GSEB 2024 Physics Question Paper with Answer and Solution

(A) The magnetic dipole moment of a current-carrying loop is given by $m = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the loop.
For the first case,$N_1 = 1$,$R_1 = R$,and $A_1 = \pi R^2$. Thus,$m = I \pi R^2$.
When the wire of length $L$ is converted into a ring of $N_2 = 2$ turns,the circumference of the new ring is $2 \times (2 \pi R_2) = L$. Since $L = 2 \pi R$,we have $4 \pi R_2 = 2 \pi R$,which gives $R_2 = \frac{R}{2}$.
The new area is $A_2 = \pi R_2^2 = \pi (\frac{R}{2})^2 = \frac{\pi R^2}{4}$.
The new magnetic dipole moment $m'$ is $m' = N_2 I A_2 = 2 \times I \times \frac{\pi R^2}{4} = \frac{I \pi R^2}{2}$.
Since $m = I \pi R^2$,we get $m' = \frac{m}{2}$.

(D) Given:
Number of turns,$N = 1000$
Area of cross-section,$A = 2 \times 10^{-4} \ m^2$
Current,$I = 5.0 \ A$
The magnetic moment $\mu$ of a solenoid is given by the formula:
$\mu = N \cdot I \cdot A$
Substituting the given values:
$\mu = 1000 \times 5.0 \times (2 \times 10^{-4})$
$\mu = 5000 \times 2 \times 10^{-4}$
$\mu = 10000 \times 10^{-4}$
$\mu = 1 \ A \ m^2$
Therefore,the correct option is $D$.

(B) The force per unit length $f$ between two parallel conductors carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the formula: $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Given values are $I_1 = I_2 = 1 \text{ mA} = 10^{-3} \text{ A}$,$d = 1 \text{ m}$,and $\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$.
Substituting these values into the formula:
$f = \frac{(4\pi \times 10^{-7}) \times (10^{-3}) \times (10^{-3})}{2 \pi \times 1}$
$f = 2 \times 10^{-7} \times 10^{-6}$
$f = 2 \times 10^{-13} \text{ N/m}$.
Thus,the correct option is $B$.

(B) The magnetic field $B$ inside a long solenoid near its centre is given by the formula $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Given:
Length of solenoid $L = 120 \ cm = 1.2 \ m$.
Number of layers $= 4$.
Turns per layer $= 400$.
Total number of turns $N = 4 \times 400 = 1600$.
Current $I = 8.0 \ A$.
Number of turns per unit length $n = \frac{N}{L} = \frac{1600}{1.2} = \frac{4000}{3} \ m^{-1}$.
Substituting the values into the formula:
$B = (4 \pi \times 10^{-7} \ T \cdot m/A) \times (\frac{4000}{3} \ m^{-1}) \times (8.0 \ A)$.
$B = 4 \pi \times 10^{-7} \times \frac{32000}{3} \ T$.
$B = \frac{128000}{3} \pi \times 10^{-7} \ T$.
$B \approx 42666.67 \pi \times 10^{-7} \ T$.
$B \approx 4.266 \pi \times 10^{-3} \ T$.
Rounding to two decimal places,we get $B \approx 4.27 \pi \times 10^{-3} \ T$.

(B) For a point inside the wire $(r < a)$:
The magnetic field is given by $B_{in} = \frac{\mu_0 I r}{2 \pi a^2}$.
Thus, $B_{in} \propto r$, which represents a straight line passing through the origin.
For a point outside the wire $(r > a)$:
The magnetic field is given by $B_{out} = \frac{\mu_0 I}{2 \pi r}$.
Thus, $B_{out} \propto \frac{1}{r}$, which represents a rectangular hyperbola.
At the surface $(r = a)$, the magnetic field is maximum: $B_{max} = \frac{\mu_0 I}{2 \pi a}$.
Therefore, the graph shows a linear increase up to $r = a$ and then a hyperbolic decrease for $r > a$.

(A) When the battery is connected across the diametrically opposite points $A$ and $B$,the ring is divided into two equal semicircular arcs,$ACB$ and $ADB$.
Since the arcs are in parallel,the potential difference across both is the same $(12 \text{ V})$.
Let $R_1$ and $R_2$ be the resistances of the two semicircular parts. Since they have equal lengths and cross-sectional areas,$R_1 = R_2 = R_{arc}$.
By Ohm's law,the current $I_1$ through arc $ACB$ and $I_2$ through arc $ADB$ will be equal because the resistances are equal $(I_1 = I_2 = I/2)$.
The magnetic field at the centre due to a semicircular arc carrying current $I'$ is $B = \frac{\mu_0 I'}{4R}$.
The magnetic field due to arc $ACB$ $(B_1)$ at the centre is $\frac{\mu_0 (I/2)}{4R}$ directed into the plane (using the right-hand rule).
The magnetic field due to arc $ADB$ $(B_2)$ at the centre is $\frac{\mu_0 (I/2)}{4R}$ directed out of the plane.
Since the magnitudes are equal and the directions are opposite,the net magnetic field at the centre is $B_{net} = B_1 - B_2 = 0$.

(D) The correct answer is $D$.
Let $m$ be the average mass of a nucleon and $A$ be the mass number.
The mass of the nucleus is given by $M = m A$.
The radius of the nucleus is given by $R = R_0 A^{1/3}$,where $R_0$ is a constant.
The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
The density of the nucleus $\rho$ is defined as the ratio of mass to volume:
$\rho = \frac{M}{V} = \frac{m A}{\frac{4}{3} \pi R_0^3 A} = \frac{m}{\frac{4}{3} \pi R_0^3}$.
Since $m$ and $R_0$ are constants,the density $\rho$ is independent of the mass number $A$.

(B) In a nuclear fission reaction,both the total mass number $(A)$ and the total atomic number $(Z)$ must be conserved.
Given reaction: ${ }_{0}^{1} n+{ }_{92}^{235} U \rightarrow{ }_{92}^{236} U \rightarrow X + { }_{41}^{99} Nb + Y{ }_{0}^{1} n$.
Conservation of atomic number $(Z)$:
$92 = Z_X + 41 \Rightarrow Z_X = 92 - 41 = 51$.
Conservation of mass number $(A)$:
$236 = A_X + 99 + Y(1)$.
For option $(B)$,$X = { }_{51}^{133} Sb$ and $Y = 4$.
$A_X = 133$.
$133 + 99 + 4 = 236$.
Since both $A$ and $Z$ are conserved,option $(B)$ is correct.

(C) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
Therefore,the ratio of the radii is given by $\frac{R_{Au}}{R_{Ag}} = \left( \frac{A_{Au}}{A_{Ag}} \right)^{1/3}$.
Given $A_{Au} = 197$ and $A_{Ag} = 107$,we have:
$\frac{R_{Au}}{R_{Ag}} = \left( \frac{197}{107} \right)^{1/3}$.
Calculating the value: $\frac{197}{107} \approx 1.841$.
Then,$(1.841)^{1/3} \approx 1.2256$.
Rounding to two decimal places,we get $1.23$.

(B) The correct answer is $B$ (isotones).
For ${ }_{92}^{238} U$,the number of neutrons $N$ is calculated as $N = A - Z = 238 - 92 = 146$.
For ${ }_{90}^{236} Th$,the number of neutrons $N$ is calculated as $N = A - Z = 236 - 90 = 146$.
Since both nuclei have the same number of neutrons $(N = 146)$,they are isotones.

(B) The average atomic mass is calculated by taking the weighted average of the isotopic masses based on their relative abundances.
Average atomic mass $= \frac{(m_1 \times p_1) + (m_2 \times p_2)}{100}$
$= \frac{(34.98 \times 75.4) + (36.98 \times 24.6)}{100}$
$= \frac{2637.492 + 909.708}{100}$
$= \frac{3547.2}{100}$
$= 35.47 \ u$

(C) According to Einstein's mass-energy equivalence relation,the energy $E$ is given by $E = mc^2$.
Given mass $m = 1 \mu g = 1 \times 10^{-6} \ kg$.
Speed of light $c = 3 \times 10^{8} \ m/s$.
Substituting the values into the formula:
$E = (1 \times 10^{-6} \ kg) \times (3 \times 10^{8} \ m/s)^2$
$E = 1 \times 10^{-6} \times 9 \times 10^{16} \ J$
$E = 9 \times 10^{10} \ J$.
Therefore,the correct option is $C$.

(D) The correct answer is $D$. Iron $(Fe)$.
The binding energy per nucleon curve shows that the binding energy per nucleon increases with the mass number $A$ for light nuclei and reaches a maximum value for nuclei with mass numbers between $40$ and $120$.
For Iron $(^{56}Fe)$,the binding energy per nucleon is approximately $8.75 \text{ MeV}$,which is the highest among the given options.
Therefore,Iron is the most stable nucleus among the choices provided.

(C) The radius of a nucleus is given by the formula $R = R_{0} A^{1/3}$,where $R_{0} = 1.2 \text{ fm} = 1.2 \times 10^{-15} \text{ m}$ and $A$ is the mass number.
For ${ }_{13}^{27} Al$,the mass number $A = 27$.
Substituting the values into the formula:
$R = 1.2 \times 10^{-15} \times (27)^{1/3} \text{ m}$
Since $(27)^{1/3} = 3$,we get:
$R = 1.2 \times 10^{-15} \times 3 \text{ m}$
$R = 3.6 \times 10^{-15} \text{ m}$
Thus,the correct option is $C$.

(B) The given nuclear fusion reaction is: ${ }_1^3 H + { }_1^3 H \rightarrow { }_2^4 He + { }_1^1 H + { }_1^1 H + Q$.
To find the energy released $(Q)$, we calculate the mass defect $(\Delta m)$:
Mass of reactants: $2 \times M({ }_1^3 H) = 2 \times 3.016049 \text{ u} = 6.032098 \text{ u}$.
Mass of products: $M({ }_2^4 He) + 2 \times M({ }_1^1 H) = 4.002603 \text{ u} + 2 \times 1.007825 \text{ u} = 4.002603 + 2.015650 = 6.018253 \text{ u}$.
Mass defect $\Delta m = 6.032098 \text{ u} - 6.018253 \text{ u} = 0.013845 \text{ u}$.
Energy released $Q = \Delta m \times 931.5 \text{ MeV/u} = 0.013845 \times 931.5 \approx 12.89 \text{ MeV}$.
Rounding to the nearest provided option, the correct value is $12.86 \text{ MeV}$.

(B) The magnifying power $m$ of a telescope in normal adjustment is given by the ratio of the focal length of the objective lens $(f_o)$ to the focal length of the eyepiece $(f_e)$.
Given:
$f_o = 140 \ cm$
$f_e = 5 \ cm$
Using the formula:
$m = \frac{f_o}{f_e}$
$m = \frac{140}{5} = 28$
Therefore,the magnifying power is $28$.

(A) The refractive index $n$ of a medium is defined as the ratio of the speed of light in vacuum $c$ to the speed of light in the medium $v$.
The formula is given by $n = \frac{c}{v}$.
Rearranging the formula to solve for $v$,we get $v = \frac{c}{n}$.
Given $c = 3 \times 10^{8} \,m \,s^{-1}$ and $n = 1.25$.
Substituting the values: $v = \frac{3 \times 10^{8}}{1.25} = 2.4 \times 10^{8} \,m \,s^{-1}$.
Therefore,the correct option is $A$.

(A) The correct answer is $GaAs$.
Semiconductors are classified into elemental and compound semiconductors.
$Si$ (Silicon),$Ge$ (Germanium),and $C$ (Carbon) are elemental semiconductors as they consist of only one type of atom.
$GaAs$ (Gallium Arsenide) is an inorganic compound semiconductor because it is formed by the combination of two different elements,Gallium $(Ga)$ and Arsenic $(As)$.

(C) The energy band gap $(E_g)$ is the energy difference between the conduction band and the valence band.
For Carbon (diamond),the band gap is approximately $5.4 \ eV$.
For Silicon,the band gap is approximately $1.1 \ eV$.
For Germanium,the band gap is approximately $0.7 \ eV$.
Comparing these values,we find that $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$.
Therefore,option $(C)$ is correct.

(D) In an $n$-type semiconductor,the material is doped with pentavalent impurity atoms (such as Phosphorus,Arsenic,or Antimony) to provide extra electrons.
Since these pentavalent atoms provide free electrons,electrons become the majority charge carriers.
Conversely,holes are generated only due to thermal excitation,making them the minority charge carriers.
Therefore,the correct statement is that holes are minority carriers and pentavalent atoms are the dopants.

(A) In the given circuit,the diode is connected in reverse bias because the positive terminal of the battery is connected to the $n$-side (cathode) and the negative terminal is connected to the $p$-side (anode) of the diode.
Since the diode has infinite reverse bias resistance,the current $I_1$ flowing through the diode branch is $I_1 = 0.0 \ A$.
The resistor of $50 \ \Omega$ is connected in parallel to the battery of $10 \ V$. Therefore,the current $I_2$ flowing through the resistor is given by Ohm's law:
$I_2 = \frac{V}{R} = \frac{10 \ V}{50 \ \Omega} = 0.2 \ A$.
Thus,$I_1 = 0.0 \ A$ and $I_2 = 0.2 \ A$.

(D) The energy gap $(E_g)$ is the energy difference between the valence band and the conduction band.
- For metals,the valence and conduction bands overlap,so $E_g = 0$.
- For semiconductors,the energy gap is small,typically $E_g < 3 \ eV$ (e.g.,Silicon is $1.1 \ eV$,Germanium is $0.7 \ eV$).
- For insulators (often referred to as non-metals in this context),the energy gap is very large,typically $E_g > 3 \ eV$.
Therefore,the correct answer is insulators or non-metals.

(B) $CdS$ (Cadmium Sulfide) is a compound semiconductor.
It is composed of elements from groups $12$ and $16$ of the periodic table.
Since it is a compound formed by inorganic elements,it is classified as an inorganic semiconductor.
Therefore,the correct option is $B$.

(B) When a $p-n$ junction is forward biased,the positive terminal of the external battery is connected to the $p$-side and the negative terminal to the $n$-side.
This external electric field opposes the internal electric field of the depletion region.
As a result,the width of the depletion layer decreases,and the potential barrier height is lowered.
Therefore,the correct option is $B$.

(D) In a $p$-type semiconductor,the material is doped with trivalent impurity atoms (such as Boron,Aluminum,or Indium).
These trivalent atoms create vacancies in the crystal lattice,which act as holes.
Therefore,holes become the majority charge carriers,while electrons are the minority charge carriers.
Thus,the correct statement is that holes are majority carriers and trivalent atoms are the dopants.

(B) According to Huygen's principle,every point on a wavefront acts as a source of secondary wavelets. The amplitude of these secondary wavelets is given by the factor $(1 + \cos \theta) / 2$,where $\theta$ is the angle between the normal to the wavefront and the direction of the secondary wavelet.
$1$. In the forward direction,$\theta = 0^\circ$,so the amplitude factor is $(1 + \cos 0^\circ) / 2 = (1 + 1) / 2 = 1$ (Maximum).
$2$. In the backward direction,$\theta = 180^\circ$,so the amplitude factor is $(1 + \cos 180^\circ) / 2 = (1 - 1) / 2 = 0$ (Zero).
Therefore,the amplitude is maximum in the forward direction and zero in the backward direction.

(C) When a plane wavefront is incident on a convex lens,the lens causes the light rays to converge at the focal point.
Since the wavefront is always perpendicular to the direction of light propagation (rays),the converging rays result in a spherical wavefront that is converging towards the focal point.
Therefore,the emerging wavefront is spherical.

(C) The resultant intensity $I$ of two superposing waves is given by the formula: $I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} \cos \phi$.
For non-coherent sources,the phase difference $\phi$ changes randomly with time.
Therefore,the average value of the interference term $\langle 2\sqrt{I_{1}I_{2}} \cos \phi \rangle$ over time is $0$ because the average value of $\cos \phi$ is $0$.
Given that $I_{1} = I_{2} = I_{0}$,the resultant average intensity is:
$I_{avg} = I_{1} + I_{2} = I_{0} + I_{0} = 2I_{0}$.

(C) According to Huygens' principle,the speed of light in a denser medium (glass) is less than in a rarer medium (air).
When a plane wavefront passes through a prism,the portion of the wavefront passing through the base of the prism travels through a greater thickness of glass compared to the portion passing through the apex.
Since the speed of light is lower in glass,the part of the wavefront traveling through the thicker part of the prism is delayed more than the part traveling through the thinner part.
This results in a tilt in the emerging wavefront,as correctly depicted in Figure $C$.