GUJCET 2013 Physics Question Paper with Answer and Solution

(B) The power consumed in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
Given: $V_{m} = 100 \ V$,$I_{m} = 1.1 \ A$,$X_{C} = 60 \ \Omega$,$R = 80 \ \Omega$.
The impedance $Z$ is calculated as $Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{80^{2} + 60^{2}} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \ \Omega$.
The power factor is $\cos \phi = \frac{R}{Z} = \frac{80}{100} = 0.8$.
The $rms$ values are $V_{rms} = \frac{V_{m}}{\sqrt{2}}$ and $I_{rms} = \frac{I_{m}}{\sqrt{2}}$.
Substituting these into the power formula:
$P = \left(\frac{100}{\sqrt{2}}\right) \times \left(\frac{1.1}{\sqrt{2}}\right) \times 0.8$
$P = \frac{100 \times 1.1}{2} \times 0.8$
$P = 50 \times 1.1 \times 0.8 = 55 \times 0.8 = 44.0 \ W$.

(A) The given $AC$ voltage source is $V = 300 \sqrt{2} \sin(\omega t)$.
The peak voltage is $V_m = 300 \sqrt{2} \text{ V}$.
The $RMS$ voltage is $V_{rms} = \frac{V_m}{\sqrt{2}} = 300 \text{ V}$.
The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $R = 30 \Omega$, $X_L = 10 \Omega$, and $X_C = 10 \Omega$.
$Z = \sqrt{30^2 + (10 - 10)^2} = \sqrt{30^2 + 0^2} = 30 \Omega$.
The $RMS$ current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{Z}$.
$I_{rms} = \frac{300}{30} = 10 \text{ A}$.

(C) The correct option is $C$.
In a $DC$ circuit,the frequency $v = 0 \ Hz$.
The angular frequency is given by $\omega = 2 \pi v = 2 \pi \times 0 = 0 \ rad/s$.
The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{\omega C}$.
Substituting the values,we get $X_C = \frac{1}{0 \times 1} = \frac{1}{0} = \infty$.
Therefore,the capacitive reactance of a capacitor in a $DC$ circuit is infinite,meaning it acts as an open circuit.

(B) The relationship between terminal potential difference $(V)$, $emf$ $(\varepsilon)$, current $(I)$, and internal resistance $(r)$ is given by: $V = \varepsilon - Ir$.
Rearranging, we get: $\varepsilon = V + Ir$.
Using the values from the first two rows of the table:
For row $1$: $\varepsilon = 1.0 + 0.08r \quad (1)$
For row $2$: $\varepsilon = 0.5 + 0.18r \quad (2)$
Equating $(1)$ and $(2)$: $1.0 + 0.08r = 0.5 + 0.18r$.
Rearranging terms: $1.0 - 0.5 = 0.18r - 0.08r$.
$0.5 = 0.1r$, which gives $r = 5 \ \Omega$.
Substituting $r = 5 \ \Omega$ into equation $(1)$: $\varepsilon = 1.0 + 0.08(5) = 1.0 + 0.4 = 1.4 \ V$.

(A) Let the resistance of each resistor be $R$ and the battery voltage be $V$.
In a series connection,the equivalent resistance is $R_s = 4R$.
The power dissipated in series is $P_s = \frac{V^2}{R_s} = \frac{V^2}{4R} = 20 \ W$.
From this,we find $\frac{V^2}{R} = 20 \times 4 = 80 \ W$.
In a parallel connection,the equivalent resistance is $R_p = \frac{R}{4}$.
The power dissipated in parallel is $P_p = \frac{V^2}{R_p} = \frac{V^2}{R/4} = 4 \times \frac{V^2}{R}$.
Substituting the value of $\frac{V^2}{R}$,we get $P_p = 4 \times 80 \ W = 320 \ W$.

(A) For $n$ resistors each of resistance $R$ connected in parallel,the minimum resistance is given by $R_{\min} = \frac{R}{n}$.
Given $R = 1 \ \Omega$ and $n = 10$,we have $R_{\min} = \frac{1}{10} \ \Omega$.
For $n$ resistors connected in series,the maximum resistance is given by $R_{\max} = nR$.
Thus,$R_{\max} = 10 \times 1 = 10 \ \Omega$.
The ratio of minimum to maximum resistance is $\frac{R_{\min}}{R_{\max}} = \frac{1/10}{10} = \frac{1}{100}$.

(D) Given: $q_1 = +16 \mu C$,$q_2 = -9 \mu C$,distance $d = 10 \ cm$.
Let the point where the resultant electric field is zero be at a distance $x$ from the $-9 \mu C$ charge. Since the charges are unlike,the null point lies outside the region between the charges,closer to the charge with the smaller magnitude.
At the null point,the magnitude of the electric field due to $q_1$ must equal the magnitude of the electric field due to $q_2$.
$E_1 = E_2$
$\frac{k |q_1|}{(d + x)^2} = \frac{k |q_2|}{x^2}$
$\frac{16}{(10 + x)^2} = \frac{9}{x^2}$
Taking the square root on both sides:
$\frac{4}{10 + x} = \frac{3}{x}$
$4x = 3(10 + x)$
$4x = 30 + 3x$
$x = 30 \ cm$
Thus,the resultant electric field is zero at a distance of $30 \ cm$ from the $-9 \mu C$ charge.

(B) Let the charge at the origin be $q = 10^{-9} \ C$ and the charge at $(2, 0, 0) \ m$ be $Q$.
Let point $P$ be $(3, 1, 1) \ m$.
The position vector of $P$ relative to the origin $O(0, 0, 0)$ is $\vec{r}_1 = (3 - 0)\hat{i} + (1 - 0)\hat{j} + (1 - 0)\hat{k} = 3\hat{i} + \hat{j} + \hat{k}$.
The magnitude is $r_1 = |\vec{r}_1| = \sqrt{3^2 + 1^2 + 1^2} = \sqrt{11} \ m$.
The position vector of $P$ relative to the charge $Q$ at $(2, 0, 0)$ is $\vec{r}_2 = (3 - 2)\hat{i} + (1 - 0)\hat{j} + (1 - 0)\hat{k} = \hat{i} + \hat{j} + \hat{k}$.
The magnitude is $r_2 = |\vec{r}_2| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \ m$.
The total electric field at $P$ is $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{kq}{r_1^3}\vec{r}_1 + \frac{kQ}{r_2^3}\vec{r}_2$.
The $Y$-component of the electric field is $E_y = \frac{kq}{r_1^3}(y_1) + \frac{kQ}{r_2^3}(y_2) = 0$.
Substituting the values: $k \left[ \frac{10^{-9} \times 1}{(\sqrt{11})^3} + \frac{Q \times 1}{(\sqrt{3})^3} \right] = 0$.
$\frac{Q}{3\sqrt{3}} = -\frac{10^{-9}}{11\sqrt{11}}$.
$Q = -10^{-9} \times \frac{3\sqrt{3}}{11\sqrt{11}} = -10^{-9} \times \left( \frac{3}{11} \right)^{1.5} \approx -0.1424 \times 10^{-9} \ C$.

(B) The force between two point charges in a medium is given by $F_{\text{medium}} = \frac{F_{\text{air}}}{K}$,where $F_{\text{air}}$ is the force between the same charges in a vacuum (or air) and $K$ is the dielectric constant of the medium.
Given that the force in the medium is $F$,we have $F = \frac{F_{\text{air}}}{K}$.
To find the force when the material is removed (i.e.,the force in air),we rearrange the equation:
$F_{\text{air}} = F \times K$.
Therefore,the force becomes $FK$.

(C) Given: Number of turns $N = 60$.
Rate of change of magnetic flux $\frac{d\phi}{dt} = 1 \text{ Wb/hour}$.
Convert the rate into $SI$ units (Weber per second): $\frac{d\phi}{dt} = \frac{1}{3600} \text{ Wb/s}$.
According to Faraday's law of electromagnetic induction,the magnitude of induced emf is given by $|\varepsilon| = N \left| \frac{d\phi}{dt} \right|$.
Substituting the values: $|\varepsilon| = 60 \times \frac{1}{3600} \text{ V}$.
$|\varepsilon| = \frac{1}{60} \text{ V}$.

(D) When we pass a current $I$ through the outer loop of radius $R$,the magnetic field at the centre is given by:
$B_{\text{center}} = \frac{\mu_0 I}{2 R}$
Since $r \ll R$,the magnetic field $B$ is approximately uniform over the area of the smaller loop $A$.
The magnetic flux $\phi$ associated with the smaller loop is:
$\phi = B \cdot A = \left( \frac{\mu_0 I}{2 R} \right) \times (\pi r^2)$
$\phi = \frac{\mu_0 I \pi r^2}{2 R}$
The mutual inductance $M$ is defined as the ratio of the magnetic flux $\phi$ through the secondary coil to the current $I$ in the primary coil:
$M = \frac{\phi}{I} = \frac{\mu_0 \pi r^2}{2 R}$

(B) The electrical power of the bulb is $P_{elec} = 800 \text{ W}$.
The efficiency of the bulb is $\eta = 3 \% = 0.03$.
The radiant power (energy emitted per second) is $P_{rad} = P_{elec} \times \eta = 800 \times 0.03 = 24 \text{ J/s}$.
The force exerted by an electromagnetic wave on a surface that absorbs it is given by $F = \frac{P_{rad}}{c}$,where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light.
Substituting the values: $F = \frac{24}{3 \times 10^8} \text{ N}$.
$F = 8 \times 10^{-8} \text{ N}$.
Thus,the correct option is $B$.

(A) The electron microscope operates on the principle of the wave nature of electrons.
According to the de Broglie hypothesis,moving electrons are associated with a wave,known as matter waves.
The wavelength of these matter waves is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.
Because the wavelength of electrons is much smaller than that of visible light,electron microscopes can achieve much higher resolution than optical microscopes.
Therefore,the correct option is $(A)$.

(D) Microwave ovens operate by using electromagnetic waves to excite water molecules in food. The standard frequency commonly used for domestic microwave ovens is $2.45 \text{ GHz}$. However,among the provided options,$0.915 \text{ GHz}$ is a frequency also allocated for industrial,scientific,and medical $(ISM)$ applications,including specific types of microwave heating. Given the choices,$0.915 \text{ GHz}$ is the correct answer.

(A) The given system consists of four plates. Let the capacitance of each pair of adjacent plates be $C = \frac{A \varepsilon_0}{d}$.
Based on the circuit diagram,plates $1$ and $3$ are connected together,and plates $2$ and $4$ are connected to terminals $A$ and $B$ respectively.
This forms two capacitors in parallel (between plates $1-2$ and $3-2$) which are then in series with a third capacitor (between plates $3-4$).
Let $C_1$ be the capacitance between plates $1$ and $2$,$C_2$ between $3$ and $2$,and $C_3$ between $3$ and $4$. Thus,$C_1 = C_2 = C_3 = C$.
The parallel combination of $C_1$ and $C_2$ gives $C' = C_1 + C_2 = C + C = 2C$.
This $C'$ is in series with $C_3$. The equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = \frac{C' \cdot C_3}{C' + C_3} = \frac{(2C) \cdot C}{2C + C} = \frac{2C^2}{3C} = \frac{2}{3}C$.
Substituting $C = \frac{A \varepsilon_0}{d}$,we get $C_{eq} = \frac{2}{3} \frac{A \varepsilon_0}{d}$.

(C) Let the potentials of the spheres with radii $a$ and $b$ be $V_a$ and $V_b$ respectively,and the charges on them be $Q_a$ and $Q_b$.
Since the two spheres are connected by a conducting wire,they must be at the same potential:
$V_a = V_b$
$\frac{k Q_a}{a} = \frac{k Q_b}{b}$
$\frac{Q_a}{Q_b} = \frac{a}{b}$
Using the property of ratios,$\frac{Q_a}{Q_a + Q_b} = \frac{a}{a + b}$.
Since the total charge $Q_a + Q_b = Q$,we have:
$Q_a = \frac{a Q}{a + b}$ and $Q_b = \frac{b Q}{a + b}$.
Now,calculating the potential $V$ of each sphere:
$V = V_a = V_b = \frac{k Q_a}{a} = \frac{k}{a} \left( \frac{a Q}{a + b} \right) = \frac{k Q}{a + b}$.

(D) The relation between electric field $\overrightarrow{E}$ and electric potential $V$ is given by $\overrightarrow{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$.
Given $V = 5x^2$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{d}{dx}(5x^2) = 10x$
$\frac{\partial V}{\partial y} = 0$
$\frac{\partial V}{\partial z} = 0$
Thus,$\overrightarrow{E} = -(10x)\hat{i} = -10x\hat{i} \text{ N/C}$.
At the point $(1, 2, 3) \text{ m}$,substituting $x = 1$:
$\overrightarrow{E} = -10(1)\hat{i} = -10\hat{i} \text{ N/C}$.
Therefore,the correct option is $D$.

(A) Magnetization $(M)$ is defined as the net magnetic moment per unit volume.
$M = \frac{M_{\text{net}}}{V}$
Given:
Number of atoms $(N)$ = $4 \times 10^{14}$
Magnetic moment of each atom $(m)$ = $16 \times 10^{-24} \text{ A m}^2$
Side length of the cube $(a)$ = $2 \mu m = 2 \times 10^{-6} \text{ m}$
Volume of the cube $(V)$ = $a^3 = (2 \times 10^{-6})^3 = 8 \times 10^{-18} \text{ m}^3$
Total magnetic moment $(M_{\text{net}})$ = $N \times m = (4 \times 10^{14}) \times (16 \times 10^{-24}) = 64 \times 10^{-10} \text{ A m}^2$
Now,calculating magnetization:
$M = \frac{64 \times 10^{-10}}{8 \times 10^{-18}}$
$M = 8 \times 10^8 \times 10^{-4} = 8 \times 10^4 \text{ A m}^{-1}$

(C) The time period $T$ of a bar magnet oscillating in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic dipole moment,and $B$ is the magnetic field strength.
Since the mass and volume are the same,the moment of inertia $I$ remains constant.
For the new magnet,the magnetic dipole moment $M' = 9M$.
The new time period $T'$ is given by: $T' = 2\pi \sqrt{\frac{I}{M'B}} = 2\pi \sqrt{\frac{I}{9MB}}$.
Simplifying this,we get: $T' = \frac{1}{\sqrt{9}} \times 2\pi \sqrt{\frac{I}{MB}} = \frac{1}{3} T$.
Therefore,the new time period is $\frac{T}{3}$.

(B) The magnetic field on the axis of a circular loop is given by: $B_{\text{axis}} = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$
The magnetic field at the centre of the loop is given by: $B_{\text{centre}} = \frac{\mu_0 I}{2a}$
Taking the ratio of the two fields:
$\frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{\mu_0 I}{2a} \times \frac{2(a^2 + x^2)^{3/2}}{\mu_0 I a^2} = \frac{(a^2 + x^2)^{3/2}}{a^3}$
Given $a = 6 \ cm$ and $x = 8 \ cm$:
$\frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{(6^2 + 8^2)^{3/2}}{6^3} = \frac{(36 + 64)^{3/2}}{216} = \frac{(100)^{3/2}}{216} = \frac{1000}{216}$
Given $B_{\text{axis}} = 216 \ \mu T$:
$B_{\text{centre}} = 216 \times \frac{1000}{216} = 1000 \ \mu T$.

(D) The magnetic force per unit length between two parallel wires is given by $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi y}$.
For wire $B$ to remain at rest,the upward magnetic force must balance the downward gravitational force (weight per unit length).
$\frac{F}{l} = \text{weight per unit length} = 40 \times 10^{-3} \text{ N/m}$.
Substituting the values: $40 \times 10^{-3} = \frac{4 \pi \times 10^{-7} \times 10 \times 20}{2 \pi \times y}$.
Simplifying the equation: $40 \times 10^{-3} = \frac{2 \times 10^{-7} \times 200}{y}$.
$40 \times 10^{-3} = \frac{400 \times 10^{-7}}{y}$.
$y = \frac{400 \times 10^{-7}}{40 \times 10^{-3}} = 10 \times 10^{-4} = 10^{-3} \text{ m}$.
Since the magnetic force must be upward (repulsive) to counteract gravity,the currents must flow in opposite directions.

(C) The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $y$ is given by the formula:
$F = \frac{\mu_0 I_1 I_2}{2 \pi y}$
Given that both wires carry the same current $I$,the initial force is:
$F = \frac{\mu_0 I^2}{2 \pi y}$
If the current in each wire is halved,the new current becomes $I' = \frac{I}{2}$.
The new force $F'$ is:
$F' = \frac{\mu_0 (I/2) (I/2)}{2 \pi y} = \frac{\mu_0 I^2}{4 \cdot 2 \pi y}$
Substituting the original force $F$ into this equation:
$F' = \frac{1}{4} F$
Therefore,the force becomes $\frac{F}{4}$.

(B) The average density of a nucleus is approximately $\rho_{n} = 2.3 \times 10^{17} \ kg/m^{3}$.
The density of water is $\rho_{w} = 10^{3} \ kg/m^{3}$.
To find the ratio,we divide the density of the nucleus by the density of water:
$\text{Ratio} = \frac{\rho_{n}}{\rho_{w}} = \frac{2.3 \times 10^{17} \ kg/m^{3}}{10^{3} \ kg/m^{3}} = 2.3 \times 10^{14}$.
Therefore,the density of the nucleus is $2.3 \times 10^{14}$ times that of water.