GUJCET 2021 Physics Question Paper with Answer and Solution

(D) The $Q$ factor of a series $LCR$ circuit is given by the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Given values are $L = 2 \ H$,$C = 18 \ \mu F = 18 \times 10^{-6} \ F$,and $R = 10 \ \Omega$.
Substituting these values into the formula:
$Q = \frac{1}{10} \sqrt{\frac{2}{18 \times 10^{-6}}}$
$Q = \frac{1}{10} \sqrt{\frac{1}{9 \times 10^{-6}}}$
$Q = \frac{1}{10} \times \frac{1}{3 \times 10^{-3}}$
$Q = \frac{1}{10} \times \frac{1000}{3}$
$Q = \frac{100}{3} \approx 33.33$
Therefore,the correct option is $D$.

(C) The resonance frequency $v_{0}$ for an $LCR$ series circuit is given by the formula:
$v_{0} = \frac{1}{2 \pi \sqrt{LC}}$
Given values are $L = 25 \times 10^{-3} \, H$, $C = 62.5 \times 10^{-6} \, F$.
Substituting these values into the formula:
$v_{0} = \frac{1}{2 \times 3.14159 \times \sqrt{25 \times 10^{-3} \times 62.5 \times 10^{-6}}}$
$v_{0} = \frac{1}{2 \times 3.14159 \times \sqrt{1562.5 \times 10^{-9}}}$
$v_{0} = \frac{1}{2 \times 3.14159 \times \sqrt{1.5625 \times 10^{-6}}}$
$v_{0} = \frac{1}{2 \times 3.14159 \times 1.25 \times 10^{-3}}$
$v_{0} = \frac{1}{7.85398 \times 10^{-3}}$
$v_{0} \approx 127.39 \, Hz$
Thus, the correct option is $C$.

(D) Given: Capacitance $C = 50 \mu F = 50 \times 10^{-6} F$,Voltage $V_{rms} = 110 V$,Frequency $\nu = 60 Hz$.
First,calculate the capacitive reactance $X_C$ using the formula $X_C = \frac{1}{2 \pi \nu C}$.
$X_C = \frac{1}{2 \times 3.1416 \times 60 \times 50 \times 10^{-6}} \approx 53.05 \Omega$.
The rms current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_C}$.
$I_{rms} = \frac{110}{53.05} \approx 2.07 A$.
Rounding to one decimal place,we get $I_{rms} = 2.1 A$.

(A) The radius of an electron orbit in a hydrogen-like atom is given by the formula: $r_n = n^2 a_0$,where $n$ is the principal quantum number and $a_0$ is the Bohr radius $(5.3 \times 10^{-11} \ m)$.
For the $n=4$ state,we substitute the values into the formula:
$r_4 = (4)^2 \times (5.3 \times 10^{-11} \ m)$
$r_4 = 16 \times 5.3 \times 10^{-11} \ m$
$r_4 = 84.8 \times 10^{-11} \ m$
$r_4 = 8.48 \times 10^{-10} \ m$
Therefore,the correct option is $A$.

(C) The total energy $E$ of an electron in a hydrogen atom is given by the sum of its kinetic energy $K$ and potential energy $U$.
For a hydrogen atom, the total energy $E$ is related to the kinetic energy $K$ by the relation $E = -K$.
Given that the ground state energy $E = -13.6 \text{ eV}$.
Therefore, the kinetic energy $K = -E = -(-13.6 \text{ eV}) = 13.6 \text{ eV}$.
Thus, the correct option is $C$.

(D) In a potentiometer,the emf $\varepsilon$ of a cell is directly proportional to the balancing length $l$ of the wire,given by $\varepsilon \propto l$ or $\varepsilon = kl$,where $k$ is the potential gradient of the wire.
For two different cells,we have the ratio: $\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}$.
Given: $\varepsilon_1 = 1.5 \ V$,$l_1 = 150 \ cm$,and $l_2 = 210 \ cm$.
Substituting the values: $\frac{1.5}{\varepsilon_2} = \frac{150}{210}$.
Solving for $\varepsilon_2$: $\varepsilon_2 = \frac{1.5 \times 210}{150}$.
$\varepsilon_2 = \frac{1.5}{150} \times 210 = 0.01 \times 210 = 2.1 \ V$.
Thus,the emf of the second cell is $2.1 \ V$.

(B) The current $I$ flowing through a conductor is given by the relation: $I = n A v_d e$,where $n$ is the number density of electrons,$A$ is the cross-sectional area,$v_d$ is the drift velocity,and $e$ is the charge of an electron.
Since the current $I$ is constant throughout the wire,we have $v_d = \frac{I}{n A e}$.
Since $I$,$n$,and $e$ are constant,$v_d \propto \frac{1}{A}$.
Since the cross-section is circular,$A = \pi r^2$,therefore $v_d \propto \frac{1}{r^2}$.
Now,calculating the ratio of drift velocities at points $A$ and $B$:
$\frac{(v_d)_A}{(v_d)_B} = \frac{r_B^2}{r_A^2} = \frac{(r)^2}{(3r)^2} = \frac{r^2}{9r^2} = \frac{1}{9}$.
Thus,the ratio is $\frac{1}{9}$.

(A) The electric field due to a thin infinite sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.
Since both plates have the same positive surface charge density $\sigma = 17.7 \times 10^{-22} \ C/m^2$,the electric fields produced by both plates in the outer region of the second plate point in the same direction (away from the plates).
Therefore,the total electric field $E$ in the outer region is the sum of the fields from both plates:
$E = E_1 + E_2 = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$.
Given $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2/(N \cdot m^2)$:
$E = \frac{17.7 \times 10^{-22}}{8.85 \times 10^{-12}} = 2 \times 10^{-10} \ N/C$.

(C) The electric field intensity due to an electric dipole on its axis at a large distance $x$ is given by $E_{\text{axis}} = \frac{2kp}{x^3}$.
The electric field intensity due to an electric dipole on its equatorial line at a large distance $y$ is given by $E_{\text{equator}} = \frac{kp}{y^3}$.
Given that the magnitudes are equal: $E_{\text{axis}} = E_{\text{equator}}$.
Substituting the formulas: $\frac{2kp}{x^3} = \frac{kp}{y^3}$.
Canceling $kp$ from both sides: $\frac{2}{x^3} = \frac{1}{y^3}$.
Rearranging the terms: $\frac{x^3}{y^3} = 2$.
Taking the cube root on both sides: $\frac{x}{y} = \sqrt[3]{2} : 1$.

(B) The force $F$ experienced by an electron of charge $e$ in an electric field $E$ is given by $F = eE$.
According to Newton's second law of motion,$F = ma$,where $m_e$ is the mass of the electron and $a$ is its acceleration.
Equating the two expressions for force: $ma = eE$.
Therefore,the acceleration $a$ is given by $a = \frac{eE}{m_e}$.
Given:
Charge of electron $e = 1.6 \times 10^{-19} \ C$
Electric field $E = 2.0 \times 10^4 \ NC^{-1}$
Mass of electron $m_e = 9.1 \times 10^{-31} \ kg$
Substituting the values:
$a = \frac{(1.6 \times 10^{-19} \ C) \times (2.0 \times 10^4 \ NC^{-1})}{9.1 \times 10^{-31} \ kg}$
$a = \frac{3.2 \times 10^{-15}}{9.1 \times 10^{-31}} \ ms^{-2}$
$a \approx 0.3516 \times 10^{16} \ ms^{-2}$
$a \approx 3.52 \times 10^{15} \ ms^{-2}$.

(B) The flux linkage $\phi$ in a coil due to current $I$ in an adjacent coil is given by $\phi = M I$,where $M$ is the mutual inductance.
The change in flux linkage $\Delta \phi$ is given by:
$\Delta \phi = M \Delta I$
Given:
Mutual inductance $M = 1.5 \ H$
Change in current $\Delta I = I_{final} - I_{initial} = 20 \ A - 0 \ A = 20 \ A$
Substituting the values:
$\Delta \phi = 1.5 \times 20$
$\Delta \phi = 30 \ Wb$
Therefore,the change in flux linkage is $30 \ Wb$.

(B) The formula for the self-inductance $L$ of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$.
From this expression,we can see that $L \propto \frac{A}{l}$.
To increase the self-inductance $L$,the numerator $A$ (area of cross-section) must increase and the denominator $l$ (length of the solenoid) must decrease.
Therefore,the correct option is $B$.

(C) The magnetic energy density $(u)$ stored in a magnetic field is given by the formula:
$u = \frac{B^2}{2\mu_0}$
Given:
$B = 0.6 \ T$
$\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A \approx 12.56 \times 10^{-7} \ T \cdot m/A$
Substituting the values:
$u = \frac{(0.6)^2}{2 \times 4 \times 3.14 \times 10^{-7}}$
$u = \frac{0.36}{25.12 \times 10^{-7}}$
$u = \frac{0.36}{2.512 \times 10^{-6}}$
$u \approx 0.1433 \times 10^6 \ J/m^3$
$u = 1.43 \times 10^5 \ J/m^3$

(B) The relationship between the magnitudes of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is given by $E = cB$,where $c$ is the speed of light in free space $(c = 3 \times 10^8 \text{ ms}^{-1})$.
Substituting the given values: $E = (3 \times 10^8 \text{ ms}^{-1}) \times (2.1 \times 10^{-8} \text{ T}) = 6.3 \text{ Vm}^{-1}$.
The direction of propagation of the electromagnetic wave is given by the direction of the vector $\overrightarrow{E} \times \overrightarrow{B}$.
Given that the wave travels along the $X$-direction $(\hat{i})$ and $\overrightarrow{B}$ is along the $Z$-direction $(\hat{k})$,we have $\overrightarrow{E} \times \hat{k} = \hat{i}$.
Since $\hat{j} \times \hat{k} = \hat{i}$,the electric field $\overrightarrow{E}$ must be in the $Y$-direction $(\hat{j})$.
Therefore,$\overrightarrow{E} = 6.3 \hat{j} \text{ Vm}^{-1}$.

(D) The standard frequency range allocated for $FM$ (Frequency Modulated) radio broadcasting is $88 \text{ MHz}$ to $108 \text{ MHz}$.
This band is part of the Very High Frequency $(VHF)$ spectrum.
Therefore,the correct option is $D$.

(B) polar molecule is one that possesses a permanent electric dipole moment due to an uneven distribution of charge,resulting from differences in electronegativity between the bonded atoms and an asymmetric molecular geometry.
$1$. $HCl$: Chlorine is more electronegative than Hydrogen,creating a dipole moment. It is a polar molecule.
$2$. $H_2O$: The bent geometry of the water molecule and the high electronegativity of Oxygen result in a net dipole moment. It is a polar molecule.
$3$. $H_2$ and $O_2$: These are homonuclear diatomic molecules with zero dipole moment. They are non-polar.
Therefore,the pair $[HCl, H_2O]$ consists of two polar molecules.

(B) The circuit consists of two parts connected in series.
First part (left): Two capacitors of capacitance $C$ are in series on the top branch,and two are in series on the bottom branch. The middle capacitor is in parallel with these branches.
Equivalent capacitance of the top branch = $C/2$.
Equivalent capacitance of the bottom branch = $C/2$.
These two branches are in parallel with the middle capacitor $C$.
So,$C_{eq1} = C/2 + C/2 + C = 2C$.
Second part (right): Two capacitors of capacitance $C$ are in series on the top branch,and two are in series on the bottom branch.
Equivalent capacitance of the top branch = $C/2$.
Equivalent capacitance of the bottom branch = $C/2$.
These two branches are in parallel.
So,$C_{eq2} = C/2 + C/2 = C$.
Now,$C_{eq1}$ and $C_{eq2}$ are in series.
$C_{eq} = \frac{C_{eq1} \times C_{eq2}}{C_{eq1} + C_{eq2}} = \frac{2C \times C}{2C + C} = \frac{2C^2}{3C} = \frac{2C}{3}$.
Given $C = 3 \mu F$,we have:
$C_{eq} = \frac{2 \times 3 \mu F}{3} = 2 \mu F$.

(A) The magnetic field $B$ on the equatorial line of a short bar magnet is given by the formula:
$B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$
Given:
Magnetic moment $m = 0.75 \ A \ m^2$
Distance $r = 75 \ cm = 0.75 \ m$
Constant $\frac{\mu_0}{4\pi} = 10^{-7} \ T \ m/A$
Substituting the values:
$B = 10^{-7} \times \frac{0.75}{(0.75)^3}$
$B = 10^{-7} \times \frac{1}{(0.75)^2}$
$B = 10^{-7} \times \frac{1}{0.5625}$
$B \approx 1.78 \times 10^{-7} \ T$

(B) The angle of dip $(I)$ is defined by the relationship between the vertical component $(Z_E)$ and the horizontal component $(H_E)$ of the earth's magnetic field as:
$\tan I = \frac{Z_E}{H_E}$
Given that the horizontal component is equal to the vertical component,we have:
$Z_E = H_E$
Substituting this into the formula:
$\tan I = \frac{Z_E}{Z_E} = 1$
Therefore,the angle of dip is:
$I = \tan^{-1}(1) = 45^{\circ}$

(A) The magnetic dipole moment $M$ of a current loop is given by $M = I A$,where $I$ is the current and $A$ is the area of the loop.
For a circular loop of radius $r$,the area is $A = \pi r^2$.
Thus,$M = I (\pi r^2) \quad \dots(1)$
The magnetic field $B$ at the centre of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
Rearranging this formula to solve for current $I$,we get $I = \frac{2rB}{\mu_0}$.
Substituting the expression for $I$ into equation $(1)$:
$M = \left( \frac{2rB}{\mu_0} \right) (\pi r^2)$
$M = \frac{2 \pi B r^3}{\mu_0}$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula: $B = \mu_0 n I$,where $n = \frac{N}{l}$ is the number of turns per unit length.
Given:
Length $l = 0.5 \ m$
Number of turns $N = 1000$
Current $I = 10 \ A$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$
Substituting the values:
$n = \frac{1000}{0.5} = 2000 \ turns/m$
$B = (4\pi \times 10^{-7}) \times 2000 \times 10$
$B = 8\pi \times 10^{-3} \ T$
$B \approx 25.12 \times 10^{-3} \ T = 2.51 \times 10^{-2} \ T$.

(B) The magnetic force on a current-carrying conductor is given by the formula $F = i l B \sin(\theta)$.
Here,the current $i = 2 \text{ A}$,the magnetic field $B = 3.0 \times 10^{-5} \text{ T}$,and the angle $\theta$ between the current (east-west) and the magnetic field (south-north) is $90^\circ$.
Since $\sin(90^\circ) = 1$,the force per unit length is given by:
$\frac{F}{l} = i B \sin(90^\circ)$
$\frac{F}{l} = 2 \times 3.0 \times 10^{-5} \times 1$
$\frac{F}{l} = 6 \times 10^{-5} \text{ N/m}$.

(C) The formula for apparent depth is given by:
$\mu = \frac{\text{Real depth}}{\text{Apparent depth}}$
Given:
Real depth $(h_2) = 16 \text{ cm}$
Refractive index $(\mu) = \frac{4}{3}$
Let the apparent depth be $h_1$.
Substituting the values in the formula:
$\frac{4}{3} = \frac{16}{h_1}$
$h_1 = \frac{16 \times 3}{4}$
$h_1 = 4 \times 3 = 12 \text{ cm}$
Therefore, the apparent depth of the needle is $12.0 \text{ cm}$.

(D) The formula for the equivalent focal length $f$ of two thin lenses in contact is given by:
$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$
Given that both lenses are convex with focal lengths $f_1 = 30 \ cm$ and $f_2 = 30 \ cm$:
$\frac{1}{f} = \frac{1}{30} + \frac{1}{30}$
$\frac{1}{f} = \frac{2}{30} = \frac{1}{15}$
Therefore,$f = 15 \ cm$.
The correct option is $D$.

(B) Using the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given:
$f = 12 \ cm$,$R_1 = 10 \ cm$,$R_2 = -15 \ cm$ (for a double convex lens,the second surface has a negative radius of curvature in the sign convention).
Substituting the values:
$\frac{1}{12} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{-15} \right)$
$\frac{1}{12} = (\mu - 1) \left( \frac{1}{10} + \frac{1}{15} \right)$
$\frac{1}{12} = (\mu - 1) \left( \frac{3 + 2}{30} \right)$
$\frac{1}{12} = (\mu - 1) \times \frac{5}{30}$
$\frac{1}{12} = (\mu - 1) \times \frac{1}{6}$
$\frac{6}{12} = \mu - 1$
$0.5 = \mu - 1$
$\mu = 1.5$
Thus,the refractive index of the material of the lens is $1.5$.

(D) The formula for the refractive index of a prism is given by $\mu = \frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \frac{A}{2}}$,where $A$ is the angle of the prism and $D_m$ is the angle of minimum deviation.
Given that $D_m = A$,we substitute this into the formula:
$\mu = \frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}$
$\mu = \frac{\sin A}{\sin \frac{A}{2}}$
Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we get:
$\mu = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}$
$\mu = 2 \cos \frac{A}{2}$
$\frac{\mu}{2} = \cos \frac{A}{2}$
$\frac{A}{2} = \cos ^{-1}\left(\frac{\mu}{2}\right)$
$A = 2 \cos ^{-1}\left(\frac{\mu}{2}\right)$

(C) The position of the $n^{th}$ bright fringe in a Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
Given:
$\lambda = 500 \, nm = 500 \times 10^{-9} \, m = 5 \times 10^{-7} \, m$
$D = 100 \, cm = 1 \, m$
$d = 1 \, mm = 10^{-3} \, m$
The distance between the fifth $(n=5)$ and third $(n=3)$ bright fringes is $\Delta y = y_5 - y_3$.
$\Delta y = \frac{5 \lambda D}{d} - \frac{3 \lambda D}{d} = \frac{2 \lambda D}{d}$.
Substituting the values:
$\Delta y = \frac{2 \times (5 \times 10^{-7} \, m) \times (1 \, m)}{10^{-3} \, m}$.
$\Delta y = 10 \times 10^{-4} \, m = 10^{-3} \, m$.
$\Delta y = 1 \, mm$.