to gravity is \(g = 10 m/s^2\).


Substitute these values into the maximum range formula:
\[ R_{max} = \frac{u^2}{g} \] \[ 6.4 = \frac{u^2}{10} \]
Multiply both sides by 10:
\[ u^2 = 6.4 \times 10 = 64 \]
Take the square root to find the initial speed \(u\):
\[ u = \sqrt{64} = 8 m/s \]

Step 4: Final Answer:

The speed of the bullets is 8 m/s. Quick Tip: If a problem mentions a fountain of water or a gun firing in "all directions," the "farthest distance" is automatically the maximum projectile range achieved at a \(45^\circ\) launch angle.

Step 1: Understanding the Concept:

Point P lies midway between two perpendicular bar magnets. For the horizontal magnet, P lies exactly on its axial line. For the vertical magnet, P lies exactly on its equatorial line. Because the magnetic fields from these two configurations are perpendicular to each other, the net magnetic field is their vector sum via the Pythagorean theorem.


Step 2: Key Formula or Approach:

Magnetic field on the axial line of a short dipole: \(B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}\)

Magnetic field on the equatorial line of a short dipole: \(B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{r^3}\)

Net Field: \(B_{net} = \sqrt{B_{axial}^2 + B_{equatorial}^2}\)


Step 3: Detailed Explanation:

Let the horizontal magnet be Magnet 1 and the vertical magnet be Magnet 2.

Distance between centers is \(10 cm\). Since P is midway, the distance from each center to P is \(r = 5 cm = 0.05 m = 5 \times 10^{-2} m\).

Magnetic moment \(M = 3\sqrt{5} J/T\).


Calculate field due to Magnet 1 (Axial):
\(B_1 = \frac{\mu_0}{4\pi} \frac{2M}{r^3}\)

Calculate field due to Magnet 2 (Equatorial):
\(B_2 = \frac{\mu_0}{4\pi} \frac{M}{r^3}\)


Since Magnet 1 is horizontal, \(B_1\) is horizontal. Magnet 2 is vertical, and its equatorial field at P is anti-parallel to its axis, meaning \(B_2\) is vertical. Thus, the vectors are at \(90^\circ\) to each other.
\[ B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{\left(\frac{\mu_0}{4\pi}\frac{2M}{r^3}\right)^2 + \left(\frac{\mu_0}{4\pi}\frac{M}{r^3}\right)^2} \] \[ B_{net} = \frac{\mu_0}{4\pi} \frac{M}{r^3} \sqrt{2^2 + 1^2} = \frac{\mu_0}{4\pi} \frac{M\sqrt{5}}{r^3} \]

Substitute the values:
\(\frac{\mu_0}{4\pi} = 10^{-7} Tm/A\).
\(M = 3\sqrt{5}\).
\(r = 5 \times 10^{-2} m \implies r^3 = 125 \times 10^{-6} m^3\).

\[ B_{net} = 10^{-7} \times \frac{(3\sqrt{5})(\sqrt{5})}{125 \times 10^{-6}} \] \[ B_{net} = 10^{-7} \times \frac{3 \times 5}{125 \times 10^{-6}} = 10^{-7} \times \frac{15}{125} \times 10^6 \] \[ B_{net} = \frac{15}{125} \times 10^{-1} = \frac{3}{25} \times 10^{-1} \]
Convert fraction to decimal: \(\frac{3}{25} = 0.12\).
\[ B_{net} = 0.12 \times 10^{-1} T = 0.012 T = 12 \times 10^{-3} T \]

We are given \(B_{net} = \alpha \times 10^{-3} T\).

Therefore, \(\alpha = 12\).


Step 4: Final Answer:

The value of \(\alpha\) is 12. Quick Tip: To avoid dragging messy numbers through squares and square roots, always factor out the common term \(\left(\frac{\mu_0}{4\pi}\frac{M}{r^3}\right)\) before evaluating the vector sum magnitude.

Step 1: Understanding the Concept:

When a current-carrying coil is placed in a magnetic field, it experiences a magnetic torque. The magnitude of this torque depends on the coil's magnetic moment, the external magnetic field, and the angle between the coil's normal vector (its axis) and the magnetic field.


Step 2: Key Formula or Approach:

Torque on a coil: \(\tau = |\vec{M} \times \vec{B}| = M B \sin \theta\)

Magnetic moment of the coil: \(M = N I A\)

Area of circular coil: \(A = \pi r^2\)

Thus, \(\tau = N I A B \sin \theta\).


Step 3: Detailed Explanation:

Identify the given values:

Number of turns, \(N = 125\)

Current, \(I = 1 A\)

Radius, \(r = 2 cm = 0.02 m\)

Magnetic field, \(B = 0.4 T\)

Angle between axis and field, \(\theta = 30^\circ\)
\(\pi = 3.14\)


First, calculate the area of the coil \(A\):
\[ A = \pi r^2 = 3.14 \times (0.02)^2 = 3.14 \times 0.0004 = 12.56 \times 10^{-4} m^2 \]

Now, substitute everything into the torque formula:
\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin \theta \] \[ \tau = 125 \times 1 \times (12.56 \times 10^{-4}) \times 0.4 \times \sin(30^\circ) \]
Since \(\sin(30^\circ) = 0.5\):
\[ \tau = 125 \times 0.4 \times 0.5 \times 12.56 \times 10^{-4} \] \[ \tau = 125 \times 0.2 \times 12.56 \times 10^{-4} \] \[ \tau = 25 \times 12.56 \times 10^{-4} \] \[ \tau = 314 \times 10^{-4} N.m \]

The problem states the torque is \(\alpha \times 10^{-4} N.m\).

Matching our result with the required format:
\(\alpha = 314\).


Step 4: Final Answer:

The value of \(\alpha\) is 314. Quick Tip: Always read carefully whether the angle given is between the magnetic field and the "plane of the coil" or the "axis of the coil". If it's the axis (normal vector), use \(\sin \theta\) directly. If it's the plane, you must use \(\cos \theta\).

Step 1: Understanding the Concept:

Introducing a transparent sheet of thickness \(t\) and refractive index \(\mu\) in front of one slit introduces an additional optical path difference of \((\mu - 1)t\). This extra path difference shifts the entire interference pattern. If the shift corresponds to a specific fringe position, we can equate the path shift to the required wavelength condition.


Step 2: Key Formula or Approach:

Shift of the fringe pattern: \(\Delta y = \frac{D}{d} (\mu - 1)t\)

Position of the \(n^{th}\) bright fringe (without sheet): \(y_n = n \frac{\lambda D}{d}\)

Equating the shift to the \(n^{th}\) bright fringe position:
\(\frac{D}{d} (\mu - 1)t = n \frac{\lambda D}{d} \implies (\mu - 1)t = n\lambda\)


Step 3: Detailed Explanation:

Given parameters:

Refractive index \(\mu = 1.56\)

Fringe shift \(n = 7\) (central fringe shifts to the 7th bright fringe position)

Wavelength \(\lambda = 450 nm = 450 \times 10^{-9} m\)


Use the derived equation:
\[ (\mu - 1)t = n\lambda \] \[ (1.56 - 1)t = 7 \times 450 \times 10^{-9} \] \[ 0.56 t = 3150 \times 10^{-9} \]

Solve for \(t\):
\[ t = \frac{3150 \times 10^{-9}}{0.56} \]
To simplify the division, multiply numerator and denominator by 100:
\[ t = \frac{315000}{56} \times 10^{-9} \]
Divide by 7:
\[ t = \frac{45000}{8} \times 10^{-9} \]
Divide by 8:
\[ 45000 / 8 = 5625 \] \[ t = 5625 \times 10^{-9} m \]

The problem states thickness is \(\alpha \times 10^{-9} m\).

Thus, \(\alpha = 5625\).


Step 4: Final Answer:

The value of \(\alpha\) is 5625. Quick Tip: The geometrical setup constants (\(D\) and \(d\)) perfectly cancel out when relating optical path shift directly to the equivalent number of wavelengths (\(\Delta path = (\mu - 1)t = n\lambda\)). You don't need \(D\) or \(d\) at all.