JEE Main 2018 Physics Question Paper with Answer and Solution

(C) According to Kepler's third law,the time period $T$ of a satellite orbiting a planet of mass $M$ at a radius $r$ is given by:
$T = 2\pi \sqrt{\frac{r^3}{GM}}$
Squaring both sides,we get:
$T^2 = \frac{4\pi^2}{GM} r^3$
Rearranging for the mass $M$:
$M = \frac{4\pi^2 r^3}{GT^2}$
Taking the natural logarithm and differentiating to find the relative uncertainty:
$\ln M = \ln(4\pi^2) + 3\ln r - \ln G - 2\ln T$
$\frac{\Delta M}{M} = 3\frac{\Delta r}{r} - 2\frac{\Delta T}{T}$
Given that the relative uncertainty in the radius $\frac{\Delta r}{r}$ is negligible (i.e.,$\frac{\Delta r}{r} = 0$):
$\left| \frac{\Delta M}{M} \right| = |-2| \frac{\Delta T}{T} = 2 \times 10^{-2}$

(B) For a mixture of gases,the adiabatic index $\gamma_{mix}$ is given by the formula:
$\gamma_{mix} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 C_{V1} + n_2 C_{V2}}$
Alternatively,using the degrees of freedom $f$:
$C_V = \frac{f}{2}R$ and $C_P = (\frac{f}{2} + 1)R$
For Helium (monoatomic),$f_1 = 3$,so $C_{V1} = \frac{3}{2}R$ and $C_{P1} = \frac{5}{2}R$.
For Hydrogen (diatomic),$f_2 = 5$,so $C_{V2} = \frac{5}{2}R$ and $C_{P2} = \frac{7}{2}R$.
Given $n_1 = 2$ and $n_2 = n$,the mixture ratio is $\frac{C_P}{C_V} = \frac{3}{2}$.
Using the relation $\frac{C_{P,mix}}{C_{V,mix}} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 C_{V1} + n_2 C_{V2}} = \frac{3}{2}$:
$\frac{2(\frac{5}{2}R) + n(\frac{7}{2}R)}{2(\frac{3}{2}R) + n(\frac{5}{2}R)} = \frac{3}{2}$
$\frac{5 + 3.5n}{3 + 2.5n} = \frac{3}{2}$
$10 + 7n = 9 + 7.5n$
$0.5n = 1$
$n = 2$.

(D) Let the equation of motion be $x = A \cos(\omega t + \phi)$. However,for simplicity,we can use $x = A \cos(\omega t)$ if the phase is zero at $t=0$. Given the positions at specific times:
$a = A \cos(\omega t_0)$
$b = A \cos(2\omega t_0)$
$c = A \cos(3\omega t_0)$
Using the trigonometric identity $\cos(3\theta) + \cos(\theta) = 2 \cos(2\theta) \cos(\theta)$:
$a + c = A \cos(3\omega t_0) + A \cos(\omega t_0) = A [2 \cos(2\omega t_0) \cos(\omega t_0)]$
Substitute $b = A \cos(2\omega t_0)$:
$a + c = 2b \cos(\omega t_0)$
$\cos(\omega t_0) = \frac{a+c}{2b}$
$\omega t_0 = \cos^{-1} \left( \frac{a+c}{2b} \right)$
Since $\omega = 2\pi f$,we have $2\pi f t_0 = \cos^{-1} \left( \frac{a+c}{2b} \right)$
$f = \frac{1}{2\pi t_0} \cos^{-1} \left( \frac{a+c}{2b} \right)$

(D) Given: Frequency of string $A$,$n_A = 425 \, Hz$. Initial beat frequency $x_1 = 5 \, Hz$.
Beat frequency is given by $|n_A - n_B| = 5 \, Hz$. This implies $n_B$ could be $420 \, Hz$ or $430 \, Hz$.
When the tension of string $B$ is increased,its frequency $n_B$ increases because $n \propto \sqrt{T}$.
Case $1$: If $n_B = 420 \, Hz$,increasing tension increases $n_B$. As $n_B$ approaches $n_A$ $(425 \, Hz)$,the beat frequency $|n_A - n_B|$ decreases. This matches the observation that the beat frequency decreased to $3 \, Hz$.
Case $2$: If $n_B = 430 \, Hz$,increasing tension increases $n_B$ further away from $n_A$ $(425 \, Hz)$,which would increase the beat frequency. This contradicts the observation.
Therefore,the original frequency of $B$ must be $420 \, Hz$.

(D) The effective acceleration due to gravity $g'$ at a latitude $\phi$ is given by the formula: $g' = g - \omega^2 R \cos^2 \phi$,where $\omega$ is the angular velocity of the earth and $R$ is the radius of the earth.
At the poles,the latitude $\phi = 90^\circ$,so $\cos 90^\circ = 0$. Thus,$g' = g$,which means there is no change in gravity at the poles regardless of the change in $\omega$.
At all other latitudes,as the angular velocity $\omega$ increases,the term $\omega^2 R \cos^2 \phi$ increases.
Since $g' = g - \omega^2 R \cos^2 \phi$,an increase in $\omega$ leads to a decrease in $g'$.
Consequently,the weight of an object,$W = mg'$,will decrease at all points on the earth except at the poles.

(A) The path $BC$ is a straight line passing through points $(V_0, 3P_0)$ and $(2V_0, P_0)$.
The slope of the line $BC$ is $m = \frac{P_0 - 3P_0}{2V_0 - V_0} = \frac{-2P_0}{V_0}$.
The equation of the line $BC$ is $P - 3P_0 = \frac{-2P_0}{V_0}(V - V_0)$,which simplifies to $P = 3P_0 - \frac{2P_0}{V_0}(V - V_0) = P_0(5 - \frac{2V}{V_0})$.
Using the ideal gas law $PV = nRT$ with $n = 1$,we have $T = \frac{PV}{R} = \frac{P_0}{R}(5V - \frac{2V^2}{V_0})$.
To find the maximum temperature,we set $\frac{dT}{dV} = 0$:
$\frac{dT}{dV} = \frac{P_0}{R}(5 - \frac{4V}{V_0}) = 0 \implies V = \frac{5}{4}V_0$.
Substituting $V = \frac{5}{4}V_0$ back into the expression for $T$:
$T_{max} = \frac{P_0}{R}(5(\frac{5}{4}V_0) - \frac{2}{V_0}(\frac{25}{16}V_0^2)) = \frac{P_0}{R}(\frac{25}{4}V_0 - \frac{25}{8}V_0) = \frac{25}{8} \frac{P_0 V_0}{R}$.

(B) Given: Battery voltage $E = 6\,V$,high resistance $R = 11\,k\Omega = 11000\,\Omega$,figure of merit $k = 60\,\mu A/\text{div}$,initial deflection $\theta = 9\,\text{div}$.
$1$. Calculate the current $I$ for full deflection $\theta = 9$:
$I = k \cdot \theta = 60 \times 10^{-6} \times 9 = 540 \times 10^{-6} = 5.4 \times 10^{-4}\,A$.
$2$. Using Ohm's law for the circuit without shunt:
$I = \frac{E}{R + G} \implies 5.4 \times 10^{-4} = \frac{6}{11000 + G}$.
Since $R \gg G$,$R + G \approx R = 11000\,\Omega$.
$G = \frac{E}{I} - R = \frac{6}{5.4 \times 10^{-4}} - 11000 = 11111 - 11000 = 111.1\,\Omega$.
$3$. In the half-deflection method,the shunt resistance $S$ is connected such that the deflection becomes $\theta/2 = 4.5\,\text{div}$.
The formula for shunt resistance in the half-deflection method is $S = \frac{G \cdot R}{R - G}$.
Given $R = 11000\,\Omega$ and $G \approx 111.1\,\Omega$:
$S = \frac{111.1 \times 11000}{11000 - 111.1} \approx \frac{111.1 \times 11000}{10888.9} \approx 112.2\,\Omega$.
Rounding to the nearest given option,the value is $110\,\Omega$.

(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.
For an electron in a hydrogen atom,the orbital radius $r_n$ is proportional to $n^2$ and the velocity $v_n$ is proportional to $1/n$.
The angular momentum quantization condition is $mvr = \frac{nh}{2\pi}$,which implies $2\pi r = n\lambda$.
Thus,the de-Broglie wavelength $\lambda$ is proportional to the circumference of the orbit,$\lambda = \frac{2\pi r_n}{n}$.
Since $r_n \propto n^2$,we have $\lambda \propto \frac{n^2}{n} = n$.
For the ground state,$n_G = 1$,so $\lambda_G \propto 1$.
For the second excited state,$n_B = 3$,so $\lambda_B \propto 3$.
Therefore,$\frac{\lambda_B}{\lambda_G} = \frac{3}{1}$,which gives $\lambda_B = 3\lambda_G$.

(A) The magnetic flux $\phi$ linked with a coil of $n$ turns rotating in a magnetic field $B$ is given by $\phi = nBA \cos(\omega t).$
According to Faraday's law of electromagnetic induction,the induced $e.m.f.$ is $e = -\frac{d\phi}{dt}.$
Substituting the expression for $\phi,$ we get $e = -\frac{d}{dt}(nBA \cos(\omega t)) = nBA\omega \sin(\omega t).$
The maximum value of the induced $e.m.f.$ $(e_0)$ occurs when $\sin(\omega t) = 1.$
Therefore,$e_0 = nBA\omega.$

(A) When unpolarized light of intensity $I$ passes through the first polarizer $A$,the intensity becomes $I_A = I/2$.
Since the emergent intensity after $B$ is $I/2$,the polarizers $A$ and $B$ must be parallel (angle $0^\circ$ between them).
When a third polarizer $C$ is placed between $A$ and $B$ at an angle $\theta$ with $A$,the angle between $C$ and $B$ is $(90^\circ - \theta)$ if $A$ and $B$ were crossed,but here they are parallel,so the angle between $C$ and $B$ is $\theta$.
Using Malus' Law: $I_{final} = I_A \cos^2 \theta \cos^2 \theta = (I/2) \cos^4 \theta$.
Given $I_{final} = I/3$,we have $(I/2) \cos^4 \theta = I/3$.
$\cos^4 \theta = 2/3$.
Therefore,$\cos \theta = (2/3)^{1/4}$.

(C) Given: Primary voltage $V_{P} = 2300\,V$,Secondary voltage $V_{S} = 230\,V$,Primary current $I_{P} = 5\,A$,Efficiency $\eta = 90\% = 0.9$.
The efficiency of a transformer is defined as the ratio of output power $(P_{S})$ to input power $(P_{P})$:
$\eta = \frac{P_{S}}{P_{P}} \Rightarrow P_{S} = \eta \times P_{P}$.
Since power $P = V \times I$,we can write:
$V_{S} \times I_{S} = 0.9 \times (V_{P} \times I_{P})$.
Substituting the given values:
$230 \times I_{S} = 0.9 \times 2300 \times 5$.
Solving for $I_{S}$:
$I_{S} = \frac{0.9 \times 2300 \times 5}{230} = 0.9 \times 10 \times 5 = 45\,A$.
Therefore,the output current is $45\,A$.

(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
From this,we have $\lambda = \frac{hc}{E}$,which implies $\frac{\lambda_N}{\lambda_A} = \frac{E_A}{E_N}$.
Here,$E_N$ is the energy of the photon emitted during nuclear de-excitation,which is typically in the order of $MeV$ $(10^6 \ eV)$.
$E_A$ is the energy of the photon emitted during atomic de-excitation,which is typically in the order of a few $eV$ (e.g.,$1-10 \ eV$).
Therefore,the ratio $\frac{\lambda_N}{\lambda_A} = \frac{E_A}{E_N} \approx \frac{1 \ eV}{10^6 \ eV} = 10^{-6}$.

(A) The circuit consists of two capacitors $C_1$ and $C_2$ connected in series with a resistor $R$ and a battery of $EMF$ $E$.
When the switch $S$ is closed at $t = 0$,the capacitors begin to charge.
The equivalent capacitance of the series combination is given as $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.
The charging equation for a series $RC$ circuit is $Q(t) = Q_0(1 - e^{-t/\tau})$,where $Q_0$ is the maximum charge and $\tau = RC_{eq}$ is the time constant.
The maximum charge on the equivalent capacitor is $Q_0 = C_{eq}E$.
Since the capacitors are in series,the charge on each capacitor is the same and equal to the charge on the equivalent capacitor.
Thus,the charge on $C_1$ as a function of time is $Q(t) = C_{eq}E[1 - \exp(-t/RC_{eq})]$.

(D) Given:
Galvanometer resistance,$R_g = 25\,\Omega$
Full-scale deflection current,$I_g = 1\,mA = 10^{-3}\,A$
Maximum current to be measured,$I = 2\,A$
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
The potential difference across the galvanometer and the shunt must be equal:
$I_g R_g = (I - I_g) S$
Substituting the values:
$10^{-3} \times 25 = (2 - 10^{-3}) S$
$0.025 = (2 - 0.001) S$
$0.025 = 1.999 S$
$S = \frac{0.025}{1.999} \approx \frac{0.025}{2} = 0.0125\,\Omega$
Thus,$S = 1.25 \times 10^{-2}\,\Omega$.

(B) The voltage across the zener diode is $V_{Z} = 10 \ V$. Since the zener diode is in parallel with resistor $R_{2} = 1500 \ \Omega$,the voltage across $R_{2}$ is $V_{R_{2}} = V_{Z} = 10 \ V$.
The current through $R_{2}$ is:
$I_{R_{2}} = \frac{V_{R_{2}}}{R_{2}} = \frac{10 \ V}{1500 \ \Omega} = \frac{1}{150} \ A \approx 6.67 \times 10^{-3} \ A = 6.67 \ mA$.
The voltage across $R_{1} = 500 \ \Omega$ is:
$V_{R_{1}} = V_{source} - V_{Z} = 15 \ V - 10 \ V = 5 \ V$.
The total current through $R_{1}$ is:
$I_{R_{1}} = \frac{V_{R_{1}}}{R_{1}} = \frac{5 \ V}{500 \ \Omega} = 0.01 \ A = 10 \ mA$.
Applying Kirchhoff's Current Law at the junction,the current through the zener diode $I_{Z}$ is:
$I_{Z} = I_{R_{1}} - I_{R_{2}} = 10 \ mA - 6.67 \ mA = 3.33 \ mA \approx 3.3 \ mA$.

(D) Let the initial charge on spheres $A$ and $B$ be $q$. The force between them is given by $F = \frac{k q^2}{r^2}$.
When sphere $C$ (uncharged) is touched to $A$,the charge redistributes equally between them. Thus,the new charge on $A$ is $q_A = \frac{q + 0}{2} = \frac{q}{2}$. The charge on $C$ becomes $\frac{q}{2}$.
Next,sphere $C$ (now with charge $\frac{q}{2}$) is touched to $B$ (with charge $q$). The total charge is $\frac{q}{2} + q = \frac{3q}{2}$. This charge is shared equally,so the new charge on $B$ is $q_B = \frac{3q/2}{2} = \frac{3q}{4}$.
The new force between $A$ and $B$ is $F' = \frac{k q_A q_B}{r^2} = \frac{k (q/2) (3q/4)}{r^2} = \frac{3}{8} \frac{k q^2}{r^2} = \frac{3}{8} F$.

(C) Given: Angle of prism,$A = 30^o$,Angle of incidence,$i = 60^o$,Angle of deviation,$\delta = 30^o$.
Using the prism formula for deviation: $\delta = i + e - A$.
Substituting the values: $30^o = 60^o + e - 30^o$.
Solving for the angle of emergence $e$: $e = 30^o + 30^o - 60^o = 0^o$.
Since the angle of emergence $e = 0^o$,the emergent ray is normal (perpendicular) to the second face of the prism.
Therefore,the angle made by the emergent ray with the second face of the prism is $90^o$.

(A) Given: Modulation index $m = 80\% = 0.8$.
Carrier wave peak voltage $E_c = 14\,V$.
We know that the modulation index $m$ is defined as the ratio of the peak voltage of the modulating signal $(E_m)$ to the peak voltage of the carrier wave $(E_c)$.
Formula: $m = \frac{E_m}{E_c}$.
Rearranging the formula to find $E_m$: $E_m = m \times E_c$.
Substituting the values: $E_m = 0.8 \times 14\,V = 11.2\,V$.
Therefore,the peak voltage of the modulating signal is $11.2\,V$.

(C) The intensity $I$ of an electromagnetic wave is related to the amplitude of the electric field $E_0$ by the formula $I = \frac{1}{2} \varepsilon_0 c E_0^2$.
Solving for $E_0$,we get $E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}}$.
The magnitude of the magnetic field is $B_0 = \frac{E_0}{c}$.
The wave propagates in the direction of $\vec E \times \vec B$. Given the propagation is along the $+Y$ direction (unit vector $\hat j$),we check the cross product for option $C$: $\hat k \times \hat i = \hat j$. This matches the propagation direction. Thus,the correct expression is $\vec E = E_0 \cos[\frac{2\pi}{\lambda}(y - ct)] \hat k$ and $\vec B = \frac{E_0}{c} \cos[\frac{2\pi}{\lambda}(y - ct)] \hat i$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.
Given that the de Broglie wavelengths are equal,$\lambda_p = \lambda_{\alpha}$.
Therefore,$\frac{h}{m_p v_p} = \frac{h}{m_{\alpha} v_{\alpha}}$.
This implies $m_p v_p = m_{\alpha} v_{\alpha}$.
We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton,so $m_{\alpha} = 4m_p$.
Substituting this into the equation: $m_p v_p = (4m_p) v_{\alpha}$.
Dividing both sides by $m_p$,we get $v_p = 4v_{\alpha}$.
Thus,the ratio of their velocities is $\frac{v_p}{v_{\alpha}} = 4:1$.