JEE Main 2021 22nd July Morning Shift Question Paper with Solutions

The vector projection of $\vec{A}$ onto $\vec{B}$ is given by the formula $\text{proj}_{\vec{B}}\vec{A} = \dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\,\vec{B}$.

First, compute the dot product $\vec{A} \cdot \vec{B} = (1)(1) + (1)(1) + (1)(0) = 2$.

Next, find the magnitude squared of $\vec{B}$: $|\vec{B}|^2 = 1^2 + 1^2 = 2$.

Therefore the projection is $\dfrac{2}{2}(\hat{i} + \hat{j}) = \hat{i} + \hat{j}$.

We apply the law of conservation of momentum. Initially, both the bullet and gun are at rest, so the total momentum is zero.

The bullet has mass $m = 4 \text{ g} = 0.004 \text{ kg}$ and muzzle speed $v = 50 \text{ ms}^{-1}$.

The impulse imparted to the gun equals the change in momentum of the bullet (Newton's third law). The magnitude of impulse is: $J = mv = 0.004 \times 50 = 0.2 \text{ kg m s}^{-1}$

By conservation of momentum, the gun (mass $M = 4 \text{ kg}$) recoils with velocity $V$ such that: $MV = mv$ $V = \frac{mv}{M} = \frac{0.004 \times 50}{4} = \frac{0.2}{4} = 0.05 \text{ m s}^{-1}$

Therefore, the impulse imparted to the gun is $0.2 \text{ kg m s}^{-1}$ and the velocity of recoil is $0.05 \text{ m s}^{-1}$.

We need to find the constants $C$ and $\phi$ when the equation of motion of a mass on a spring is represented in the cosine form, given its initial position $x(0)$ and initial velocity $v(0)$ at $t = 0$.

1. Understand the Two Representations of the Motion

The displacement of the mass can be written in two equivalent ways:

$\text{Form 1: } x(t) = A \sin \omega t + B \cos \omega t \quad \text{--- (Equation 1)}$

$\text{Form 2: } x(t) = C \cos(\omega t - \phi) \quad \text{--- (Equation 2)}$

---

2. Expand the Cosine Form

Using the trigonometric identity $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$, we can expand Equation 2:

$x(t) = C (\cos \omega t \cos \phi + \sin \omega t \sin \phi)$

$x(t) = (C \sin \phi) \sin \omega t + (C \cos \phi) \cos \omega t \quad \text{--- (Equation 3)}$

Comparing the coefficients of $\sin \omega t$ and $\cos \omega t$ between Equation 1 and Equation 3, we get:

$A = C \sin \phi \quad \text{--- (Equation 4)}$

$B = C \cos \phi \quad \text{--- (Equation 5)}$

---

3. Relate Constants to Initial Conditions ($t = 0$)

Let's find the values of $A$ and $B$ using the position and velocity at $t = 0$:

• For Position ($x(0)$): Substitute $t = 0$ into Equation 1:

  $x(0) = A \sin(0) + B \cos(0) \implies B = x(0)$

• For Velocity ($v(0)$): First, differentiate Equation 1 with respect to time $t$ to find the velocity equation $v(t)$:

  $v(t) = \frac{dx}{dt} = A \omega \cos \omega t - B \omega \sin \omega t$

  Now, substitute $t = 0$ into this velocity equation:

  $v(0) = A \omega \cos(0) - B \omega \sin(0) \implies v(0) = A \omega \implies A = \frac{v(0)}{\omega}$

---

4. Solve for Amplitude ($C$) and Phase Constant ($\phi$)

Now substitute the expressions for $A$ and $B$ back into our coefficient relations (Equation 4 and Equation 5):

$C \sin \phi = \frac{v(0)}{\omega}$

$C \cos \phi = x(0)$

• To find $C$: Square both equations and add them together ($\sin^2 \phi + \cos^2 \phi = 1$):

  $C^2 \sin^2 \phi + C^2 \cos^2 \phi = \left(\frac{v(0)}{\omega}\right)^2 + [x(0)]^2$

  $C^2 = \frac{v(0)^2}{\omega^2} + x(0)^2 \implies C = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}$

• To find $\phi$: Divide the sine equation by the cosine equation ($\frac{\sin \phi}{\cos \phi} = \tan \phi$):

  $\tan \phi = \frac{C \sin \phi}{C \cos \phi} = \frac{\frac{v(0)}{\omega}}{x(0)} = \frac{v(0)}{x(0)\omega}$

  $\phi = \tan^{-1}\left(\frac{v(0)}{x(0)\omega}\right)$

Final Answer: Option D: $C = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}$, $\phi = \tan^{-1}\left(\frac{v(0)}{x(0)\omega}\right)$

The porter lowers the suitcase with constant velocity, meaning the net force is zero. We need to find the work done by the porter (i.e., the work done by the force applied by the porter).

The suitcase has mass $m = 80 \text{ kg}$ and is lowered by a distance $d = 80 \text{ cm} = 0.80 \text{ m}$.

Since the suitcase moves down with constant velocity, the porter applies an upward force equal in magnitude to the weight: $F_{\text{porter}} = mg = 80 \times 9.8 = 784 \text{ N (upward)}$

The displacement is downward ($d = 0.80 \text{ m}$ downward), while the force is upward. Since force and displacement are in opposite directions, the work done by the porter is negative: $W_{\text{porter}} = -F \cdot d = -(784)(0.80) = -627.2 \text{ J}$

The work done by the porter in lowering the suitcase is $-627.2 \text{ J}$.

When a rigid body rolls down an inclined plane without slipping, we use energy conservation. Starting from rest at height $h$, all potential energy converts to translational and rotational kinetic energy:

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Using the rolling condition $\omega = v/R$ and writing $I = kmR^2$ where $k$ is a constant depending on the body's geometry:

$mgh = \frac{1}{2}mv^2(1 + k)$ $v = \sqrt{\frac{2gh}{1+k}}$

The moment of inertia factors are: for a ring, $I = mR^2$ so $k = 1$; for a solid cylinder, $I = \frac{1}{2}mR^2$ so $k = \frac{1}{2}$; for a solid sphere, $I = \frac{2}{5}mR^2$ so $k = \frac{2}{5}$.

The velocity at the bottom is: $v_{\text{ring}} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$ $v_{\text{cylinder}} = \sqrt{\frac{2gh}{1+\frac{1}{2}}} = \sqrt{\frac{4gh}{3}}$ $v_{\text{sphere}} = \sqrt{\frac{2gh}{1+\frac{2}{5}}} = \sqrt{\frac{10gh}{7}}$

Comparing: $\frac{10}{7} > \frac{4}{3} > 1$, so $v_{\text{sphere}} > v_{\text{cylinder}} > v_{\text{ring}}$. The sphere has the greatest and the ring has the least velocity at the bottom.

A body projected with escape velocity satisfies the energy equation at any height $r$ from the Earth's center: $\frac{1}{2}mv^2 - \frac{GMm}{r} = 0$ so $v = \sqrt{\frac{2GM}{r}}$.

Using $GM = gR_e^2$, we get $v = \frac{dr}{dt} = \sqrt{\frac{2gR_e^2}{r}}$.

Separating variables: $\sqrt{r}\, dr = \sqrt{2gR_e^2}\, dt = R_e\sqrt{2g}\, dt$

Integrating from $r = R_e$ to $r = R_e + h$: $\int_{R_e}^{R_e+h} r^{1/2}\, dr = R_e\sqrt{2g} \int_0^t dt$

$\left[\frac{2}{3}r^{3/2}\right]_{R_e}^{R_e+h} = R_e\sqrt{2g}\cdot t$

$\frac{2}{3}\left[(R_e+h)^{3/2} - R_e^{3/2}\right] = R_e\sqrt{2g}\cdot t$

$t = \frac{2}{3R_e\sqrt{2g}}\left[(R_e+h)^{3/2} - R_e^{3/2}\right]$

Factoring out $R_e^{3/2}$: $t = \frac{2R_e^{3/2}}{3R_e\sqrt{2g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2} - 1\right] = \frac{2\sqrt{R_e}}{3\sqrt{2g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2} - 1\right]$

Simplifying: $t = \frac{1}{3}\sqrt{\frac{2R_e}{g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2} - 1\right]$

This matches option 4.

According to the equipartition theorem, each degree of freedom of a gas molecule contributes $\frac{1}{2}k_BT$ to the average energy.

A monoatomic gas (like He, Ne, Ar) has only translational degrees of freedom. Since there are 3 translational degrees of freedom (motion along $x$, $y$, and $z$ axes), the average kinetic energy per molecule is:

$\langle E \rangle = 3 \times \frac{1}{2}k_BT = \frac{3}{2}k_BT$

This is the well-known result for the average thermal energy of a monoatomic ideal gas molecule at temperature $T$.

The time period of a simple pendulum is given by: $T = 2\pi\sqrt{\frac{L}{g}}$

Initially, the time period is $T_0 = 2\pi\sqrt{\frac{L}{g}}$.

When the length is reduced to $\frac{1}{16}$ of its original value, the new length is $L' = \frac{L}{16}$.

The new time period is: $T' = 2\pi\sqrt{\frac{L'}{g}} = 2\pi\sqrt{\frac{L/16}{g}} = 2\pi\sqrt{\frac{L}{16g}} = \frac{1}{4} \cdot 2\pi\sqrt{\frac{L}{g}} = \frac{T_0}{4}$

Therefore, the modified time period is $\frac{T_0}{4}$.

The line charge lies along the $z$-axis, so the electric field at any point in the $x$-$y$ plane is radial and has magnitude

$E(r)=\frac{\lambda}{2\pi\varepsilon_0\,r}\qquad -(1)$
where $\lambda=3.0\times10^{-6}\,{\rm C\,m^{-1}}$ and $r$ is the perpendicular distance from the line.

The dipole is on the $x$-axis.
Positive charge $+q$ is at $x_1=10\;{\rm mm}=0.01\;{\rm m}$.
Negative charge $-q$ is at $x_2=12\;{\rm mm}=0.012\;{\rm m}$.

Using $-(1)$, the fields at the two charges are

$E_1=\frac{\lambda}{2\pi\varepsilon_0\,x_1},\qquad E_2=\frac{\lambda}{2\pi\varepsilon_0\,x_2}\qquad -(2)$

Direction of the field is $+x$ (away from the line) at both points.
Hence the individual forces are

$F_+=+qE_1\quad\text{(on }+q\text{, toward }+x),$ $F_-=-qE_2\quad\text{(on }-q\text{, toward }-x)$

The net force on the dipole is therefore

$F=F_+ + F_- = q\bigl(E_1-E_2\bigr)\qquad -(3)$

Substituting $-(2)$ into $-(3)$ gives

$F = q\,\frac{\lambda}{2\pi\varepsilon_0}\left(\frac1{x_1}-\frac1{x_2}\right)\qquad -(4)$

Given $F=4\;{\rm N}$, solve $-(4)$ for $q$:

First evaluate the bracket:
$\frac1{x_1}-\frac1{x_2}= \frac1{0.01}-\frac1{0.012} =100-83.333=16.6667\;{\rm m^{-1}}\qquad -(5)$

Next compute the constant factor:
$\frac{\lambda}{2\pi\varepsilon_0}= \frac{3.0\times10^{-6}}{2\pi(8.854\times10^{-12})} =5.39\times10^{4}\;{\rm N\,C^{-1}}\qquad -(6)$

Hence the field difference is
$E_1-E_2 = (5.39\times10^{4})(16.6667) =8.98\times10^{5}\;{\rm N\,C^{-1}}\qquad -(7)$

Finally, from $-(4)$,
$q=\frac{F}{E_1-E_2}= \frac{4}{8.98\times10^{5}} =4.46\times10^{-6}\;{\rm C}\qquad -(8)$

$4.46\times10^{-6}\;{\rm C}=4.46\;\mu{\rm C}$

The magnitude of each charge in the dipole is therefore approximately $4.44\;\mu{\rm C}$.

Answer: Option D - $4.44\;\mu{\rm C}$

We need to find the total equivalent resistance of a parallel combination of a Copper ($\text{Cu}$) rod and an Aluminium ($\text{Al}$) rod between points $A$ and $B$.

1. Extract Given Values and Convert to SI Units

• Length of each rod ($L$): $25\text{ cm} = 0.25\text{ m}$
• Cross-sectional area ($A_{\text{area}}$): $3\text{ mm}^2 = 3 \times 10^{-6}\text{ m}^2$
• Resistivity of Copper ($\rho_{\text{Cu}}$): $1.7 \times 10^{-8}\ \Omega\cdot\text{m}$
• Resistivity of Aluminium ($\rho_{\text{Al}}$): $2.6 \times 10^{-8}\ \Omega\cdot\text{m}$

---

2. Calculate the Resistance of Each Individual Rod

The resistance of a conductor is given by the formula:

$R = \rho \frac{L}{A_{\text{area}}}$

For the Copper Rod ($R_{\text{Cu}}$):

$R_{\text{Cu}} = (1.7 \times 10^{-8}) \times \frac{0.25}{3 \times 10^{-6}}$

$R_{\text{Cu}} = \frac{0.425 \times 10^{-2}}{3} = \frac{4.25}{3} \times 10^{-3}\ \Omega \approx 1.4167\text{ m}\Omega$

For the Aluminium Rod ($R_{\text{Al}}$):

$R_{\text{Al}} = (2.6 \times 10^{-8}) \times \frac{0.25}{3 \times 10^{-6}}$

$R_{\text{Al}} = \frac{0.65 \times 10^{-2}}{3} = \frac{6.5}{3} \times 10^{-3}\ \Omega \approx 2.1667\text{ m}\Omega$

---

3. Calculate the Equivalent Resistance ($R_{AB}$) in Parallel

Since the two rods are connected side-by-side in a parallel configuration across terminals $A$ and $B$, the equivalent resistance is calculated as:

$R_{AB} = \frac{R_{\text{Cu}} \times R_{\text{Al}}}{R_{\text{Cu}} + R_{\text{Al}}}$

Let's factor out the common term $10^{-3}$ to simplify calculation with fractions:

$R_{AB} = \left( \frac{\frac{4.25}{3} \times \frac{6.5}{3}}{\frac{4.25}{3} + \frac{6.5}{3}} \right) \times 10^{-3}\ \Omega$

$R_{AB} = \left( \frac{\frac{27.625}{9}}{\frac{10.75}{3}} \right) \times 10^{-3}\ \Omega$

$R_{AB} = \left( \frac{27.625}{9} \times \frac{3}{10.75} \right) \times 10^{-3}\ \Omega$

$R_{AB} = \left( \frac{27.625}{3 \times 10.75} \right) \times 10^{-3}\ \Omega = \frac{27.625}{32.25} \times 10^{-3}\ \Omega$

$R_{AB} \approx 0.8566 \times 10^{-3}\ \Omega \approx 0.858\text{ m}\Omega$

Final Answer: 0.858 m$\Omega$

Statement I states that ferromagnetic property depends on temperature, and above a certain temperature (the Curie temperature), a ferromagnet becomes a paramagnet. This is correct — it is the well-established Curie temperature phenomenon.

Statement II states that at high temperature, the domain wall area of a ferromagnetic substance increases. This is incorrect. At high temperature, the domains actually break down or become disordered. The thermal energy disrupts the alignment of magnetic moments within domains. Above the Curie temperature, domain structure itself ceases to exist as the material becomes paramagnetic — it is not that domain walls increase in area. In fact, as temperature increases, thermal agitation destroys the long-range magnetic order, causing the domains to shrink and eventually disappear rather than domain walls expanding.

Therefore, Statement I is true but Statement II is false.

We need to determine the correct relationship between the true dip ($\delta$) and the apparent dip ($\delta'$) in Earth's magnetism.

1. Understand the Concepts of Dip

• Magnetic Meridian: The vertical plane passing through the magnetic axis of Earth. The actual magnetic field ($\vec{B}$) of Earth lies in this plane.
• True Dip ($\delta$): The angle made by the total magnetic field of Earth with the horizontal direction in the true magnetic meridian.
• Apparent Dip ($\delta'$): The angle of dip measured in any vertical plane other than the true magnetic meridian.

---

2. Set Up the Mathematical Relations

In the true magnetic meridian, let the horizontal component of Earth's magnetic field be $B_H$ and the vertical component be $B_V$. The true dip angle $\delta$ is given by:

$\tan \delta = \frac{B_V}{B_H} \quad \text{--- (Equation 1)}$

Now, consider another vertical plane that makes an angle $\alpha$ with the true magnetic meridian. In this new plane:

• The vertical component remains unchanged because it acts straight down: $B_V' = B_V$.
• The horizontal component is the projection of $B_H$ onto this new plane: $B_H' = B_H \cos \alpha$.

The apparent dip $\delta'$ measured in this plane is given by:

$\tan \delta' = \frac{B_V'}{B_H'} = \frac{B_V}{B_H \cos \alpha}$

---

3. Compare True Dip and Apparent Dip

Using the value from Equation 1 ($\frac{B_V}{B_H} = \tan \delta$), we can rewrite the equation for apparent dip as:

$\tan \delta' = \frac{\tan \delta}{\cos \alpha}$

$\tan \delta = \tan \delta' \cdot \cos \alpha$

For any plane other than the true magnetic meridian, the angle $\alpha$ is between $0^\circ$ and $90^\circ$ ($0^\circ < \alpha < 90^\circ$). In this range, the value of cosine is always a fraction less than $1$:

$\cos \alpha < 1$

Therefore, multiplying $\tan \delta'$ by a fraction less than $1$ makes it equal to $\tan \delta$:

$\tan \delta < \tan \delta' \implies \delta < \delta'$

---

4. Conclusion

Since the angle of true dip is smaller than the angle of apparent dip, it means that the True dip is less than apparent dip.

Final Answer: Option B (True dip is less than apparent dip.)

In a purely capacitive circuit with $E_g = E_{g0}\sin\omega t$, the current leads the voltage by $90°$. This is because in a capacitor, charge builds up as voltage rises, and maximum current flows when the voltage is zero (rate of change is maximum), while zero current flows when the voltage is at its peak.

Specifically, the current is: $I_C = I_0\sin\left(\omega t + \frac{\pi}{2}\right) = I_0\cos(\omega t)$

while the voltage across the capacitor is $V_C = V_0\sin(\omega t)$ (in phase with the applied emf).

In the phasor diagram, $V_C$ (or the emf $E_g$) is taken as the reference phasor, and $I_C$ leads $V_C$ by $90°$ (i.e., $I_C$ is drawn $90°$ ahead of $V_C$ in the direction of rotation). The correct phasor diagram shows $I_C$ perpendicular to and leading $V_C$, which corresponds to phasor diagram (3).

For a series $RLC$ circuit the impedance is $Z = \sqrt{R^{2}+\left(\,\omega L-\frac{1}{\omega C}\right)^{2}}$ and the phase angle $\phi$ between the current $I$ and the applied emf $E$ is given by

$\tan \phi = \frac{\omega L-\dfrac{1}{\omega C}}{R}$

Sign of $\left(\omega L-\dfrac{1}{\omega C}\right)$ decides whether the circuit behaves inductively or capacitively.

Case a: $\omega L \gt \dfrac{1}{\omega C}$ ⇒ $\tan\phi \gt 0$, so $\phi$ is positive. A positive phase angle means the current lags behind the emf.
Hence (a) matches with (ii) “Current lags behind the applied emf”.

Case b: $\omega L = \dfrac{1}{\omega C}$ ⇒ $\tan\phi = 0$, so $\phi = 0$. The current is exactly in phase with the emf and the circuit is at resonance.
Hence (b) matches with (i) “Current is in phase with emf”.

Case c: $\omega L \lt \dfrac{1}{\omega C}$ ⇒ $\tan\phi \lt 0$, so $\phi$ is negative. A negative phase angle means the current leads the emf.
Hence (c) matches with (iv) “Current leads the emf”.

Case d: “Resonant frequency” refers to the condition $\omega L = \dfrac{1}{\omega C}$, where the impedance is minimum ($Z = R$) and the circuit admits the largest current.
Hence (d) matches with (iii) “Maximum current occurs”.

Collecting all matches:
(a) - (ii); (b) - (i); (c) - (iv); (d) - (iii)

Therefore the correct option is Option A.

The intensity of an electromagnetic wave is related to the peak electric field by: $I = \frac{1}{2}\varepsilon_0 c E_0^2$

And the peak magnetic field is related to the peak electric field by: $B_0 = \frac{E_0}{c}$

Combining these: $I = \frac{1}{2}\varepsilon_0 c \cdot (B_0 c)^2 = \frac{1}{2}\varepsilon_0 c^3 B_0^2$

Solving for $B_0$: $B_0 = \sqrt{\frac{2I}{\varepsilon_0 c^3}}$

Substituting the values: $I = 0.092 \text{ W m}^{-2}$, $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{N}^{-1}\text{m}^{-2}$, $c = 3 \times 10^8 \text{ m s}^{-1}$: $B_0 = \sqrt{\frac{2 \times 0.092}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}}$

Computing the denominator: $(3\times10^8)^3 = 27\times10^{24}$ $8.85\times10^{-12} \times 27\times10^{24} = 238.95\times10^{12} = 2.3895\times10^{14}$

$B_0 = \sqrt{\frac{0.184}{2.3895\times10^{14}}} = \sqrt{7.699\times10^{-16}} = 2.775\times10^{-8} \text{ T} \approx 2.77\times10^{-8} \text{ T}$

A ray travels from denser medium to rarer medium at angle of incidence $i$. The angle of reflection equals the angle of incidence, so $r = i$.

The reflected and refracted rays are perpendicular to each other, so: $r + r' = 90°$ $r' = 90° - r = 90° - i$

By Snell's law at the interface (denser to rarer): $n_1 \sin i = n_2 \sin r'$

where $n_1$ is the refractive index of the denser medium and $n_2$ is that of the rarer medium. Assuming the rarer medium is air ($n_2 = 1$): $n_1 \sin i = \sin r' = \sin(90° - i) = \cos i$ $n_1 = \frac{\cos i}{\sin i} = \cot i = \cot r$

The critical angle $\theta_c$ is defined by $\sin\theta_c = \frac{n_2}{n_1} = \frac{1}{n_1}$, so: $\sin\theta_c = \frac{1}{\cot r} = \tan r$ $\theta_c = \sin^{-1}(\tan r)$

We need to find the ratio of the de-Broglie wavelength associated with an electron to that with a proton when both are accelerated through the same potential difference.

1. Understand the de-Broglie Wavelength Formula

According to de-Broglie's hypothesis, a moving particle behaves as a wave, and its wavelength ($\lambda$) is given by:

$\lambda = \frac{h}{p}$

Where:

• $h$ = Planck's constant
• $p$ = Momentum of the particle

---

2. Relate Momentum to Potential Difference

When a particle of charge $q$ and mass $m$ is accelerated from rest through a potential difference $V$, the kinetic energy ($K$) gained by the particle is:

$K = qV$

We also know that kinetic energy is related to momentum ($p$) by the formula:

$K = \frac{p^2}{2m} \implies p = \sqrt{2mK}$

Substituting the kinetic energy equation ($K = qV$) into the momentum formula gives:

$p = \sqrt{2mqV}$

---

3. Substitute Momentum into the Wavelength Equation

Now, substitute the expression for momentum ($p$) back into the de-Broglie wavelength formula:

$\lambda = \frac{h}{\sqrt{2mqV}}$

---

4. Find the Ratio for Electron and Proton

Let's write out the wavelengths for both the electron and the proton:

• For the electron (mass = $m_e$, charge = $e$):

  $\lambda_e = \frac{h}{\sqrt{2m_e eV}}$

• For the proton (mass = $m_p$, charge = $e$):

  $\lambda_p = \frac{h}{\sqrt{2m_p eV}}$

(Note: Both the electron and proton have the same magnitude of charge $q = e$ and are accelerated through the same potential difference $V$.)

Taking the ratio of the de-Broglie wavelength of the electron to that of the proton:

$\frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_e eV}}}{\frac{h}{\sqrt{2m_p eV}}}$

Canceling out the common terms ($h$, $2$, $e$, and $V$), the equation simplifies inversely to the square root of their masses:

$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$

Final Answer: Option C ($\sqrt{\frac{m_p}{m_e}}$)

The nucleus has mass number $A = 184$ and is initially at rest. It emits an alpha particle (mass number 4), leaving a daughter nucleus with mass number $A - 4 = 180$.

By conservation of momentum (initial momentum = 0): $m_\alpha v_\alpha = M_D v_D$ $v_D = \frac{m_\alpha v_\alpha}{M_D} = \frac{4}{180}v_\alpha$

The total kinetic energy released equals the Q-value: $K_\alpha + K_D = Q$ $\frac{1}{2}m_\alpha v_\alpha^2 + \frac{1}{2}M_D v_D^2 = Q$

Substituting $v_D = \frac{4}{180}v_\alpha$: $K_D = \frac{1}{2}M_D\left(\frac{4}{180}\right)^2 v_\alpha^2 = \frac{4}{180} \cdot \frac{1}{2}m_\alpha v_\alpha^2 = \frac{4}{180}K_\alpha$

Therefore: $K_\alpha + \frac{4}{180}K_\alpha = Q$ $K_\alpha\left(1 + \frac{4}{180}\right) = Q$ $K_\alpha \cdot \frac{184}{180} = 5.5 \text{ MeV}$ $K_\alpha = 5.5 \times \frac{180}{184} = 5.5 \times 0.9783 = 5.38 \text{ MeV}$

The kinetic energy of the $\alpha$-particle is approximately $5.38 \text{ MeV}$.

When light of wavelength $\lambda \leq 621 \text{ nm}$ falls on the reverse-biased P-N junction, it generates hole-electron pairs, increasing the reverse current. This means the photon energy must be at least equal to the band gap energy.

The minimum photon energy (at maximum wavelength $\lambda = 621 \text{ nm}$) equals the band gap: $E_g = \frac{hc}{\lambda}$

Using $h = 6.626 \times 10^{-34} \text{ J s}$, $c = 3 \times 10^8 \text{ m s}^{-1}$, and $\lambda = 621 \times 10^{-9} \text{ m}$: $E_g = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{621 \times 10^{-9}}$

$E_g = \frac{1.988 \times 10^{-25}}{6.21 \times 10^{-7}} = 3.20 \times 10^{-19} \text{ J}$

Converting to electron volts: $E_g = \frac{3.20 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0 \text{ eV}$

The band gap of the semiconductor is approximately $2 \text{ eV}$.