GUJCET 2022 Physics Question Paper with Answer and Solution

(C) The power rating of the bulb is $P = 200 \text{ W}$ and the voltage supply is $V = 220 \text{ V}$.
Using the power formula $P = \frac{V^2}{R}$,we can rearrange it to solve for resistance $R$:
$R = \frac{V^2}{P}$
Substituting the given values:
$R = \frac{(220)^2}{200} = \frac{48400}{200} = 242 \ \Omega$.
Therefore,the resistance of the bulb is $242 \ \Omega$.

(D) The angular frequency of free oscillations for an $LC$ circuit is given by the formula $\omega_{0} = \frac{1}{\sqrt{LC}}$.
Given values are $L = 16 \text{ mH} = 16 \times 10^{-3} \text{ H}$ and $C = 10 \mu F = 10 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$\omega_{0} = \frac{1}{\sqrt{16 \times 10^{-3} \times 10 \times 10^{-6}}}$
$\omega_{0} = \frac{1}{\sqrt{16 \times 10^{-8}}}$
$\omega_{0} = \frac{1}{4 \times 10^{-4}}$
$\omega_{0} = \frac{10^{4}}{4} = 2500 \text{ rad s}^{-1}$.

(B) For an ideal transformer,the relationship between the voltage and the number of turns in the primary and secondary coils is given by the transformer equation: $\frac{V_S}{V_P} = \frac{N_S}{N_P}$.
Given that $N_S > N_P$,it follows that $\frac{N_S}{N_P} > 1$.
Therefore,$\frac{V_S}{V_P} > 1$,which implies $V_S > V_P$.
This type of transformer is known as a step-up transformer.

(D) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{n^2} \right]$,where $n = 4, 5, 6, \dots$
To find the shortest wavelength,we consider the limit where $n = \infty$.
Substituting $n = \infty$ into the formula:
$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{9} - 0 \right] = \frac{R}{9}$.
Thus,$\lambda_{\min} = \frac{9}{R}$.
Given the Rydberg constant $R \approx 1.097 \times 10^7 \ m^{-1}$:
$\lambda_{\min} = \frac{9}{1.097 \times 10^7} \approx 8.204 \times 10^{-7} \ m$.
Converting to nanometers:
$\lambda_{\min} \approx 820.4 \ nm \approx 820 \ nm$.

(B) The energy difference between the two levels is given by $\Delta E = 5.4 \ eV$.
Converting this energy into Joules: $\Delta E = 5.4 \times 1.6 \times 10^{-19} \ J = 8.64 \times 10^{-19} \ J$.
The energy of the emitted photon is related to the frequency $\nu$ by the equation $\Delta E = h\nu$.
Therefore,the frequency $\nu = \frac{\Delta E}{h}$.
Substituting the values: $\nu = \frac{8.64 \times 10^{-19}}{6.625 \times 10^{-34}} \ Hz$.
Calculating the result: $\nu \approx 1.304 \times 10^{15} \ Hz$.

(B) The relationship between current $I$ and drift velocity $v_d$ is given by $I = n A v_d e$,where $n$ is the number density,$A$ is the cross-sectional area,and $e$ is the elementary charge.
Since $v_d = \frac{l}{t}$,where $l$ is the length of the wire and $t$ is the time taken,we can write:
$I = n A \left( \frac{l}{t} \right) e$
Rearranging for $t$:
$t = \frac{n A l e}{I}$
Given values:
$n = 8.5 \times 10^{28} \ m^{-3}$
$A = 1.0 \times 10^{-6} \ m^2$
$l = 6 \ m$
$e = 1.6 \times 10^{-19} \ C$
$I = 1.5 \ A$
Substituting these values:
$t = \frac{8.5 \times 10^{28} \times 1.0 \times 10^{-6} \times 6 \times 1.6 \times 10^{-19}}{1.5}$
$t = \frac{81.6 \times 10^3}{1.5} = 54.4 \times 10^3 \ s$
$t = 5.44 \times 10^4 \ s$
Thus,the correct option is $B$.

(C) The power dissipated in the circuit is given by the formula $P = \frac{V^2}{R_{eq}}$.
Given $P = 150 \text{ W}$ and $V = 15 \text{ V}$.
Substituting the values:
$150 = \frac{(15)^2}{R_{eq}}$
$150 = \frac{225}{R_{eq}}$
$R_{eq} = \frac{225}{150} = 1.5 \Omega$.
The two resistors $R$ and $2 \Omega$ are connected in parallel.
Therefore,$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{2}$.
$\frac{1}{1.5} = \frac{1}{R} + \frac{1}{2}$.
$\frac{1}{R} = \frac{1}{1.5} - \frac{1}{2} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$.
Thus,$R = 6 \Omega$.

(A) The relationship between resistance and temperature is given by the formula: $R_\theta = R_0 [1 + \alpha (\theta - \theta_0)]$
Given values are: $R_0 = 100 \ \Omega$,$\theta_0 = 27^\circ \text{C}$,$R_\theta = 137 \ \Omega$,and $\alpha = 1.35 \times 10^{-4} \ ^\circ \text{C}^{-1}$.
Substituting these values into the equation:
$137 = 100 [1 + 1.35 \times 10^{-4} (\theta - 27)]$
$1.37 = 1 + 1.35 \times 10^{-4} (\theta - 27)$
$0.37 = 1.35 \times 10^{-4} (\theta - 27)$
$\theta - 27 = \frac{0.37}{1.35 \times 10^{-4}}$
$\theta - 27 \approx 2740.74$
$\theta \approx 2740.74 + 27 = 2767.74^\circ \text{C}$
Rounding to the nearest integer,we get $\theta \approx 2767^\circ \text{C}$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$.
Given values are $h = 6.63 \times 10^{-34} \ J \cdot s$,$m_{e} = 9.11 \times 10^{-31} \ kg$,and $v = 6.4 \times 10^{6} \ m/s$.
Substituting these values into the formula:
$\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 6.4 \times 10^{6}}$
$\lambda = \frac{6.63 \times 10^{-34}}{58.304 \times 10^{-25}}$
$\lambda \approx 0.1137 \times 10^{-9} \ m$
Since $1 \ nm = 10^{-9} \ m$,we get $\lambda \approx 0.114 \ nm$.

(A) Given:
$E_1 = 9 \times 10^4 \text{ NC}^{-1}$
$r_1 = 2 \text{ cm}$
$r_2 = 3 \text{ cm}$
The electric field $E$ due to an infinite line charge is given by the formula:
$E = \frac{\lambda}{2 \pi \varepsilon_0 r}$
From this relation,we can see that $E \propto \frac{1}{r}$.
Therefore,the ratio of the fields at two different distances is:
$\frac{E_2}{E_1} = \frac{r_1}{r_2}$
Substituting the given values:
$\frac{E_2}{9 \times 10^4} = \frac{2 \text{ cm}}{3 \text{ cm}}$
Solving for $E_2$:
$E_2 = \frac{9 \times 10^4 \times 2}{3}$
$E_2 = 3 \times 10^4 \times 2$
$E_2 = 6 \times 10^4 \text{ NC}^{-1}$
Thus,the electric field at a distance of $3 \text{ cm}$ is $6 \times 10^4 \text{ NC}^{-1}$.

(B) The torque $\tau$ acting on an electric dipole in a uniform electric field is given by the formula: $\tau = pE \sin \theta$.
Given values are:
Dipole moment $p = 4 \times 10^{-9} \text{ Cm}$
Electric field $E = 5 \times 10^4 \text{ NC}^{-1}$
Angle $\theta = 60^{\circ}$
Substituting these values into the formula:
$\tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 60^{\circ}$
$\tau = 20 \times 10^{-5} \times \frac{\sqrt{3}}{2}$
$\tau = 10 \times 10^{-5} \times 1.732$
$\tau = 1.732 \times 10^{-4} \text{ Nm}$
Thus,the magnitude of the torque is $1.73 \times 10^{-4} \text{ Nm}$.

(A) The magnetic field inside an air-cored solenoid is given by $B = \frac{\mu_0 NI}{l}$.
Initial magnetic flux linkage $\phi_1 = N B A = N \left( \frac{\mu_0 NI}{l} \right) A = \frac{\mu_0 N^2 AI}{l}$.
Given values: $N = 500$,$I = 2.5 \ A$,$l = 0.3 \ m$,$A = 25 \times 10^{-4} \ m^2$,$\Delta t = 10^{-3} \ s$,and $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
When the current is switched off,the final flux $\phi_2 = 0$.
The magnitude of induced emf is $|\varepsilon| = \frac{|\Delta \phi|}{\Delta t} = \frac{|\phi_2 - \phi_1|}{\Delta t} = \frac{\mu_0 N^2 AI}{l \Delta t}$.
Substituting the values:
$|\varepsilon| = \frac{4\pi \times 10^{-7} \times (500)^2 \times 25 \times 10^{-4} \times 2.5}{0.3 \times 10^{-3}}$.
$|\varepsilon| = \frac{4 \times 3.1416 \times 10^{-7} \times 250000 \times 25 \times 10^{-4} \times 2.5}{3 \times 10^{-4}}$.
$|\varepsilon| = \frac{1.9635 \times 10^{-3}}{3 \times 10^{-4}} \approx 6.54 \ V$.

(A) The magnetic flux linkage $\phi$ in the second coil is related to the current $I$ in the first coil by the relation $\phi = M I$,where $M$ is the mutual inductance.
For a change in current $\Delta I$,the change in flux linkage $\Delta \phi$ is given by:
$\Delta \phi = M \Delta I$
Given:
$M = 1.5 \ H$
$\Delta I = I_f - I_i = 10 \ A - 0 \ A = 10 \ A$
Substituting the values:
$\Delta \phi = 1.5 \ H \times 10 \ A$
$\Delta \phi = 15 \ Wb$
Therefore,the change in flux linkage is $15 \ Wb$.

(D) The correct answer is $D$.
According to Lenz's law, the direction of the induced current in a circuit is always such that it opposes the change in magnetic flux that produces it. This law is a direct consequence of the law of conservation of energy.

(A) The relationship between speed of light $(c)$, frequency $(\nu)$, and wavelength $(\lambda)$ is given by $c = \nu \lambda$.
Therefore, the wavelength is $\lambda = \frac{c}{\nu}$.
For the lower frequency limit $\nu_1 = 6 \text{ MHz} = 6 \times 10^6 \text{ Hz}$, the wavelength is:
$\lambda_1 = \frac{3 \times 10^8}{6 \times 10^6} = \frac{300}{6} = 50 \text{ m}$.
For the upper frequency limit $\nu_2 = 12 \text{ MHz} = 12 \times 10^6 \text{ Hz}$, the wavelength is:
$\lambda_2 = \frac{3 \times 10^8}{12 \times 10^6} = \frac{300}{12} = 25 \text{ m}$.
Thus, the corresponding wavelength band is $25 \text{ m}$ to $50 \text{ m}$.

(A) According to the theory of electromagnetic waves, an oscillating charge is a source of electromagnetic radiation.
When a charged particle oscillates with a frequency $f$, it produces electromagnetic waves of the same frequency $f$.
Given that the frequency of the oscillating particle is $10^{9} \,Hz$, the frequency of the electromagnetic waves produced will also be $10^{9} \,Hz$.
Therefore, the correct option is $A$.

(B) The dipole moment of a single molecule is $p_0 = 10^{-29} \text{ C m}$.
Since $1 \text{ mole}$ contains $6 \times 10^{23}$ molecules,the total number of molecules in $2 \text{ moles}$ is $N = 2 \times 6 \times 10^{23} = 12 \times 10^{23}$ molecules.
The total dipole moment $P$ of the substance when all dipoles are aligned with the field is $P = N \times p_0 = 12 \times 10^{23} \times 10^{-29} = 12 \times 10^{-6} \text{ C m}$.
The potential energy $U$ of a dipole in an external electric field $E$ is given by $U = -P E \cos \theta$. For maximum polarization at low temperature,the dipoles align with the field,so $\theta = 0^{\circ}$.
Substituting the values: $U = -(12 \times 10^{-6} \text{ C m}) \times (10^6 \text{ V m}^{-1}) \times \cos(0^{\circ}) = -12 \times 1 \times 1 = -12 \text{ J}$.

(C) In the absence of a dielectric,the electric field between the two plates is $E_0 = \frac{V_0}{d}$.
When the dielectric slab is inserted,the electric field inside the dielectric becomes $E = \frac{E_0}{K}$,where $K=3$ is the dielectric constant.
The thickness of the dielectric slab is $t = \frac{3}{4}d$,and the thickness of the air gap is $d - t = d - \frac{3}{4}d = \frac{1}{4}d$.
The new potential difference $V$ is the sum of the potential drops across the air gap and the dielectric slab:
$V = E_0 \left(\frac{1}{4}d\right) + E \left(\frac{3}{4}d\right)$
$V = E_0 \left(\frac{1}{4}d\right) + \frac{E_0}{K} \left(\frac{3}{4}d\right)$
$V = E_0 d \left[ \frac{1}{4} + \frac{3}{4K} \right]$
Since $E_0 d = V_0$ and $K = 3$:
$V = V_0 \left[ \frac{1}{4} + \frac{3}{4 \times 3} \right]$
$V = V_0 \left[ \frac{1}{4} + \frac{1}{4} \right]$
$V = V_0 \left( \frac{2}{4} \right) = \frac{V_0}{2}$

(A) The correct configuration is shown in the figure,where two capacitors ($C_2$ and $C_3$) are connected in parallel,and this combination is connected in series with two other capacitors ($C_1$ and $C_4$).
Given: $C_1 = C_2 = C_3 = C_4 = 4 \mu F$.
First,calculate the equivalent capacitance of the parallel combination $(C_p)$:
$C_p = C_2 + C_3 = 4 \mu F + 4 \mu F = 8 \mu F$.
Now,the total equivalent capacitance $(C_{eq})$ is the series combination of $C_1$,$C_p$,and $C_4$:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_p} + \frac{1}{C_4} = \frac{1}{4} + \frac{1}{8} + \frac{1}{4}$.
$\frac{1}{C_{eq}} = \frac{2 + 1 + 2}{8} = \frac{5}{8} \mu F^{-1}$.
$C_{eq} = \frac{8}{5} \mu F = 1.6 \mu F$.

(C) The magnetic field inside a solenoid with a magnetic core is given by the formula:
$B = \mu_0 \mu_r n I$
Where:
$\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$
$\mu_r = 400$
$n = 1000 \text{ turns/m}$
$I = 1 \text{ A}$
Substituting the values:
$B = (4\pi \times 10^{-7}) \times 400 \times 1000 \times 1$
$B = 4\pi \times 4 \times 10^5 \times 10^{-7}$
$B = 16\pi \times 10^{-2} \text{ T}$

(B) The force per unit length between two parallel current-carrying wires is given by the formula: $F = \frac{\mu_0 I_1 I_2 l}{2 \pi d}$.
Given:
$I_1 = 10 \ A$
$I_2 = 4 \ A$
$d = 2 \ cm = 2 \times 10^{-2} \ m$
$l = 4 \ cm = 4 \times 10^{-2} \ m$
$\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
Substituting the values into the formula:
$F = \frac{(4 \pi \times 10^{-7}) \times 10 \times 4 \times (4 \times 10^{-2})}{2 \pi \times (2 \times 10^{-2})}$
$F = \frac{2 \times 10^{-7} \times 40 \times 4 \times 10^{-2}}{2 \times 10^{-2}}$
$F = 2 \times 10^{-7} \times 40 \times 2$
$F = 160 \times 10^{-7} \ N = 1.6 \times 10^{-5} \ N$.
Thus,the correct option is $B$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$, where $n = N/l$ is the number of turns per unit length.
Given:
Length $l = 0.25 \ m$
Number of turns $N = 500$
Current $I = 2.5 \ A$
Permeability $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
Calculation:
$n = \frac{500}{0.25} = 2000 \ turns/m$
$B = (4 \pi \times 10^{-7}) \times 2000 \times 2.5$
$B = 4 \times 3.14 \times 10^{-7} \times 5000$
$B = 12.56 \times 10^{-4} \ T = 6.28 \times 2 \times 10^{-4} \ T = 6.28 \times 10^{-3} \ T$.

(A) The focal length of a convex lens is $f_1 = +30 \ cm$.
The focal length of a concave lens is $f_2 = -10 \ cm$.
When two thin lenses are placed in contact,the equivalent focal length $f$ is given by the formula:
$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$
Substituting the given values:
$\frac{1}{f} = \frac{1}{30} + \frac{1}{-10}$
$\frac{1}{f} = \frac{1 - 3}{30}$
$\frac{1}{f} = \frac{-2}{30}$
$\frac{1}{f} = \frac{-1}{15}$
Therefore,$f = -15 \ cm$.

(C) Given:
Focal length $f = +20 \ cm$
Refractive index $n = 1.55$
For a double convex lens with equal radii of curvature,$R_1 = +R$ and $R_2 = -R$.
Using the lens maker's formula:
$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Substituting the values:
$\frac{1}{20} = (1.55 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$
$\frac{1}{20} = 0.55 \times \left( \frac{2}{R} \right)$
$\frac{1}{20} = \frac{1.1}{R}$
$R = 20 \times 1.1 = 22 \ cm$
Therefore,the required radius of curvature is $22 \ cm$.

(D) Given: Refractive index of air $n_1 = 1$,refractive index of glass $n_2 = 1.5$,radius of curvature $R = +20 \ cm$,and object distance $u = -100 \ cm$.
Using the formula for refraction at a spherical surface:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Substituting the values:
$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$
$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}$
$\frac{1.5}{v} = \frac{0.5}{20} - \frac{1}{100}$
$\frac{1.5}{v} = \frac{2.5 - 1}{100}$
$\frac{1.5}{v} = \frac{1.5}{100}$
$v = +100 \ cm$
Thus,the image is formed at a distance of $100 \ cm$ from the surface.