GUJCET 2025 Physics Question Paper with Answer and Solution

(D) The relationship between maximum (peak) voltage $V_0$ and root mean square (rms) voltage $V_{rms}$ for a sinusoidal $AC$ circuit is given by $V_{rms} = \frac{V_0}{\sqrt{2}}$.
Rearranging this formula,we get the ratio of maximum voltage to $rms$ voltage as $\frac{V_0}{V_{rms}} = \sqrt{2}$.
Given that $\sqrt{2} \approx 1.414$,we convert this ratio into a percentage by multiplying by $100\%$.
Therefore,the percentage value is $1.414 \times 100\% = 141.4\%$.

(A) In an $LCR$ series circuit,resonance occurs when the inductive reactance $X_L$ equals the capacitive reactance $X_C$ $(X_L = X_C)$.
At this condition,the total impedance $Z$ is equal to the resistance $R$,meaning the circuit behaves purely resistively.
The power factor is given by $\cos \phi = \frac{R}{Z}$.
At resonance,$Z = R$,so $\cos \phi = \frac{R}{R} = 1$.

(D) The power $P$ is given by the formula $P = V_{rms} \times I_{rms}$.
Given $P = 12 \text{ W}$ and $V_{rms} = 24 \text{ V}$,we can calculate the root mean square current $I_{rms}$ as:
$I_{rms} = \frac{P}{V_{rms}} = \frac{12}{24} = 0.5 \text{ A}$.
The peak current $I_0$ is related to the $I_{rms}$ by the formula $I_0 = I_{rms} \times \sqrt{2}$.
Substituting the value of $I_{rms}$,we get:
$I_0 = 0.5 \times \sqrt{2} = \frac{1}{2} \times \sqrt{2} = \frac{1}{\sqrt{2}} \text{ A}$.

(B) Gamma rays possess high energy and high penetrating power. They are widely used in medical treatments such as radiotherapy to destroy or inhibit the growth of cancer cells by damaging their $DNA$.

(C) The speed of light in a vacuum is $c = 3 \times 10^8 \text{ m/s}$.
The speed of light in the medium is $v = 200 \times 10^8 \text{ cm/s}$.
First,convert the speed in the medium to $\text{m/s}$:
$v = 200 \times 10^8 \times 10^{-2} \text{ m/s} = 2 \times 10^8 \text{ m/s}$.
The refractive index $n$ is given by the formula $n = \frac{c}{v}$.
Substituting the values: $n = \frac{3 \times 10^8 \text{ m/s}}{2 \times 10^8 \text{ m/s}} = 1.5$.

(A) The power of a lens is given by the formula $P = \frac{1}{f(m)}$,where $f$ is the focal length in meters.
For a convex lens,the focal length is positive,so $f_1 = +0.25 \text{ m}$. Thus,$P_1 = \frac{1}{0.25} = +4 \text{ D}$.
For a concave lens,the focal length is negative,so $f_2 = -0.25 \text{ m}$. Thus,$P_2 = \frac{1}{-0.25} = -4 \text{ D}$.
The power of the combination of lenses in contact is the algebraic sum of their individual powers: $P = P_1 + P_2$.
Substituting the values,$P = 4 \text{ D} + (-4 \text{ D}) = 0 \text{ D}$.

(D) For the condition of minimum deviation in an equilateral prism,the angle of incidence $i$ is equal to the angle of emergence $e$.
The relationship is given by $i = e = \frac{A + \delta_m}{2}$.
For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given the angle of minimum deviation $\delta_m = 46^{\circ}$.
Substituting these values into the formula:
$i = \frac{60^{\circ} + 46^{\circ}}{2} = \frac{106^{\circ}}{2} = 53^{\circ}$.
Therefore,the angle of incidence is $53^{\circ}$.

(B) The magnification $(M)$ of a compound microscope is given by the formula $M \approx \frac{L}{f_o} \times \frac{D}{f_e}$,where $L$ is the length of the tube,$f_o$ is the focal length of the objective lens,$f_e$ is the focal length of the eyepiece,and $D$ is the least distance of distinct vision.
Since $M$ is directly proportional to the tube length $L$ $(M \propto L)$,increasing the tube length $L$ results in an increase in the magnification of the compound microscope.

(C) The resultant intensity $I_R$ for two waves of intensity $I_0$ with a phase difference $\phi$ is given by the formula:
$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$
Since $I_1 = I_2 = I_0$,we substitute these values:
$I_R = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos \phi$
$I_R = 2I_0 + 2I_0 \cos \phi = 2I_0(1 + \cos \phi)$
Using the trigonometric identity $1 + \cos \phi = 2 \cos^2(\frac{\phi}{2})$:
$I_R = 2I_0 \cdot 2 \cos^2(\frac{\phi}{2}) = 4I_0 \cos^2(\frac{\phi}{2})$
Therefore,the resultant intensity $I_R$ is directly proportional to $\cos^2(\frac{\phi}{2})$.

(D) When light diverges from a point source in a homogeneous isotropic medium,the energy spreads out uniformly in all directions. The locus of points having the same phase forms a sphere centered at the source. Hence,the wavefront is spherical.

(B) Given: Slit separation $d = 0.54$ mm $= 0.54 \times 10^{-3}$ m.
Distance of screen $D = 1.8$ m.
Distance of the $n^{th}$ bright fringe from the central fringe is given by $y_n = \frac{n\lambda D}{d}$.
For the $6^{th}$ bright fringe,$n = 6$ and $y_6 = 1.2$ cm $= 1.2 \times 10^{-2}$ m.
Rearranging the formula for wavelength $\lambda$: $\lambda = \frac{y_n d}{n D}$.
Substituting the values: $\lambda = \frac{1.2 \times 10^{-2} \times 0.54 \times 10^{-3}}{6 \times 1.8}$.
$\lambda = \frac{0.648 \times 10^{-5}}{10.8} = 0.06 \times 10^{-5}$ m.
$\lambda = 600 \times 10^{-9}$ m $= 600$ nm.

(B) The electric field required to pull out an electron from a metal surface (a process known as field emission) is typically of the order of $10^6 \text{ V/cm}$.
This value is derived from the electrostatic force required to overcome the work function of the metal,allowing electrons to tunnel through the potential barrier.

(A) The power $P$ emitted by a source is given by the product of the number of photons emitted per second $(n)$ and the energy of a single photon $(E = h\nu)$.
Given: Power $P = 4 \times 10^{-3} \text{ W}$,frequency $\nu = 6 \times 10^{14} \text{ Hz}$,and Planck's constant $h = 6.63 \times 10^{-34} \text{ Js}$.
The formula is $P = n \cdot h\nu$.
Rearranging for $n$: $n = \frac{P}{h\nu}$.
Substituting the values: $n = \frac{4 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}}$.
$n = \frac{4 \times 10^{-3}}{39.78 \times 10^{-20}} = \frac{4}{39.78} \times 10^{17} \approx 0.10055 \times 10^{17} = 1.0055 \times 10^{16}$.
Rounding to the nearest significant figure,we get $n \approx 1 \times 10^{16} \text{ photons per second}$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
Mass $m = 0.033 \text{ kg}$
Velocity $v = 1 \text{ km/s} = 1000 \text{ m/s} = 10^3 \text{ m/s}$
Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$
Substituting the values into the formula:
$\lambda = \frac{6.6 \times 10^{-34}}{0.033 \times 1000}$
$\lambda = \frac{6.6 \times 10^{-34}}{33}$
$\lambda = 0.2 \times 10^{-34} \text{ m}$
$\lambda = 2 \times 10^{-35} \text{ m}$.

(A) According to Bohr's model,the orbital angular momentum $L$ is given by the formula $L = n \frac{h}{2\pi}$.
The ground state corresponds to $n = 1$.
The first excited state is $n = 2$,the second excited state is $n = 3$,and the third excited state is $n = 4$.
Substituting $n = 4$ into the formula:
$L = 4 \times \frac{h}{2\pi} = 2 \times \frac{h}{\pi}$.
Given $h = 6.63 \times 10^{-34} \text{ Js}$ and $\pi \approx 3.14$:
$L = 2 \times \frac{6.63 \times 10^{-34}}{3.14} \approx 2 \times 2.111 \times 10^{-34} = 4.222 \times 10^{-34} \text{ kg m}^2\text{s}^{-1}$.

(D) For a hydrogen atom in the ground state $(n=1)$,Bohr's quantization condition for angular momentum is given by $mvr = \frac{nh}{2\pi}$.
Here,$m$ is the mass of the electron $(9.1 \times 10^{-31} \text{ kg})$,$v$ is the velocity,$r$ is the orbital radius $(5.3 \times 10^{-11} \text{ m})$,and $h$ is Planck's constant $(6.63 \times 10^{-34} \text{ Js})$.
Rearranging the formula for velocity $v$,we get $v = \frac{h}{2\pi mr}$.
Substituting the values: $v = \frac{6.63 \times 10^{-34}}{2 \times 3.14159 \times 9.1 \times 10^{-31} \times 5.3 \times 10^{-11}}$.
Calculating this,$v \approx 2.18 \times 10^6 \text{ ms}^{-1}$.
Rounding to two significant figures,we get $v \approx 2.2 \times 10^6 \text{ ms}^{-1}$.

(C) In the Bohr model,the Total Energy $(TE)$ of an electron in the ground state is $-13.6 \text{ eV}$.
Kinetic Energy $(KE)$ is given by the relation $KE = -TE$. Therefore,$KE = -(-13.6 \text{ eV}) = +13.6 \text{ eV}$.
Potential Energy $(PE)$ is given by the relation $PE = 2 \times TE$. Therefore,$PE = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$.
Thus,the potential energy is $-27.2 \text{ eV}$ and the kinetic energy is $+13.6 \text{ eV}$.

(D) The potential energy of the barrier at the point of contact is given by $U = k \frac{q_1 q_2}{r}$.
Here,the charge of each deuteron is $q_1 = q_2 = e = 1.6 \times 10^{-19}$ $C$.
The distance between the centers of the two deuterons at the point of contact is $r = R_1 + R_2 = 2 \text{ fm} + 2 \text{ fm} = 4 \text{ fm} = 4 \times 10^{-15}$ m.
Substituting the values into the formula:
$U = \frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2) \times (1.6 \times 10^{-19} \text{ C})^2}{4 \times 10^{-15} \text{ m}}$
$U = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{4 \times 10^{-15}}$
$U = \frac{23.04 \times 10^{-29}}{4 \times 10^{-15}}$
$U = 5.76 \times 10^{-14}$ $J$.

(B) The proton-proton chain reaction,which is the primary fusion process in the Sun,can be represented by the net equation:
$4_1^1H + 2e^- \rightarrow ^4_2He + 2\nu_e + 26.7 \text{ MeV}$.
In this process,four hydrogen nuclei (protons) fuse to form one helium nucleus $(^4_2He)$,releasing two neutrinos $(
u_e)$ and a total energy of approximately $26.7 \text{ MeV}$.

(A) Isotones are atoms of different elements that have the same number of neutrons.
To find the number of neutrons $(N)$,we use the formula $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For $^{198}_{80}\text{Hg}$: $N = 198 - 80 = 118$.
For $^{197}_{79}\text{Au}$: $N = 197 - 79 = 118$.
Since both atoms have the same number of neutrons $(118)$,they are isotones.
For other options:
- $^3_1\text{H}$ $(N=2)$ and $^3_2\text{He}$ $(N=1)$ are not isotones.
- $^{214}_{82}\text{Pb}$ $(N=132)$ and $^{214}_{83}\text{Bi}$ $(N=131)$ are not isotones.
- $^{12}_6\text{C}$ $(N=6)$ and $^{14}_6\text{C}$ $(N=8)$ are isotopes,not isotones.

(D) The diode is ideal and forward-biased because the potential at the anode $(+6 \text{ V})$ is greater than the potential at the cathode $(+5 \text{ V})$.
In forward bias,an ideal diode acts as a short circuit (zero resistance).
Thus,the current $I$ flowing through the circuit is determined by Ohm's law:
$I = \frac{V_{\text{applied}}}{R} = \frac{6 \text{ V} - 5 \text{ V}}{100 \text{ } \Omega} = \frac{1 \text{ V}}{100 \text{ } \Omega} = 0.01 \text{ A} = 10 \text{ mA}$.
Therefore,the correct option is $(D)$.

(D) When a reverse bias is applied to a $p-n$ junction,the negative terminal of the external battery is connected to the $p$-region and the positive terminal to the $n$-region.
This configuration increases the width of the depletion layer.
As the depletion layer widens,the potential barrier at the junction increases,which makes it more difficult for the majority charge carriers to cross the junction.
Therefore,the correct option is $D$.

(B) The time constant $\tau$ for an $RC$ circuit is defined as the product of resistance $R$ and capacitance $C$: $\tau = R \times C$.
Given values are $R = 200 \Omega$ and $C = 15 \mu\text{F} = 15 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$\tau = 200 \Omega \times 15 \times 10^{-6} \text{ F}$
$\tau = 3000 \times 10^{-6} \text{ s}$
$\tau = 3 \times 10^{-3} \text{ s} = 3 \text{ ms}$.
Therefore,the correct option is $(B)$.

(B) For a point charge $2q$ at distance $r$,the electric field is $E = k \frac{2q}{r^2}$.
For a thin spherical shell of charge $q$ and radius $R$,the electric field at a distance $r' = \frac{r}{2}$ from the center is considered. Since $r \gg R$,the condition $r' > R$ is satisfied,meaning the shell acts as a point charge at its center.
Thus,$E' = k \frac{q}{(r/2)^2} = k \frac{q}{r^2/4} = 4k \frac{q}{r^2}$.
From the first equation,$k \frac{q}{r^2} = \frac{E}{2}$.
Substituting this into the expression for $E'$,we get $E' = 4 \times \frac{E}{2} = 2E$.
Therefore,the correct option is $B$.

(A) The charges are placed at $0.5R$ intervals along the $X$-axis. Given that one charge is at the origin (center of the sphere),the positions of the charges are $x = 0, \pm 0.5R, \pm 1.0R, \pm 1.5R, \dots$.
For a spherical surface of radius $1.5R$,the charges enclosed are those located at $x = 0$,$x = +0.5R$,$x = -0.5R$,$x = +1.0R$,and $x = -1.0R$.
The total number of enclosed charges is $1 + 2 + 2 = 5$.
According to Gauss's Law,the electric flux $\phi$ is given by $\phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$.
Substituting the value of enclosed charge,we get $\phi = \frac{5q}{\epsilon_0}$.
Therefore,option $(A)$ is correct.

(C) The force on the charge is $F = qE$.
Using Newton's second law,the acceleration $a$ is given by $a = \frac{F}{M} = \frac{qE}{2m}$.
Starting from rest $(u = 0)$,the speed $v$ after $n$ seconds is calculated using the equation of motion $v = u + at$.
Substituting the values,we get $v = 0 + (\frac{qE}{2m}) \times n = \frac{qEn}{2m}$.
Therefore,the correct option is $C$.

(D) Point '$A$' is at a distance $l$ from the top $+q$ charge and the bottom $+q$ charge.
The distance from point '$A$' to both $-q$ charges is $\sqrt{(2l)^2 + l^2} = \sqrt{4l^2 + l^2} = \sqrt{5l^2} = l\sqrt{5}$.
The electric potential $V$ at point '$A$' due to all four charges is given by the sum of potentials from each charge:
$V_A = k(\frac{q}{l} + \frac{q}{l} - \frac{q}{l\sqrt{5}} - \frac{q}{l\sqrt{5}})$
$V_A = k(\frac{2q}{l} - \frac{2q}{l\sqrt{5}})$
$V_A = \frac{2kq}{l}(1 - \frac{1}{\sqrt{5}})$
Therefore,option $(D)$ is correct.

(B) The electric potential $V$ inside a spherical conductor is constant and equal to the potential on its surface,given by $V = \frac{kq}{R}$.
The electric field $E$ on the surface of the spherical conductor is given by $E = \frac{kq}{R^2}$.
To find the ratio of the electric potential inside to the electric field on the surface,we calculate:
$\frac{V}{E} = \frac{kq/R}{kq/R^2} = \frac{kq}{R} \times \frac{R^2}{kq} = R$.
Therefore,the ratio is $R$. Option $(B)$ is correct.

(A) In a parallel plate capacitor,the electric field between the plates is uniform.
Since the force $F = eE$,where $e$ is the charge of the electron and $E$ is the uniform electric field,the force experienced by an electron at any point between the plates is constant.
Thus,the forces acting on the electrons at both points $P$ and $Q$ are the same.
Therefore,option $(A)$ is correct.

(D) The relation between current $I$ and drift velocity $v_d$ is given by the formula $I = n e A v_d$,where $n$ is the electron density and $e$ is the charge of an electron.
From this,the drift velocity is expressed as $v_d = \frac{I}{neA}$.
Given that the new current $I' = 2I$ and the new area of cross-section $A' = 2A$,the new drift velocity $v'_d$ is calculated as:
$v'_d = \frac{I'}{neA'} = \frac{2I}{ne(2A)} = \frac{I}{neA} = v_d$.
Therefore,the drift velocity remains unchanged. Option $(D)$ is correct.

(D) In a balanced Wheatstone bridge,the ratio of resistances in opposite arms is equal.
From the figure,the arms of the bridge are $15 \Omega$,$10 \Omega$,$r \Omega$,and the parallel combination of $r \Omega$ and $n \Omega$.
Let $R_1 = 15 \Omega$,$R_2 = 10 \Omega$,$R_3 = r \Omega$,and $R_4 = \frac{r \cdot n}{r + n} \Omega$.
The condition for a balanced Wheatstone bridge is $\frac{R_1}{R_3} = \frac{R_2}{R_4}$.
Substituting the values: $\frac{15}{r} = \frac{10}{\frac{rn}{r+n}}$.
This simplifies to $\frac{15}{r} = \frac{10(r+n)}{rn}$.
Canceling $r$ from both sides: $15 = \frac{10(r+n)}{n}$.
$15n = 10r + 10n \implies 5n = 10r \implies n = 2r$.
Assuming the question implies a specific ratio or standard values where $r$ is given as $1.25 \Omega$ (or similar context from standard problems),if $r = 1.25 \Omega$,then $n = 2(1.25) = 2.5 = \frac{5}{2} \Omega$.
Thus,the correct option is $D$.

(B) The circuit is a voltage divider with two resistors in series,$R_1 = 64 \, \Omega$ and $R_2 = 32 \, \Omega$,connected across a potential difference of $60 \text{ V} - 0 \text{ V} = 60 \text{ V}$.
The total resistance of the circuit is $R_{eq} = R_1 + R_2 = 64 \, \Omega + 32 \, \Omega = 96 \, \Omega$.
The current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{60 \text{ V}}{96 \, \Omega} = \frac{5}{8} \text{ A}$.
The potential difference between points $P$ and $Q$ is the voltage across the $32 \, \Omega$ resistor.
$V_{PQ} = I \cdot R_2 = \left( \frac{5}{8} \text{ A} \right) \times 32 \, \Omega = 5 \times 4 = 20 \text{ V}$.

(A) The magnetic field at the centre of a circular ring carrying current $I$ is given by $B_c = \frac{\mu_0 I}{2R}$.
The magnetic field at a point on the axis of the ring at a distance $x$ from the centre is given by $B_a = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$.
Given $x = 2\sqrt{2}R$,we calculate the term $(R^2+x^2)$:
$R^2 + x^2 = R^2 + (2\sqrt{2}R)^2 = R^2 + 8R^2 = 9R^2$.
Substituting this into the expression for $B_a$:
$B_a = \frac{\mu_0 I R^2}{2(9R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(27R^3)} = \frac{\mu_0 I}{54R}$.
Now,the ratio $B_c/B_a$ is:
$\frac{B_c}{B_a} = \frac{\mu_0 I}{2R} \div \frac{\mu_0 I}{54R} = \frac{54}{2} = 27$.
Therefore,the ratio is $27:1$.

(C) Current sensitivity $I_s$ is defined as the deflection per unit current,given by $I_s = \frac{\theta}{I} = \frac{NAB}{k}$.
Here,$N$ is the number of turns (dimensionless),$A$ is the area $[L^2]$,$B$ is the magnetic field,and $k$ is the restoring torque per unit twist.
The dimensions of magnetic field $B$ are $[MT^{-2}A^{-1}]$.
The dimensions of restoring torque constant $k$ are the same as torque,which is $[ML^2T^{-2}]$.
Substituting these dimensions into the formula: $[I_s] = \frac{[L^2] \cdot [MT^{-2}A^{-1}]}{[ML^2T^{-2}]}$.
Simplifying the expression: $[I_s] = \frac{L^2 MT^{-2}A^{-1}}{ML^2T^{-2}} = [A^{-1}]$.
Thus,the dimensional formula of current sensitivity is $[A^{-1}]$.

(A) The force $\vec{F}$ on a conductor of length $L$ carrying current $I$ in a magnetic field $\vec{B}$ is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
The magnitude of the force per unit length $f$ is given by $f = I B \sin\theta$,where $\theta$ is the angle between the direction of the current and the magnetic field.
In this case,the magnetic field $\vec{B}$ is directed from South to North,and the current $I$ is also flowing from South to North.
Therefore,the angle $\theta$ between the current and the magnetic field is $0^\circ$.
Since $\sin(0^\circ) = 0$,the force per unit length $f = I \times B \times \sin(0^\circ) = 0 \text{ N/m}$.

(B) For a long cylindrical wire of radius '$a$' carrying current '$I$':
$1$. Inside the wire $(r < a)$,the magnetic field is given by $B = \frac{\mu_0 I r}{2\pi a^2}$. This shows that $B$ is directly proportional to $r$ $(B \propto r)$,resulting in a linear increase from the center to the surface.
$2$. Outside the wire $(r > a)$,the magnetic field is given by $B = \frac{\mu_0 I}{2\pi r}$. This shows that $B$ is inversely proportional to $r$ $(B \propto 1/r)$,resulting in a hyperbolic decrease as distance increases.
$3$. Combining these,the graph shows a linear increase up to $r = a$ and a hyperbolic decrease for $r > a$. This corresponds to Graph $B$.

(B) Paramagnetic substances possess a small positive magnetic susceptibility. When placed in a non-uniform magnetic field,they experience a net force directed towards the region of higher magnetic field intensity. Therefore,a paramagnetic substance tends to move from a region of weak magnetic field to a region of strong magnetic field.

According to Lenz's law, the induced current creates a magnetic field that opposes the change in magnetic flux.
In figure (a), the magnetic field is directed into the page $(\times)$ and is increasing. To oppose this increase, the induced current must create a magnetic field directed out of the page $(\bullet)$. According to the right-hand rule, an outward magnetic field corresponds to an anticlockwise current.
In figure (b), the magnetic field is directed out of the page $(\bullet)$ and is decreasing. To oppose this decrease, the induced current must create a magnetic field into the page $(\times)$. According to the right-hand rule, this corresponds to a clockwise current.
Therefore, the direction of current in ring (a) is anticlockwise and in ring (b) is clockwise.

(C) The magnetic flux linkage $\phi$ in the secondary coil is related to the current $I$ in the primary coil by the formula $\phi = M I$,where $M$ is the mutual inductance.
To find the change in flux linkage $\Delta \phi$,we use the formula $\Delta \phi = M \Delta I$.
Given:
Mutual inductance $M = 2 \text{H}$.
Change in current $\Delta I = I_f - I_i = 30 \text{A} - 0 \text{A} = 30 \text{A}$.
Substituting these values into the formula:
$\Delta \phi = 2 \text{H} \times 30 \text{A} = 60 \text{ Wb}$.
Thus,the change in flux linkage with the other coil is $60 \text{ Wb}$.

(D) The induced emf in an $AC$ generator is given by the equation $\varepsilon = \varepsilon_0 \sin(\omega t)$.
At $t = 0$,$\varepsilon = \varepsilon_0 \sin(0) = 0$.
The emf reaches its maximum value when $\sin(\omega t) = 1$.
This occurs when $\omega t = \frac{\pi}{2}$.
Solving for $t$,we get $t = \frac{\pi}{2\omega}$.
Therefore,the induced emf is maximum at time $t = \frac{\pi}{2\omega}$.