GUJCET 2008 Physics Question Paper with Answer and Solution

(B) The instantaneous voltage of an $AC$ generator is given by $V = V_m \sin(\omega t)$, where $V_m$ is the peak voltage and $\omega = 2 \pi \nu$ is the angular frequency.
Given: $V_m = 100 \text{ V}$, $V = 2 \text{ V}$ at $t = \frac{1}{100 \pi} \text{ s}$.
Substituting these values into the equation:
$2 = 100 \sin\left(2 \pi \nu \times \frac{1}{100 \pi}\right)$
$2 = 100 \sin\left(\frac{\nu}{50}\right)$
$\frac{2}{100} = \sin\left(\frac{\nu}{50}\right)$
Since the angle $\frac{\nu}{50}$ is very small, we can use the approximation $\sin(\theta) \approx \theta$:
$\frac{1}{50} = \frac{\nu}{50}$
$\nu = 1 \text{ Hz}$.

(C) In an $RL$ series circuit,the phase angle $\phi$ between the voltage and the current is given by the formula: $\tan \phi = \frac{X_L}{R}$,where $X_L = \omega L$ is the inductive reactance.
Given that the phase angle $\phi = 45^{\circ}$.
Substituting the value into the formula: $\tan 45^{\circ} = \frac{X_L}{R}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{X_L}{R}$.
Therefore,$X_L = R$.

(C) The power consumed by a bulb with resistance $R$ connected to a potential difference $V$ is given by $P = \frac{V^2}{R}$.
For the first bulb,$P_1 = \frac{V^2}{R_1}$,which implies $R_1 = \frac{V^2}{P_1}$.
For the second bulb,$P_2 = \frac{V^2}{R_2}$,which implies $R_2 = \frac{V^2}{P_2}$.
When the bulbs are connected in series,the total resistance is $R_{eq} = R_1 + R_2$.
The total power consumed in series is $P_{series} = \frac{V^2}{R_{eq}} = \frac{V^2}{R_1 + R_2}$.
Substituting the values of $R_1$ and $R_2$,we get $P_{series} = \frac{V^2}{\frac{V^2}{P_1} + \frac{V^2}{P_2}} = \frac{1}{\frac{1}{P_1} + \frac{1}{P_2}}$.
Simplifying this,we get $P_{series} = \frac{P_1 P_2}{P_1 + P_2}$.

(D) The heat produced per second in a conductor is equal to the power dissipated,given by $P = I^2 R$.
Since the wires are made of the same material and have the same length,the resistance $R$ is given by $R = \rho \frac{l}{A}$.
Substituting this into the power formula,we get $P = I^2 \left( \frac{\rho l}{A} \right)$.
Since $I$,$\rho$,and $l$ are constant for both wires,we have $P \propto \frac{1}{A}$.
Therefore,the ratio of heat produced is $\frac{P_1}{P_2} = \frac{A_2}{A_1}$.
Given the ratio of areas $\frac{A_1}{A_2} = \frac{1}{2}$,we have $\frac{A_2}{A_1} = \frac{2}{1}$.
Thus,the ratio of heat produced per second is $2 : 1$.

(D) Given: Electromotive force $\varepsilon = 4 \text{ V}$, internal resistance $r = 0.1 \ \Omega$, and external resistance $R = 3.9 \ \Omega$.
First, calculate the current $I$ flowing through the circuit using Ohm's law:
$I = \frac{\varepsilon}{R + r} = \frac{4}{3.9 + 0.1} = \frac{4}{4} = 1 \text{ A}$.
The terminal voltage $V$ across the cell is given by the potential difference across the external resistor $R$:
$V = I \times R$
$V = 1 \text{ A} \times 3.9 \ \Omega = 3.9 \text{ V}$.
Alternatively, using the formula $V = \varepsilon - Ir$:
$V = 4 - (1 \times 0.1) = 4 - 0.1 = 3.9 \text{ V}$.
Therefore, the correct option is $D$.

(A) Given: Area $A = (2 \ mm)^2 = (2 \times 10^{-3} \ m)^2 = 4 \times 10^{-6} \ m^2$. Current $I = 8 \ A$. Number density $n = 8 \times 10^{28} \ m^{-3}$. Charge of electron $e = 1.6 \times 10^{-19} \ C$.
Using the formula for drift velocity: $I = n A v_d e$.
Rearranging for $v_d$: $v_d = \frac{I}{n A e}$.
Substituting the values: $v_d = \frac{8}{8 \times 10^{28} \times 4 \times 10^{-6} \times 1.6 \times 10^{-19}}$.
$v_d = \frac{8}{51.2 \times 10^3} = \frac{8}{51200} \approx 1.56 \times 10^{-4} \ ms^{-1}$.

(A) As shown in the figure,let us consider an element of angular width $d \theta$ at an angle $\theta$.
The length of this element is $dl = a d \theta$.
The charge on this element is $dq = \lambda dl = (\lambda_0 \cos^2 \theta) (a d \theta)$.
The total charge $Q$ on the circle is obtained by integrating $dq$ over the entire circumference from $0$ to $2 \pi$:
$Q = \int_{0}^{2 \pi} \lambda_0 \cos^2 \theta a d \theta$
$Q = a \lambda_0 \int_{0}^{2 \pi} \cos^2 \theta d \theta$
Using the identity $\cos^2 \theta = \frac{1 + \cos 2 \theta}{2}$:
$Q = a \lambda_0 \int_{0}^{2 \pi} \frac{1 + \cos 2 \theta}{2} d \theta$
$Q = \frac{a \lambda_0}{2} [\theta + \frac{\sin 2 \theta}{2}]_{0}^{2 \pi}$
$Q = \frac{a \lambda_0}{2} [(2 \pi + 0) - (0 + 0)]$
$Q = \frac{a \lambda_0}{2} (2 \pi) = \pi a \lambda_0$
Thus,the total charge is $\pi a \lambda_0$.

(C) Let the two point charges be $q_1$ and $q_2$,separated by a distance $r$. The initial force is given by Coulomb's Law: $F = \frac{k q_1 q_2}{r^2} = 200 \ N$.
When the charge $q_1$ increases by $10 \ \%$,the new charge is $q_1' = q_1 + 0.1 q_1 = 1.1 q_1$.
When the charge $q_2$ decreases by $10 \ \%$,the new charge is $q_2' = q_2 - 0.1 q_2 = 0.9 q_2$.
The new electrical force $F'$ at the same distance $r$ is: $F' = \frac{k q_1' q_2'}{r^2} = \frac{k (1.1 q_1) (0.9 q_2)}{r^2}$.
Substituting the initial force value: $F' = (1.1 \times 0.9) \times \frac{k q_1 q_2}{r^2} = 0.99 \times F$.
Therefore,$F' = 0.99 \times 200 \ N = 198 \ N$.

(C) According to Faraday's law of electromagnetic induction,the induced $emf$ $(\varepsilon)$ is given by the negative rate of change of magnetic flux $(\phi)$ through the loop:
$\varepsilon = -\frac{d\phi}{dt}$
The magnetic flux $\phi$ through the ring of area $A = \pi r^2$ is given by $\phi = B \cdot A = (B_0 + \alpha t) \cdot \pi r^2$.
Substituting this into the $emf$ formula:
$\varepsilon = -\frac{d}{dt} [\pi r^2 (B_0 + \alpha t)]$
Since $\pi$ and $r$ are constants:
$\varepsilon = -\pi r^2 \frac{d}{dt} (B_0 + \alpha t)$
Calculating the derivative with respect to time $t$:
$\frac{d}{dt} (B_0 + \alpha t) = 0 + \alpha = \alpha$
Therefore,the induced $emf$ is:
$\varepsilon = -\pi r^2 \alpha$

(B) The mutual inductance $M$ between two coils is defined by the relationship between the change in magnetic flux $\Delta \phi$ linked with one coil and the change in current $\Delta I$ in the other coil.
The formula is given by:
$M = \frac{\Delta \phi}{\Delta I}$
Given values:
Change in current in coil $X$,$\Delta I = 4 \ A$
Change in magnetic flux in coil $Y$,$\Delta \phi = 0.4 \ Wb$
Substituting these values into the formula:
$M = \frac{0.4 \ Wb}{4 \ A}$
$M = 0.1 \ H$
Therefore,the mutual inductance of the system is $0.1 \ H$.

(A) According to Lenz's law,the direction of the induced current is such that it opposes the cause that produces it.
As the north pole $(N)$ of the magnet moves towards the metallic ring,the magnetic flux linked with the ring increases.
To oppose this increase in magnetic flux,the upper face of the ring must behave like a north pole $(N)$ to repel the approaching magnet.
$A$ face of a current-carrying loop behaves as a north pole when the current flows in an anticlockwise direction as viewed from that side.
Therefore,when looked at from above,the induced current in the ring will be anticlockwise.

(B) The work done $W$ in moving a charge $q$ between two points with potential difference $\Delta V$ is given by the formula: $W = q \Delta V$.
Here,$W = 20 \ J$,$q = 4 \ C$,initial potential $V_i = 10 \ V$,and final potential $V_f = V$.
The potential difference is $\Delta V = V_f - V_i = V - 10$.
Substituting the values into the formula:
$20 = 4(V - 10)$
Divide both sides by $4$:
$5 = V - 10$
Adding $10$ to both sides:
$V = 15 \ V$.

(A) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.
Since the volume remains constant,the volume of $64$ small drops equals the volume of the big drop:
$64 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$R^3 = 64 r^3$
$R = 4r$
The capacitance of a spherical drop is given by $C = 4 \pi \epsilon_0 r$.
Therefore,the capacitance of the small drop is $C_{\text{small}} = 4 \pi \epsilon_0 r$ and the capacitance of the big drop is $C_{\text{big}} = 4 \pi \epsilon_0 R$.
The ratio of the capacitance of the big drop to the small drop is:
$\frac{C_{\text{big}}}{C_{\text{small}}} = \frac{4 \pi \epsilon_0 R}{4 \pi \epsilon_0 r} = \frac{R}{r} = \frac{4r}{r} = 4$.
Thus,the ratio is $4: 1$.

(D) The given circuit can be redrawn as a balanced Wheatstone bridge.
In a balanced Wheatstone bridge,the potential difference across the central capacitor is zero,so it can be removed from the circuit.
After removing the central capacitor,the two upper capacitors ($4 \mu F$ and $4 \mu F$) are in series,and the two lower capacitors ($4 \mu F$ and $4 \mu F$) are in series.
The equivalent capacitance of the upper branch is $C_1 = \frac{4 \times 4}{4 + 4} = 2 \mu F$.
The equivalent capacitance of the lower branch is $C_2 = \frac{4 \times 4}{4 + 4} = 2 \mu F$.
These two branches are in parallel,so the total equivalent capacitance is $C_{eq} = C_1 + C_2 = 2 \mu F + 2 \mu F = 4 \mu F$.

(D) The length of the body diagonal of a cube with side $a$ is $\sqrt{3} a$. The distance $r$ of each vertex from the centre $O$ of the cube is half of the body diagonal: $r = \frac{\sqrt{3} a}{2}$.
Since there are $8$ identical charges $q$ at the vertices,the total electric potential $V$ at the centre $O$ is the sum of the potentials due to each individual charge:
$V = 8 \times \frac{1}{4 \pi \varepsilon_0} \times \frac{q}{r}$
Substituting the value of $r$:
$V = 8 \times \frac{1}{4 \pi \varepsilon_0} \times \frac{q}{\frac{\sqrt{3} a}{2}}$
$V = 8 \times \frac{1}{4 \pi \varepsilon_0} \times \frac{2q}{\sqrt{3} a}$
$V = \frac{4 q}{\sqrt{3} \pi \varepsilon_0 a}$

(B) The maximum magnetization $(M_{\max})$ occurs when all molecular dipoles are aligned in the direction of the external magnetic field.
It is calculated by the product of the total number of dipoles $(n)$ and the magnetic dipole moment of each molecule $(m)$.
Given:
$n = 2.0 \times 10^{24}$
$m = 1.5 \times 10^{-23} \text{ A m}^2$
Formula:
$M_{\max} = n \times m$
Calculation:
$M_{\max} = (2.0 \times 10^{24}) \times (1.5 \times 10^{-23})$
$M_{\max} = 3.0 \times 10^{1} = 30 \text{ A m}^2$
Therefore,the correct option is $B$.

(D) The angle between the Geographic Meridian and the Magnetic Meridian at any point on the Earth's surface is known as the Magnetic Declination or simply Declination.
It represents the deviation of the magnetic compass needle from the true geographic north.

(B) The deflection $\theta$ in a galvanometer is proportional to the current $I$ flowing through it,so $\theta \propto I$.
Initially,the current is $I$ and the deflection is $\theta_1 = 50$ divisions.
When a shunt resistance $S$ is connected in parallel,the current $I$ divides such that the galvanometer current $I_g$ corresponds to the new deflection $\theta_2 = 10$ divisions.
The ratio of the deflections is $n = \frac{\theta_1}{\theta_2} = \frac{50}{10} = 5$.
The formula relating the shunt resistance $S$,galvanometer resistance $G$,and the ratio $n$ is $S = \frac{G}{n-1}$.
Rearranging for $G$,we get $G = S(n-1)$.
Substituting the given values $S = 12 \Omega$ and $n = 5$:
$G = 12 \times (5 - 1) = 12 \times 4 = 48 \Omega$.
Thus,the resistance of the galvanometer coil is $48 \Omega$.

(C) To find $X$,we apply the law of conservation of mass number and atomic number.
For mass number (top indices): $9 + 4 = 12 + A \implies 13 = 12 + A \implies A = 1$.
For atomic number (bottom indices): $4 + 2 = 6 + Z \implies 6 = 6 + Z \implies Z = 0$.
$A$ particle with mass number $1$ and atomic number $0$ is a neutron,denoted as ${ }_{0}^{1} n$.
Therefore,$X$ represents a neutron $(n)$.