GSEB 2023 Physics Question Paper with Answer and Solution

(B) Given: $I_{rms} = 5 \text{ A}$,frequency $f = 50 \text{ Hz}$.
At $t=0$,$I=0$,so the equation for current is $I = I_0 \sin(\omega t)$.
First,calculate the peak current $I_0 = \sqrt{2} \times I_{rms} = 5\sqrt{2} \text{ A}$.
Next,calculate the angular frequency $\omega = 2\pi f = 2\pi \times 50 = 100\pi \text{ rad/s}$.
Now,substitute $t = \frac{1}{300} \text{ s}$ into the equation:
$I = 5\sqrt{2} \sin(100\pi \times \frac{1}{300})$
$I = 5\sqrt{2} \sin(\frac{\pi}{3})$
Since $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$,we get:
$I = 5\sqrt{2} \times \frac{\sqrt{3}}{2} = 5 \sqrt{\frac{2 \times 3}{4}} = 5 \sqrt{\frac{3}{2}} \text{ A}$.

(D) The power transmitted in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.
From this,the current is $I_{rms} = \frac{P}{V_{rms} \cos \phi}$.
For a fixed power $P$ and voltage $V_{rms}$,if the power factor $\cos \phi$ is low,the current $I_{rms}$ must be large to transmit the same amount of power.
The power loss in the transmission lines is given by $P_{loss} = I_{rms}^2 R$.
Since $I_{rms}$ is large,the power loss $I_{rms}^2 R$ will be large.
Therefore,a low power factor implies a large power loss in transmission.

(C) The transformer equation is given by $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
Given:
Primary voltage $V_p = 2300 \ V$
Secondary voltage $V_s = 230 \ V$
Primary turns $N_p = 4000$
Substituting these values into the equation:
$\frac{230}{2300} = \frac{N_s}{4000}$
$N_s = \frac{230 \times 4000}{2300}$
$N_s = \frac{1}{10} \times 4000 = 400$.
Therefore,the number of turns in the secondary winding should be $400$.

(A) The correct answer is $A$. Inductive reactance $(X_L)$ is defined as the opposition offered by an inductor to the flow of alternating current.
$X_L = \omega L = 2 \pi f L$
where $f$ is the frequency of the $AC$ source and $L$ is the inductance.
Since $X_L$ is directly proportional to the frequency $(f)$,it effectively limits the flow of alternating current in a circuit.
For $DC$ circuits,the frequency $f = 0$,which means $X_L = 0$. Therefore,a pure inductor offers no resistance to $DC$ current.

(C) The total energy of an electron in a hydrogen atom is given by the formula:
$E_{n} = \frac{-13.6}{n^{2}} \text{ eV}$
From this,we can see that the energy is inversely proportional to the square of the principal quantum number:
$E_{n} \propto \frac{1}{n^{2}}$
The first excited state corresponds to $n = 2$,and the third excited state corresponds to $n = 4$.
Therefore,the ratio of the energy in the first excited state $(E_{2})$ to the energy in the third excited state $(E_{4})$ is:
$\frac{E_{2}}{E_{4}} = \frac{n_{4}^{2}}{n_{2}^{2}} = \left(\frac{4}{2}\right)^{2}$
$\frac{E_{2}}{E_{4}} = (2)^{2} = \frac{4}{1}$
Thus,the ratio is $4: 1$.

(B) According to Bohr's postulate,the angular momentum $L$ of an electron in an orbit $n$ is given by the formula:
$L = \frac{nh}{2\pi}$
Given that $n = 5$ and Planck's constant $h \approx 6.63 \times 10^{-34} \text{ J s}$:
$L = \frac{5 \times 6.63 \times 10^{-34}}{2 \times 3.14}$
$L = \frac{33.15 \times 10^{-34}}{6.28}$
$L \approx 5.278 \times 10^{-34} \text{ J s}$
Rounding to two significant figures,we get $L \approx 5.3 \times 10^{-34} \text{ J s}$.

(C) The correct option is $C$.
When a steady current flows through a metallic conductor of non-uniform cross-section,the current $I$ remains constant throughout the conductor due to the principle of conservation of charge.
From the relation for drift velocity,$v_d = \frac{I}{neA}$,since $n$,$e$,and $I$ are constant,$v_d$ depends on the cross-sectional area $A$.
From the relation for current density,$J = \frac{I}{A}$,since $I$ is constant,$J$ depends on $A$.
From the relation for electric field,$E = \rho J = \frac{\rho I}{A}$,since $\rho$ and $I$ are constant,$E$ depends on $A$.
Therefore,only the current $I$ remains constant along the conductor.

(D) Kirchhoff's loop rule (also known as Kirchhoff's second law or the voltage law) states that the algebraic sum of changes in potential around any closed loop in a circuit is zero.
This is based on the fact that electrostatic force is a conservative force,meaning the work done in moving a unit charge around a closed path is zero.
Therefore,the loop rule is a direct consequence of the law of conservation of energy.
Thus,the correct option is $D$.

(D) For two batteries connected in parallel,the equivalent emf $\varepsilon_{eq}$ and equivalent internal resistance $r_{eq}$ are given by the formulas:
$\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2}$
and
$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \implies r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$
Substituting $r_{eq}$ into the first equation:
$\varepsilon_{eq} = \left( \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2} \right) \times \left( \frac{r_1 r_2}{r_1 + r_2} \right)$
$\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$
Since $\varepsilon_2 > \varepsilon_1$,the equivalent emf $\varepsilon_{eq}$ is a weighted average of $\varepsilon_1$ and $\varepsilon_2$,which implies that $\varepsilon_1 < \varepsilon_{eq} < \varepsilon_2$.

(D) The energy of a photon is given by the formula $E = h \nu$,where $h$ is Planck's constant $(6.63 \times 10^{-34} \,J \cdot s)$ and $\nu$ is the frequency of the light.
Given: $\nu = 6 \times 10^{14} \,Hz$.
Substituting the values: $E = (6.63 \times 10^{-34} \,J \cdot s) \times (6 \times 10^{14} \,Hz) = 3.978 \times 10^{-19} \,J$.
To convert the energy from Joules to electron-volts $(eV)$,we divide by the charge of an electron $(1.6 \times 10^{-19} \,C)$:
$E = \frac{3.978 \times 10^{-19} \,J}{1.6 \times 10^{-19} \,J/eV} \approx 2.486 \,eV$.
Rounding to the nearest value,we get $E \approx 2.5 \,eV$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h\nu - \phi_{0}$.
Since the stopping potential $V_{0}$ is related to the maximum kinetic energy by $K_{max} = eV_{0}$,we can write $eV_{0} = h\nu - \phi_{0}$.
Dividing by the elementary charge $e$,we get $V_{0} = \frac{h}{e}\nu - \frac{\phi_{0}}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V_{0}$,$x = \nu$,and $c = -\frac{\phi_{0}}{e}$,the slope $m$ is equal to $\frac{h}{e}$.

(C) The electric charge of an $\alpha$-particle is $q = 2e = 2 \times 1.6 \times 10^{-19} \ C = 3.2 \times 10^{-19} \ C$.
Given distance $r = 3 \ cm = 3 \times 10^{-2} \ m$.
The Coulombian repulsive force is given by $F = \frac{k q_1 q_2}{r^2}$.
Since $q_1 = q_2 = q$,we have $F = \frac{k q^2}{r^2}$.
Substituting the values:
$F = \frac{9 \times 10^9 \times (3.2 \times 10^{-19})^2}{(3 \times 10^{-2})^2}$
$F = \frac{9 \times 10^9 \times 10.24 \times 10^{-38}}{9 \times 10^{-4}}$
$F = 10.24 \times 10^{9 - 38 + 4}$
$F = 10.24 \times 10^{-25} \ N = 1.024 \times 10^{-24} \ N$.

(B) The surface charge density $\sigma$ is defined as the charge $Q$ per unit area $A$ of the surface.
Given: Radius $r = 25 \ cm = 0.25 \ m = \frac{1}{4} \ m$.
Surface charge density $\sigma = \frac{3}{\pi} \ \mu C/m^2$.
The surface area of a spherical shell is $A = 4 \pi r^2$.
Using the formula $\sigma = \frac{Q}{A}$,we get $Q = \sigma \times A = \sigma \times 4 \pi r^2$.
Substituting the values:
$Q = \left( \frac{3}{\pi} \ \mu C/m^2 \right) \times 4 \pi \times (0.25 \ m)^2$
$Q = 3 \times 4 \times (0.0625) \ \mu C$
$Q = 12 \times 0.0625 \ \mu C = 0.75 \ \mu C$.
Therefore,the charge required is $0.75 \ \mu C$.

(D) The charge on the plastic rod is $q = -8 \times 10^{-7} \ C$. Since the rod becomes negatively charged,it has gained electrons.
Using the quantization of charge formula,$q = ne$,where $e = 1.6 \times 10^{-19} \ C$ is the magnitude of the charge of an electron.
$n = \frac{|q|}{e} = \frac{8 \times 10^{-7}}{1.6 \times 10^{-19}}$
$n = 5 \times 10^{12}$
Since the plastic rod gained a negative charge,electrons must have been transferred from the wool to the plastic rod.

(B) Electric flux is defined by the formula $\Phi = E \cdot A$.
The dimensional formula of the Electric Field $(E)$ is $[M^1 L^1 T^{-3} A^{-1}]$.
The dimensional formula of Area $(A)$ is $[L^2]$.
Thus,the dimensional formula of electric flux is $[M^1 L^1 T^{-3} A^{-1}] \times [L^2] = [M^1 L^3 T^{-3} A^{-1}]$.

(B) The electric field lines are closer together on the left side,indicating a stronger electric field $(E_L > E_R)$.
For a dipole,the force on the negative charge $(-q)$ is in the direction opposite to the electric field,and the force on the positive charge $(+q)$ is in the direction of the electric field.
Let $E_L$ be the electric field at the position of $-q$ and $E_R$ be the electric field at the position of $+q$.
The force on $-q$ is $F_L = q E_L$ directed towards the right.
The force on $+q$ is $F_R = q E_R$ directed towards the left.
Since the field lines are denser on the left,$E_L > E_R$,therefore $F_L > F_R$.
The net force is $F_{net} = F_L - F_R$,which is directed towards the right.

(B) According to Faraday's law of electromagnetic induction,the induced electromotive force (emf) $\varepsilon$ is given by the rate of change of magnetic flux: $\varepsilon = |\frac{d\phi}{dt}|$.
Given $\phi(t) = 2t^2 + 2t + 1$,we differentiate with respect to $t$:
$\varepsilon = |\frac{d}{dt}(2t^2 + 2t + 1)| = |4t + 2|$.
At $t = 2 \ s$,the induced emf is:
$\varepsilon = 4(2) + 2 = 10 \ V$.
Using Ohm's law,the current $i$ is given by $i = \frac{\varepsilon}{R}$.
Given resistance $R = 10 \ \Omega$,the current is:
$i = \frac{10 \ V}{10 \ \Omega} = 1 \ A$.

(A) The self-inductance $L$ of an inductor 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 inductor.
This formula shows that the self-inductance $L$ depends only on the geometric properties and the core material of the inductor.
It is independent of the current $I$ flowing through it.
Therefore,if the current is doubled,the self-inductance remains unchanged at $4 \ H$.

(B) The magnetic flux $\phi$ through a surface is given by the dot product of the magnetic field vector $\overrightarrow{B}$ and the area vector $\overrightarrow{A}$.
Since the square of side $L$ lies in the $x-y$ plane,its area vector $\overrightarrow{A}$ is directed along the $z$-axis.
Therefore,$\overrightarrow{A} = L^2 \hat{k}$.
The magnetic field is $\overrightarrow{B} = B_0(2 \hat{i} + 4 \hat{j} + 3 \hat{k})$.
The magnetic flux is calculated as:
$\phi = \overrightarrow{B} \cdot \overrightarrow{A}$
$\phi = [B_0(2 \hat{i} + 4 \hat{j} + 3 \hat{k})] \cdot [L^2 \hat{k}]$
$\phi = B_0 L^2 (2 \hat{i} \cdot \hat{k} + 4 \hat{j} \cdot \hat{k} + 3 \hat{k} \cdot \hat{k})$
Since $\hat{i} \cdot \hat{k} = 0$,$\hat{j} \cdot \hat{k} = 0$,and $\hat{k} \cdot \hat{k} = 1$,we get:
$\phi = B_0 L^2 (0 + 0 + 3(1)) = 3 B_0 L^2 \text{ Wb}$.

(D) The relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is given by the equation: $E/B = c$,where $c$ is the speed of light in vacuum.
Given: $E = 6.6 \,V \,m^{-1}$ and $c = 3 \times 10^8 \,m \,s^{-1}$.
Rearranging the formula to solve for $B$: $B = E/c$.
Substituting the values: $B = 6.6 / (3 \times 10^8) \,T$.
Calculating the result: $B = 2.2 \times 10^{-8} \,T$.
Therefore,the correct option is $D$.

(C) $TV$ waves,which are a part of the radio wave spectrum used for television broadcasting,typically operate in the frequency range of $54 MHz$ to $890 MHz$. This range covers both Very High Frequency $(VHF)$ and Ultra High Frequency $(UHF)$ bands used for television signals. Therefore,the correct option is $C$.

(B) The correct option is $(B)$.
For a collection of charges with a non-zero total sum $(Q \neq 0)$,the system behaves like a point charge when viewed from a very large distance $(r \to \infty)$.
The electric potential $V$ due to a point charge $q$ is given by the formula:
$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$
This equation shows that the potential $V$ depends only on the distance $r$ from the charge. Therefore,all points at a constant distance $r$ from the charge have the same potential.
The locus of points at a constant distance $r$ from a point source in three-dimensional space forms the surface of a sphere.
Thus,at a very large distance,the equipotential surfaces of such a charge distribution are approximately spheres.

(B) The electric potential $V$ at a point is defined as the electric potential energy $U$ per unit charge $q$.
Formula: $V = \frac{U}{q}$
Given:
Charge $q = 2 \mu C = 2 \times 10^{-6} \text{ C}$
Potential Energy $U = 3 \times 10^{-5} \text{ J}$
Calculation:
$V = \frac{3 \times 10^{-5}}{2 \times 10^{-6}}$
$V = 1.5 \times 10^1$
$V = 15 \text{ V}$
Therefore,the electric potential at the point is $15 \text{ V}$.

(D) The initial energy stored in the capacitor is $U = \frac{Q^2}{2C}$,where $Q$ is the initial charge and $C$ is the capacitance.
When the charged capacitor is disconnected from the battery and connected in parallel to two other identical uncharged capacitors,the total charge $Q$ is shared equally among the three capacitors because they are identical.
Therefore,the charge on each capacitor becomes $Q' = \frac{Q}{3}$.
The new energy $U'$ stored in each capacitor is given by $U' = \frac{(Q')^2}{2C}$.
Substituting $Q' = \frac{Q}{3}$ into the equation:
$U' = \frac{(\frac{Q}{3})^2}{2C} = \frac{Q^2}{9 \times 2C} = \frac{1}{9} \times \frac{Q^2}{2C}$.
Since $U = \frac{Q^2}{2C}$,we get $U' = \frac{U}{9}$.

(C) For a positive charge $+q$,the electric potential at a distance $r$ is given by $V = \frac{kq}{r}$. Since $V \propto \frac{1}{r}$,the potential is higher closer to the charge.
For the positive charge,point $Q$ is closer to the charge than point $P$,so $r_Q < r_P$. This implies $V_Q > V_P$,therefore $V_Q - V_P > 0$ (positive).
For a negative charge $-q$,the electric potential is given by $V = \frac{k(-q)}{r} = -\frac{kq}{r}$. Here,the potential is more negative (lower) closer to the charge.
For the negative charge,point $A$ is closer to the charge than point $B$,so $r_A < r_B$. This implies $V_A < V_B$ (since $V_A$ is more negative than $V_B$). Therefore,$V_B - V_A > 0$ (positive).
Thus,both potential differences are positive.

(A) Ferromagnetic materials are characterized by their ability to be strongly magnetized in an external magnetic field.
They possess a very high relative permeability $(\mu_r \gg 1)$, which allows them to concentrate magnetic field lines within themselves.
Additionally, they exhibit high retentivity, meaning they retain a significant amount of magnetization even after the external magnetic field is removed.
Therefore, the correct description is high permeability and high retentivity.
However, based on the provided options, the intended answer is $A$.

(B) The magnetic induction $B$ is defined by the force $F$ on a current-carrying wire of length $l$ carrying current $I$ as $F = BIl \sin \theta$.
Thus,the $SI$ unit of magnetic induction is $\frac{\text{Newton}}{\text{Ampere-meter}}$.
Since $1 \text{ Tesla} = 1 \frac{\text{Weber}}{\text{m}^2} = 1 \frac{\text{Newton}}{\text{Ampere-meter}}$,options $A$,$C$,and $D$ are all valid units of magnetic induction.
Option $B$,$\frac{\text{Newton-meter}}{\text{Ampere}}$,is not a unit of magnetic induction.

(C) Given:
Resistance of galvanometer $(G)$ = $10 \ \Omega$
Full-scale deflection current $(I_g)$ = $3 \ \text{mA} = 3 \times 10^{-3} \ \text{A}$
Range of ammeter $(I)$ = $10 \ \text{A}$
The formula for shunt resistance $(S)$ required to convert a galvanometer into an ammeter is given by:
$S = \frac{I_g \cdot G}{I - I_g}$
Substituting the values:
$S = \frac{3 \times 10^{-3} \times 10}{10 - 3 \times 10^{-3}}$
Since $3 \times 10^{-3} = 0.003 \ \text{A}$,the denominator is $10 - 0.003 = 9.997 \ \text{A}$.
$S = \frac{0.03}{9.997} \approx \frac{0.03}{10} = 0.003 \ \Omega$
Therefore,$S = 3 \times 10^{-3} \ \Omega$.

(D) The magnetic field is given by $\vec{B} = 0.3 \hat{k} \text{ T}$.
The area vector $\vec{A}$ of the loop placed in the $xy$-plane is perpendicular to the plane,so $\vec{A} = (10 \times 10^{-2} \text{ m} \times 5 \times 10^{-2} \text{ m}) \hat{k} = 50 \times 10^{-4} \hat{k} \text{ m}^2 = 5 \times 10^{-3} \hat{k} \text{ m}^2$.
The magnetic moment $\vec{m}$ is given by $\vec{m} = I \vec{A} = 12 \times 5 \times 10^{-3} \hat{k} = 60 \times 10^{-3} \hat{k} = 0.06 \hat{k} \text{ A m}^2$.
The torque $\vec{\tau}$ acting on the loop is given by $\vec{\tau} = \vec{m} \times \vec{B}$.
Substituting the values,$\vec{\tau} = (0.06 \hat{k}) \times (0.3 \hat{k})$.
Since the cross product of a vector with itself is zero $(\hat{k} \times \hat{k} = 0)$,the torque $\vec{\tau} = 0$.

(A) The mass of ${ }_{1}^{2} H$ is $2.014102 \ u$.
The mass of ${ }_{1}^{1} H$ is $1.007825 \ u$.
The mass of ${ }_{2}^{3} He$ is $3.016029 \ u$.
The mass defect $\Delta m = (m({ }_{1}^{2} H) + m({ }_{1}^{1} H)) - m({ }_{2}^{3} He)$.
$\Delta m = (2.014102 + 1.007825) - 3.016029 = 3.021927 - 3.016029 = 0.005898 \ u$.
The energy released $Q = \Delta m \times 931.5 \ MeV/u$.
$Q = 0.005898 \times 931.5 \approx 5.49 \ MeV$.

(D) According to the mass-energy equivalence relation:
$E = mc^{2}$
Therefore,$m = \frac{E}{c^{2}}$
Substituting the given values:
$m = \frac{9 \times 10^{13}}{(3 \times 10^{8})^{2}}$
$m = \frac{9 \times 10^{13}}{9 \times 10^{16}}$
$m = 10^{-3} \text{ kg}$
Since $1 \text{ kg} = 1000 \text{ g}$,we have:
$m = 10^{-3} \times 10^{3} \text{ g} = 1 \text{ g}$
Thus,the correct option is $D$.

(B) mirror works on the principle of the reflection of light,where light rays bounce off the surface of the mirror. $A$ lens works on the principle of the refraction of light,where light rays pass through the medium and change their direction due to a change in speed. Therefore,the correct phenomena are reflection and refraction respectively.

(B) The focal length of a plane mirror is $f = \infty$.
The power $P$ of a mirror is defined as the reciprocal of its focal length in meters,given by $P = \frac{1}{f}$.
Substituting the value of $f$:
$P = \frac{1}{\infty} = 0 \text{ dioptre}$.
Therefore,the power of a plane mirror is $0$.

(B) The apparent depth of an object viewed through a glass slab is given by the formula:
$\text{Apparent depth} = \frac{\text{Real depth}}{\mu}$
Given, real depth $(t)$ = $9 \ cm$ and refractive index $(\mu)$ = $1.5$.
$\text{Apparent depth} = \frac{9}{1.5} = 6 \ cm$
The shift or the distance by which the pin appears to be raised is given by:
$\text{Shift} = \text{Real depth} - \text{Apparent depth}$
$\text{Shift} = 9 \ cm - 6 \ cm = 3 \ cm$
Therefore, the pin appears to be raised by $3 \ cm$.

(D) When unpolarized light of intensity $I_1$ is incident on a polaroid,the intensity of the emergent polarized light $I_2$ is given by Malus's Law for unpolarized light.
According to the law,the intensity of light transmitted through a polarizer is half the intensity of the incident unpolarized light.
Therefore,$I_2 = \frac{I_1}{2}$.
Rearranging this equation,we get $I_1 = 2I_2$.

(A) The distance between any two consecutive bright fringes is known as fringe width,denoted by $\beta$.
The formula for fringe width is $\beta = \frac{\lambda D}{d}$.
Given values are:
Wavelength $\lambda = 600 \ nm = 600 \times 10^{-9} \ m$.
Distance between slits and screen $D = 1.5 \ m$.
Distance between slits $d = 0.2 \ mm = 0.2 \times 10^{-3} \ m$.
Substituting these values into the formula:
$\beta = \frac{600 \times 10^{-9} \times 1.5}{0.2 \times 10^{-3}}$
$\beta = \frac{900 \times 10^{-9}}{0.2 \times 10^{-3}}$
$\beta = 4500 \times 10^{-6} \ m$
$\beta = 4.5 \times 10^{-3} \ m = 4.5 \ mm$.
Thus,the distance between any two consecutive bright fringes is $4.5 \ mm$.

(C) The refractive index $n$ of a medium is defined as the ratio of the speed of light in a vacuum $c$ to the speed of light in the medium $v$,given by $n = \frac{c}{v}$.
Given $n = \frac{3}{2}$ and $c = 3 \times 10^{8} \,ms^{-1}$.
Rearranging the formula to solve for $v$,we get $v = \frac{c}{n}$.
Substituting the values,$v = \frac{3 \times 10^{8}}{3/2} = 3 \times 10^{8} \times \frac{2}{3} = 2 \times 10^{8} \,ms^{-1}$.
Therefore,the correct option is $C$.

(B) In single-slit diffraction,the condition for the first minimum (where the light is diffracted) is given by the formula $a \sin \theta = n \lambda$.
For the first minimum,we take $n = 1$,so $a \sin \theta = \lambda$.
Since the angle $\theta$ is very small,we can approximate $\sin \theta \approx \theta$.
Therefore,$a \theta \approx \lambda$,which gives $\theta \approx \frac{\lambda}{a}$.
Thus,the angle of diffraction is approximately $\frac{\lambda}{a}$.