GUJCET 2016 Physics Question Paper with Answer and Solution

(C) The $Q$-factor (Quality factor) of an $L-C-R$ circuit is defined as the ratio of the resonant frequency to the bandwidth.
$Q = \frac{f_0}{f_2 - f_1}$
Given:
Resonant frequency $f_0 = 5000 \ Hz$
Half-power frequencies $f_1 = 4950 \ Hz$ and $f_2 = 5050 \ Hz$
Bandwidth $\Delta f = f_2 - f_1 = 5050 \ Hz - 4950 \ Hz = 100 \ Hz$
Substituting the values:
$Q = \frac{5000}{100} = 50$
Therefore,the $Q$-factor is $50$.

(B) step-down transformer is designed to decrease the voltage, meaning the output voltage is less than the input voltage $(V_s < V_p)$.
In an ideal transformer, the output power equals the input power $(P_{out} = P_{in})$.
However, in a real-world transformer, there are energy losses due to factors like resistance of windings (copper loss), eddy currents, and hysteresis.
Therefore, the output power is always slightly less than the input power $(P_{out} < P_{in})$.

(B) The distance of closest approach $r_0$ is the point where the initial kinetic energy of the $\alpha$-particle is completely converted into electrostatic potential energy.
$r_0 = \frac{k q_1 q_2}{K}$
Here,$q_1 = 2e$ (charge of $\alpha$-particle),$q_2 = Ze$ (charge of nucleus),and $K = 5 \text{ MeV} = 5 \times 10^6 \times 1.6 \times 10^{-19} \text{ J}$.
$r_0 = \frac{9 \times 10^9 \times (2e) \times (Ze)}{K} = \frac{9 \times 10^9 \times 2 \times 50 \times e^2}{5 \times 10^6 \times 1.6 \times 10^{-19}}$
$r_0 = \frac{9 \times 10^9 \times 100 \times (1.6 \times 10^{-19})^2}{5 \times 10^6 \times 1.6 \times 10^{-19}}$
$r_0 = \frac{9 \times 10^9 \times 100 \times 1.6 \times 10^{-19}}{5 \times 10^6} = \frac{1440 \times 10^{-10}}{5 \times 10^6} = 288 \times 10^{-16} \text{ m} = 2.88 \times 10^{-14} \text{ m}$.
Thus,the distance is $2.88 \times 10^{-14} \text{ m}$.

(B) The number of spectral lines emitted when an electron transitions from an excited state $n_2$ to a lower state $n_1$ is given by the formula:
$N = \frac{(n_2 - n_1 + 1)(n_2 - n_1)}{2}$
Here,the ground state is $n_1 = 1$ and the excited state is $n_2 = 4$.
Substituting these values into the formula:
$N = \frac{(4 - 1 + 1)(4 - 1)}{2}$
$N = \frac{(4)(3)}{2}$
$N = \frac{12}{2} = 6$
Thus,the total number of spectral lines observed is $6$.

(B) The heat energy $H$ required to boil a given quantity of water is constant.
Using the formula for heat energy $H = \frac{V^2}{R} t$,where $V$ is the voltage,$R$ is the resistance,and $t$ is the time.
Since $H$ and $R$ are constant,we have $H_1 = H_2$.
$\frac{V_1^2}{R} t_1 = \frac{V_2^2}{R} t_2$
Given $V_1 = V$,$t_1 = 5 \text{ min}$,and $V_2 = \frac{V}{2}$.
Substituting these values: $\frac{V^2}{R} (5) = \frac{(\frac{V}{2})^2}{R} t_2$
$5 = \frac{V^2}{4R} \cdot \frac{R}{V^2} \cdot t_2$
$5 = \frac{t_2}{4}$
$t_2 = 20 \text{ min}$.

(D) The correct option is $D$.
From the circuit diagram,the $6 \Omega$ and $2 \Omega$ resistors are connected in parallel. Their equivalent resistance $R_p$ is given by:
$R_p = \frac{6 \times 2}{6 + 2} = \frac{12}{8} = 1.5 \Omega$
Now,this equivalent resistance $R_p = 1.5 \Omega$ is in series with the $1.5 \Omega$ resistor and the $3 \Omega$ resistor.
Therefore,the total equivalent resistance $R_{eq}$ of the circuit is:
$R_{eq} = 1.5 \Omega + 1.5 \Omega + 3 \Omega = 6 \Omega$
Using Ohm's law,the total current $I$ supplied by the battery is:
$I = \frac{V}{R_{eq}} = \frac{9 \text{ V}}{6 \Omega} = 1.5 \text{ A}$

(A) The total number of batteries is $n = 4$. Each battery has an $EMF$ $\varepsilon = 1.5 \ V$ and internal resistance $r = 0.1 \ \Omega$.
When batteries are connected in series,if one battery is connected in reverse,its $EMF$ opposes the others.
Resultant $EMF$ $\varepsilon' = (\varepsilon + \varepsilon + \varepsilon) - \varepsilon = 2\varepsilon$.
$\varepsilon' = 2 \times 1.5 \ V = 3 \ V$.
Since internal resistances are always in series regardless of the polarity of the $EMF$,the resultant internal resistance $r'$ is the sum of all individual internal resistances.
$r' = r + r + r + r = 4r$.
$r' = 4 \times 0.1 \ \Omega = 0.4 \ \Omega$.
Therefore,the resultant $EMF$ is $3 \ V$ and the resultant internal resistance is $0.4 \ \Omega$.

(A) From Coulomb's law,the force $F$ between two charges $q_1$ and $q_2$ separated by distance $r$ is given by $F = k \frac{q_1 q_2}{r^2}$.
Rearranging for $k$,we get $k = \frac{F r^2}{q_1 q_2}$.
The $SI$ unit of $k$ is $\frac{N \cdot m^2}{C^2}$.
Since $1 \ C = 1 \ A \cdot s$,the unit can be written as $\frac{N \cdot m^2}{A^2 \cdot s^2}$.
The dimensions are: Force $[F] = M^1 L^1 T^{-2}$,distance $[r] = L^1$,current $[I] = I^1$,and time $[t] = T^1$.
Substituting these into the formula for $k$:
$[k] = \frac{[M^1 L^1 T^{-2}] [L^2]}{[I^2] [T^2]} = \frac{M^1 L^3 T^{-2}}{I^2 T^2} = M^1 L^3 T^{-4} I^{-2}$.

(D) Given that the electrostatic repulsive force is equal to the weight of the particle:
$F_e = F_g$
$\frac{k q^2}{r^2} = mg$
Rearranging for $r^2$:
$r^2 = \frac{k q^2}{mg}$
Substituting the given values:
$r^2 = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{1.66 \times 10^{-27} \times 10}$
$r^2 = \frac{9 \times 2.56 \times 10^{9} \times 10^{-38}}{1.66 \times 10^{-26}}$
$r^2 = \frac{23.04 \times 10^{-29}}{1.66 \times 10^{-26}}$
$r^2 = 13.8795 \times 10^{-3} = 1.38795 \times 10^{-2}$
Taking the square root:
$r = \sqrt{1.38795} \times 10^{-1} \ m$
$r \approx 1.178 \times 10^{-1} \ m$
Rounding to two decimal places,we get $r \approx 1.18 \times 10^{-1} \ m$.

(A) The self-inductance $L$ is defined by the relation $\phi = LI$,where $\phi$ is magnetic flux and $I$ is current. Thus,$L = \frac{\phi}{I}$.
Units of $L$ are $\frac{\text{weber}}{\text{ampere}}$ (Henry).
Since $\text{weber} = \text{volt} \cdot \text{second}$,the unit is also $\frac{\text{volt} \cdot \text{second}}{\text{ampere}}$.
Since $\text{volt} = \text{ampere} \cdot \text{ohm}$,the unit is $\frac{(\text{ampere} \cdot \text{ohm}) \cdot \text{second}}{\text{ampere}} = \text{ohm} \cdot \text{second}$.
'mho-second' is not a unit of self-inductance,as 'mho' is the unit of conductance $(1/\text{ohm})$.

(B) The electromagnetic spectrum is ordered by wavelength as follows: Gamma rays < $X$-rays < Ultraviolet < Visible < Infrared < Microwaves < Radio waves.
Given that $\lambda_r$ is the wavelength of Gamma rays,$\lambda_x$ is the wavelength of $X$-rays,and $\lambda_m$ is the wavelength of microwaves.
According to the electromagnetic spectrum,the wavelength of Gamma rays is the shortest and the wavelength of microwaves is much longer than that of $X$-rays.
Therefore,the correct order is $\lambda_r < \lambda_x < \lambda_m$.

(C) Initial energy of the capacitor is $U = \frac{Q^2}{2C}$.
When the capacitor is disconnected from the battery and connected in parallel to an uncharged capacitor of capacitance $2C$,the total charge $Q$ is redistributed such that both capacitors have the same potential difference $V'$.
Since they are in parallel,$V' = \frac{Q_1}{C} = \frac{Q_2}{2C}$.
This implies $Q_2 = 2Q_1$.
Since the total charge is conserved,$Q = Q_1 + Q_2 = Q_1 + 2Q_1 = 3Q_1$.
Therefore,$Q_1 = \frac{Q}{3}$ and $Q_2 = \frac{2Q}{3}$.
The new energy of the first capacitor is $U_1 = \frac{Q_1^2}{2C} = \frac{(Q/3)^2}{2C} = \frac{1}{9} \left(\frac{Q^2}{2C}\right) = \frac{1}{9} U$.
The new energy of the second capacitor is $U_2 = \frac{Q_2^2}{2(2C)} = \frac{(2Q/3)^2}{4C} = \frac{4Q^2/9}{4C} = \frac{1}{9} \left(\frac{Q^2}{C}\right) = \frac{2}{9} \left(\frac{Q^2}{2C}\right) = \frac{2}{9} U$.
Thus,the energies are $\frac{1}{9} U$ and $\frac{2}{9} U$ respectively.

(B) The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = \frac{k q_1 q_2}{r}$,where $k$ is Coulomb's constant.
For a proton $(+e)$ and an electron $(-e)$,the potential energy is $U = \frac{k(e)(-e)}{r} = -\frac{k e^2}{r}$.
As the proton moves away from the electron,the separation distance $r$ increases.
Since $r$ is in the denominator and the potential energy is negative,as $r$ increases,the value of $-\frac{k e^2}{r}$ becomes less negative (i.e.,it approaches zero from the negative side).
Therefore,the potential energy of the system increases.

(D) The area enclosed by a hysteresis loop ($B-H$ curve) represents the energy dissipated as heat per unit volume during one complete cycle of magnetization and demagnetization of a ferromagnetic material.
Therefore,the correct option is $D$.

(D) The centripetal force required for circular motion is provided by the magnetic Lorentz force.
$F_c = F_m$
$\frac{m v^2}{r} = q_{\alpha} v B$
Since the charge of an $\alpha$-particle is $q_{\alpha} = 2e$,we have:
$\frac{m v}{r} = (2e) B$
We know that the velocity $v$ is related to the time period $T$ by $v = \frac{2 \pi r}{T}$. Substituting this into the equation:
$\frac{m}{r} \left( \frac{2 \pi r}{T} \right) = 2e B$
$\frac{2 \pi m}{T} = 2e B$
Solving for $T$:
$T = \frac{\pi m}{e B}$
Thus,the correct answer is $\frac{\pi m}{e B}$.

(C) The correct option is $C$.
The magnetic force acting on a charged particle is given by $\overrightarrow{F} = q(\vec{v} \times \overrightarrow{B})$.
Since the magnetic force $\overrightarrow{F}$ is always perpendicular to the velocity vector $\vec{v}$,the work done by the magnetic force is zero $(W = \overrightarrow{F} \cdot \vec{d} = 0)$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done. Since the work done is zero,the kinetic energy remains constant.
However,because the force acts perpendicular to the velocity,it changes the direction of the velocity vector. Since momentum $\vec{p} = m\vec{v}$,a change in the direction of velocity implies a change in momentum.
Therefore,the momentum changes,but the kinetic energy remains the same.

(B) $1$. Two parallel wires carrying currents in the same direction attract each other,while those carrying currents in opposite directions repel each other.
$2$. In the figure,the current in wire $P$ is downward $(20 \ A)$,while the current in wire $Q$ is upward $(40 \ A)$. Since these currents are anti-parallel,wire $P$ exerts a repulsive force on wire $Q$,pushing it to the right.
$3$. The current in wire $Q$ is upward $(40 \ A)$,and the current in wire $R$ is also upward $(60 \ A)$. Since these currents are parallel,wire $R$ exerts an attractive force on wire $Q$,pulling it to the right.
$4$. Since both forces (repulsion from $P$ and attraction from $R$) act towards the right,the resultant force on wire $Q$ is directed to the right.

(C) The atomic mass number $(A)$ is defined as the sum of the number of protons $(Z)$ and the number of neutrons $(N)$ in the nucleus,so $A = Z + N$.
Since the number of neutrons $(N)$ is always greater than or equal to zero,the atomic mass number $(A)$ must be greater than or equal to the atomic number $(Z)$.
For the simplest element,Hydrogen $(_{1}^{1}H)$,the number of neutrons is $0$,so $A = Z = 1$.
For all other elements,the number of neutrons is greater than zero,making $A > Z$.
Therefore,the atomic mass number is always equal to or greater than the atomic number.

(B) According to the Lens Maker's Formula,the focal length $f$ of a lens is given by $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
This implies that $f \propto \frac{1}{n - 1}$,where $n$ is the refractive index of the material of the lens.
According to Cauchy's dispersion formula,the refractive index of a material is higher for light of shorter wavelengths (violet) and lower for light of longer wavelengths (red).
Therefore,$n_{V} > n_{R}$.
Since $f$ is inversely proportional to $(n - 1)$,a higher refractive index results in a shorter focal length.
Thus,$f_{V} < f_{R}$ or $f_{R} > f_{V}$.

(B) The correct answer is $B$.
$A$ solar cell is a photovoltaic device that converts light energy directly into electrical energy.
Unlike a photodiode or a varactor diode,which require an external bias voltage to function in their specific modes (reverse bias),a solar cell operates in the fourth quadrant of the $I-V$ characteristic curve without any external bias voltage.
It generates its own electromotive force when exposed to light.

(D) For the $n$-th order secondary maxima in single-slit diffraction,the condition is given by:
$a \sin \theta = (n + \frac{1}{2}) \lambda$
Given:
Slit width $a = 0.01 \ cm = 10^{-4} \ m$
Wavelength $\lambda = 5000 \ \mathring{A} = 5 \times 10^{-7} \ m$
Order of maxima $n = 2$
Substituting the values:
$\sin \theta = (2 + \frac{1}{2}) \frac{\lambda}{a}$
$\sin \theta = \frac{2.5 \times 5 \times 10^{-7}}{10^{-4}}$
$\sin \theta = 12.5 \times 10^{-3} = 0.0125$
Since $\theta$ is very small,$\sin \theta \approx \theta$.
Therefore,$\theta = 0.0125 \ \text{rad}$.

(A) The correct answer is $A$.
In a single-slit diffraction pattern, the angular width of the central maximum is given by $\theta = \frac{2\lambda}{a}$, where $\lambda$ is the wavelength of light and $a$ is the slit width.
This shows that the diffraction effect is directly proportional to the wavelength $(\text{diffraction} \propto \lambda)$.
Since the wavelength of yellow light $(\lambda_{Y})$ is greater than the wavelength of blue light $(\lambda_{B})$, i.e., $\lambda_{Y} > \lambda_{B}$, the diffraction pattern produced by yellow light will be more spread out.
Therefore, the maxima and minima are broadened and become more distant from each other.