GUJCET 2026 Physics Question Paper with Answer and Solution

(A) The quantity $\frac{B^2}{2\mu_0}$ represents the magnetic energy density $(u_B)$,which is defined as the energy stored per unit volume in a magnetic field.
Energy $(E)$ has the dimensional formula $[ML^2T^{-2}]$.
Volume $(V)$ has the dimensional formula $[L^3]$.
Therefore,the dimensional formula for magnetic energy density is $\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Thus,the correct option is $A$.

(C) Planck's constant $(h)$ has the dimensions of action,which is defined as energy multiplied by time $(E \times t)$.
The $SI$ unit of energy is Joule $(J)$ and time is second $(s)$,so the unit of Planck's constant is $J \cdot s$.
Angular momentum $(L)$ is defined as the product of moment of inertia $(I)$ and angular velocity $(\omega)$,or $L = mvr$.
The dimensions of angular momentum are $[M L^2 T^{-1}]$,which are identical to the dimensions of Planck's constant.
Therefore,both Planck's constant and angular momentum have the same $SI$ unit,which is $kg \cdot m^2/s$ or $J \cdot s$.
Thus,the correct option is $C$.

(B) To determine the relationship,we calculate the number of neutrons $(N)$ for each nuclide using the formula $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For $^{197}_{79} \text{Au}$: $A = 197$,$Z = 79$. Thus,$N = 197 - 79 = 118$.
For $^{198}_{80} \text{Hg}$: $A = 198$,$Z = 80$. Thus,$N = 198 - 80 = 118$.
Since both nuclides have the same number of neutrons $(N = 118)$,they are classified as isotones.

(C) An oscillating charge produces electromagnetic waves of the same frequency as the frequency of oscillation of the charge.
Since the frequency of the oscillating particle is $8 \times 10^9 \text{ Hz}$,the frequency of the electromagnetic waves produced will also be $8 \times 10^9 \text{ Hz}$.

(D) In an ideal transformer,the input power is equal to the output power,so $V_p I_p = V_s I_s$.
For a transformer,the relationship between voltage,current,and number of turns is given by $\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$.
Given values are $N_p = 100$,$N_s = 200$,and $I_s = 5 \text{ A}$.
Using the relation $\frac{I_p}{I_s} = \frac{N_s}{N_p}$,we get $I_p = I_s \times \frac{N_s}{N_p}$.
Substituting the values: $I_p = 5 \times \frac{200}{100} = 5 \times 2 = 10 \text{ A}$.
Therefore,the input current is $10 \text{ A}$.

(B) The angular frequency $\omega$ of free oscillations in an $LC$ circuit is given by the formula $\omega = \frac{1}{\sqrt{LC}}$.
Given values are $L = 27 \text{ mH} = 27 \times 10^{-3} \text{ H}$ and $C = 30 \mu\text{F} = 30 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$\omega = \frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}}$
$\omega = \frac{1}{\sqrt{810 \times 10^{-9}}}$
$\omega = \frac{1}{\sqrt{81 \times 10^{-8}}}$
$\omega = \frac{1}{9 \times 10^{-4}}$
$\omega = \frac{10^4}{9} \approx 1111.11 \text{ rad/s}$.
The closest option is $1100 \text{ rad/s}$.

(B) The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2\pi f C}$.
Given values are frequency $f = 50 \text{ Hz}$ and capacitance $C = 15.0 \mu\text{F} = 15.0 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$X_C = \frac{1}{2 \times 3.14159 \times 50 \times 15.0 \times 10^{-6}}$
$X_C = \frac{1}{314.159 \times 15.0 \times 10^{-6}}$
$X_C = \frac{1}{4712.385 \times 10^{-6}}$
$X_C = \frac{10^6}{4712.385} \approx 212.2 \Omega$.
Rounding to the nearest integer,the capacitive reactance is $212 \Omega$.

(C) The self-inductance $L$ of a coil is a property that depends solely on its physical geometry,such as the number of turns $N$,the cross-sectional area,the length of the coil,and the magnetic permeability of the core material.
It does not depend on the magnitude of the current $I$ flowing through the coil.
Therefore,if the current is doubled,the self-inductance remains unchanged.
Thus,the new self-inductance is $L$.

(C) The magnetic field $B$ at the center of a large circular coil of radius $r_2$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r_2}$.
Since $r_1 \ll r_2$,the magnetic field produced by the larger coil is approximately uniform over the area of the smaller coil.
The magnetic flux $\phi$ linked with the smaller coil of radius $r_1$ is $\phi = B \times A = \frac{\mu_0 I}{2r_2} \times (\pi r_1^2)$.
By definition,the mutual inductance $M$ is given by $M = \frac{\phi}{I}$.
Substituting the expression for $\phi$,we get $M = \frac{\mu_0 \pi r_1^2}{2r_2}$.
Therefore,the mutual inductance $M$ is proportional to $\frac{r_1^2}{r_2}$.

(D) Superconductors exhibit the Meissner effect,which means they are perfect diamagnetic materials.
For a perfect diamagnetic material,the internal magnetic field $B$ is equal to $0$.
Since $B = \mu_0(H + M) = 0$,we have $M = -H$.
The magnetic susceptibility $\chi$ is defined as the ratio of magnetization $M$ to the magnetic field intensity $H$,i.e.,$\chi = M/H$.
Substituting $M = -H$,we get $\chi = -H/H = -1$.

(B) The magnetic moment $m$ of a solenoid is given by the formula $m = N \cdot I \cdot A$.
Here,the number of turns $N = 800$,the current $I = 3.0 \text{ A}$,and the area of cross-section $A = 2.5 \times 10^{-4} \text{ m}^2$.
Substituting these values into the formula:
$m = 800 \times 3.0 \times 2.5 \times 10^{-4}$
$m = 2400 \times 2.5 \times 10^{-4}$
$m = 6000 \times 10^{-4}$
$m = 0.6 \text{ JT}^{-1}$.
Thus,the magnetic moment associated with the solenoid is $0.60 \text{ JT}^{-1}$.

(B) The magnetic Lorentz force experienced by a charged particle moving in a magnetic field is $F = qvB$.
This magnetic force provides the necessary centripetal force for circular motion: $qvB = \frac{mv^2}{R}$.
Solving for the radius $R$,we get $R = \frac{mv}{qB}$.
The period of revolution $T$ is defined as the time taken to complete one full circle,given by $T = \frac{2\pi R}{v}$.
Substituting the expression for $R$ into the formula for $T$,we get $T = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}$.
Since $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$,it is independent of both the linear speed $v$ and the radius $R$.

(A) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula $B = \frac{\mu_0 I}{2\pi r}$.
Given values are current $I = 90 \text{ A}$ and distance $r = 1.5 \text{ m}$.
The permeability of free space is $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Substituting these values into the formula:
$B = \frac{4\pi \times 10^{-7} \times 90}{2\pi \times 1.5}$
$B = \frac{2 \times 10^{-7} \times 90}{1.5}$
$B = 2 \times 10^{-7} \times 60 = 120 \times 10^{-7} = 1.2 \times 10^{-5} \text{ T}$.
According to the right-hand thumb rule,if the current flows from east to west,the magnetic field at a point above the wire points towards the north.

(C) Current sensitivity is defined as $I_s = \frac{NAB}{k}$,where $N$ is the number of turns,$A$ is the area,$B$ is the magnetic field,and $k$ is the torsional constant.
If $N$ is doubled,$I_s$ becomes $2 \times I_s$,meaning it is doubled.
Voltage sensitivity is defined as $V_s = \frac{I_s}{R} = \frac{NAB}{kR}$,where $R$ is the resistance of the coil.
Since the resistance $R$ is directly proportional to the length of the wire,and the length is proportional to the number of turns $(R \propto N)$,doubling $N$ also doubles $R$.
Substituting these into the formula: $V_s = \frac{(2N)AB}{k(2R)} = \frac{NAB}{kR}$.
Thus,the voltage sensitivity remains unchanged.

(D) For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal,i.e.,$\frac{P}{Q} = \frac{R}{S}$.
From the given figure,the resistances are $P = 2\Omega$,$Q = 2\Omega$,$R = 6\Omega$,and the arm $S$ consists of a resistor $X$ in parallel with a $6\Omega$ resistor.
The equivalent resistance of the parallel combination in arm $S$ is $S = \frac{X \cdot 6}{X + 6}$.
Applying the balanced bridge condition: $\frac{2}{2} = \frac{6}{\left(\frac{6X}{X + 6}\right)}$.
This simplifies to $1 = \frac{6(X + 6)}{6X}$.
$1 = \frac{X + 6}{X}$.
$X = X + 6$ is not possible,let us re-examine the circuit. The arm $CD$ has $X$ and $6\Omega$ in parallel. The condition is $\frac{AB}{BC} = \frac{AD}{CD_{eq}}$.
Here $AB=2\Omega, BC=2\Omega, AD=2\Omega$. So,$CD_{eq} = 2\Omega$.
Since $CD_{eq} = \frac{X \cdot 6}{X + 6} = 2$,we have $6X = 2(X + 6)$.
$6X = 2X + 12$.
$4X = 12$,which gives $X = 3\Omega$.

(C) The maximum power is drawn from a battery when the external resistance $R$ is equal to the internal resistance $r$ of the battery $(R = r)$.
The power delivered to the external resistor is given by the formula: $P = I^2 R = \left(\frac{E}{R+r}\right)^2 \cdot R$.
Substituting $R = r$ for maximum power,we get:
$P_{max} = \frac{E^2 r}{(r+r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}$.
Given values are $E = 6.0 \ V$ and $r = 0.2 \ \Omega$.
Substituting these values into the formula:
$P_{max} = \frac{(6.0)^2}{4 \times 0.2} = \frac{36}{0.8} = 45 \ W$.
Therefore,the maximum power drawn is $45 \ W$.

(A) The resistance of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length of the conductor in the direction of current flow,and $A$ is the cross-sectional area perpendicular to the current flow.
To maximize the resistance $R$,we need to maximize the length $l$ and minimize the cross-sectional area $A$.
When the battery is connected across the $1 \text{ cm} \times 0.5 \text{ cm}$ faces,the current flows along the length of $10 \text{ cm}$. Thus,$l = 10 \text{ cm}$ and $A = 1 \text{ cm} \times 0.5 \text{ cm} = 0.5 \text{ cm}^2$.
The resistance is $R = \rho \frac{10}{0.5} = 20\rho$.
For the other two orientations,the length $l$ will be either $1 \text{ cm}$ or $0.5 \text{ cm}$,and the area $A$ will be larger,which results in a smaller resistance value.
Therefore,the resistance is maximum when the battery is connected across the $1 \text{ cm} \times 0.5 \text{ cm}$ faces.

(C) The work done $W$ in rotating an electric dipole in a uniform electric field is given by the formula: $W = PE(\cos \theta_1 - \cos \theta_2)$.
Initially,the dipole is parallel to the electric field,so the initial angle $\theta_1 = 0^\circ$.
Finally,the dipole is rotated by an angle of $90^\circ$,so the final angle $\theta_2 = 90^\circ$.
Substituting these values into the formula:
$W = PE(\cos 0^\circ - \cos 90^\circ)$
Since $\cos 0^\circ = 1$ and $\cos 90^\circ = 0$,we get:
$W = PE(1 - 0) = PE$.

(A) For a uniformly charged spherical shell,the electric potential at any point inside the shell is constant and equal to the potential at its surface.
The potential at the surface of the shell is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}$.
Since the distance $r = R/2$ is less than the radius $R$ (i.e.,the point is inside the shell),the electric potential at this point is the same as the potential at the surface.
Therefore,the potential at $r = R/2$ is $V = \frac{Q}{4\pi \epsilon_0 R}$.

(D) The initial capacitance of the parallel plate capacitor with air is given by $C_0 = \frac{\epsilon_0 A}{d} = 1.0 \text{ pF}$.
When the distance between the plates is doubled $(d' = 2d)$ and the space is filled with a dielectric of constant $K$,the new capacitance is $C = \frac{K \epsilon_0 A}{d'}$.
Substituting $d' = 2d$ into the equation,we get $C = \frac{K \epsilon_0 A}{2d} = \frac{K}{2} \times C_0$.
Given that $C = 2.0 \text{ pF}$ and $C_0 = 1.0 \text{ pF}$,we have $2.0 = \frac{K}{2} \times 1.0$.
Solving for $K$,we get $K = 2.0 \times 2 = 4.0$.

(A) The electric field due to an infinitely long line charge at a distance $r$ is given by $E = \frac{\lambda}{2\pi \epsilon_0 r}$.
At the midpoint between the two wires,the distance from each wire is $r = R$.
The electric field due to the positively charged wire $(+\lambda)$ points away from it,and the electric field due to the negatively charged wire $(-\lambda)$ points towards it.
Since the midpoint is between the wires,both electric field vectors point in the same direction (from the positive wire towards the negative wire).
Therefore,the total electric field $E_{total} = E_1 + E_2 = \frac{\lambda}{2\pi \epsilon_0 R} + \frac{\lambda}{2\pi \epsilon_0 R} = \frac{2\lambda}{2\pi \epsilon_0 R} = \frac{\lambda}{\pi \epsilon_0 R}$.

(B) According to Gauss's law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\varepsilon_0}$.
When a charge $q$ is placed at a vertex of a cube,the cube occupies only $\frac{1}{8}$ of the total space surrounding the charge because $8$ identical cubes are required to enclose the charge completely at the center of a larger cube.
Therefore,the electric flux passing through the cube is $\Phi = \frac{1}{8} \times \frac{q}{\varepsilon_0} = \frac{q}{8\varepsilon_0}$.

(C) The force on a charged particle in an electric field is given by $F = qE$.
According to Newton's second law of motion,$F = ma$,therefore $a = \frac{qE}{m}$.
For a proton,the charge is $q_p = e$ and the mass is $m_p$. Thus,$a_p = \frac{eE}{m_p}$.
For an $\alpha$-particle,the charge is $q_\alpha = 2e$ and the mass is $m_\alpha = 4m_p$.
Substituting these values,$a_\alpha = \frac{2eE}{4m_p} = \frac{1}{2} \frac{eE}{m_p} = \frac{a_p}{2}$.
Therefore,the ratio $\frac{a_p}{a_\alpha} = 2$,which means $a_p : a_\alpha = 2 : 1$.

(B) Trivalent impurities have $3$ valence electrons and belong to Group $13$ of the periodic table (e.g.,$B, Al, Ga, In$).
Pentavalent impurities have $5$ valence electrons and belong to Group $15$ of the periodic table (e.g.,$N, P, As, Sb$).
Antimony $(Sb)$ belongs to Group $15$ and is therefore a pentavalent impurity,not a trivalent one.

(A) For a full-wave rectifier,the output frequency $(f_{out})$ is twice the input frequency $(f_{in})$.
The relationship is given by the formula: $f_{out} = 2 \times f_{in}$.
Given that the output frequency $(f_{out})$ is $100 \text{ Hz}$.
Substituting the value into the formula: $100 \text{ Hz} = 2 \times f_{in}$.
Therefore,$f_{in} = \frac{100 \text{ Hz}}{2} = 50 \text{ Hz}$.

(B) In forward biasing,the positive terminal of the external battery is connected to the $p$-type region and the negative terminal to the $n$-type region.
This configuration pushes the majority charge carriers toward the junction.
As a result,the width of the depletion region decreases,which effectively lowers the potential barrier across the $p-n$ junction.

(A) For a concave mirror,the focal length $f = -15 \text{ cm}$ and the object distance $u = -10 \text{ cm}$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-10} = -\frac{1}{15} + \frac{1}{10} = \frac{-2 + 3}{30} = \frac{1}{30}$.
Thus,$v = +30 \text{ cm}$.
Since $v$ is positive,the image is formed behind the mirror,which means it is virtual and erect.
The magnification $m = -\frac{v}{u} = -\frac{30}{-10} = +3$.
Since the magnification $m$ is positive and $|m| > 1$,the image is virtual,erect,and magnified.

(B) The magnification $M$ of a compound microscope is given by the formula $M \approx \frac{L}{f_o} \times \frac{D}{f_e}$.
Here,$L = 20 \text{ cm}$ (tube length),$f_o = 1.0 \text{ cm}$ (focal length of objective),$f_e = 2.0 \text{ cm}$ (focal length of eyepiece),and $D = 25 \text{ cm}$ (near point distance).
Substituting these values into the formula:
$M = \frac{20}{1.0} \times \frac{25}{2.0}$
$M = 20 \times 12.5$
$M = 250$.
Therefore,the magnification is $250$.

(B) When two thin lenses are placed in contact,the equivalent focal length $F$ is given by the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
By simplifying this,we get $\frac{1}{F} = \frac{f_1 + f_2}{f_1 f_2}$.
The power $P$ of a lens is defined as the reciprocal of its focal length in meters,i.e.,$P = \frac{1}{F}$.
Therefore,the equivalent power $P$ of the combination is $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} = \frac{f_1 + f_2}{f_1 f_2}$.

(A) $1$. From the figure,the incident ray strikes the first surface of the prism normally. Therefore,the angle of incidence $i = 0^\circ$,and the ray passes undeviated into the prism.
$2$. The prism is a right-angled triangle with one angle $60^\circ$. The angle at the top vertex is $180^\circ - 90^\circ - 60^\circ = 30^\circ$.
$3$. Inside the prism,the ray hits the second surface (the hypotenuse). The angle of incidence at this surface $(r_2)$ is the angle between the normal and the ray. Since the ray is perpendicular to the first surface,the angle of incidence at the second surface is $r_2 = 30^\circ$.
$4$. Applying Snell's law at the second surface: $n_1 \sin(r_2) = n_2 \sin(e)$,where $n_1 = \sqrt{3}$,$n_2 = 1$,and $r_2 = 30^\circ$.
$5$. $\sqrt{3} \sin(30^\circ) = 1 \times \sin(e)$.
$6$. $\sqrt{3} \times \frac{1}{2} = \sin(e) \implies \sin(e) = \frac{\sqrt{3}}{2}$.
$7$. Therefore,$e = 60^\circ$.

(C) wave front is defined as the locus of all points that oscillate in phase.
Since every point on the same wave front vibrates in phase,the phase difference between any two particles on the same wave front is $0$.

(A) The electromagnetic spectrum is arranged in order of increasing frequency or decreasing wavelength.
The general order of wavelengths for these radiations is: $\lambda_{\text{microwave}} > \lambda_{\text{visible}} > \lambda_{\text{X-rays}}$.
Given that $\lambda_m$ represents the wavelength of microwaves,$\lambda_v$ represents the wavelength of visible rays,and $\lambda_x$ represents the wavelength of $X$-rays.
Therefore,the correct relation is $\lambda_m > \lambda_v > \lambda_x$.
Thus,the correct option is $A$.

(B) Given: Slit separation $d = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}$.
Distance to screen $D = 2.0 \text{ m}$.
Order of fringe $n = 3$.
Distance of $n^{th}$ bright fringe $y_n = 1.5 \text{ cm} = 1.5 \times 10^{-2} \text{ m}$.
The formula for the position of the $n^{th}$ bright fringe is $y_n = \frac{n \lambda D}{d}$.
Rearranging for wavelength $\lambda$: $\lambda = \frac{y_n d}{n D}$.
Substituting the values: $\lambda = \frac{1.5 \times 10^{-2} \times 2 \times 10^{-4}}{3 \times 2.0}$.
$\lambda = \frac{3 \times 10^{-6}}{6} = 0.5 \times 10^{-6} \text{ m}$.
Converting to $ \mathring{A}$s: $\lambda = 0.5 \times 10^{-6} \times 10^{10} \mathring{A} = 5000 \mathring{A}$.
Thus,the wavelength of light used is $5000 \mathring{A}$.

(D) Einstein's photoelectric equation is given by $eV_0 = h\nu - \Phi$,where $\Phi$ is the work function.
Rearranging the equation to the form $y = mx + c$,we get $V_0 = (\frac{h}{e})\nu - \frac{\Phi}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V_0$ and $x = \nu$,the slope $m$ is equal to $\frac{h}{e}$.
Therefore,the correct option is $D$.

(B) The maximum kinetic energy $(K_{max})$ of photoelectrons is related to the stopping potential $(V_0)$ by the equation $K_{max} = eV_0$.
Given that the cut-off voltage (stopping potential) $V_0 = 1.5 \text{ V}$.
Substituting the value into the equation,we get $K_{max} = e \times 1.5 \text{ V} = 1.5 \text{ eV}$.
Therefore,the maximum kinetic energy is $1.5 \text{ eV}$.

(B) For a photon,the momentum $p$ is given by $p = \frac{h}{\lambda}$,where $h$ is Planck's constant.
According to the de Broglie relation,the wavelength $\lambda'$ associated with a particle of momentum $p$ is $\lambda' = \frac{h}{p}$.
Substituting the expression for $p$ into the de Broglie equation,we get $\lambda' = \frac{h}{h/\lambda} = \lambda$.
Therefore,the de Broglie wavelength of a photon is equal to the wavelength of the electromagnetic radiation associated with it.
Thus,the correct option is $B$.

(A) In a hydrogen-like atom,the total energy $E$ of an electron is related to its kinetic energy $K$ and potential energy $U$ by the following relations:
$E = -K$
$U = 2E$
Given that the total energy $E = -13.6 \text{ eV}$,we can find $K$ and $U$:
$K = -E = -(-13.6 \text{ eV}) = 13.6 \text{ eV}$
$U = 2E = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{13.6 \text{ eV}}{-27.2 \text{ eV}} = -\frac{1}{2}$.
Thus,the correct option is $A$.

(D) The radius of the $n$-th orbit in a hydrogen atom is given by the formula $r_n = n^2 r_1$,where $r_1$ is the radius of the first orbit (Bohr radius).
Given $r_1 = 5.3 \times 10^{-11} \text{ m}$ and $n = 3$.
Substituting these values into the formula:
$r_3 = 3^2 \times (5.3 \times 10^{-11} \text{ m})$
$r_3 = 9 \times 5.3 \times 10^{-11} \text{ m}$
$r_3 = 47.7 \times 10^{-11} \text{ m}$
$r_3 = 4.77 \times 10^{-10} \text{ m}$.
Thus,the radius of the $n = 3$ orbit is $4.77 \times 10^{-10} \text{ m}$.
Therefore,the correct option is $D$.

(C) According to Einstein's mass-energy equivalence principle,the energy $E$ is given by the formula $E = mc^2$.
Here,$m$ is the mass of the substance and $c$ is the speed of light in a vacuum.
Given:
Mass $m = 1.0 \text{ kg}$
Speed of light $c = 3 \times 10^8 \text{ m/s}$
Substituting these values into the formula:
$E = 1.0 \text{ kg} \times (3 \times 10^8 \text{ m/s})^2$
$E = 1.0 \times 9 \times 10^{16} \text{ J}$
$E = 9 \times 10^{16} \text{ J}$
Therefore,the energy equivalent of $1.0 \text{ kg}$ of substance is $9 \times 10^{16} \text{ J}$.
The correct option is $C$.

(B) The nuclear radius $R$ is calculated using the formula $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 \text{ fm}$ and $A$ is the mass number of the nucleus.
For a hydrogen atom,the nucleus consists of a single proton,so the mass number $A = 1$.
Substituting the value of $A$ into the formula:
$R = 1.2 \times (1)^{1/3} \text{ fm}$
$R = 1.2 \times 1 \text{ fm} = 1.2 \text{ fm}$.
Therefore,the nuclear radius of a hydrogen atom is $1.2 \text{ fm}$.
The correct option is $B$.