GUJCET 2015 Physics Question Paper with Answer and Solution

(B) Given: $\theta = 45^{\circ}$,$d = 5.60 \ m$,$E = 100 \ V \ m^{-1}$,$m = 1 \ kg$,$\mu = 0.1$,$q = 10^{-2} \ C$,$v_0 = 0$.
From the free body diagram,the normal force $N$ is:
$N = mg \cos 45^{\circ} + qE \sin 45^{\circ}$
$N = (1 \times 10 \times \frac{1}{\sqrt{2}}) + (10^{-2} \times 100 \times \frac{1}{\sqrt{2}})$
$N = \frac{10}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.778 \ N$
The net force $F$ acting down the incline is:
$F = mg \sin 45^{\circ} - qE \cos 45^{\circ} - \mu N$
$F = (1 \times 10 \times \frac{1}{\sqrt{2}}) - (10^{-2} \times 100 \times \frac{1}{\sqrt{2}}) - (0.1 \times 7.778)$
$F = \frac{10}{\sqrt{2}} - \frac{1}{\sqrt{2}} - 0.7778 = \frac{9}{\sqrt{2}} - 0.7778 \approx 6.364 - 0.778 = 5.586 \ N$
Acceleration $a = \frac{F}{m} = \frac{5.586}{1} = 5.586 \ m \ s^{-2}$.
Using $d = v_0 t + \frac{1}{2} a t^2$:
$5.60 = 0 + \frac{1}{2} \times 5.586 \times t^2$
$t^2 = \frac{2 \times 5.60}{5.586} \approx 2$
$t = \sqrt{2} \approx 1.41 \ s$.

(A) The given alternating voltage is $V = 100 \sqrt{2} \sin(100t) \text{ V}$.
Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the peak voltage $V_m = 100 \sqrt{2} \text{ V}$ and angular frequency $\omega = 100 \text{ rad/s}$.
The capacitive reactance is given by $X_C = \frac{1}{\omega C}$.
Substituting the values,$X_C = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \text{ } \Omega$.
The root mean square $(RMS)$ voltage is $V_{\text{rms}} = \frac{V_m}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 \text{ V}$.
The $RMS$ current $I_{\text{rms}}$ is given by $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}$.
$I_{\text{rms}} = \frac{100}{10^4} = 10^{-2} \text{ A}$.
Converting to milliamperes,$I_{\text{rms}} = 10^{-2} \times 10^3 \text{ mA} = 10 \text{ mA}$.

(B) In a purely capacitive $AC$ circuit,the voltage $V$ is given by $V = V_m \sin(\omega t)$ and the current $I$ is given by $I = I_m \sin(\omega t + \frac{\pi}{2})$.
This shows that the current leads the voltage by a phase angle of $\frac{\pi}{2}$ radians.
Therefore,the correct option is $B$.

(D) For an $RL$ series circuit,the phase difference $\phi$ between voltage and current is given by the formula: $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.
Given: Frequency $\nu = \frac{25}{\pi} \text{ Hz}$,Resistance $R = 100 \ \Omega$,Inductance $L = 2 \text{ H}$.
Since $\omega = 2 \pi \nu$,we have:
$\tan \phi = \frac{2 \pi \nu L}{R}$
$\tan \phi = \frac{2 \pi \times (\frac{25}{\pi}) \times 2}{100}$
$\tan \phi = \frac{2 \times 25 \times 2}{100} = \frac{100}{100} = 1$.
Therefore,$\phi = \tan^{-1}(1) = 45^{\circ}$.

(B) The distance of the closest approach $r_{0}$ for an alpha particle of kinetic energy $K$ is given by the formula:
$r_{0} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2Ze^{2}}{K}$
From this expression,we can see that the distance of the closest approach is inversely proportional to the kinetic energy:
$r_{0} \propto \frac{1}{K}$
Let $r_{01} = r_{0}$ when kinetic energy $K_{1} = K$.
When the kinetic energy is increased to $K_{2} = 2K$,let the new distance be $r_{02}$.
Using the proportionality $r_{01} K_{1} = r_{02} K_{2}$,we get:
$r_{0} \cdot K = r_{02} \cdot (2K)$
$r_{02} = \frac{r_{0} \cdot K}{2K} = \frac{r_{0}}{2}$
Therefore,the new distance of the closest approach is $\frac{r_{0}}{2}$.

(D) In a hydrogen atom, an electron can exist in any of the infinite number of energy levels $(n = 1, 2, 3, \dots, \infty)$.
When an electron transitions from a higher energy level to a lower energy level, it emits a photon corresponding to a spectral line.
Since there are an infinite number of possible energy levels, there are an infinite number of possible transitions between these levels.
Therefore, the total number of spectral lines in a hydrogen atom is infinite $(\infty)$.

(B) The dimensional formulas are derived as follows:
$(A)$ Electrical resistance $(R)$: $R = V/I$. Potential $(V)$ is $M^1 L^2 T^{-3} A^{-1}$ and Current $(I)$ is $A^1$. Thus, $R = [M^1 L^2 T^{-3} A^{-1}] / [A^1] = M^1 L^2 T^{-3} A^{-2}$. This matches $(Q)$.
$(B)$ Electrical potential $(V)$: $V = W/q$. Work $(W)$ is $M^1 L^2 T^{-2}$ and Charge $(q)$ is $A^1 T^1$. Thus, $V = [M^1 L^2 T^{-2}] / [A^1 T^1] = M^1 L^2 T^{-3} A^{-1}$. This matches $(R)$.
$(C)$ Specific resistance (Resistivity, $\rho$): $\rho = R \cdot A / l$. Dimensions are $[M^1 L^2 T^{-3} A^{-2}] \cdot [L^2] / [L] = M^1 L^3 T^{-3} A^{-2}$. This matches $(P)$.
$(D)$ Specific conductance ($\sigma$): $\sigma = 1 / \rho$. Dimensions are $[M^{-1} L^{-3} T^3 A^2]$. This is not listed in the options, so it matches $(S)$.
Therefore, the correct matching is $A-Q, B-R, C-P, D-S$.

(D) Let the total resistance of the ring be $R$. The total length of the ring is $2 \pi r$. The resistance per unit length is $\lambda = \frac{R}{2 \pi r}$.
The length of the minor arc $AB$ is $l_1 = r \theta$. The resistance of this part is $R_1 = \lambda l_1 = \left(\frac{R}{2 \pi r}\right) (r \theta) = \frac{R \theta}{2 \pi}$.
The length of the major arc $AB$ is $l_2 = r(2 \pi - \theta)$. The resistance of this part is $R_2 = \lambda l_2 = \left(\frac{R}{2 \pi r}\right) (r(2 \pi - \theta)) = \frac{R(2 \pi - \theta)}{2 \pi}$.
Since the two arcs are connected in parallel between points $A$ and $B$,the equivalent resistance $R_{AB}$ is given by:
$R_{AB} = \frac{R_1 R_2}{R_1 + R_2}$
$R_{AB} = \frac{\left(\frac{R \theta}{2 \pi}\right) \left(\frac{R(2 \pi - \theta)}{2 \pi}\right)}{\frac{R \theta}{2 \pi} + \frac{R(2 \pi - \theta)}{2 \pi}}$
$R_{AB} = \frac{\frac{R^2 \theta (2 \pi - \theta)}{4 \pi^2}}{\frac{R}{2 \pi} (\theta + 2 \pi - \theta)}$
$R_{AB} = \frac{\frac{R^2 \theta (2 \pi - \theta)}{4 \pi^2}}{\frac{R}{2 \pi} (2 \pi)}$
$R_{AB} = \frac{R^2 \theta (2 \pi - \theta)}{4 \pi^2} \cdot \frac{1}{R} = \frac{R \theta (2 \pi - \theta)}{4 \pi^2}$

(B) When two wires are connected in series,the total resistance $R$ is the sum of individual resistances $R_1$ and $R_2$.
$R = R_1 + R_2$
Using the formula $R = \frac{\rho L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area:
$\frac{\rho(2l)}{A} = \frac{\rho_1 l}{A} + \frac{\rho_2 l}{A}$
Dividing both sides by $\frac{l}{A}$:
$2\rho = \rho_1 + \rho_2$
$\therefore \rho = \frac{\rho_1 + \rho_2}{2}$

(A) The two $8 \ \Omega$ resistors are connected in parallel to each other.
Their equivalent resistance $R_{eq}$ is given by:
$R_{eq} = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 \ \Omega$
The total resistance of the circuit is:
$R_{total} = 6 \ \Omega + 4 \ \Omega = 10 \ \Omega$
The current $I$ flowing through the circuit is:
$I = \frac{V_{source}}{R_{total}} = \frac{10 \ V}{10 \ \Omega} = 1 \ A$
The voltmeter is connected across the $6 \ \Omega$ resistor. Since the voltmeter has a very high resistance,it does not draw significant current,and the voltage across the $6 \ \Omega$ resistor is:
$V = I \times R = 1 \ A \times 6 \ \Omega = 6 \ V$

(D) As shown in the figure,consider an elementary charge $dq$ having length $dx$ at a distance $x$ from the point charge $q$. The linear charge density is $\lambda = \frac{Q}{L}$.
Then,$dq = \lambda dx = \frac{Q}{L} dx$.
By Coulomb's law,the force $dF$ between the elementary charge $dq$ and the point charge $q$ is:
$dF = \frac{k q dq}{x^2} = \frac{k q Q dx}{L x^2}$.
To find the total force $F$,integrate $dF$ from $x = r$ to $x = r + L$:
$F = \int_r^{r+L} \frac{k Q q}{L x^2} dx = \frac{k Q q}{L} \int_r^{r+L} x^{-2} dx$.
$F = \frac{k Q q}{L} \left[ -\frac{1}{x} \right]_r^{r+L} = \frac{k Q q}{L} \left( -\frac{1}{r+L} - (-\frac{1}{r}) \right)$.
$F = \frac{k Q q}{L} \left( \frac{1}{r} - \frac{1}{r+L} \right) = \frac{k Q q}{L} \left( \frac{r+L-r}{r(r+L)} \right)$.
$F = \frac{k Q q}{L} \left( \frac{L}{r(r+L)} \right) = \frac{k Q q}{r(r+L)}$.

(B) The correct option is $B$.
According to the quantization of charge,the total charge $Q$ on a body is given by the formula $Q = n \times e$,where $n$ is the number of electrons transferred and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \ C)$.
Given:
$n = 10^{19}$
$e = 1.6 \times 10^{-19} \ C$
Calculation:
$Q = 10^{19} \times 1.6 \times 10^{-19} \ C$
$Q = 1.6 \times 10^{(19-19)} \ C$
$Q = 1.6 \times 10^0 \ C$
$Q = 1.6 \ C$
Since electrons (which are negatively charged) are removed from the neutral plate,the plate develops a deficiency of electrons,resulting in a net positive charge.
Therefore,the charge on the plate becomes $+1.6 \ C$.

(A) The induced $emf$ $(\varepsilon)$ in a rod of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $\varepsilon = B l v$.
To find the rate of increase of induced $emf$, we differentiate with respect to time $t$:
$\frac{d\varepsilon}{dt} = B l \frac{dv}{dt}$.
Since $\frac{dv}{dt} = a$ (acceleration), we have $\frac{d\varepsilon}{dt} = B l a$.
Given: $B = 5 \times 10^{-4} \ Wb/m^2$, $l = 10 \ cm = 0.1 \ m$, $a = 5 \ m/s^2$.
Substituting the values:
$\frac{d\varepsilon}{dt} = (5 \times 10^{-4}) \times (0.1) \times (5) = 25 \times 10^{-4} \times 0.1 = 2.5 \times 10^{-4} \ V/s$.

(B) Given: Current $I = 2 \ A$,rate of change of current $\frac{dI}{dt} = -10^2 \ A \ s^{-1}$ (since it is decreasing),resistance $R = 2 \ \Omega$,$EMF$ $\varepsilon = 12 \ V$,and inductance $L = 5 \ mH = 5 \times 10^{-3} \ H$.
Applying Kirchhoff's voltage law from point $A$ to $B$:
$V_A - IR + \varepsilon - L \frac{dI}{dt} = V_B$
$V_B - V_A = -IR + \varepsilon - L \frac{dI}{dt}$
Substituting the values:
$V_B - V_A = -(2 \times 2) + 12 - (5 \times 10^{-3} \times (-10^2))$
$V_B - V_A = -4 + 12 + 0.5$
$V_B - V_A = 8.5 \ V$

(B) The number of waves $N$ in a given length $L$ is calculated by the formula: $N = \frac{L}{\lambda}$.
Given:
Length $L = 1 \text{ mm} = 10^{-3} \text{ m}$.
Wavelength $\lambda = 4000 \text{ Å} = 4000 \times 10^{-10} \text{ m} = 4 \times 10^{-7} \text{ m}$.
Substituting the values:
$N = \frac{10^{-3}}{4 \times 10^{-7}}$
$N = \frac{1}{4} \times 10^{-3 - (-7)}$
$N = 0.25 \times 10^4$
$N = 2500$.
Therefore, the number of waves is $2500$.

(A) The electromagnetic spectrum in increasing order of frequency is: Radio waves < Microwaves < Infrared < Visible < Ultraviolet < $X-$rays < $\gamma-$rays.
Given frequencies are:
$p$ (frequency of $X-$rays)
$q$ (frequency of $\gamma-$rays)
$r$ (frequency of ultraviolet rays)
Comparing their positions in the spectrum:
$1$. $\gamma-$rays have the highest frequency,so $q > p$.
$2$. $X-$rays have a higher frequency than ultraviolet rays,so $p > r$.
Combining these,we get $q > p > r$.
Looking at the options provided,we evaluate the relations:
For $p < q$ and $q > r$,this is consistent with $q > p$ and $q > r$.
Thus,the correct relation is $p < q$ and $q > r$.

(A) The length of the diagonal of a square with side $a$ is $a \sqrt{2}$.
Given side $a = 2 \sqrt{2} \text{ m}$,the diagonal length is $d = (2 \sqrt{2}) \times \sqrt{2} = 4 \text{ m}$.
The distance $r$ of each vertex from the center (intersection of diagonals) is half the diagonal length:
$r = \frac{d}{2} = \frac{4}{2} = 2 \text{ m}$.
The electric potential $V$ at the center due to four identical charges $q = 1 \mu C = 10^{-6} \text{ C}$ at distance $r$ is given by:
$V = 4 \times \frac{k q}{r}$,where $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.
Substituting the values:
$V = 4 \times \frac{9 \times 10^9 \times 10^{-6}}{2}$
$V = 2 \times 9 \times 10^3 = 18 \times 10^3 \text{ V}$.

(C) The correct option is $C$.
When a moving electron approaches another stationary electron,the electrostatic force of repulsion between them acts in the direction opposite to the motion of the moving electron.
This repulsive force does negative work on the moving electron,which causes its kinetic energy to decrease.
According to the law of conservation of energy,the total energy of the system remains constant.
Since the kinetic energy of the moving electron decreases,its potential energy must increase as the distance between the two electrons decreases.

(A) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of a small drop is given by $V = \frac{kq}{r} = 10 \ V$.
When $n = 27$ drops combine to form a large drop of radius $R$,the volume remains conserved: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which implies $R = n^{1/3}r$.
For $n = 27$,$R = (27)^{1/3}r = 3r$.
The total charge on the large drop is $Q = nq = 27q$.
The potential of the large drop is $V^{\prime} = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \left(\frac{kq}{r}\right) = n^{2/3}V$.
Substituting the values: $V^{\prime} = (27)^{2/3} \times 10 = (3^3)^{2/3} \times 10 = 3^2 \times 10 = 9 \times 10 = 90 \ V$.

(B) The angle of dip $(I)$ is defined by the relationship between the vertical component $(Z_E)$ and the horizontal component $(H_E)$ of the Earth's magnetic field as:
$\tan I = \frac{Z_E}{H_E}$
Given that the vertical component is $\sqrt{3}$ times the horizontal component:
$Z_E = \sqrt{3} H_E$
Substituting this into the formula:
$\tan I = \frac{\sqrt{3} H_E}{H_E} = \sqrt{3}$
Since $\tan 60^{\circ} = \sqrt{3}$,we have:
$I = 60^{\circ}$

(D) The current $I$ flowing through the galvanometer is proportional to the deflection $\phi$ in divisions,i.e.,$I \propto \phi$.
Using Ohm's law,$I = \frac{V}{R + G}$,where $V = 8 \ V$,$G = 50 \ \Omega$,and $R$ is the series resistance.
For the first case: $I_1 = \frac{V}{R_1 + G}$ with $\phi_1 = 30$ divisions and $R_1 = 3950 \ \Omega$.
For the second case: $I_2 = \frac{V}{R_2 + G}$ with $\phi_2 = 15$ divisions.
Taking the ratio: $\frac{I_1}{I_2} = \frac{\phi_1}{\phi_2} \implies \frac{R_2 + G}{R_1 + G} = \frac{30}{15} = 2$.
Substituting the values: $\frac{R_2 + 50}{3950 + 50} = 2$.
$\frac{R_2 + 50}{4000} = 2$.
$R_2 + 50 = 8000$.
$R_2 = 7950 \ \Omega$.

(B) When two long parallel wires carry currents in the same direction,each wire produces a magnetic field at the location of the other wire.
According to the right-hand rule,the magnetic field produced by one wire exerts a Lorentz force on the current-carrying second wire.
For currents in the same direction,the force between the wires is attractive.
Therefore,the wires will attract each other.

(B) The power $P$ produced is given by the product of the number of fissions per second $(n)$ and the energy released per fission $(E)$.
$P = n \times E$
Given:
$P = 6.4 \text{ W} = 6.4 \text{ J/s}$
$E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$
Rearranging for $n$:
$n = \frac{P}{E}$
$n = \frac{6.4}{3.2 \times 10^{-11}}$
$n = 2 \times 10^{11} \text{ fissions/second}$
Thus,the correct option is $B$.

(A) The correct answer is $A$.
The velocity of a wave is given by the relation $v = \nu \lambda$,where $\nu$ is the frequency and $\lambda$ is the wavelength.
When light travels from one medium to another,its frequency $\nu$ remains constant because it depends on the source of the light.
Since $v = \nu \lambda$ and $\nu$ is constant,the velocity $v$ is directly proportional to the wavelength $\lambda$ $(v \propto \lambda)$.
Therefore,when the velocity of light decreases upon entering a denser medium,the wavelength must also decrease.

(D) The critical angle $i_c$ for a ray of light passing from a denser medium (refractive index $n_1$) to a rarer medium (refractive index $n_2$) is given by the formula:
$\sin i_c = \frac{n_2}{n_1}$
Given:
$n_1 = 1.6$ (Refractive index of medium $A$)
$n_2 = 1.5$ (Refractive index of medium $B$)
Substituting the values:
$\sin i_c = \frac{1.5}{1.6}$
$\sin i_c = \frac{15}{16}$
Therefore,the critical angle is:
$i_c = \sin^{-1} \left(\frac{15}{16}\right)$
Thus,the correct option is $D$.

(D) The refractive index $n$ of a prism is given by the formula: $n = \frac{\sin((A + D_m)/2)}{\sin(A/2)}$.
Given that the angle of minimum deviation $D_m$ is equal to the angle of the prism $A$,we substitute $D_m = A$ into the formula:
$1.5 = \frac{\sin((A + A)/2)}{\sin(A/2)}$
$1.5 = \frac{\sin(A)}{\sin(A/2)}$
Using the trigonometric identity $\sin(A) = 2 \sin(A/2) \cos(A/2)$,we get:
$1.5 = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)}$
$1.5 = 2 \cos(A/2)$
$\cos(A/2) = 1.5 / 2 = 0.75$
Since $\sin(48^{\circ} 36^{\prime}) = 0.75$,we know that $\cos(90^{\circ} - 48^{\circ} 36^{\prime}) = 0.75$,which is $\cos(41^{\circ} 24^{\prime}) = 0.75$.
Therefore,$A/2 = 41^{\circ} 24^{\prime} = 41.4^{\circ}$.
Thus,$A = 2 \times 41.4^{\circ} = 82.8^{\circ}$.

(C) Let the initial intensity of the light be $I_{0}$.
According to Malus's Law,the final intensity $I$ of the light after passing through the tourmaline plate is given by:
$I = I_{0} \cos^{2} \theta$
Given that the angle $\theta = 60^{\circ}$,we have:
$I = I_{0} \cos^{2} 60^{\circ} = I_{0} \left(\frac{1}{2}\right)^{2} = \frac{I_{0}}{4}$
The percentage difference between the initial and final intensities is calculated as:
$\text{Percentage Difference} = \left(\frac{I_{0} - I}{I_{0}}\right) \times 100$
Substituting the value of $I$:
$\text{Percentage Difference} = \left(\frac{I_{0} - \frac{I_{0}}{4}}{I_{0}}\right) \times 100 = \left(\frac{\frac{3I_{0}}{4}}{I_{0}}\right) \times 100 = \frac{3}{4} \times 100 = 75 \%$

(B) The linear width of the central maximum in a single-slit diffraction pattern is given by $\beta = \frac{2 \lambda D}{a}$.
According to the problem,the linear width of the central maximum is equal to the width of the slit,so $\beta = a$.
Equating the two expressions:
$a = \frac{2 \lambda D}{a}$
Solving for $D$:
$D = \frac{a^{2}}{2 \lambda}$