GUJCET 2018 Physics Question Paper with Answer and Solution

(A) The speed of light in vacuum is given by the relation $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
Squaring both sides,we get $c^{2} = \frac{1}{\mu_{0} \varepsilon_{0}}$.
Therefore,$\mu_{0} \varepsilon_{0} = \frac{1}{c^{2}}$.
The dimensional formula for the speed of light $c$ is $[M^{0} L^{1} T^{-1}]$.
Substituting this into the expression,we get $\mu_{0} \varepsilon_{0} = \frac{1}{(M^{0} L^{1} T^{-1})^{2}}$.
$\mu_{0} \varepsilon_{0} = \frac{1}{M^{0} L^{2} T^{-2}} = M^{0} L^{-2} T^{2}$.

(D) The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2 \pi \nu C}$,where $\nu$ is the frequency of the source.
For a $DC$ source,the frequency $\nu = 0 \ Hz$.
Substituting this value into the formula,we get $X_C = \frac{1}{2 \pi (0) C} = \frac{1}{0}$.
Therefore,the reactance of the capacitor in a $DC$ circuit is infinite.

(B) The power consumed in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\phi$ is the phase difference between voltage and current.
Maximum power $P_{max}$ occurs when $\cos \phi = 1$,so $P_{max} = V_{rms} I_{rms}$.
Given that the lamp consumes $50 \%$ of the maximum power:
$P = 0.5 \times P_{max}$
$V_{rms} I_{rms} \cos \phi = 0.5 \times V_{rms} I_{rms}$
$\cos \phi = 0.5 = \frac{1}{2}$
$\phi = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} \ rad$.

(D) Mobility $\mu$ is defined as the ratio of drift velocity $v_d$ to the applied electric field $E$: $\mu = \frac{v_d}{E}$.
The dimensional formula for drift velocity $v_d$ is $[L^1 T^{-1}]$.
The dimensional formula for electric field $E$ is $[M^1 L^1 T^{-3} A^{-1}]$.
Substituting these into the formula: $\mu = \frac{[L^1 T^{-1}]}{[M^1 L^1 T^{-3} A^{-1}]}$.
Simplifying the expression: $\mu = [M^{-1} L^{1-1} T^{-1 - (-3)} A^1] = [M^{-1} L^0 T^2 A^1]$.

(C) To obtain the minimum resistance,the resistors must be connected in parallel.
$R_{\min} = \frac{R}{n} = \frac{2}{10} = 0.2 \ \Omega$
To obtain the maximum resistance,the resistors must be connected in series.
$R_{\max} = n \times R = 10 \times 2 = 20 \ \Omega$
The ratio of maximum to minimum resistance is:
$\frac{R_{\max}}{R_{\min}} = \frac{20}{0.2} = 100$

(A) The ratio of conductivity $(\sigma)$ to resistivity $(\rho)$ is given by $\frac{\sigma}{\rho}$.
Since $\sigma = \frac{1}{\rho}$, we can write the ratio as $\frac{1/\rho}{\rho} = \frac{1}{\rho^2}$.
For a conductor, as the temperature increases, the resistivity $(\rho)$ increases due to increased collisions of electrons with the lattice ions.
Since $\rho$ is in the denominator and is squared, as $\rho$ increases, the value of $\frac{1}{\rho^2}$ decreases.
Therefore, the ratio of conductivity to resistivity decreases.

(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Since $h$ and $c$ are constants,the energy $E$ is inversely proportional to the wavelength $\lambda$,i.e.,$E \propto \frac{1}{\lambda}$.
Therefore,we can write the ratio as $\frac{E_{2}}{E_{1}} = \frac{\lambda_{1}}{\lambda_{2}}$.
Given: $E_{1} = 3.2 \times 10^{-19} \ J$,$\lambda_{1} = 6000 \mathring{A}$,and $\lambda_{2} = 4000 \mathring{A}$.
Substituting the values: $\frac{E_{2}}{3.2 \times 10^{-19}} = \frac{6000}{4000} = \frac{6}{4} = 1.5$.
$E_{2} = 1.5 \times 3.2 \times 10^{-19} \ J$.
$E_{2} = 4.80 \times 10^{-19} \ J$.

(A) Given:
Energy $E = 35 \text{ keV} = 35 \times 10^{3} \times 1.6 \times 10^{-19} \text{ J} = 56 \times 10^{-16} \text{ J}$.
Using the formula for the wavelength of a photon:
$E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$
Substituting the values:
$\lambda = \frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{35 \times 10^{3} \times 1.6 \times 10^{-19}}$
$\lambda = \frac{19.875 \times 10^{-26}}{56 \times 10^{-16}}$
$\lambda \approx 0.3549 \times 10^{-10} \text{ m} \approx 0.355 \times 10^{-10} \text{ m} = 35.5 \times 10^{-12} \text{ m}$.
Rounding to the nearest provided option,the correct value is $35 \times 10^{-12} \text{ m}$.

(D) The electric force between two point charges in vacuum is given by $F_{\text{air}} = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2} = 16 \ N$.
When the same charges are placed in a medium with dielectric constant $K$,the force is given by $F_{\text{medium}} = \frac{F_{\text{air}}}{K}$.
Given $F_{\text{air}} = 16 \ N$ and $K = 8$.
Therefore,$F_{\text{medium}} = \frac{16}{8} = 2 \ N$.
Thus,the correct option is $D$.

(C) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.
Initially,the charge enclosed is $q = 10 \mu C$,so $\phi = \frac{q}{\varepsilon_0} \quad ... (1)$.
When another $10 \mu C$ charge is placed inside,the total charge enclosed becomes $Q = 10 \mu C + 10 \mu C = 20 \mu C = 2q$.
The new flux $\phi'$ passing through the surface is $\phi' = \frac{Q}{\varepsilon_0} = \frac{2q}{\varepsilon_0}$.
Substituting from equation $(1)$,we get $\phi' = 2 \phi$.

(A) Let the side length of the square be $r$. The charges at vertices $1, 2,$ and $3$ are identical,so $q_1 = q_2 = q_3 = q$.
The distance between charges $q_1$ and $q_2$ is the side of the square,$r_{12} = r$.
According to Coulomb's Law,the force $F_{12}$ is:
$F_{12} = \frac{k q_1 q_2}{r_{12}^2} = \frac{k q^2}{r^2} \quad \dots (1)$
The distance between charges $q_1$ and $q_3$ is the diagonal of the square,$r_{13} = \sqrt{r^2 + r^2} = r\sqrt{2}$.
According to Coulomb's Law,the force $F_{13}$ is:
$F_{13} = \frac{k q_1 q_3}{r_{13}^2} = \frac{k q^2}{(r\sqrt{2})^2} = \frac{k q^2}{2r^2} \quad \dots (2)$
Taking the ratio of equation $(2)$ to equation $(1)$:
$\frac{F_{13}}{F_{12}} = \frac{\frac{k q^2}{2r^2}}{\frac{k q^2}{r^2}} = \frac{1}{2}$
Thus,the correct option is $A$.

(D) The equivalent inductance of two inductors $L$ connected in parallel is given by:
$L_p = \frac{L \times L}{L + L} = \frac{L^2}{2L} = \frac{L}{2}$
This parallel combination is connected in series with an inductor of $5 \text{ mH}$. The total effective inductance $L_{eq}$ is:
$L_{eq} = L_p + 5 \text{ mH}$
Given $L_{eq} = 15 \text{ mH}$,we have:
$15 = \frac{L}{2} + 5$
$10 = \frac{L}{2}$
$L = 20 \text{ mH}$
Therefore,the correct option is $D$.

(B) According to Faraday's law of electromagnetic induction,the magnitude of induced electromotive force (emf) is given by:
$|\varepsilon| = N \frac{\Delta \phi}{\Delta t}$
Since the magnetic field is perpendicular to the area,the magnetic flux $\phi = B \cdot A$. Thus,the change in flux is $\Delta \phi = A(B_2 - B_1)$.
Substituting the given values:
$N = 200$,$A = 0.15 \ m^2$,$B_1 = 0.2 \ T$,$B_2 = 0.6 \ T$,$\Delta t = 0.4 \ s$
$|\varepsilon| = \frac{N \cdot A \cdot (B_2 - B_1)}{\Delta t}$
$|\varepsilon| = \frac{200 \times 0.15 \times (0.6 - 0.2)}{0.4}$
$|\varepsilon| = \frac{200 \times 0.15 \times 0.4}{0.4}$
$|\varepsilon| = 200 \times 0.15 = 30 \ V$
Therefore,the induced emf is $30 \ V$.

(B) The correct order is $v_r < v_v < v_{uv}$.
In the electromagnetic spectrum,waves are arranged in increasing order of frequency as follows:
Radio Waves $\rightarrow$ Microwaves $\rightarrow$ Infrared rays $\rightarrow$ Visible light $\rightarrow$ Ultraviolet rays $\rightarrow$ $X$-rays $\rightarrow$ Gamma rays.
Since Radio waves have the lowest frequency and Ultraviolet waves have a higher frequency than Visible light,the correct relationship is $v_r < v_v < v_{uv}$.

(B) The arrangement consists of three parallel plate capacitors connected in parallel.
Let the area of each plate be $A = l^2$.
The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
From the figure,we identify three capacitors:
$C_1$ formed by plates $1$ and $2$ with separation $3d$: $C_1 = \frac{\varepsilon_0 l^2}{3d}$.
$C_2$ formed by plates $3$ and $4$ with separation $d$: $C_2 = \frac{\varepsilon_0 l^2}{d}$.
$C_3$ formed by plates $5$ and $6$ with separation $3d$: $C_3 = \frac{\varepsilon_0 l^2}{3d}$.
Since these capacitors are connected in parallel,the equivalent capacitance $C_{eq}$ is:
$C_{eq} = C_1 + C_2 + C_3$
$C_{eq} = \frac{\varepsilon_0 l^2}{3d} + \frac{\varepsilon_0 l^2}{d} + \frac{\varepsilon_0 l^2}{3d}$
$C_{eq} = \frac{\varepsilon_0 l^2}{d} (\frac{1}{3} + 1 + \frac{1}{3}) = \frac{\varepsilon_0 l^2}{d} (\frac{1+3+1}{3}) = \frac{5 \varepsilon_0 l^2}{3d}$.
Thus,the correct option is $(B)$.

(D) The electric potential $V$ due to an electric dipole at a point $(r, \theta)$ is given by the formula:
$V(r, \theta) = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos \theta}{r^2}$
$1$. On the axis of the dipole,the angle $\theta$ is either $0$ or $\pi$ radians. Substituting this into the formula:
$V = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos(0)}{r^2} = \frac{p}{4 \pi \varepsilon_0 r^2}$ or $V = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos(\pi)}{r^2} = -\frac{p}{4 \pi \varepsilon_0 r^2}$
Thus,on the axis,$V \neq 0$.
$2$. On the equatorial plane of the dipole,the angle $\theta = \frac{\pi}{2}$ radians. Substituting this into the formula:
$V = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos(\pi/2)}{r^2} = 0$ (since $\cos(\pi/2) = 0$)
Thus,on the equator,$V = 0$.
Therefore,the correct option is $D$.

(A) The induced dipole moment $\vec{p}$ is proportional to the applied electric field $\vec{E_0}$,given by the relation $\vec{p} = \alpha \vec{E_0}$,where $\alpha$ is the polarizability of the molecule.
From this,we can express polarizability as $\alpha = \frac{\vec{p}}{\vec{E_0}}$.
The unit of dipole moment $\vec{p}$ is Coulomb-meter $(C \cdot m)$.
The unit of electric field $\vec{E_0}$ is Newton per Coulomb ($N/C$ or $N \cdot C^{-1}$).
Therefore,the unit of $\alpha = \frac{C \cdot m}{N \cdot C^{-1}} = \frac{C^2 \cdot m}{N} = C^2 \cdot m \cdot N^{-1}$.

(A) In the context of the Earth's magnetic field,$B_E$ represents the total magnetic field intensity,$H_E$ represents the horizontal component,and $Z_E$ represents the vertical component.
According to the vector resolution of the Earth's magnetic field,the total magnetic field $B_E$ is the vector sum of its horizontal and vertical components.
Since these components are perpendicular to each other,the magnitude is given by the Pythagorean theorem: $B_E = \sqrt{Z_E^2 + H_E^2}$.

(C) substance that experiences a weak force towards the region of a stronger magnetic field when placed in a non-uniform magnetic field is classified as a $Paramagnetic$ substance.
$1$. $Diamagnetic$ substances are weakly repelled by a magnetic field and move towards the weaker field region.
$2$. $Paramagnetic$ substances are weakly attracted by a magnetic field and move towards the stronger field region.
$3$. $Ferromagnetic$ substances are strongly attracted by a magnetic field.
Therefore,the correct option is $C$.

(D) For a long straight wire of radius $r$ carrying a uniform current $I$,the magnetic field $B$ at a distance $a$ from the axis inside the wire $(a < r)$ is given by Ampere's Law.
Using Ampere's circuital law: $\oint B \cdot dl = \mu_0 I_{enclosed}$.
For a point inside the wire,the enclosed current $I_{enclosed} = I \times (\frac{\pi a^2}{\pi r^2}) = I \frac{a^2}{r^2}$.
Thus,$B(2 \pi a) = \mu_0 I \frac{a^2}{r^2}$.
Solving for $B$,we get $B = \frac{\mu_0 I a}{2 \pi r^2}$.
Since $\mu_0$,$I$,and $r$ are constants,$B \propto a$.

(B) The force per unit length between two parallel current-carrying wires is given by the formula: $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Given: $I_1 = I_2 = 5 \text{ A}$,$d = 1 \text{ m}$,and $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Substituting the values: $\frac{F}{l} = \frac{4\pi \times 10^{-7} \times 5 \times 5}{2 \pi \times 1}$.
Simplifying the expression: $\frac{F}{l} = 2 \times 10^{-7} \times 25 = 50 \times 10^{-7} \text{ N/m}$.
Thus,$\frac{F}{l} = 5 \times 10^{-6} \text{ N/m}$.
Since the currents are in the same direction,the force between the wires is attractive.

(A) To convert a galvanometer into a voltmeter,a high resistance $R_s$ must be connected in series with the galvanometer.
The formula for the series resistance is given by $R_s = \frac{V}{I_G} - G$,where $V$ is the required voltage range,$I_G$ is the full-scale deflection current,and $G$ is the resistance of the galvanometer.
Given values: $V = 100 \ V$,$I_G = 10 \ mA = 10 \times 10^{-3} \ A$,and $G = 50 \ \Omega$.
Substituting these values into the formula:
$R_s = \frac{100}{10 \times 10^{-3}} - 50$
$R_s = 10000 - 50$
$R_s = 9950 \ \Omega$.

(C) The radius $r$ of the circular path of a charged particle in a magnetic field is given by the formula: $r = \frac{mv}{Bq}$.
Rearranging the formula to solve for the magnetic field intensity $B$,we get: $B = \frac{mv}{qr}$.
Given values are:
Mass $m = 9.1 \times 10^{-31} \ kg$
Velocity $v = 10^6 \ ms^{-1}$
Charge $q = 1.6 \times 10^{-19} \ C$
Radius $r = 0.2 \ m$
Substituting these values into the equation:
$B = \frac{9.1 \times 10^{-31} \times 10^6}{1.6 \times 10^{-19} \times 0.2}$
$B = \frac{9.1 \times 10^{-25}}{0.32 \times 10^{-19}}$
$B = 28.4375 \times 10^{-6} \ T = 2.84 \times 10^{-5} \ T$.
Thus,the intensity of the magnetic field is $2.84 \times 10^{-5} \ T$.

(A) In a nuclear fission reaction,such as the fission of $U^{235}$ by thermal neutrons,the neutrons released are known as fast neutrons.
These neutrons possess high kinetic energy,which is typically around $2 \text{ MeV}$.
These fast neutrons must be slowed down to thermal energies (approximately $0.025 \text{ eV}$) by a moderator to sustain the chain reaction.
Therefore,the correct option is $A$.

(B) Using the lens maker's formula:
$\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
For a plano-convex lens,one surface is plane $(R_1 = \infty)$ and the other is convex ($R_2 = -60 \ cm$ according to sign convention).
Given $n = 1.5$.
Substituting the values:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{-60} \right)$
$\frac{1}{f} = 0.5 \times \left( 0 + \frac{1}{60} \right)$
$\frac{1}{f} = \frac{0.5}{60} = \frac{1}{120}$
Therefore,$f = 120 \ cm$.

(B) For a small angled prism,the angle of deviation $\delta$ is given by the formula: $\delta = A(n - 1)$,where $A$ is the angle of the prism and $n$ is the refractive index of the material of the prism.
Given:
Refractive index $n = 1.6$
Angle of deviation $\delta = 3.6^{\circ}$
Substituting the values in the formula:
$3.6^{\circ} = A(1.6 - 1)$
$3.6^{\circ} = A(0.6)$
$A = \frac{3.6^{\circ}}{0.6}$
$A = 6^{\circ}$
Therefore,the angle of the prism is $6^{\circ}$.

(D) When two thin lenses of focal lengths $f_{1}$ and $f_{2}$ are placed in contact,the equivalent focal length $f$ of the combination is given by the formula: $\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$.
Since the power of a lens is defined as $P = \frac{1}{f}$ (where $f$ is in meters),the power of the combination $P$ is the sum of the individual powers: $P = P_{1} + P_{2}$.
Substituting the expressions for power,we get: $P = \frac{1}{f_{1}} + \frac{1}{f_{2}}$.
Taking the common denominator,we obtain: $P = \frac{f_{1} + f_{2}}{f_{1} f_{2}}$.
Thus,the correct option is $D$.

(B) In an ideal Young's Double Slit Experiment $(YDSE)$,the intensity of all bright fringes is considered equal. However,in real-world interference patterns,the intensity of bright fringes decreases as we move away from the central maximum due to the diffraction effect caused by the finite width of the slits. Therefore,the statement that 'all bright fringes are equally bright' is technically incorrect in a practical physical context compared to the other fundamental properties of interference.