GUJCET 2011 Physics Question Paper with Answer and Solution

(A) The $rms$ value of the alternating current is given by $I_{rms} = \frac{I_m}{\sqrt{2}}$.
Given $I_m = 5 \ A$,we have $I_{rms} = \frac{5}{1.414} \approx 3.536 \ A$.
The time taken to reach the peak value from zero is one-fourth of the time period $T$.
Since $T = \frac{1}{f} = \frac{1}{60} \ s$,the time $t = \frac{T}{4} = \frac{1}{4 \times 60} \ s$.
$t = \frac{1}{240} \ s \approx 0.004167 \ s = 4.167 \ ms$.

(B) The resistance $R$ of a wire is given by $R = \frac{\rho l}{A}$.
Since volume $V = A \times l$,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \frac{\rho l^2}{V}$.
Since density $d = \frac{m}{V}$,we have $V = \frac{m}{d}$.
Thus,$R = \frac{\rho l^2 d}{m}$.
For wires of the same material,$\rho$ and $d$ are constant,so $R \propto \frac{l^2}{m}$.
Given the ratios $m_1:m_2:m_3 = 1:3:5$ and $l_1:l_2:l_3 = 5:3:1$,the ratio of resistances is:
$R_1: R_2: R_3 = \frac{l_1^2}{m_1} : \frac{l_2^2}{m_2} : \frac{l_3^2}{m_3}$
$R_1: R_2: R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5}$
$R_1: R_2: R_3 = 25 : 3 : 0.2$
Multiplying by $5$ to clear the fraction:
$R_1: R_2: R_3 = 125 : 15 : 1$.

(D) The given circuit can be analyzed by identifying the Wheatstone bridge structure. The resistors form a bridge between points $A$ and $B$ with nodes $C$ and $D$.
Since all resistors are $3 \ \Omega$,the ratio of resistances in the arms is $\frac{3}{3} = \frac{3}{3}$,which satisfies the balanced Wheatstone bridge condition.
Therefore,no current flows through the central resistor connected between $C$ and $D$. We can remove this resistor from the circuit.
After removing the central resistor,the circuit simplifies to two parallel branches connected between $A$ and $B$:
$1$. The upper branch consists of two $3 \ \Omega$ resistors in series: $R_{ACB} = 3 + 3 = 6 \ \Omega$.
$2$. The lower branch consists of two $3 \ \Omega$ resistors in series: $R_{ADB} = 3 + 3 = 6 \ \Omega$.
$3$. There is also a direct $3 \ \Omega$ resistor connected in parallel between $A$ and $B$.
Now,we have three resistors of $6 \ \Omega, 6 \ \Omega$,and $3 \ \Omega$ connected in parallel.
The equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{3}$
$\frac{1}{R_{eq}} = \frac{1 + 1 + 2}{6} = \frac{4}{6} = \frac{2}{3}$
$R_{eq} = \frac{3}{2} = 1.5 \ \Omega$.

(D) The power $P$ dissipated in an external resistance $R$ connected to a cell of emf $\varepsilon$ and internal resistance $r$ is given by the formula:
$P = I^2 R = \left( \frac{\varepsilon}{R+r} \right)^2 R$
To find the maximum power,we differentiate $P$ with respect to $R$ and set it to zero:
$\frac{dP}{dR} = \varepsilon^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} \right] = 0$
$(R+r)^2 - 2R(R+r) = 0$
$(R+r)(R+r - 2R) = 0$
$r - R = 0 \implies R = r$
Thus,the power dissipated is maximum when the external resistance equals the internal resistance $(R = r)$.
Substituting $R = r$ into the power equation:
$P_{\max} = \left( \frac{\varepsilon}{r+r} \right)^2 r = \left( \frac{\varepsilon}{2r} \right)^2 r = \frac{\varepsilon^2}{4r^2} \times r = \frac{\varepsilon^2}{4r}$

(A) For a charge $q_2$ to revolve in a circular orbit,the electrostatic force between $q_1$ and $q_2$ provides the necessary centripetal force.
The centripetal force is given by $F_c = \frac{m v^2}{r}$,where $v$ is the orbital velocity.
The electrostatic force is given by Coulomb's law as $F_e = \frac{k q_1 q_2}{r^2}$.
Equating the two forces: $\frac{m v^2}{r} = \frac{k q_1 q_2}{r^2}$.
Since $v = r \omega$,where $\omega$ is the angular velocity,we have $\frac{m (r \omega)^2}{r} = \frac{k q_1 q_2}{r^2}$.
Simplifying,$m r \omega^2 = \frac{k q_1 q_2}{r^2}$,which gives $\omega^2 = \frac{k q_1 q_2}{m r^3}$.
Since $\omega = \frac{2 \pi}{T}$,where $T$ is the periodic time,we have $\left(\frac{2 \pi}{T}\right)^2 = \frac{k q_1 q_2}{m r^3}$.
$\frac{4 \pi^2}{T^2} = \frac{k q_1 q_2}{m r^3}$.
Therefore,$T^2 = \frac{4 \pi^2 m r^3}{k q_1 q_2}$.
Taking the square root,$T = \left[\frac{4 \pi^2 m r^3}{k q_1 q_2}\right]^{\frac{1}{2}}$.

(C) The correct answer is $C$.
To calculate the electric field using Gauss's Law,we choose a Gaussian surface such that the electric field is constant at every point on the surface and the angle between the electric field vector and the area vector is constant.
This is achieved by selecting a symmetrical closed surface that reflects the symmetry of the charge distribution,ensuring that the electric field magnitude is uniform across the surface.

(D) The electric field is given as $\vec{E} = 1 \hat{j} \ N/C$.
Since the square is placed in the $XY$-plane,its area vector $\vec{A}$ is perpendicular to the $XY$-plane,which is along the $Z$-axis.
Therefore,$\vec{A} = 1 \hat{k} \ m^2$.
The electric flux $\phi$ is defined as the dot product of the electric field and the area vector:
$\phi = \vec{E} \cdot \vec{A}$
$\phi = (1 \hat{j}) \cdot (1 \hat{k})$
Since the dot product of orthogonal unit vectors $\hat{j} \cdot \hat{k} = 0$,the flux is:
$\phi = 0 \ Nm^2/C$.

(A) The formula for the self-inductance $(L)$ of a solenoid is given by:
$L = \frac{\mu_0 N^2 A}{l}$
Given:
Permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
Number of turns $N = 10^3$
Area of cross-section $A = 10^{-3} \ m^2$
Length $l = 31.4 \ cm = 31.4 \times 10^{-2} \ m$
Substituting the values:
$L = \frac{4 \times 3.14 \times 10^{-7} \times (10^3)^2 \times 10^{-3}}{31.4 \times 10^{-2}}$
$L = \frac{4 \times 3.14 \times 10^{-7} \times 10^6 \times 10^{-3}}{31.4 \times 10^{-2}}$
$L = \frac{12.56 \times 10^{-4}}{31.4 \times 10^{-2}} = \frac{12.56}{31.4} \times 10^{-2} = 0.4 \times 10^{-2} \ H$
$L = 4 \times 10^{-3} \ H = 4 \ mH$

(B) The magnetic flux $\phi$ linked with a coil is given by the relation $\phi = L \cdot I$,where $L$ is the self-inductance and $I$ is the current flowing through the coil.
Given:
$\phi = 10 \ \mu Wb = 10 \times 10^{-6} \ Wb$
$I = 2 \ mA = 2 \times 10^{-3} \ A$
Rearranging the formula for self-inductance $L$:
$L = \frac{\phi}{I}$
Substituting the values:
$L = \frac{10 \times 10^{-6}}{2 \times 10^{-3}}$
$L = 5 \times 10^{-3} \ H$
$L = 5 \ mH$
Therefore,the self-inductance of the coil is $5 \ mH$.

(A) In Hertz's experiment,the two metal rods connected to the high-voltage induction coil act as the plates of a capacitor. When the potential difference between the rods becomes sufficiently high,the air between them breaks down,causing a spark to jump across the gap. This oscillation of charge creates electromagnetic waves.

(C) The line integral of the electric field $\vec{E}$ along a path is defined as the potential difference $V$ between two points.
Mathematically,$V = -\int \vec{E} \cdot d\vec{l}$.
The $SI$ unit of electric potential $V$ is the volt $(V)$,which is defined as the work done per unit charge.
Therefore,$1 \ V = 1 \ J C^{-1}$.
Thus,the unit of the physical quantity obtained by the line integral of the electric field is $J C^{-1}$ (or Volts).

(C) The work done $W$ to change the separation between two point charges is equal to the change in their electrostatic potential energy $\Delta U$.
The potential energy of a system of two charges is given by $U = \frac{k q_1 q_2}{r}$.
Given: $q_1 = 5 \times 10^{-6} \ C$,$q_2 = 10 \times 10^{-6} \ C$,$r_1 = 1 \ m$,$r_2 = 0.5 \ m$,and $k = 9 \times 10^9 \ N \cdot m^2/C^2$.
Work done $W = U_2 - U_1 = k q_1 q_2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$.
Substituting the values:
$W = (9 \times 10^9) \times (5 \times 10^{-6}) \times (10 \times 10^{-6}) \times \left( \frac{1}{0.5} - \frac{1}{1} \right)$
$W = (9 \times 10^9) \times (50 \times 10^{-12}) \times (2 - 1)$
$W = 450 \times 10^{-3} \ J = 0.45 \ J = 45 \times 10^{-2} \ J$.

(A) For a charged hollow metal sphere,the electric field inside the sphere is zero $(E = 0)$.
Since the electric field is the negative gradient of the potential $(E = -dV/dr)$,if $E = 0$,then the potential $V$ must be constant throughout the interior of the sphere.
Therefore,the potential at any point inside the sphere,including the centre,is equal to the potential on its surface.
Given that the potential on the surface is $80 \ V$,the potential at the centre of the sphere is $80 \ V$.

(A) Diamagnetic materials are weakly repelled by magnetic fields. When placed in a non-uniform magnetic field,the induced magnetic moment is in the direction opposite to the applied field. Consequently,the material experiences a force that pushes it from the region of higher magnetic field intensity (stronger part) to the region of lower magnetic field intensity (weaker part). Therefore,the correct option is $A$.

(B) The magnetic field $B$ on the axial line of a bar magnet at a distance $Z$ from its centre is given by the formula $B = \frac{\mu_0}{4 \pi} \frac{2MZ}{(Z^2 - l^2)^2}$.
For a short bar magnet,the length $2l$ is very small compared to the distance $Z$,i.e.,$Z \gg l$.
Therefore,we can approximate $(Z^2 - l^2)^2 \approx (Z^2)^2 = Z^4$.
Substituting this into the formula,we get $B = \frac{\mu_0}{4 \pi} \frac{2MZ}{Z^4} = \frac{\mu_0}{4 \pi} \frac{2M}{Z^3}$.
In vector form,this is expressed as $\vec{B} = \frac{2 \mu_0}{4 \pi} \frac{\vec{M}}{Z^3}$.

(C) An ideal voltmeter is designed to measure the potential difference between two points in a circuit without drawing any current from the circuit.
To ensure that no current flows through the voltmeter,its resistance must be infinitely high.
Therefore,the resistance of an ideal voltmeter is infinite.

(B) The formula for the shunt resistance $S$ required to send a fraction of the total current $I$ through a galvanometer of resistance $G$ is given by:
$S = \frac{G I_G}{I - I_G}$
Given that the galvanometer current $I_G = 10 \% I = 0.1 I$ and the galvanometer resistance $G = 99 \Omega$.
Substituting these values into the formula:
$S = \frac{99 \times 0.1 I}{I - 0.1 I}$
$S = \frac{9.9 I}{0.9 I}$
$S = \frac{9.9}{0.9} = 11 \Omega$
Therefore,the required shunt resistance is $11 \Omega$.