GUJCET 2010 Physics Question Paper with Answer and Solution

(B) The speed of light in vacuum is given by the relation $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
Squaring both sides,we get $c^{2} = \frac{1}{\mu_{0} \varepsilon_{0}}$.
Since $c$ represents velocity,its dimensional formula is $[M^{0} L^{1} T^{-1}]$.
Therefore,the dimensional formula of $\frac{1}{\mu_{0} \varepsilon_{0}}$ is the square of the dimensional formula of velocity:
$\frac{1}{\mu_{0} \varepsilon_{0}} = [M^{0} L^{1} T^{-1}]^{2} = M^{0} L^{2} T^{-2}$.
Thus,the correct option is $B$.

(C) The instantaneous value of an alternating voltage is given by $V(t) = V_{\max} \sin(\omega t)$.
To find the average value over a complete cycle (from $t = 0$ to $t = T$),we integrate the function over the period $T$ and divide by the period:
$V_{\text{avg}} = \frac{1}{T} \int_{0}^{T} V_{\max} \sin(\omega t) dt$.
Since $\omega = \frac{2\pi}{T}$,the integral becomes:
$V_{\text{avg}} = \frac{V_{\max}}{T} \int_{0}^{T} \sin\left(\frac{2\pi}{T} t\right) dt$.
The integral of the sine function over a full period is zero because the positive area of the first half-cycle exactly cancels out the negative area of the second half-cycle.
Therefore,$V_{\text{avg}} = 0$.

(D) The $Q$ factor (Quality factor) of a series $LCR$ circuit is given by the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Given values are $R = 6 \Omega$,$L = 1 \text{ H}$,and $C = 17.36 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$Q = \frac{1}{6} \sqrt{\frac{1}{17.36 \times 10^{-6}}}$
$Q = \frac{1}{6} \sqrt{\frac{10^6}{17.36}}$
$Q = \frac{1}{6} \times \sqrt{57603.68}$
$Q \approx \frac{1}{6} \times 240$
$Q \approx 40$
Therefore,the correct option is $D$.

(A) In an ideal transformer,the primary and secondary coils are linked by a common magnetic flux that oscillates at the same rate as the input alternating current source.
Since the frequency of the induced electromotive force $(EMF)$ in the secondary coil depends solely on the frequency of the magnetic flux variation,which is determined by the input source,the frequency remains constant.
Therefore,the frequency of the output voltage is the same as the frequency of the input voltage.

(D) For series connection:
The total electromotive force is $2\varepsilon$ and the total internal resistance is $2r$.
The current flowing through the circuit is given by:
$I_1 = \frac{2\varepsilon}{R + 2r} = \frac{2\varepsilon}{2 + 2r}$
For parallel connection:
The total electromotive force is $\varepsilon$ and the total internal resistance is $\frac{r}{2}$.
The current flowing through the circuit is given by:
$I_2 = \frac{\varepsilon}{R + \frac{r}{2}} = \frac{\varepsilon}{2 + \frac{r}{2}} = \frac{2\varepsilon}{4 + r}$
Since the current is the same in both connections,$I_1 = I_2$:
$\frac{2\varepsilon}{2 + 2r} = \frac{2\varepsilon}{4 + r}$
$4 + r = 2 + 2r$
$r = 2 \ \Omega$
Thus,the internal resistance of each cell is $2 \ \Omega$.

(C) First,calculate the resistance and rated current capacity for each bulb.
For the $25 W$ bulb: $R_1 = \frac{V^2}{P_1} = \frac{220^2}{25} = 1936 \Omega$. The rated current is $I_1 = \frac{P_1}{V} = \frac{25}{220} \approx 0.114 \ A$.
For the $100 W$ bulb: $R_2 = \frac{V^2}{P_2} = \frac{220^2}{100} = 484 \Omega$. The rated current is $I_2 = \frac{P_2}{V} = \frac{100}{220} \approx 0.454 \ A$.
When connected in series to a $440 \ V$ supply,the total resistance is $R_{eq} = R_1 + R_2 = 1936 + 484 = 2420 \ \Omega$.
The current flowing through the series circuit is $I = \frac{V_{total}}{R_{eq}} = \frac{440}{2420} \approx 0.181 \ A$.
Comparing the circuit current $I$ with the rated capacities: Since $I (0.181 \ A) > I_1 (0.114 \ A)$,the $25 \ W$ bulb will exceed its rated current capacity and fuse. Since $I (0.181 \ A) < I_2 (0.454 \ A)$,the $100 \ W$ bulb will remain safe.

(B) The circuit can be simplified step-by-step by identifying series and parallel combinations.
$1$. The resistors of $2 \ \Omega$ and $2 \ \Omega$ in branch $ACD$ are in series. Their equivalent resistance is $R_1 = 2 + 2 = 4 \ \Omega$.
$2$. This $R_1 = 4 \ \Omega$ is in parallel with the $4 \ \Omega$ resistor connected between $A$ and $D$. The equivalent resistance $R_2$ is given by $\frac{1}{R_2} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$,so $R_2 = 2 \ \Omega$.
$3$. Now,$R_2 = 2 \ \Omega$ is in series with the $2 \ \Omega$ resistor in branch $DE$. Their equivalent resistance is $R_3 = 2 + 2 = 4 \ \Omega$.
$4$. This $R_3 = 4 \ \Omega$ is in parallel with the $4 \ \Omega$ resistor connected between $A$ and $E$. The equivalent resistance $R_4$ is given by $\frac{1}{R_4} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$,so $R_4 = 2 \ \Omega$.
$5$. Finally,$R_4 = 2 \ \Omega$ is in series with the $2 \ \Omega$ resistor in branch $EB$. Their equivalent resistance is $R_5 = 2 + 2 = 4 \ \Omega$.
$6$. This $R_5 = 4 \ \Omega$ is in parallel with the $8 \ \Omega$ resistor connected directly between $A$ and $B$. The total equivalent resistance $R_{AB}$ is $\frac{1}{R_{AB}} = \frac{1}{4} + \frac{1}{8} = \frac{2 + 1}{8} = \frac{3}{8}$.
$7$. Therefore,$R_{AB} = \frac{8}{3} \ \Omega$.

(B) The electric field $E$ on the axial line of an electric dipole at a distance $r$ from its centre is given by the formula:
$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2pr}{(r^2 - a^2)^2}$
where $p$ is the dipole moment and $2a$ is the distance between the charges.
For a short dipole where $r \gg a$,we can neglect $a^2$ in the denominator:
$E \approx \frac{1}{4\pi\epsilon_0} \cdot \frac{2pr}{r^4}$
$E \approx \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}$
Therefore,$E \propto \frac{1}{r^3}$.

(A) The energy $U$ stored in an inductor with self-inductance $L$ carrying a current $I$ is given by the formula:
$U = \frac{1}{2} LI^2$
Given:
$L = 200 \ mH = 200 \times 10^{-3} \ H = 0.2 \ H$
$I = 4 \ A$
Substituting the values into the formula:
$U = \frac{1}{2} \times 0.2 \times (4)^2$
$U = 0.1 \times 16$
$U = 1.6 \ J$
Therefore,the energy required is $1.6 \ J$.

(B) The relationship between the speed of light $(c)$,frequency $(v)$,and wavelength $(\lambda)$ is given by the formula: $c = v \lambda$.
Given that the speed of light $c = 3 \times 10^{8} \ m/s$ and the frequency $v = 8.196 \times 10^{6} \ Hz$.
Rearranging the formula to solve for wavelength: $\lambda = \frac{c}{v}$.
Substituting the values: $\lambda = \frac{3 \times 10^{8}}{8.196 \times 10^{6}}$.
$\lambda = \frac{3}{8.196} \times 10^{2} \ m$.
$\lambda \approx 0.3660 \times 10^{2} \ m$.
$\lambda = 36.60 \ m$.
Converting meters to centimeters: $36.60 \ m = 3660 \ cm$.
Therefore,the correct option is $B$.

(D) For a conducting sphere,the electric potential $V$ at the surface is given by $V = \frac{kQ}{R}$,where $k$ is Coulomb's constant,$Q$ is the charge on the sphere,and $R$ is the radius of the sphere.
Since both spheres are made of copper (a conductor) and have the same size,they have the same radius $R$.
Given that both spheres are charged to the same potential $V$,we have $V_x = V_y = V$.
Using the formula $V = \frac{kQ}{R}$,we get $V = \frac{kQ_x}{R}$ and $V = \frac{kQ_y}{R}$.
Equating the two,we find $\frac{kQ_x}{R} = \frac{kQ_y}{R}$,which implies $Q_x = Q_y$.
Therefore,the charge on both spheres is equal,regardless of whether the sphere is hollow or solid,because the charge on a conductor resides entirely on its outer surface.

(D) Let the initial charge on the capacitor be $Q$ and its capacitance be $C$.
The initial energy stored in the capacitor is $U = \frac{Q^2}{2C}$.
When the battery is disconnected and an identical uncharged capacitor is connected in parallel,the total charge $Q$ is shared equally between the two capacitors because they are identical.
Thus,the charge on each capacitor becomes $Q' = \frac{Q}{2}$.
The energy stored in each capacitor is $U' = \frac{(Q')^2}{2C} = \frac{(Q/2)^2}{2C} = \frac{Q^2}{8C} = \frac{1}{4} \left( \frac{Q^2}{2C} \right) = \frac{U}{4}$.
The total energy of the system is the sum of the energies of the two capacitors:
$U_{total} = U' + U' = \frac{U}{4} + \frac{U}{4} = \frac{U}{2}$.

(C) The electric field $E$ between two parallel plates is given by the formula $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates.
Given:
Potential difference $V = 10 \ V$
Distance $d = 20 \ cm = 20 \times 10^{-2} \ m = 0.2 \ m$
Substituting these values into the formula:
$E = \frac{10}{0.2} = 50 \ Vm^{-1}$
Therefore,the electric field between the plates is $50 \ Vm^{-1}$.

(B) The capacitance $C$ of an isolated conducting sphere of radius $R$ is given by the formula:
$C = 4 \pi \varepsilon_0 R$
Since $4$,$\pi$,and $\varepsilon_0$ are constants,the capacitance is directly proportional to the radius $R$.
Therefore,$C \propto R$.
Thus,the correct option is $B$.

(C) The torque $\tau$ experienced by a magnetic dipole in a uniform magnetic field $B$ is given by $\tau = mB \sin \theta$,where $m$ is the magnetic moment and $\theta$ is the angle between the magnetic moment vector and the magnetic field vector.
Assuming the maximum torque condition (where $\sin \theta = 1$),we have $\tau = mB$.
Given $\tau = 0.64 \ J$ and $B = 0.32 \ T$.
Substituting the values: $0.64 = m \times 0.32$.
Therefore,$m = \frac{0.64}{0.32} = 2 \ Am^2$.
Thus,the magnetic moment of the magnet is $2 \ Am^2$.

(C) The magnetic field on the axial line of a short bar magnet is given by the formula:
$B = \frac{\mu_0}{4\pi} \frac{2M}{z^3}$
Given:
Magnetic moment $M = 10 \text{ Am}^2$
Distance $z = 0.1 \text{ m}$
Permeability constant $\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A}$
Substituting the values:
$B = 10^{-7} \times \frac{2 \times 10}{(0.1)^3}$
$B = 10^{-7} \times \frac{20}{0.001}$
$B = 10^{-7} \times 20000$
$B = 2 \times 10^{-3} \text{ T}$

(D) The magnetic moment $M$ of a current loop is given by $M = I \cdot A$,where $I$ is the current and $A$ is the area. The $SI$ unit is $A \cdot m^2$.
Alternatively,for a magnetic dipole in an external magnetic field $B$,the potential energy $U$ is given by $U = -M \cdot B$.
Thus,$M = U / B$.
The unit of energy $U$ is Joule $(J)$ and the unit of magnetic field $B$ is Tesla $(T)$.
Therefore,the unit of magnetic moment is $J \cdot T^{-1}$.

(C) Let the total length of the wire be $L$.
For a single-turn coil $(n_1 = 1)$,the circumference is $L = 2\pi R_1$,so $R_1 = \frac{L}{2\pi}$.
The magnetic field at the centre is $B_1 = \frac{\mu_0 I}{2R_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L} = B$.
For a two-turn coil $(n_2 = 2)$,the total length $L = n_2(2\pi R_2) = 2(2\pi R_2) = 4\pi R_2$.
Thus,the new radius is $R_2 = \frac{L}{4\pi} = \frac{R_1}{2}$.
The magnetic field at the centre for $n$ turns is $B_n = \frac{n \mu_0 I}{2R_n}$.
For $n=2$,$B_2 = \frac{2 \mu_0 I}{2R_2} = \frac{\mu_0 I}{R_2}$.
Substituting $R_2 = \frac{R_1}{2}$,we get $B_2 = \frac{\mu_0 I}{R_1/2} = \frac{2 \mu_0 I}{R_1}$.
Since $B = \frac{\mu_0 I}{2R_1}$,we have $\frac{\mu_0 I}{R_1} = 2B$.
Therefore,$B_2 = 2(2B) = 4B$.

(B) The correct answer is $B$.
An atom bomb operates on the principle of uncontrolled nuclear fission.
In this process,a heavy nucleus (such as $U^{235}$ or $Pu^{239}$) splits into smaller nuclei when bombarded with neutrons,releasing a tremendous amount of energy along with more neutrons,which sustain the chain reaction.

(A) In the ${ }_{92}^{235} U$ nucleus,the atomic number $Z$ (number of protons) is $92$.
The mass number $A$ is $235$.
The number of neutrons $N$ is given by $N = A - Z$.
$N = 235 - 92 = 143$.
The difference between the number of neutrons and protons is $N - Z$.
$N - Z = 143 - 92 = 51$.
Therefore,there are $51$ more neutrons than protons in the nucleus.