KCET 2019 Physics Question Paper with Answer and Solution

(A) In any collision,if there are no external forces acting on the system,the total linear momentum of the system remains conserved according to the law of conservation of linear momentum.
This principle applies to both elastic and inelastic collisions.
However,total kinetic energy is only conserved in perfectly elastic collisions.
In inelastic collisions,some kinetic energy is transformed into other forms of energy (such as heat,sound,or deformation energy),so it is not conserved.
Therefore,total linear momentum is the quantity that always remains conserved.

(D) The center of mass velocity is given by $V_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$.
Since the particles are initially at rest,the initial momentum of the system is $P_{initial} = 0$.
There are no external forces acting on the system (only mutual internal attraction),so the net external force $F_{ext} = 0$.
According to the law of conservation of linear momentum,the total momentum of the system remains constant.
Therefore,$P_{final} = P_{initial} = 0$.
Since $P_{final} = M_{total} \times V_{cm}$,and the total mass $M_{total}$ is non-zero,the velocity of the center of mass $V_{cm}$ must be $0$ at all times.

(B) The orbital velocity of a satellite orbiting close to the Earth is given by $v_0 = \sqrt{\frac{GM}{R}}$.
The kinetic energy of the satellite in this orbit is $K = \frac{1}{2}mv_0^2 = \frac{GMm}{2R}$.
To escape the gravitational pull of the Earth,the satellite must reach the escape velocity,which is $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2}v_0$.
The required kinetic energy to escape is $K_e = \frac{1}{2}mv_e^2 = \frac{1}{2}m(2v_0^2) = 2(\frac{1}{2}mv_0^2) = 2K$.
The extra kinetic energy required is $\Delta K = K_e - K = 2K - K = K$.

(C) For an ideal gas at constant pressure,Charles's Law states that $ V \propto T $.
Since the volume is doubled $( V_2 = 2V_1 )$,the temperature must also double.
The initial temperature is $ T_1 = 27^{\circ}C = 27 + 273 = 300 \text{ K} $.
The final temperature is $ T_2 = 2 \times 300 \text{ K} = 600 \text{ K} $.
The change in temperature is $ \Delta T = 600 \text{ K} - 300 \text{ K} = 300 \text{ K} $.
For a diatomic gas,the molar heat capacity at constant pressure is $ C_p = \frac{7}{2}R $.
The heat added is given by $ Q = n C_p \Delta T $.
Substituting the values: $ Q = 1 \times \frac{7}{2} R \times 300 = 1050 R $.

(D) Given: Mass $m = 5 \ kg$,Force $\vec{F} = (-3 \hat{i} + 4 \hat{j}) \ N$,Initial velocity $\vec{u} = (6 \hat{i} - 12 \hat{j}) \ ms^{-1}$.
Using Newton's second law,$\vec{a} = \frac{\vec{F}}{m} = \left(\frac{-3}{5} \hat{i} + \frac{4}{5} \hat{j}\right) \ ms^{-2}$.
The velocity at any time $t$ is given by $\vec{v} = \vec{u} + \vec{a}t$.
$\vec{v} = (6 \hat{i} - 12 \hat{j}) + \left(-\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}\right)t$.
$\vec{v} = \left(6 - \frac{3}{5}t\right) \hat{i} + \left(-12 + \frac{4}{5}t\right) \hat{j}$.
For the velocity to be along the $y$-axis,the $x$-component of the velocity must be zero:
$6 - \frac{3}{5}t = 0$.
$\frac{3}{5}t = 6$.
$t = \frac{6 \times 5}{3} = 10 \ s$.

(B) The buoyant force $F_b$ acting on a submerged object is given by Archimedes' principle: $F_b = V \rho g$,where $V$ is the volume of the displaced liquid,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
The density of water is maximum at $4 \ ^\circ C$ $(\rho_4 \approx 1000 \ kg/m^3)$ and is lower at $0 \ ^\circ C$ $(\rho_0 \approx 999.8 \ kg/m^3)$.
Since the volume of the aluminium sphere $V$ remains constant,the buoyancy at $0 \ ^\circ C$ is $F_b = V \rho_0 g$ and at $4 \ ^\circ C$ is $F'_b = V \rho_4 g$.
Comparing the two,since $\rho_0 < \rho_4$,it follows that $F_b < F'_b$.
Therefore,the buoyancy is less in water at $0 \ ^\circ C$ than in water at $4 \ ^\circ C$.

(B) According to Torricelli's law,the velocity of efflux $v$ is given by $v = \sqrt{2gh}$,where $h$ is the depth of the hole from the free surface.
Given $h = 2 \text{ cm}$,so $v = \sqrt{2 \times g \times 2} = 2\sqrt{g}$.
The time $t$ taken for the water to reach the ground from a height $H = 8 \text{ cm}$ is given by $H = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 8}{g}} = \sqrt{\frac{16}{g}} = \frac{4}{\sqrt{g}}$.
The horizontal range $R$ is the product of the velocity of efflux and the time of flight:
$R = v \times t = (\sqrt{2gh}) \times \sqrt{\frac{2H}{g}} = 2\sqrt{hH}$.
Substituting the values $h = 2 \text{ cm}$ and $H = 8 \text{ cm}$:
$R = 2 \times \sqrt{2 \times 8} = 2 \times \sqrt{16} = 2 \times 4 = 8 \text{ cm}$.

(B) The Poisson's ratio $\sigma$ is defined as the negative ratio of transverse strain to longitudinal strain: $\sigma = -\frac{\Delta D/D}{\Delta L/L}$.
For a wire of length $L$ and diameter $D$,the volume $V = \frac{\pi D^2 L}{4}$.
Since the volume $V$ remains constant,$\frac{\Delta V}{V} = 2\frac{\Delta D}{D} + \frac{\Delta L}{L} = 0$.
This implies $2\frac{\Delta D}{D} = -\frac{\Delta L}{L}$,or $\frac{\Delta D/D}{\Delta L/L} = -0.5$.
Substituting this into the definition of Poisson's ratio: $\sigma = -(-0.5) = 0.5$.
Therefore,for a material with constant volume under stretching,the Poisson's ratio is $0.5$.
Thus,option $(B)$ is correct.

(A) The given graph is a straight line with a positive intercept on the $v$-axis and a negative slope.
Its equation can be written as:
$v = -mx + v_0 \dots(1)$
where $m = \tan \theta = \frac{v_0}{x_0}$ is the magnitude of the slope.
The acceleration $a$ is given by $a = v \frac{dv}{dx}$.
From equation $(1)$,differentiating $v$ with respect to $x$ gives:
$\frac{dv}{dx} = -m$
Substituting this into the expression for acceleration:
$a = v(-m) = (-mx + v_0)(-m)$
$a = m^2x - mv_0$
This is the equation of a straight line with a positive slope $(m^2)$ and a negative intercept $(-mv_0)$ on the acceleration axis.
Comparing this with the given options,graph $(A)$ represents a straight line with a negative intercept and a positive slope.
Therefore,option $(A)$ is correct.

(D) The standard equation of the trajectory of a projectile is given by $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing this with the given equation $y = x - \frac{2x^2}{5}$,we get:
$\tan \theta = 1 \implies \theta = 45^{\circ}$.
Also,$\frac{g}{2u^2 \cos^2 \theta} = \frac{2}{5}$.
Substituting $g = 10 \ m/s^2$ and $\cos^2 45^{\circ} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$:
$\frac{10}{2u^2 (1/2)} = \frac{2}{5}$.
$\frac{10}{u^2} = \frac{2}{5}$.
$2u^2 = 50 \implies u^2 = 25$.
$u = 5 \ m/s$.

(B) Given frequency $f = 0.5 \,Hz$.
Angular frequency $\omega = 2 \pi f = 2 \pi (0.5) = \pi \,rad/s$.
For the block to remain in contact with the piston, the downward acceleration of the piston at the extreme position must not exceed the acceleration due to gravity $g$.
The condition for the block to just lose contact is $a_{max} = g$.
Since $a_{max} = \omega^2 A$, we have $\omega^2 A = g$.
Substituting the values, $\pi^2 A = 10$.
Taking $\pi^2 \approx 10$, we get $10 A = 10$.
Therefore, $A = 1 \,m$.

(A) The angular momentum $L$ of a particle about a point $O$ is given by the cross product $L = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
The magnitude of angular momentum is $L = mvr \sin(\theta)$,where $r$ is the position vector of the particle from the origin $O$,$v$ is the velocity,and $\theta$ is the angle between $\vec{r}$ and $\vec{v}$.
This can also be written as $L = mv d$,where $d = r \sin(\theta)$ is the perpendicular distance from the origin $O$ to the line of motion of the particle.
Since the particle is moving along a straight line,the perpendicular distance $d$ from the origin $O$ to the line of motion remains constant throughout the motion.
Since the mass $m$ and the speed $v$ are also constant (uniform motion),the product $mvd$ remains constant.
Therefore,the angular momentum of the particle about $O$ remains constant during the motion from $A$ to $B$.

(A) The work done by a system in a cyclic process is equal to the area enclosed by the $P-V$ diagram.
For a clockwise cycle,the work done is positive,and for a counter-clockwise cycle,the work done is negative.
In the given diagram,the cycle $A \rightarrow B \rightarrow C \rightarrow A$ is counter-clockwise.
Area of the triangle $ABC = \frac{1}{2} \times \text{base} \times \text{height}$.
Base $= (V_C - V_B) = (10 - 5) \ m^3 = 5 \ m^3$.
Height $= (P_A - P_B) = (400 - 100) \ N/m^2 = 300 \ N/m^2$.
Area $= \frac{1}{2} \times 5 \times 300 = 750 \ J$.
Since the cycle is counter-clockwise,the work done by the system is $W = -750 \ J$.

(B) According to the principle of homogeneity of dimensions,only physical quantities with the same dimensions can be added or subtracted.
$(A)$ $PQ/R$: Multiplication and division of physical quantities with different dimensions are allowed.
$(B)$ $(P-Q)/R$: Since $P$ and $Q$ have different dimensions,the subtraction $(P-Q)$ is physically meaningless.
$(C)$ $(PR-Q^2)/R$: Here,$PR$ and $Q^2$ must have the same dimensions for the subtraction to be valid. While $P, Q, R$ have different dimensions,specific combinations might be invalid,but $(P-Q)$ is fundamentally invalid regardless of the denominator.
$(D)$ $PQ-R$: Since $PQ$ and $R$ have different dimensions,the subtraction $(PQ-R)$ is physically meaningless.
However,in the context of standard multiple-choice questions,$(P-Q)$ is the most direct violation of the principle of homogeneity.

(B) The given equation of the stationary wave is $ y = 2 \sin \left( \frac{\pi x}{15} \right) \cos (48 \pi t) $.
Comparing this with the standard equation $ y = A \sin(kx) \cos(\omega t) $, we get the wave number $ k = \frac{\pi}{15} $.
We know that $ k = \frac{2 \pi}{\lambda} $, so $ \frac{2 \pi}{\lambda} = \frac{\pi}{15} $.
Solving for wavelength $ \lambda $, we get $ \lambda = 30 $ units.
The distance between a node and its consecutive antinode is given by $ d = \frac{\lambda}{4} $.
Substituting the value of $ \lambda $, we get $ d = \frac{30}{4} = 7.5 $ units.

(C) The force experienced by a charge $q$ in a uniform electric field $E$ is given by $F = qE$.
Since the particle starts from rest and moves a distance $y$ in the direction of the force,the work done by the electric field on the particle is $W = F \times y = (qE) \times y = qEy$.
According to the work-energy theorem,the work done by the electric field is equal to the change in kinetic energy of the particle.
Since the initial kinetic energy is $0$,the final kinetic energy attained by the particle is $K = qEy$.

(D) The impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L$.
Thus,$Z = \sqrt{R^2 + \omega^2 L^2}$.
The $RMS$ current $I$ in the circuit is $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + \omega^2 L^2}}$.
The power dissipated in the circuit is given by $P = I^2 R$.
Substituting the value of $I$,we get $P = \left( \frac{V}{\sqrt{R^2 + \omega^2 L^2}} \right)^2 R = \frac{V^2 R}{R^2 + \omega^2 L^2}$.

(D) The instantaneous current is given by $ I = I_0 \sin(\omega t) $.
At the $ rms $ value, $ I = I_{rms} = \frac{I_0}{\sqrt{2}} $.
So, $ \frac{I_0}{\sqrt{2}} = I_0 \sin(\omega t_1) \Rightarrow \sin(\omega t_1) = \frac{1}{\sqrt{2}} $.
This gives $ \omega t_1 = \frac{\pi}{4} $.
Since $ \omega = \frac{2\pi}{T} $, we have $ \frac{2\pi}{T} t_1 = \frac{\pi}{4} \Rightarrow t_1 = \frac{T}{8} $.
The peak value occurs at $ t_2 = \frac{T}{4} $.
The time taken to reach the peak value from the $ rms $ value is $ \Delta t = t_2 - t_1 = \frac{T}{4} - \frac{T}{8} = \frac{T}{8} $.
Given frequency $ f = 50 \,Hz $, the time period is $ T = \frac{1}{f} = \frac{1}{50} \,s = 0.02 \,s $.
Therefore, $ \Delta t = \frac{0.02}{8} = 0.0025 \,s = 2.5 \times 10^{-3} \,s $.
Thus, option $ D $ is correct.

(B) The impact parameter $b$ is defined as the perpendicular distance of the initial velocity vector of the $\alpha$-particle from the center of the nucleus.
For a head-on collision,the $\alpha$-particle travels directly towards the center of the nucleus and retraces its path after the collision.
In this case,the perpendicular distance between the path of the $\alpha$-particle and the center of the nucleus is zero.
Therefore,the impact parameter for a head-on collision is $0$.

(C) According to Bohr's model,the velocity of an electron in the $n^{th}$ orbit is $v \propto 1/n$.
The radius of the $n^{th}$ orbit is $r \propto n^2$.
The frequency of revolution $f$ is given by $f = v / (2 \pi r)$.
Substituting the proportionalities: $f \propto (1/n) / n^2 = 1/n^3$.
Therefore,the frequency of revolution is proportional to $1/n^3$.

(B) The energy of the $ n $-th orbit in a hydrogen atom is given by $ E_n = -\frac{13.6}{n^2} \text{ eV} $.
For the ground state $( n_1 = 1 )$,$ E_1 = -13.6 \text{ eV} $.
When the atom absorbs $ 10.2 \text{ eV} $,the new energy is $ E_2 = -13.6 + 10.2 = -3.4 \text{ eV} $.
Since $ E_2 = -\frac{13.6}{n_2^2} = -3.4 \text{ eV} $,we find $ n_2^2 = 4 $,so $ n_2 = 2 $.
The orbital angular momentum is given by $ L = \frac{nh}{2\pi} $.
The increase in angular momentum is $ \Delta L = L_2 - L_1 = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{(n_2 - n_1)h}{2\pi} $.
Substituting $ n_2 = 2 $,$ n_1 = 1 $,and $ h = 6.626 \times 10^{-34} \text{ Js} $:
$ \Delta L = \frac{(2 - 1) \times 6.626 \times 10^{-34}}{2 \times 3.14159} \approx 1.054 \times 10^{-34} \text{ Js} $.
Thus,the increase is $ 1.05 \times 10^{-34} \text{ Js} $.

(A) When two capacitors are connected in parallel,they reach a common potential $V_c$.
The total charge $Q_{total} = Q + 0 = Q$.
The total capacitance $C_{total} = C + 2C = 3C$.
The common potential is given by $V_c = \frac{Q_{total}}{C_{total}} = \frac{Q}{3C}$.
The final charge on the first capacitor is $Q_1 = C \cdot V_c = C \cdot \frac{Q}{3C} = \frac{Q}{3}$.
The final charge on the second capacitor is $Q_2 = 2C \cdot V_c = 2C \cdot \frac{Q}{3C} = \frac{2Q}{3}$.
Thus,the final charges are $\frac{Q}{3}$ and $\frac{2Q}{3}$.

(D) From the circuit diagram,we can identify the arrangement of capacitors:
$1$. The two $ 100 pF $ capacitors are connected in series. Their equivalent capacitance $ C_1 $ is given by: $\frac{1}{C_1} = \frac{1}{100} + \frac{1}{100} = \frac{2}{100} \implies C_1 = 50 pF$.
$2$. This $ C_1 = 50 pF $ is in parallel with the top $ 50 pF $ capacitor. Let this parallel combination be $ C_2 $. $ C_2 = 50 pF + 50 pF = 100 pF$.
$3$. Finally,this $ C_2 = 100 pF $ is in series with the bottom $ 50 pF $ capacitor connected to terminal $ B $. The total equivalent capacitance $ C_{AB} $ is: $\frac{1}{C_{AB}} = \frac{1}{100} + \frac{1}{50} = \frac{1+2}{100} = \frac{3}{100}$.
$4$. Therefore,$ C_{AB} = \frac{100}{3} pF $.

(B) The wavelength $\lambda$ of the electromagnetic wave is given by the formula $\lambda = \frac{c}{f}$,where $c$ is the speed of light $(3 \times 10^8 \text{ m/s})$ and $f$ is the frequency $(5 \times 10^6 \text{ Hz})$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{5 \times 10^6} = 0.6 \times 10^2 = 60 \text{ m}$.
For an antenna to work efficiently,its length should be at least $\frac{\lambda}{4}$.
Therefore,the required size of the antenna is $\frac{60}{4} = 15 \text{ m}$.

(D) The reflection of radio waves by the ionosphere occurs because the refractive index of the ionosphere decreases with altitude. As radio waves travel into the ionosphere,they undergo continuous refraction until the angle of incidence exceeds the critical angle,leading to total internal reflection. This process is analogous to the total internal reflection of light that occurs in the atmosphere during the formation of a mirage. Therefore,the correct option is $D$.

(D) To find the current through the $2 \Omega$ resistor, we first simplify the circuit.
$1$. The rightmost part of the circuit consists of two $1 \Omega$ resistors in series, which are in parallel with another $1 \Omega$ resistor. The equivalent resistance of this part is $R_1 = \frac{(1+1) \times 1}{(1+1) + 1} = \frac{2}{3} \Omega$.
$2$. This $R_1$ is in series with the $1 \Omega$ resistor, giving $R_2 = 1 + \frac{2}{3} = \frac{5}{3} \Omega$.
$3$. This $R_2$ is in parallel with the $3 \Omega$ resistor. The equivalent resistance is $R_3 = \frac{3 \times (5/3)}{3 + (5/3)} = \frac{5}{14/3} = \frac{15}{14} \Omega$.
$4$. Finally, this $R_3$ is in series with the $1 \Omega$ and $2 \Omega$ resistors. The total equivalent resistance is $R_{eq} = 1 + 2 + \frac{15}{14} = 3 + \frac{15}{14} = \frac{42+15}{14} = \frac{57}{14} \Omega$.
$5$. The total current from the battery is $I = \frac{V}{R_{eq}} = \frac{1.2}{57/14} = \frac{1.2 \times 14}{57} \approx 0.2947 \text{ A}$.
$6$. Since the $2 \Omega$ resistor is in series with the battery, the total current flows through it. Rounding to the nearest option, the current is approximately $0.3 \text{ A}$.

(D) The terminal potential difference $V$ of a cell is given by the equation: $V = E - Ir$,where $E$ is the $EMF$ and $r$ is the internal resistance.
This equation is in the form of a straight line $y = mx + c$,where $V$ is on the y-axis and $I$ is on the x-axis.
$1$. When the current $I = 0$,the terminal potential difference $V = E$. From the graph,at $I = 0$,$V = 3 \text{ V}$. Therefore,$E = 3 \text{ V}$.
$2$. When the terminal potential difference $V = 0$,the current is $I = 6 \text{ A}$. Substituting these values into the equation: $0 = E - Ir \implies 0 = 3 - (6)r$.
$3$. Solving for $r$: $6r = 3 \implies r = \frac{3}{6} = 0.5 \ \Omega$.
Thus,the $EMF$ is $3 \text{ V}$ and the internal resistance is $0.5 \ \Omega$.

(B) The current $I$ in a conductor is given by the formula $I = neAv_d$, where $n$ is the electron number density, $e$ is the charge of an electron, $A$ is the cross-sectional area, and $v_d$ is the drift velocity.
Even though the drift velocity $v_d$ is very small (typically $10^{-4} \, m/s$) and the charge $e$ is small $(1.6 \times 10^{-19} \, C)$, the number density $n$ of free electrons in a conductor is extremely large, typically of the order of $10^{28} \, m^{-3}$.
Therefore, the product $neAv_d$ results in an appreciably large current.

(D) The current sensitivity $(I_s)$ of a galvanometer is given by $I_s = \frac{\theta}{I} = \frac{NAB}{k}$,where $N$ is the number of turns,$A$ is the area,$B$ is the magnetic field,and $k$ is the restoring torque constant per unit twist.
If $N$ is tripled $(N' = 3N)$,then $I_s' = \frac{(3N)AB}{k} = 3 I_s$. Thus,current sensitivity increases $3$ times.
The voltage sensitivity $(V_s)$ is given by $V_s = \frac{\theta}{V} = \frac{\theta}{IR} = \frac{I_s}{R}$.
Since the resistance $R$ is proportional to the length of the wire,and the length of the wire is proportional to the number of turns $(R' = 3R)$,the new voltage sensitivity is $V_s' = \frac{3I_s}{3R} = \frac{I_s}{R} = V_s$.
Therefore,voltage sensitivity remains constant and current sensitivity increases $3$ times.

(C) In the given $ LCR $ series circuit, the voltage across the inductor is $ V_L = 50 \, V $ and the voltage across the capacitor is $ V_C = 50 \, V $.
Since $ V_L = V_C $, the circuit is in resonance.
In a resonant $ LCR $ circuit, the net reactance is zero $( X_L - X_C = 0 )$, meaning the circuit behaves as a purely resistive circuit.
Therefore, the entire source voltage drops across the resistor $ R $.
Thus, the voltmeter reading is $ V_R = V_{source} = 220 \, V $.
The current in the circuit is given by $ I = \frac{V_R}{R} = \frac{220 \, V}{100 \, \Omega} = 2.2 \, A $.
Hence, the ammeter reading is $ 2.2 \, A $ and the voltmeter reading is $ 220 \, V $.

(D) The internal resistance $r$ of a cell is given by the formula: $r = R \left( \frac{l_1 - l_2}{l_2} \right)$.
Here,$l_1 = 240 \ cm$ is the balancing length when the cell is in an open circuit.
$l_2 = 120 \ cm$ is the balancing length when the cell is shunted with an external resistance $R = 2 \ \Omega$.
Substituting the values into the formula:
$r = 2 \left( \frac{240 - 120}{120} \right) \ \Omega$.
$r = 2 \left( \frac{120}{120} \right) \ \Omega$.
$r = 2 \times 1 \ \Omega = 2 \ \Omega$.
Therefore,the internal resistance of the cell is $2 \ \Omega$.

(C) Kirchhoff's junction rule,also known as Kirchhoff's Current Law $(KCL)$,states that the algebraic sum of currents meeting at any junction in a circuit is zero.
This implies that the total charge entering a junction per unit time must equal the total charge leaving the junction per unit time.
Since charge is neither created nor destroyed at the junction,this rule is a direct consequence of the law of conservation of charges.
Therefore,the correct option is $C$.

(C) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Since $A = \frac{V}{l} = \frac{m}{l \cdot d}$ (where $m$ is mass and $d$ is density),we can write:
$R = \rho \frac{l^2 \cdot d}{m}$
Since $\rho$ and $d$ are constant for copper wires,$R \propto \frac{l^2}{m}$.
Given ratios: $m_1:m_2:m_3 = 1:3:5$ and $l_1:l_2:l_3 = 5:3:1$.
Calculating the ratio of resistances:
$R_1:R_2:R_3 = \frac{l_1^2}{m_1} : \frac{l_2^2}{m_2} : \frac{l_3^2}{m_3}$
$R_1:R_2:R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5}$
$R_1:R_2:R_3 = 25 : 3 : \frac{1}{5}$
Multiplying by $5$ to clear the fraction:
$R_1:R_2:R_3 = 125 : 15 : 1$.

(D) The photoelectric current is directly proportional to the intensity of the incident light,provided the incident frequency is greater than the threshold frequency.
Since the saturation current depends only on the number of photoelectrons emitted per second,which is directly proportional to the intensity of the incident light,doubling the intensity will double the saturation current.
The frequency of the incident light affects the kinetic energy of the emitted photoelectrons,not the number of photoelectrons (saturation current),as long as it remains above the threshold frequency.
Therefore,when the intensity is doubled,the photoelectric saturation current doubles.

(D) The motional electromotive force (emf) induced in a conductor moving in a magnetic field is given by $e = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$.
For a conductor moving with velocity $v$ perpendicular to a uniform magnetic field $B$,the induced emf is $e = B v L_{eff}$,where $L_{eff}$ is the effective length of the conductor perpendicular to the velocity vector.
The effective length $L_{eff}$ is the straight-line distance between the two ends of the wire that are in contact with the rails.
When the straight wire $AB$ is replaced by a semicircular wire with the same endpoints $A$ and $B$,the effective length $L_{eff}$ (the distance between $A$ and $B$) remains the same.
Since the induced emf $e = B v L_{eff}$ depends only on the distance between the contact points on the rails,the emf remains unchanged.
Assuming the resistance $R$ of the circuit remains constant,the induced current $i = e/R$ will also remain the same.

(C) The electric field is given by $\vec{E} = E_{0} \sin(kx - \omega t) \hat{j}$.
The wave travels in the $+x$-direction,so the direction of propagation is $\hat{i}$.
The relation between the magnitudes of the electric and magnetic fields is $E_{0} = C B_{0}$,which implies $B_{0} = \frac{E_{0}}{C}$.
The direction of the magnetic field $\vec{B}$ is given by the direction of $\vec{c} \times \vec{E}$,where $\vec{c}$ is the direction of wave propagation.
Here,$\hat{i} \times \hat{j} = \hat{k}$.
Therefore,the magnetic field vector is $\vec{B}_{z} = \frac{E_{0}}{C} \sin(kx - \omega t) \hat{k}$.

(A) Let the total charge be $2Q$. The two parts are $q_{1}$ and $q_{2}$,such that $q_{1} + q_{2} = 2Q$. Thus,$q_{2} = 2Q - q_{1}$.
According to Coulomb's law,the force of interaction $F$ between the charges separated by a distance $r$ is given by $F = k \frac{q_{1}q_{2}}{r^2}$.
Substituting $q_{2}$,we get $F = \frac{k}{r^2} (q_{1})(2Q - q_{1}) = \frac{k}{r^2} (2Qq_{1} - q_{1}^2)$.
For the force $F$ to be maximum,the derivative of $F$ with respect to $q_{1}$ must be zero: $\frac{dF}{dq_{1}} = 0$.
$\frac{d}{dq_{1}} [\frac{k}{r^2} (2Qq_{1} - q_{1}^2)] = \frac{k}{r^2} (2Q - 2q_{1}) = 0$.
This implies $2Q - 2q_{1} = 0$,so $q_{1} = Q$.
Therefore,the ratio $\frac{Q}{q_{1}} = \frac{Q}{Q} = 1$.

(B) In a non-uniform electric field,the electric field intensity varies at different points in space.
Since the two charges of the dipole ($+q$ and $-q$) are separated by a small distance,they experience different electric field strengths at their respective positions.
Because the forces $(F = qE)$ acting on the two charges are unequal in magnitude or direction,the net force on the dipole is non-zero.
Additionally,since the forces act at different points,they create a net torque about the center of the dipole.
Therefore,an electric dipole placed in a non-uniform electric field generally experiences both a force and a torque.

(B) The electric field $E$ between two parallel plates is given by the formula $E = \frac{|\Delta V|}{d}$,where $\Delta V$ is the potential difference and $d$ is the distance between the plates.
Given:
Potential difference $\Delta V = V_2 - V_1 = 30 \,V - (-10 \,V) = 40 \,V$.
Distance $d = 2 \,cm = 2 \times 10^{-2} \,m$.
Substituting the values:
$E = \frac{40 \,V}{2 \times 10^{-2} \,m} = 20 \times 10^2 \,V/m = 2000 \,V/m$.
Therefore,the correct option is $B$.

(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$. The magnitude of the electric field is $|E| = |\frac{dV}{dx}|$,which represents the slope of the $V-x$ graph.
In region $A$ ($x=0$ to $1$ m),$V$ is constant ($2$ $V$),so the slope $\frac{dV}{dx} = 0$. Thus,$E_{A} = 0$.
In region $B$ ($x=1$ to $2$ m),the slope is $\frac{4-2}{2-1} = 2$ $V$/m. Thus,$|E_{B}| = 2$ $V$/m.
In region $C$ ($x=2$ to $4$ m),$V$ is constant ($4$ $V$),so the slope $\frac{dV}{dx} = 0$. Thus,$E_{C} = 0$.
In region $D$ ($x=4$ to $5$ m),the slope is $\frac{0-4}{5-4} = -4$ $V$/m. Thus,$|E_{D}| = |-4| = 4$ $V$/m.
Comparing the magnitudes: $E_{A} = E_{C} = 0$ and $E_{B} = 2$ $V$/m,$E_{D} = 4$ $V$/m. Therefore,$E_{A} = E_{C}$ and $E_{B} < E_{D}$.

(A) The electrical potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}$.
Case $1$: If the charges are like (both positive or both negative),then $q_1 q_2 > 0$. As the distance $r$ increases,the value of $U$ decreases.
Case $2$: If the charges are unlike (one positive and one negative),then $q_1 q_2 < 0$. As the distance $r$ increases,the magnitude $|U|$ decreases,but since $U$ is negative,the value of $U$ increases (becomes less negative,moving closer to zero).
Therefore,the potential energy may increase or decrease depending on the nature of the charges.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $v$ is its speed.
When an electron moves in a uniform magnetic field $\vec{B}$ with a velocity $\vec{V}$ perpendicular to the field,it experiences a magnetic Lorentz force $\vec{F} = q(\vec{V} \times \vec{B})$.
This force acts as a centripetal force,causing the electron to move in a circular path.
Since the magnetic force is always perpendicular to the velocity,it does no work on the electron $(W = \vec{F} \cdot d\vec{s} = 0)$.
According to the work-energy theorem,the kinetic energy and consequently the speed $v$ of the electron remain constant throughout its motion.
Since $h$,$m$,and $v$ are all constant,the de Broglie wavelength $\lambda$ remains constant.

(B) The magnetic field $B$ inside a toroid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.
Given: $n = 500 \text{ m}^{-1}$, $I = 2 \text{ A}$, $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
The magnetic energy density $u_B$ is given by $u_B = \frac{B^2}{2\mu_0}$.
Substituting $B = \mu_0 n I$ into the formula, we get $u_B = \frac{(\mu_0 n I)^2}{2\mu_0} = \frac{\mu_0 n^2 I^2}{2}$.
Calculating the value: $u_B = \frac{(4\pi \times 10^{-7}) \times (500)^2 \times (2)^2}{2}$.
$u_B = \frac{(4\pi \times 10^{-7}) \times 250000 \times 4}{2} = 2\pi \times 10^{-7} \times 10^6 = 2\pi \times 10^{-1} = 0.2 \times 3.14 = 0.628 \text{ J/m}^3$.

(D) The magnetic field at the centre of a circular arc of radius $R$ subtending an angle $\theta$ (in radians) at the centre is given by $B = \frac{\mu_0 I \theta}{4 \pi R}$.
In the given figure,the angle subtended by the arc at the centre is $\theta = 360^{\circ} - 60^{\circ} = 300^{\circ}$.
Converting the angle to radians: $\theta = 300^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{5\pi}{3} \text{ radians}$.
Substituting this into the formula:
$B = \frac{\mu_0 I}{4 \pi R} \times \left( \frac{5\pi}{3} \right)$
$B = \frac{5 \mu_0 I}{12 R}$

(C) The potential energy $U$ of a magnetic dipole in a magnetic field is given by $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$,where $\theta$ is the angle between the magnetic moment vector $\vec{M}$ and the magnetic field vector $\vec{B}$.
When the loop is rotated about an axis perpendicular to its plane,the direction of the magnetic moment vector $\vec{M}$ (which is always perpendicular to the plane of the loop) remains unchanged relative to the magnetic field $\vec{B}$.
Since the angle $\theta$ between $\vec{M}$ and $\vec{B}$ does not change,the potential energy of the loop remains constant.
Therefore,the work done $W = \Delta U = 0$.

(B) The time period of oscillation for a magnetic needle in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.
Given: $T = 2 \text{ s}$,$I = 8 \times 10^{-6} \text{ kg m}^2$,$M = 5 \times 10^{-2} \text{ A m}^2$.
Substituting the values into the formula: $2 = 2\pi \sqrt{\frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B}}$.
Dividing by $2$: $1 = \pi \sqrt{\frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B}}$.
Squaring both sides: $1 = \pi^2 \left( \frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B} \right)$.
Rearranging for $B$: $B = \frac{\pi^2 \times 8 \times 10^{-6}}{5 \times 10^{-2}}$.
Using $\pi^2 \approx 9.86$ (or approximately $10$): $B = \frac{9.86 \times 8 \times 10^{-6}}{5 \times 10^{-2}} \approx 1.577 \times 10^{-4} \text{ T}$.
Rounding to the nearest option,$B \approx 1.6 \times 10^{-4} \text{ T}$.

(B) In a cyclotron,the charged particle is accelerated by an electric field in the gap between the two dees.
Within the dees,the particle moves in a circular path due to the perpendicular magnetic field,where its speed remains constant.
Since the particle is constantly changing its direction of motion while moving in a circle,it experiences centripetal acceleration at all times.
Therefore,the particle undergoes acceleration throughout its motion.

(D) permanent magnet is made of ferromagnetic material. In a permanent magnet,the magnetic domains are aligned in a specific direction due to the manufacturing process (like cooling in an external magnetic field). Therefore,at room temperature,the domains are all perfectly aligned to produce a net magnetic moment.

(A) The coercivity $H$ represents the magnetic field intensity required to demagnetize a ferromagnetic material.
For a solenoid,the magnetic field intensity $H$ is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current.
Given:
Coercivity $H = 3 \times 10^{3} \text{ A m}^{-1}$
Number of turns per unit length $n = 1000 \text{ turns m}^{-1} = 10^{3} \text{ m}^{-1}$
Using the formula $I = \frac{H}{n}$:
$I = \frac{3 \times 10^{3}}{10^{3}} = 3 \text{ A}$
Therefore,the minimum current required is $3 \text{ A}$.

(D) The nuclear force is a short-range force that acts effectively only within a range of approximately $1 \ fm$ to $3 \ fm$ $(1 \ fm = 10^{-15} \ m)$.
Given the separation between the two protons is $10 \ nm = 10 \times 10^{-9} \ m = 10^{-8} \ m$.
Since $10^{-8} \ m$ is much larger than the range of the nuclear force $(10^{-15} \ m)$,the nuclear force $F_n$ is negligible at this distance.
However,the electromagnetic force $F_e$ (Coulomb force) follows an inverse-square law and acts over long distances.
Therefore,at a separation of $10 \ nm$,the electromagnetic force is significantly greater than the nuclear force,i.e.,$F_e \gg F_n$.

(B) The activity of a radioactive sample is given by $R = |dN/dt|$.
From the graph,the decay curve for $A$ falls most rapidly,which means it has the highest decay constant $\lambda$.
The decay constant $\lambda$ and mean life $\tau$ are related by $\tau = 1/\lambda$.
Therefore,a higher decay constant $\lambda$ corresponds to a shorter mean life $\tau$.
Since curve $A$ has the steepest slope (highest activity),it has the largest $\lambda$ and thus the shortest mean life.
Hence,the correct option is $B$.

(D) Let the number of $ \alpha $-particles emitted be $ x $ and the number of $ \beta $-particles emitted be $ y $.
The decay reaction is: $ { }_{90} Th^{232} \rightarrow { }_{82} Pb^{208} + x({ }_{2} He^{4}) + y({ }_{-1} e^{0}) $.
Equating the mass numbers: $ 232 = 208 + 4x \implies 4x = 24 \implies x = 6 $.
Equating the atomic numbers: $ 90 = 82 + 2x - y $.
Substituting $ x = 6 $: $ 90 = 82 + 2(6) - y \implies 90 = 82 + 12 - y \implies 90 = 94 - y \implies y = 4 $.
Thus,$ 6 $ $ \alpha $-particles and $ 4 $ $ \beta $-particles are emitted.

(C) Given: Focal length $f = -25 \ cm$. Initial object distance $u_i = -50 \ cm$. Speed of object $v_o = 1 \ ms^{-1} = 100 \ cm s^{-1}$.
At $t=0$,$u_i = -50 \ cm$. Using mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v_i} + \frac{1}{-50} = \frac{1}{-25} \implies \frac{1}{v_i} = -\frac{1}{25} + \frac{1}{50} = -\frac{1}{50} \implies v_i = -50 \ cm$.
At $t=0.25 \ s$,the distance moved by the object is $d = v_o \times t = 100 \ cm s^{-1} \times 0.25 \ s = 25 \ cm$.
New object distance $u_f = -50 \ cm + 25 \ cm = -25 \ cm$.
Since the object is at the focus,the image is formed at infinity,i.e.,$v_f = \infty$.
The average velocity of the image is $\langle v \rangle = \frac{v_f - v_i}{\Delta t} = \frac{\infty - (-50)}{0.25} = \infty$.

(A) For a prism,the deviation $D$ is given by $D = (i_1 + i_2) - A$,where $i_1$ and $i_2$ are the angles of incidence and emergence,and $A$ is the prism angle.
Given $D = 44^{\circ}$ for $i_1 = 42^{\circ}$ and $i_2 = 62^{\circ}$ (or vice versa),we have:
$44^{\circ} = (42^{\circ} + 62^{\circ}) - A$
$44^{\circ} = 104^{\circ} - A$
$A = 104^{\circ} - 44^{\circ} = 60^{\circ}$.
At minimum deviation $(D_m = 38^{\circ})$,the angle of incidence $i$ is given by $i = \frac{A + D_m}{2}$.
$i = \frac{60^{\circ} + 38^{\circ}}{2} = \frac{98^{\circ}}{2} = 49^{\circ}$.

(A) The frequency response curve of a transistor amplifier shows that the voltage gain is not uniform across all frequencies.
At low frequencies,the coupling and bypass capacitors have high reactance,which reduces the gain.
At high frequencies,the internal junction capacitances of the transistor and stray capacitances become significant,providing a low-impedance path that shunts the signal,thereby reducing the gain.
In the mid-frequency range,these effects are negligible,and the amplifier provides a stable and maximum voltage gain.
Therefore,the voltage gain is low at both low and high frequencies and remains constant in the mid-frequency range.

(A) The circuit consists of two cross-coupled $NOR$ gates forming an $SR$ latch.
For a $NOR$ gate,the output $Y = \overline{A+B}$.
Let the top gate be Gate $1$ and the bottom gate be Gate $2$.
The inputs to Gate $1$ are $1$ and $Q$. Thus,$P = \overline{1+Q} = 0$.
The inputs to Gate $2$ are $0$ and $P$. Thus,$Q = \overline{0+P}$.
Substituting $P=0$ into the equation for $Q$,we get $Q = \overline{0+0} = \overline{0} = 1$.
Therefore,$P=0$ and $Q=1$.

(C) The conductivity of a semiconductor is given by $\sigma = e(n_e \mu_e + n_h \mu_h)$.
As the temperature increases,the number density of charge carriers ($n_e$ and $n_h$) increases exponentially according to the relation $n = C T^{3/2} \exp(-E_g / 2kT)$.
Although the relaxation time $(\tau)$ decreases with an increase in temperature due to increased scattering,the exponential increase in the number density of charge carriers far outweighs the decrease in mobility caused by the reduction in relaxation time.
Therefore,the net effect is an increase in the conductivity of the semiconductor.

(A) According to Brewster's law,the refractive index $n$ of the medium is given by $n = \tan \theta_{p}$,where $\theta_{p}$ is the Brewster angle.
From Snell's law,the refractive index is $n = \frac{\sin i}{\sin r}$.
From the given graph,the slope is $\frac{\sin r}{\sin i} = \tan 30^{\circ}$.
Therefore,$\frac{\sin i}{\sin r} = \frac{1}{\tan 30^{\circ}} = \frac{1}{1/\sqrt{3}} = \sqrt{3}$.
Thus,$n = \sqrt{3}$.
Equating the two expressions for $n$,we get $\tan \theta_{p} = \sqrt{3}$.
Therefore,$\theta_{p} = \tan^{-1}(\sqrt{3}) = 60^{\circ}$.

(D) The intensity $ I $ at any point is given by $ I = I_{max} \cos^2 \left( \frac{\phi}{2} \right) $, where $ \phi $ is the phase difference.
Given that at path difference $ \Delta x = \lambda $, the intensity is $ K $. Since $ \Delta x = \lambda $ corresponds to a phase difference $ \phi = 2\pi $, we have $ K = I_{max} \cos^2 \left( \frac{2\pi}{2} \right) = I_{max} \cos^2(\pi) = I_{max} $.
Now, for a path difference $ \Delta x = \frac{\lambda}{3} $, the phase difference is $ \phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} $.
The intensity at this point is $ I = I_{max} \cos^2 \left( \frac{\phi}{2} \right) = K \cos^2 \left( \frac{2\pi/3}{2} \right) $.
$ I = K \cos^2 \left( \frac{\pi}{3} \right) = K \left( \frac{1}{2} \right)^2 = \frac{K}{4} $.

(A) The formula for the Doppler shift in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$, where $v$ is the velocity of the source, $c$ is the speed of light, $\Delta \lambda$ is the shift in wavelength, and $\lambda$ is the original wavelength.
Given: $\Delta \lambda = 0.1 \text{ Å}$, $\lambda = 6000 \text{ Å}$, and $c = 3 \times 10^8 \text{ m/s}$.
Rearranging the formula to solve for $v$: $v = \frac{\Delta \lambda}{\lambda} \times c$.
Substituting the values: $v = \frac{0.1}{6000} \times 3 \times 10^8 \text{ m/s}$.
$v = \frac{1}{60000} \times 3 \times 10^8 \text{ m/s} = \frac{3 \times 10^8}{6 \times 10^4} \text{ m/s} = 0.5 \times 10^4 \text{ m/s} = 5000 \text{ m/s}$.
Converting to $\text{km/s}$: $v = 5 \text{ km/s}$.