GUJCET 2024 Physics Question Paper with Answer and Solution

(D) The correct option is $D$.
Angular momentum is defined as $L = r \times p$.
The $SI$ unit of angular momentum is $kg \cdot m^2/s$,which is equivalent to $J \cdot s$.
According to Bohr's quantization condition,$L = \frac{n h}{2 \pi}$.
Since $n$ and $2 \pi$ are dimensionless,the unit of angular momentum $L$ is the same as the unit of Planck's constant $h$,which is $J \cdot s$.

(C) Step $1$: Given values:
$V = 220 \ V$
$L = 50 \ mH = 50 \times 10^{-3} \ H$
$\nu = 50 \ Hz$
Step $2$: The inductive reactance $X_L$ is given by $X_L = \omega L = 2 \pi \nu L$.
Step $3$: The rms current $I$ is given by $I = \frac{V}{X_L} = \frac{V}{2 \pi \nu L}$.
Step $4$: Substituting the values:
$I = \frac{220}{2 \times 3.14159 \times 50 \times 50 \times 10^{-3}}$
$I = \frac{220}{15.70795} \approx 14.005 \ A$.
Thus,the rms current is approximately $14 \ A$.

(C) In the classic Geiger-Marsden experiment (also known as the Rutherford gold foil experiment),Hans Geiger and Ernest Marsden used a very thin gold foil to observe the scattering of alpha particles.
The thickness of the gold foil used in this experiment was approximately $2.1 \times 10^{-7} \ m$ (or $2.1 \times 10^{-5} \ cm$).
Therefore,the correct option is $C$.

(D) The radius of the $n^{th}$ orbit is given by $r_{n} \propto \frac{n^{2}}{Z}$.
For an $H$-atom,the atomic number $Z = 1$.
Therefore,the radius is proportional to the square of the principal quantum number: $r_{n} \propto n^{2}$.
For the second orbit $(n_{2} = 2)$ and the third orbit $(n_{3} = 3)$:
$\frac{r_{2}}{r_{3}} = \left(\frac{n_{2}}{n_{3}}\right)^{2} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9}$.
Thus,the ratio is $4: 9$.

(B) For a balanced Wheatstone bridge,the condition is $\frac{R_1}{R_3} = \frac{R_2}{R_4}$. Here,$\frac{2}{4} = \frac{4}{8} = \frac{1}{2}$.
Since the bridge is balanced,no current flows through the galvanometer.
The circuit simplifies to two parallel branches:
Branch $1$: $R_{s1} = R_1 + R_2 = 2 \Omega + 4 \Omega = 6 \Omega$.
Branch $2$: $R_{s2} = R_3 + R_4 = 4 \Omega + 8 \Omega = 12 \Omega$.
The equivalent resistance $R_{eq}$ of the parallel combination is:
$R_{eq} = \frac{R_{s1} \times R_{s2}}{R_{s1} + R_{s2}} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \Omega$.
The total current $I$ flowing through the circuit is:
$I = \frac{V}{R_{eq}} = \frac{10 \text{ V}}{4 \Omega} = 2.5 \text{ A}$.

(C) The drift velocity $v_d$ of a charge carrier is directly proportional to the applied electric field $E$,given by $v_d = \mu E$.
Here,$\mu$ is defined as the mobility of the charge carrier.
Therefore,mobility $\mu = \frac{v_d}{E}$.
It represents the magnitude of the drift velocity per unit electric field.

(D) Current density $(J)$ is defined as the current $(I)$ flowing per unit area $(A)$ perpendicular to the direction of flow.
Mathematically,$J = \frac{I}{A}$.
The $SI$ unit of current $(I)$ is Ampere $(A)$ and the $SI$ unit of area $(A)$ is square meter $(m^{2})$.
Therefore,the $SI$ unit of current density is $\frac{A}{m^{2}} = A \ m^{-2}$.
Thus,the correct option is $(D)$.

(A) Given:
Resistance at $T_0 = 27.5^{\circ} C$ is $R_0 = 2.1 \Omega$.
Resistance at $T = 100^{\circ} C$ is $R = 2.7 \Omega$.
The formula for temperature dependence of resistance is $R = R_0[1 + \alpha(T - T_0)]$.
Substituting the values:
$2.7 = 2.1[1 + \alpha(100 - 27.5)]$
$2.7 = 2.1[1 + \alpha(72.5)]$
$\frac{2.7}{2.1} = 1 + \alpha(72.5)$
$\frac{9}{7} = 1 + \alpha(72.5)$
$\frac{9}{7} - 1 = \alpha(72.5)$
$\frac{2}{7} = \alpha(72.5)$
$\alpha = \frac{2}{7 \times 72.5} = \frac{2}{507.5} \approx 0.00394 \approx 3.9 \times 10^{-3} {}^{\circ} C^{-1}$.
Thus,the correct option is $A$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
Mass $m = 0.12 \ kg$
Velocity $v = 20 \ m \ s^{-1}$
Planck's constant $h = 6.63 \times 10^{-34} \ J \ s$
Substituting the values into the formula:
$\lambda = \frac{6.63 \times 10^{-34}}{0.12 \times 20}$
$\lambda = \frac{6.63 \times 10^{-34}}{2.4}$
$\lambda = 2.7625 \times 10^{-34} \ m$
Rounding to two decimal places,we get $\lambda = 2.76 \times 10^{-34} \ m$.

(B) The process of emitting electrons from a metal surface by applying a strong external electric field is known as field emission. For a typical metal, the work function is a few electron volts $(eV)$. To overcome this potential barrier and pull an electron out of the metal surface, an electric field of the order of $10^{8} \,V \,m^{-1}$ is required. Therefore, the correct option is $B$.

(D) Given:
Electric field $E = 9 \times 10^4 \text{ N/C}$
Distance $r = 2 \text{ cm} = 2 \times 10^{-2} \text{ m}$
Coulomb constant $k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$
The electric field due to an infinite line charge is given by the formula:
$E = \frac{\lambda}{2 \pi \varepsilon_0 r} = \frac{2k\lambda}{r}$
Rearranging for linear charge density $\lambda$:
$\lambda = \frac{E \cdot r}{2k}$
Substituting the values:
$\lambda = \frac{(9 \times 10^4) \times (2 \times 10^{-2})}{2 \times (9 \times 10^9)}$
$\lambda = \frac{18 \times 10^2}{18 \times 10^9}$
$\lambda = 10^{-7} \text{ C/m}$
Converting to $\mu\text{C/m}$:
$\lambda = 0.1 \times 10^{-6} \text{ C/m} = 0.1 \mu\text{C/m}$
Therefore,the correct option is $D$.

(B) The electric dipole moment $\vec{p}$ is directed from the negative charge $(-q)$ to the positive charge $(+q)$.
On the equatorial plane of an electric dipole,the net electric field $\vec{E}_{net}$ is directed anti-parallel to the dipole moment vector $\vec{p}$.
Mathematically,the electric field on the equatorial plane is given by $\vec{E} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3}$.
Since the electric field vector is in the opposite direction to the dipole moment vector,the angle between them is $180^{\circ}$.
Therefore,the correct option is $B$.

(D) The self-inductance $L$ of a coil is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $\mu_0$ is the permeability of free space,$N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.
From this formula,it is evident that the self-inductance $L$ depends only on the geometric parameters of the coil (number of turns,area,and length) and the core material.
It is independent of the current $I$ flowing through the coil.
Therefore,if the current changes from $I$ to $5I$,the self-inductance remains unchanged.
Thus,the new self-inductance is $L$ $H$.

(A) Given:
Side of square loop,$l = 10 \ cm = 0.1 \ m$
Area of loop,$A = l^2 = (0.1 \ m)^2 = 0.01 \ m^2 = 100 \ cm^2$
Resistance,$R = 0.5 \ \Omega$
Initial magnetic field,$B_1 = 0.10 \ T$
Final magnetic field,$B_2 = 0 \ T$
Time interval,$\Delta t = 0.70 \ s$
Angle between the area vector (normal to the loop) and the magnetic field vector,$\theta = 45^{\circ}$
The magnetic flux $\phi$ is given by $\phi = BA \cos \theta$.
Initial flux,$\phi_1 = B_1 A \cos 45^{\circ} = 0.10 \times 0.01 \times \frac{1}{\sqrt{2}} \ Wb$.
Final flux,$\phi_2 = 0 \ Wb$.
Change in flux,$\Delta \phi = \phi_2 - \phi_1 = -\phi_1 = -\frac{0.001}{\sqrt{2}} \ Wb$.
Induced emf,$\varepsilon = -\frac{\Delta \phi}{\Delta t} = -\left( -\frac{0.001}{\sqrt{2} \times 0.70} \right) = \frac{0.001}{0.70 \times 1.414} \approx 1.01 \times 10^{-3} \ V$.
Using the provided approximation in the original problem: $\varepsilon = \frac{0.1 \times 100 \times 10^{-4} \times \cos 45^{\circ}}{0.70} = \frac{0.001 \times 0.707}{0.70} \approx 10^{-3} \ V$.
Induced current,$I = \frac{\varepsilon}{R} = \frac{10^{-3} \ V}{0.5 \ \Omega} = 2 \times 10^{-3} \ A$.

(A) Cellular phones operate by transmitting and receiving signals within the Ultra High Frequency $(UHF)$ band,which typically ranges from $300 \ MHz$ to $3 \ GHz$. This frequency range is ideal for mobile communication because it allows for high data rates and supports the compact antenna sizes required for portable devices. Therefore,the correct option is $A$.

(C) Let the potential at point $B$ be $V$.
Since the capacitors $C_1 = 2 \ \mu F$ and $C_2 = 3 \ \mu F$ are in series,the charge $Q$ on both capacitors is the same.
The potential difference across $C_1$ is $V_A - V_B = 40 - V$.
Thus,$Q = C_1(V_A - V_B) = 2(40 - V)$.
The potential difference across $C_2$ is $V_B - V_C = V - 10$.
Thus,$Q = C_2(V_B - V_C) = 3(V - 10)$.
Since the charges are equal:
$2(40 - V) = 3(V - 10)$
$80 - 2V = 3V - 30$
$5V = 110$
$V = 22 \ V$.

(D) The initial capacitance of the parallel plate capacitor with air is given by:
$C_0 = \frac{A \varepsilon_0}{d} = 4 \ pF$
When the distance between the plates is reduced by half,the new distance is $d' = \frac{d}{2}$.
When the space is filled with a dielectric of constant $K = 6$,the new capacitance $C$ is given by:
$C = K \frac{A \varepsilon_0}{d'}$
Substituting $d' = \frac{d}{2}$ and $K = 6$:
$C = 6 \times \frac{A \varepsilon_0}{d/2} = 12 \times \frac{A \varepsilon_0}{d}$
Since $\frac{A \varepsilon_0}{d} = C_0 = 4 \ pF$,we have:
$C = 12 \times 4 \ pF = 48 \ pF$.

(A) For a spherical charged shell,the electric potential at any point inside the shell is constant and equal to the potential at its surface.
Given that the radius of the shell is $R = 10 \ cm$ and the potential on the surface is $V_{surface} = 100 \ V$.
Since $2 \ cm < 10 \ cm$,the point is inside the shell.
Therefore,the potential at $2 \ cm$ from the centre is the same as the potential on the surface,which is $100 \ V$.

(C) The energy gained by a charged particle accelerated through a potential difference $V$ is given by the formula: $E = qV$.
Here,the charge of an electron is $q = e = 1.6 \times 10^{-19} \ C$ and the potential difference is $V = 2.5 \ V$.
Therefore,the energy gained is $E = e \times 2.5 \ V = 2.5 \ eV$.
Since $1 \ eV$ is the energy gained by an electron when accelerated through a potential difference of $1 \ V$,the energy gained is $2.5 \ eV$.

(D) The torque $\tau$ acting on a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $\tau = m B \sin \theta$.
Given:
Torque $\tau = 4.5 \times 10^{-2} \ J$
Magnetic field $B = 0.5 \ T$
Angle $\theta = 30^{\circ}$
Substituting the values into the formula:
$4.5 \times 10^{-2} = m \times 0.5 \times \sin 30^{\circ}$
Since $\sin 30^{\circ} = 0.5$:
$4.5 \times 10^{-2} = m \times 0.5 \times 0.5$
$4.5 \times 10^{-2} = m \times 0.25$
$m = \frac{4.5 \times 10^{-2}}{0.25}$
$m = 18 \times 10^{-2} \ J \ T^{-1}$.

(B) The magnetic intensity $H$ inside a solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.
Given:
Number of turns per unit length $n = 1000 \text{ m}^{-1}$
Current $I = 2 \text{ A}$
Relative permeability $\mu_r = 400$ (Note: This value is not required to calculate magnetic intensity $H$,as $H$ depends only on the geometry and current).
Substituting the values into the formula:
$H = 1000 \times 2$
$H = 2000 \text{ A m}^{-1}$
$H = 2 \times 10^3 \text{ A m}^{-1}$
Therefore,the correct option is $B$.

(D) An ideal ammeter is designed to measure current without affecting the circuit,which requires it to have zero resistance. Thus,the resistance of an ideal ammeter is $0 \ \Omega$.
An ideal voltmeter is designed to measure potential difference without drawing any current from the circuit,which requires it to have infinite resistance. Thus,the resistance of an ideal voltmeter is $\infty \ \Omega$.
Therefore,the correct option is $(0, \infty)$.

(B) The unit of magnetic permeability of free space,$\mu_0$,is given by the formula $B = \frac{\mu_0 I}{2\pi r}$.
Rearranging for $\mu_0$,we get $\mu_0 = \frac{B \cdot 2\pi r}{I}$.
The unit of magnetic field $B$ is Tesla $(T)$,which is equivalent to $\frac{V \cdot s}{m^2}$.
Substituting this into the expression for $\mu_0$,the units are $\frac{(V \cdot s / m^2) \cdot m}{A} = \frac{V \cdot s}{A \cdot m}$.
Therefore,$\frac{V \cdot s}{A \cdot m}$ is the unit of $\mu_0$ (permeability of free space).

(B) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ are conserved.
< strong>Conservation of atomic number $(Z)$:
$92 + 0 = x + 41 + 4(0)$
$92 = x + 41 \Rightarrow x = 51$.
< strong>Conservation of mass number $(A)$:
$235 + 1 = 133 + y + 4(1)$
$236 = 137 + y \Rightarrow y = 99$.
Therefore,the values are $x = 51$ and $y = 99$.

(D) Atoms of the same element have the same atomic number $(Z)$ but different mass numbers $(A)$. These atoms are defined as isotopes. For example,hydrogen has three naturally occurring isotopes: protium $(^1H_1)$,deuterium $(^2H_1)$,and tritium $(^3H_1)$.

(B) The refractive index $n$ is defined as the ratio of the speed of light in vacuum $c$ to the speed of light in the medium $v$.
$n = \frac{c}{v}$
Given:
$n = 1.6$
$c = 3 \times 10^{8} \,m \,s^{-1}$
Rearranging the formula to solve for $v$:
$v = \frac{c}{n}$
$v = \frac{3 \times 10^{8}}{1.6}$
$v = 1.875 \times 10^{8} \,m \,s^{-1} \approx 1.88 \times 10^{8} \,m \,s^{-1}$
Thus,the correct option is $B$.

(D) The magnifying power $(m)$ of a refracting telescope is given by the formula $m = \frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.
Given:
Focal length of objective $f_o = 1 \ m = 100 \ cm$.
Focal length of eyepiece $f_e = 1 \ cm$.
Substituting these values into the formula:
$m = \frac{100 \ cm}{1 \ cm} = 100$.
Therefore,the magnifying power of the telescope is $100$.

(C) For a thin prism,the angle of minimum deviation $(D_{m})$ is given by the formula:
$D_{m} = A(n - 1)$
where $A$ is the angle of the prism and $n$ is the refractive index.
Given:
$A = 4^{\circ}$
$n = 1.6$
Substituting the values into the formula:
$D_{m} = 4^{\circ}(1.6 - 1)$
$D_{m} = 4^{\circ}(0.6)$
$D_{m} = 2.4^{\circ}$
Therefore,the angle of minimum deviation is $2.4^{\circ}$.

(C) The formula for refraction at a spherical surface is given by $\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2}-n_{1}}{R}$.
Given that the object is at infinity,$u = \infty$.
The refractive index of air is $n_{1} = 1$ and the refractive index of glass is $n_{2} = 1.5$.
Substituting these values into the formula:
$\frac{1.5}{v} - \frac{1}{\infty} = \frac{1.5-1}{R}$
Since $\frac{1}{\infty} = 0$,we have:
$\frac{1.5}{v} = \frac{0.5}{R}$
$v = \frac{1.5 R}{0.5} = 3 R$.
Thus,the distance of the image is $3 R$.

(D) The focal length $f$ of a spherical mirror is related to its radius of curvature $R$ by the formula $f = R/2$. $A$ plane mirror can be considered as a spherical mirror with an infinite radius of curvature $(R = \infty)$. Therefore,the focal length of a plane mirror is $f = \infty/2 = \infty$. Thus,the correct option is $D$.

(A) The forbidden energy gap $(E_g)$ represents the minimum energy required for an electron to jump from the valence band to the conduction band.
For Germanium $(Ge)$ at room temperature $(300 \ K)$,the forbidden energy gap is approximately $0.72 \ eV$.
For Silicon $(Si)$ at room temperature $(300 \ K)$,the forbidden energy gap is approximately $1.1 \ eV$.
Therefore,the correct value for Germanium is $0.72 \ eV$.

(C) The correct answer is $C$. Indium.
To obtain a $p$-type semiconductor,a trivalent impurity (an element from group $13$ of the periodic table) must be added to a pure (intrinsic) semiconductor like Silicon or Germanium.
Trivalent impurities include Boron $(B)$,Gallium $(Ga)$,Indium $(In)$,and Aluminium $(Al)$.
Antimony $(Sb)$,Arsenic $(As)$,and Phosphorus $(P)$ are pentavalent impurities,which are used to create $n$-type semiconductors.

(C) Given:
Slit separation $d = 0.28 \ mm = 0.28 \times 10^{-3} \ m$
Distance to screen $D = 1.4 \ m$
Distance of $n^{th}$ bright fringe $x_n = 1.2 \ cm = 1.2 \times 10^{-2} \ m$
Order of fringe $n = 4$
The formula for the position of the $n^{th}$ bright fringe is given by:
$x_n = \frac{n \lambda D}{d}$
Rearranging for wavelength $\lambda$:
$\lambda = \frac{x_n d}{n D}$
Substituting the values:
$\lambda = \frac{(1.2 \times 10^{-2} \ m) \times (0.28 \times 10^{-3} \ m)}{4 \times 1.4 \ m}$
$\lambda = \frac{0.336 \times 10^{-5}}{5.6} \ m$
$\lambda = 0.06 \times 10^{-5} \ m = 6 \times 10^{-7} \ m$
Converting to nanometers $(1 \ nm = 10^{-9} \ m)$:
$\lambda = 600 \times 10^{-9} \ m = 600 \ \text{nm}$
Therefore,the correct option is $C$.

(A) wavefront is defined as the locus of all points that are in the same phase of vibration. Since all particles on a given wavefront oscillate in phase,the phase difference between any two particles on the same wavefront is $0 \ rad$.