GUJCET 2014 Physics Question Paper with Answer and Solution

(C) The correct answer is $C$.
In a practical transformer,there are various energy losses such as hysteresis loss,eddy current loss,copper loss,and flux leakage.
Due to these energy losses,the output power is always less than the input power.
Therefore,the efficiency of a practical transformer is always less than unity $(1)$.

(B) The resultant current is given by $I = 8 + 6 \sin \omega t$.
To find the $rms$ value,we first calculate the mean square value $\langle I^2 \rangle$.
$\langle I^2 \rangle = \langle (8 + 6 \sin \omega t)^2 \rangle$
$\langle I^2 \rangle = \langle 64 + 96 \sin \omega t + 36 \sin^2 \omega t \rangle$
Using the properties of average values over a full cycle: $\langle \sin \omega t \rangle = 0$ and $\langle \sin^2 \omega t \rangle = \frac{1}{2}$.
$\langle I^2 \rangle = 64 + 96(0) + 36(\frac{1}{2})$
$\langle I^2 \rangle = 64 + 18 = 82 \ A^2$
$I_{rms} = \sqrt{\langle I^2 \rangle} = \sqrt{82} \approx 9.05 \ A$.

(B) Since both wires are connected in parallel to the same potential source,the potential difference across each wire is the same.
$V_1 = V_2 = 6 \ V$
Using Ohm's Law,$V = IR$,we have $I_1 R_1 = I_2 R_2$,which implies $\frac{I_1}{I_2} = \frac{R_2}{R_1}$.
The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
Since the material is the same,the resistivity $\rho$ is the same for both wires.
Thus,$\frac{R_2}{R_1} = \frac{\rho \frac{l_2}{\pi r_2^2}}{\rho \frac{l_1}{\pi r_1^2}} = \frac{l_2}{l_1} \times \left(\frac{r_1}{r_2}\right)^2$.
Given $\frac{l_1}{l_2} = \frac{3}{4}$ and $\frac{r_1}{r_2} = \frac{3}{2}$.
Substituting these values: $\frac{I_1}{I_2} = \frac{4}{3} \times \left(\frac{3}{2}\right)^2 = \frac{4}{3} \times \frac{9}{4} = \frac{3}{1}$.
Therefore,the ratio $I_1:I_2 = 3:1$.

(B) The color code for orange is $3$.
Since there are three orange bands,the first two bands represent the digits $3$ and $3$,and the third band represents the multiplier $10^3$.
As there is no fourth band,the tolerance is assumed to be $\pm 20\%$.
The resistance value is $R = (33 \times 10^3 \pm 20\%) \Omega = (33 \text{ k}\Omega \pm 20\%)$.
The maximum value of resistance is calculated as:
$R_{\text{max}} = 33 \text{ k}\Omega + 20\% \text{ of } 33 \text{ k}\Omega$
$R_{\text{max}} = 33000 + 0.20 \times 33000$
$R_{\text{max}} = 33000 + 6600 = 39600 \Omega$
$R_{\text{max}} = 39.6 \text{ k}\Omega$.

(A) The total length of the wire is $L = 2 \pi r = 2 \pi (2) = 4 \pi \ m$.
The total resistance of the wire is $R_{total} = (\text{resistance per unit length}) \times L = \frac{1}{\pi} \times 4 \pi = 4 \ \Omega$.
Since $\angle AOB = 90^{\circ}$,the wire is divided into two arcs: a minor arc $AB$ and a major arc $AB$.
The length of the minor arc is $L_1 = \frac{90^{\circ}}{360^{\circ}} \times (2 \pi r) = \frac{1}{4} \times 4 \pi = \pi \ m$.
The resistance of the minor arc is $R_1 = \frac{1}{\pi} \times \pi = 1 \ \Omega$.
The length of the major arc is $L_2 = L - L_1 = 4 \pi - \pi = 3 \pi \ m$.
The resistance of the major arc is $R_2 = \frac{1}{\pi} \times 3 \pi = 3 \ \Omega$.
These two resistances are connected in parallel between points $A$ and $B$.
The equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{1} + \frac{1}{3} = \frac{4}{3} \ \Omega^{-1}$,so $R_p = \frac{3}{4} \ \Omega$.
The current flowing through the battery is $I = \frac{V}{R_p} = \frac{6}{3/4} = 6 \times \frac{4}{3} = 8 \ A$.

(A) The electric field $E$ at a perpendicular distance $r$ from an infinitely long straight uniformly charged wire is given by the formula:
$E = \frac{\lambda}{2 \pi \varepsilon_0 r} = \frac{2 k \lambda}{r}$
Given:
$E = 3 \times 10^8 \ N C^{-1}$
$r = 2 \ cm = 2 \times 10^{-2} \ m$
$k = 9 \times 10^9 \ N m^2 C^{-2}$
Rearranging the formula to solve for the linear charge density $\lambda$:
$\lambda = \frac{E \cdot r}{2k}$
Substituting the values:
$\lambda = \frac{(3 \times 10^8) \times (2 \times 10^{-2})}{2 \times (9 \times 10^9)}$
$\lambda = \frac{6 \times 10^6}{18 \times 10^9}$
$\lambda = \frac{1}{3} \times 10^{-3} \ C/m$
$\lambda = 0.3333 \times 10^{-3} \ C/m = 333 \times 10^{-6} \ C/m$
$\lambda = 333 \ \mu C/m$
Therefore,the correct option is $A$.

(B) Let $l = 10 \ cm = 0.1 \ m$ be the length of the string. The angle between the strings is $60^{\circ}$,so the angle with the vertical is $\theta = 30^{\circ}$.
In equilibrium,the forces acting on each sphere are tension $T$,gravitational force $mg$,and electrostatic force $F_e = \frac{kq^2}{x^2}$,where $x$ is the distance between the centers of the spheres.
From the geometry,$x = 2l \sin \theta = 2(0.1) \sin 30^{\circ} = 0.2 \times 0.5 = 0.1 \ m$.
Resolving forces:
$T \sin \theta = F_e$
$T \cos \theta = mg$
Dividing the two equations: $\tan \theta = \frac{F_e}{mg} = \frac{kq^2}{x^2 mg}$.
Rearranging for mass $m$: $m = \frac{kq^2}{x^2 g \tan \theta}$.
Substituting the values:
$m = \frac{(9 \times 10^9) \times (2 \times 10^{-6})^2}{(0.1)^2 \times 10 \times \tan 30^{\circ}}$
$m = \frac{9 \times 10^9 \times 4 \times 10^{-12}}{0.01 \times 10 \times (1/\sqrt{3})}$
$m = \frac{36 \times 10^{-3}}{0.1 \times (1/\sqrt{3})} = 36 \times 10^{-2} \times \sqrt{3} \approx 0.36 \times 1.732 = 0.6235 \ kg$.

(D) The force acting on the particle due to the electric field is given by $F = qE$.
Since the particle starts from rest and moves in the direction of the force,the work done by the electric field on the particle over a distance $x$ is $W = F \cdot x = (qE) \cdot x = qEx$.
According to the work-energy theorem,the work done on a particle is equal to the change in its kinetic energy.
Since the initial kinetic energy is $0$,the final kinetic energy is equal to the work done,which is $qEx$.

(D) The induced electromotive force (emf) in a coil is given by Faraday's law of induction: $|\varepsilon| = N \frac{|\Delta \phi|}{\Delta t}$.
Here,$N = 500$,$A = 0.15 \ m^2$,$B_1 = 0.2 \ T$,$B_2 = 1.0 \ T$,and $\Delta t = 0.4 \ s$.
Since the magnetic field is perpendicular to the area,the magnetic flux $\phi = BA$.
Therefore,$|\varepsilon| = N \frac{A(B_2 - B_1)}{\Delta t}$.
Substituting the values:
$|\varepsilon| = \frac{500 \times 0.15 \times (1.0 - 0.2)}{0.4}$.
$|\varepsilon| = \frac{500 \times 0.15 \times 0.8}{0.4}$.
$|\varepsilon| = 500 \times 0.15 \times 2 = 150 \ V$.

(B) The magnetic force per unit length between two parallel current-carrying wires is given by the formula: $F/L = \frac{\mu_{0} I_{1} I_{2}}{2 \pi d}$.
Rearranging for $\mu_{0}$,we get: $\mu_{0} = \frac{(F/L) \cdot 2 \pi d}{I_{1} I_{2}}$.
The $SI$ unit of force $(F)$ is Newton $(N)$,length $(L)$ is meter $(m)$,distance $(d)$ is meter $(m)$,and current $(I)$ is Ampere $(A)$.
Substituting the units: $\text{Unit of } \mu_{0} = \frac{(N/m) \cdot m}{A \cdot A} = \frac{N}{A^{2}}$.
Alternatively,since $1 \text{ Tesla} = 1 \text{ N}/(A \cdot m)$,the unit can also be expressed as $T \cdot m/A$ or $Wb/(A \cdot m)$.

(A) The electromagnetic spectrum in increasing order of frequency (or decreasing order of wavelength) is: Radio waves > Microwaves > Infrared > Visible > Ultraviolet > $X$-rays > Gamma rays.
Given:
$\lambda_{1}$ = Short radio waves
$\lambda_{2}$ = $X$-rays
$\lambda_{3}$ = Ultraviolet waves
Comparing their positions in the spectrum,the wavelength order is: $\lambda_{1} > \lambda_{3} > \lambda_{2}$.
Therefore,the correct decreasing order is $\lambda_{1}, \lambda_{3}, \lambda_{2}$.

(C) The electrical potential energy $U$ of a charged sphere can be calculated by considering the work done in bringing small charge elements $dq$ from infinity to the surface of the sphere.
Alternatively,we can use the average potential method.
Initially,when there is no charge on the sphere,the potential $V_1 = 0$.
When the total charge $Q$ is on the sphere,the potential $V_2 = \frac{k Q}{R}$.
The average potential $V$ during the charging process is $V = \frac{V_1 + V_2}{2} = \frac{0 + \frac{k Q}{R}}{2} = \frac{k Q}{2 R}$.
The total potential energy $U$ is the work done to charge the sphere,which is given by $U = \int_0^Q V(q) dq = \int_0^Q \frac{k q}{R} dq = \frac{k}{R} \left[ \frac{q^2}{2} \right]_0^Q = \frac{1}{2} \frac{k Q^2}{R}$.

(B) Let the charges on the rings be $Q_1 = 10 \text{ C}$ and $Q_2 = 5 \text{ C}$. The distance between the centres $O$ and $O'$ is $R$.
The potential at the centre $O$ of the first ring is due to its own charge $Q_1$ and the charge $Q_2$ on the second ring:
$V_O = \frac{k Q_1}{R} + \frac{k Q_2}{\sqrt{R^2 + R^2}} = \frac{k Q_1}{R} + \frac{k Q_2}{\sqrt{2}R} = \frac{k}{R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} \right)$
The potential at the centre $O'$ of the second ring is due to its own charge $Q_2$ and the charge $Q_1$ on the first ring:
$V_{O'} = \frac{k Q_2}{R} + \frac{k Q_1}{\sqrt{R^2 + R^2}} = \frac{k Q_2}{R} + \frac{k Q_1}{\sqrt{2}R} = \frac{k}{R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} \right)$
The potential difference is $\Delta V = V_O - V_{O'}$:
$\Delta V = \frac{k}{R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} - Q_2 - \frac{Q_1}{\sqrt{2}} \right) = \frac{k}{R} \left( Q_1(1 - \frac{1}{\sqrt{2}}) - Q_2(1 - \frac{1}{\sqrt{2}}) \right)$
$\Delta V = \frac{k}{R} (Q_1 - Q_2) \left( 1 - \frac{1}{\sqrt{2}} \right)$
Substituting $Q_1 = 10 \text{ C}$,$Q_2 = 5 \text{ C}$,and $k = \frac{1}{4 \pi \varepsilon_0}$:
$W = q \Delta V = \frac{q}{4 \pi \varepsilon_0 R} (10 - 5) \left( 1 - \frac{1}{\sqrt{2}} \right) = \frac{5 q}{4 \pi \varepsilon_0 R} \left( 1 - \frac{1}{\sqrt{2}} \right) \text{ J}$.

(B) The magnetic field $B$ at the magnetic equator of the Earth is given by the formula:
$B = \frac{\mu_0}{4\pi} \frac{m}{R^3}$
where $m$ is the magnetic dipole moment and $R$ is the radius of the Earth.
Given values:
$B = 0.5 \times 10^{-4} \text{ T}$
$R = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}$
$\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m A}^{-1}$
Substituting these values into the formula:
$0.5 \times 10^{-4} = 10^{-7} \times \frac{m}{(6.4 \times 10^6)^3}$
$m = \frac{0.5 \times 10^{-4} \times (6.4 \times 10^6)^3}{10^{-7}}$
$m = 0.5 \times 10^3 \times (6.4)^3 \times 10^{18}$
$m = 0.5 \times 262.144 \times 10^{21}$
$m = 131.072 \times 10^{21} = 1.31 \times 10^{23} \text{ Am}^2$
Thus,the magnetic dipole moment of the Earth is $1.31 \times 10^{23} \text{ Am}^2$.

(C) According to Curie's law for a paramagnetic substance,the magnetic susceptibility $\chi_m$ is inversely proportional to the absolute temperature $T$,i.e.,$\chi_m \propto \frac{1}{T}$.
Given:
$\chi_{m_1} = 1.0 \times 10^{-5}$
$T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$
$\chi_{m_2} = 1.5 \times 10^{-5}$
Using the relation $\frac{\chi_{m_1}}{\chi_{m_2}} = \frac{T_2}{T_1}$:
$\frac{1.0 \times 10^{-5}}{1.5 \times 10^{-5}} = \frac{T_2}{300}$
$\frac{1}{1.5} = \frac{T_2}{300}$
$T_2 = \frac{300}{1.5} = 200 \ K$
Converting back to Celsius: $T_2(^{\circ} C) = 200 - 273 = -73^{\circ} C$.

(A) The time period $T$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula: $T = \frac{2 \pi m}{q B}$.
Since $2, \pi,$ and $B$ are constants,the time period is proportional to the mass-to-charge ratio: $T \propto \frac{m}{q}$.
For an $\alpha$-particle,the mass $m_{\alpha} = 4 m_p$ and the charge $q_{\alpha} = 2 q_p$,where $m_p$ and $q_p$ are the mass and charge of a proton,respectively.
Therefore,the ratio of the time periods is: $\frac{T_{\alpha}}{T_p} = \frac{m_{\alpha}}{q_{\alpha}} \times \frac{q_p}{m_p}$.
Substituting the values: $\frac{T_{\alpha}}{T_p} = \frac{4 m_p}{2 q_p} \times \frac{q_p}{m_p} = \frac{4}{2} = 2: 1$.

(D) The magnetic field at the center of a circular coil with $N$ turns and radius $a$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2a}$.
Since the currents are in mutually opposite directions,the net magnetic field at the center is the difference between the fields produced by the two rings: $B = |B_1 - B_2|$.
Given: $N = 25$,$I_1 = 0.1 \text{ A}$,$a_1 = 0.5 \text{ m}$,$I_2 = 0.2 \text{ A}$,$a_2 = 2.0 \text{ m}$.
Substituting the values:
$B = \frac{\mu_0 N}{2} \left| \frac{I_1}{a_1} - \frac{I_2}{a_2} \right|$
$B = \frac{\mu_0 \times 25}{2} \left| \frac{0.1}{0.5} - \frac{0.2}{2.0} \right|$
$B = \frac{25 \mu_0}{2} \left| 0.2 - 0.1 \right|$
$B = \frac{25 \mu_0}{2} \times 0.1 = \frac{2.5 \mu_0}{2} = \frac{5}{4} \mu_0 \text{ T}$.

(A) The time period $T$ of a charged particle moving in a uniform magnetic field is given by the formula: $T = \frac{2 \pi m}{q B}$.
From this expression,we can see that $T \propto \frac{m}{q}$.
For an $\alpha$-particle,the mass $m_\alpha = 4 m_p$ and the charge $q_\alpha = 2 q_p$,where $m_p$ and $q_p$ are the mass and charge of a proton,respectively.
Now,calculating the ratio of the time periods:
$\frac{T_\alpha}{T_p} = \frac{m_\alpha}{q_\alpha} \times \frac{q_p}{m_p}$
Substituting the values:
$\frac{T_\alpha}{T_p} = \frac{4 m_p}{2 q_p} \times \frac{q_p}{m_p} = \frac{4}{2} = \frac{2}{1}$.
Therefore,the ratio is $2: 1$.

(B) To convert a galvanometer into a voltmeter,a high resistance (multiplier) must be connected in series with the galvanometer coil.
This is done to ensure that the voltmeter draws a negligible current from the circuit,thereby maintaining the potential difference across the points to be measured.
By increasing the total resistance of the device,the current flowing through the galvanometer is limited,preventing damage and allowing it to measure higher voltages.

(C) The reaction for removing one neutron from ${ }_{8}^{17}O$ is given by: ${ }_{8}^{17}O \rightarrow { }_{8}^{16}O + { }_{0}^{1}n$.
To find the energy required,we calculate the difference between the total binding energy of the products and the reactant.
Total binding energy of ${ }_{8}^{17}O = 17 \times 7.75 \text{ MeV} = 131.75 \text{ MeV}$.
Total binding energy of ${ }_{8}^{16}O = 16 \times 7.97 \text{ MeV} = 127.52 \text{ MeV}$.
The energy required to remove the neutron is the difference in total binding energies:
$E = 131.75 \text{ MeV} - 127.52 \text{ MeV} = 4.23 \text{ MeV}$.

(B) The electric field intensity $E$ is given by the formula $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the width of the depletion region.
Given: $V = 3 \ V$ and $d = 300 \ \mathring{A} = 300 \times 10^{-8} \ cm$.
Substituting the values:
$E = \frac{3}{300 \times 10^{-8}} \ V/cm$
$E = \frac{3}{3 \times 10^{-6}} \ V/cm$
$E = 10^{6} \ V/cm$.
Therefore,the correct option is $B$.