KCET 2014 Physics Question Paper with Answer and Solution

(B) The maximum speed $v_{max}$ to avoid slipping on a level circular path is given by the condition $v_{max} = \sqrt{\mu_s rg}$.
Squaring both sides,we get $v_{max}^2 = \mu_s rg$.
Given values are $\mu_s = 0.1$,$r = 5 \text{ m}$,and $g = 10 \text{ ms}^{-2}$.
Substituting these values: $v_{max}^2 = 0.1 \times 5 \times 10 = 5 \text{ m}^2\text{s}^{-2}$.
For the vehicle to not slip,the condition is $v^2 \leq v_{max}^2$,which means $v^2 \leq 5 \text{ m}^2\text{s}^{-2}$.
If the square of the actual speed $v^2$ exceeds this value,the vehicle will slip.
Therefore,the person will slip if $v^2 > 5 \text{ m}^2\text{s}^{-2}$.

(C) Let the total time be $t$ and acceleration be $a$. The initial velocity $u = 0$.
For the first half of the time,$t_{1} = t/2$:
$x_{1} = u t_{1} + \frac{1}{2} a t_{1}^{2} = 0 + \frac{1}{2} a (t/2)^{2} = \frac{1}{8} a t^{2}$.
For the total time $t$,the total distance $x_{total} = x_{1} + x_{2} = \frac{1}{2} a t^{2}$.
Substituting $x_{1} = \frac{1}{8} a t^{2}$ into the total distance equation:
$\frac{1}{8} a t^{2} + x_{2} = \frac{1}{2} a t^{2}$.
$x_{2} = \frac{1}{2} a t^{2} - \frac{1}{8} a t^{2} = \frac{4-1}{8} a t^{2} = \frac{3}{8} a t^{2}$.
Comparing $x_{1}$ and $x_{2}$:
$x_{2} = 3 \times (\frac{1}{8} a t^{2}) = 3 x_{1}$.

(C) Given the relation: $Q = \frac{x^3 y^2}{z}$.
Using the formula for propagation of errors,the relative error in $Q$ is given by: $\frac{\Delta Q}{Q} = 3 \frac{\Delta x}{x} + 2 \frac{\Delta y}{y} + \frac{\Delta z}{z}$.
Given percentage errors:
$\frac{\Delta x}{x} \times 100 = 1\%$
$\frac{\Delta y}{y} \times 100 = 2\%$
$\frac{\Delta z}{z} \times 100 = 4\%$
Substituting these values into the error equation:
$\frac{\Delta Q}{Q} \times 100 = 3(1\%) + 2(2\%) + 1(4\%)$
$= 3\% + 4\% + 4\% = 11\%$.
Therefore,the percentage error in $Q$ is $11\%$.

(C) Initial angular speed $\omega_{1} = 1800 \ rpm = \frac{1800 \times 2\pi}{60} \ rad \ s^{-1} = 60\pi \ rad \ s^{-1}$.
Final angular speed $\omega_{2} = 3000 \ rpm = \frac{3000 \times 2\pi}{60} \ rad \ s^{-1} = 100\pi \ rad \ s^{-1}$.
Time interval $t = 20 \ s$.
Angular acceleration $\alpha = \frac{\omega_{2} - \omega_{1}}{t}$.
$\alpha = \frac{100\pi - 60\pi}{20} = \frac{40\pi}{20} = 2\pi \ rad \ s^{-2}$.

(A) Given: mass $m = 10 \,kg$, spring constant $k = 1000 \,N/m$, and amplitude $A = 10 \,cm = 0.1 \,m$.
In simple harmonic motion, the restoring force is $F = -kx$.
The maximum force occurs at maximum displacement (amplitude), so $F_{max} = kA$.
Using Newton's second law, $F_{max} = m a_{max}$.
Therefore, $m a_{max} = kA$.
$a_{max} = \frac{kA}{m} = \frac{1000 \,N/m \times 0.1 \,m}{10 \,kg} = \frac{100}{10} = 10 \,m/s^2$.
The maximum acceleration of the block is $10 \,m/s^2$.

(D) vector quantity is defined as a physical quantity that possesses both magnitude and direction.
Conversely,a scalar quantity is defined as a physical quantity that possesses only magnitude and no direction.
$1$. Weight is the force of gravity acting on an object,which has both magnitude and direction (downward),making it a vector.
$2$. Nuclear spin is an intrinsic angular momentum of a nucleus,which is a vector quantity.
$3$. Momentum is defined as the product of mass and velocity $(p = mv)$. Since velocity is a vector,momentum is also a vector.
$4$. Potential energy is a scalar quantity because it represents the energy stored in an object due to its position or configuration,which does not have a specific direction.
Therefore,potential energy is not a vector quantity.

(A) Given: $\vec{F} = 5\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$.
Torque $\vec{\tau}$ is defined as the cross product of the position vector and the force vector: $\vec{\tau} = \vec{r} \times \vec{F}$.
Using the determinant method:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 5 & 2 & -5 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}((-2)(-5) - (1)(2)) - \hat{j}((1)(-5) - (1)(5)) + \hat{k}((1)(2) - (-2)(5))$
$\vec{\tau} = \hat{i}(10 - 2) - \hat{j}(-5 - 5) + \hat{k}(2 + 10)$
$\vec{\tau} = 8\hat{i} - (-10)\hat{j} + 12\hat{k}$
$\vec{\tau} = 8\hat{i} + 10\hat{j} + 12\hat{k}$.

(A) Given, speed of aeroplane $v = 720 \text{ km/h} = 720 \times \frac{5}{18} \text{ m/s} = 200 \text{ m/s}$.
Angle of banking $\theta = 45^{\circ}$.
Acceleration due to gravity $g = 10 \text{ m/s}^2$.
The formula for the radius of a horizontal loop is given by $\tan \theta = \frac{v^2}{rg}$.
Rearranging for $r$, we get $r = \frac{v^2}{g \tan \theta}$.
Substituting the values: $r = \frac{(200)^2}{10 \times \tan 45^{\circ}}$.
Since $\tan 45^{\circ} = 1$, we have $r = \frac{40000}{10 \times 1} = 4000 \text{ m}$.
Converting to kilometers, $r = 4 \text{ km}$.

(A) Anomalous expansion of water is a unique property where water contracts instead of expanding when heated between $ 0^{\circ} C $ and $ 4^{\circ} C $.
Since density is defined as mass per unit volume $( \rho = m/V )$, as the volume decreases while the mass remains constant, the density increases.
At $ 4^{\circ} C $, the volume of a given mass of water is minimum.
Therefore, the density of water is maximum at $ 4^{\circ} C $.

(A) Given: Moment of inertia $I = 3 \ kg \ m^{2}$,angular velocity $\omega = 3 \ rad \ s^{-1}$,and mass $m = 27 \ kg$.
The rotational kinetic energy of the body is given by $K_{rot} = \frac{1}{2} I \omega^{2}$.
The translational kinetic energy of the second body is given by $K_{trans} = \frac{1}{2} m v^{2}$.
According to the problem,$K_{rot} = K_{trans}$.
Therefore,$\frac{1}{2} I \omega^{2} = \frac{1}{2} m v^{2}$.
$I \omega^{2} = m v^{2}$.
$v^{2} = \frac{I \omega^{2}}{m}$.
$v = \omega \sqrt{\frac{I}{m}}$.
Substituting the values: $v = 3 \times \sqrt{\frac{3}{27}} = 3 \times \sqrt{\frac{1}{9}} = 3 \times \frac{1}{3} = 1 \ m \ s^{-1}$.

(D) Given,efficiency,$\eta = 70 \% = 0.7$.
Sink temperature,$T_2 = 27^{\circ}C = 273 + 27 = 300 \ K$.
The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.
Substituting the given values: $0.7 = 1 - \frac{300}{T_1}$.
Rearranging the terms: $\frac{300}{T_1} = 1 - 0.7 = 0.3$.
Solving for source temperature $T_1$: $T_1 = \frac{300}{0.3} = 1000 \ K$.
Converting the temperature back to Celsius: $T_1 = 1000 - 273 = 727^{\circ}C$.
Therefore,the source temperature required is $727^{\circ}C$.

$(A)$ The Reynolds number $(Re)$ is a dimensionless quantity that represents the ratio of inertial forces to viscous forces in a fluid flow.
It is used to predict the flow regime of a fluid.
For a flow in a pipe, if the Reynolds number is less than $2000$, the flow is considered to be streamline or laminar.
If the Reynolds number is between $2000$ and $4000$, the flow is in a transition state.
If the Reynolds number is greater than $4000$, the flow is turbulent.
Among the given options, the condition for streamline flow is satisfied by the range less than $1000$.

(B) Let the distance between $A$ and $B$ be $d$.
Time taken to travel from $A$ to $B$ is $t_{1} = \frac{d}{v_{1}} = \frac{d}{30}$.
Time taken to travel from $B$ to $A$ is $t_{2} = \frac{d}{v_{2}} = \frac{d}{20}$.
Average speed is defined as the total distance divided by the total time.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{d + d}{t_{1} + t_{2}} = \frac{2d}{\frac{d}{30} + \frac{d}{20}}$.
Average speed $= \frac{2d}{d(\frac{20 + 30}{600})} = \frac{2 \times 600}{50} = \frac{1200}{50} = 24 \ km/h$.

(B) According to the Doppler effect, the apparent frequency $f^{\prime}$ heard by an observer when the source is moving towards a stationary observer is given by the formula:
$f^{\prime} = \left( \frac{v}{v - v_{s}} \right) f$
Where:
$f = 340 \,Hz$ (frequency of the source)
$v = 340 \,ms^{-1}$ (speed of sound)
$v_{s} = 10 \,ms^{-1}$ (speed of the source/train)
Substituting the values into the formula:
$f^{\prime} = \left( \frac{340}{340 - 10} \right) \times 340$
$f^{\prime} = \left( \frac{340}{330} \right) \times 340$
$f^{\prime} = \frac{115600}{330} \approx 350.3 \,Hz$
Rounding to the nearest integer, the frequency heard is $350 \,Hz$.

(B) Given: Initial speed $u = 30 \,ms^{-1}$, angle of projection $\theta = 45^{\circ}$, and acceleration due to gravity $g = 10 \,ms^{-2}$.
The formula for the maximum height $H$ reached by a projectile is given by:
$H = \frac{u^2 \sin^2 \theta}{2g}$
Substituting the given values into the formula:
$H = \frac{(30)^2 \times (\sin 45^{\circ})^2}{2 \times 10}$
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, then $\sin^2 45^{\circ} = \frac{1}{2}$.
$H = \frac{900 \times \frac{1}{2}}{20} = \frac{450}{20} = 22.5 \,m$.
Therefore, the maximum height reached by the stone is $22.5 \,m$.

(D) Given: Length of pipe $L = 30 \text{ cm} = 0.3 \text{ m}$.
Frequency $f = 1.1 \text{ kHz} = 1100 \text{ Hz}$.
Speed of sound $v = 330 \text{ ms}^{-1}$.
For an open pipe,the frequency of the $n^{th}$ harmonic is given by $f_n = n \times \frac{v}{2L}$.
Substituting the values: $1100 = n \times \frac{330}{2 \times 0.3}$.
$1100 = n \times \frac{330}{0.6}$.
$1100 = n \times 550$.
$n = \frac{1100}{550} = 2$.
Therefore,the pipe resonates in the second harmonic mode.

(B) When a cycle tyre bursts suddenly,the process is adiabatic.
This is because the expansion of air occurs very rapidly,leaving no time for heat exchange between the system (the air inside the tyre) and the surroundings.
Since $dQ = 0$,the process satisfies the condition for an adiabatic process.

(C) The period of revolution $T$ of a satellite orbiting very close to the Earth's surface is given by the formula:
$T = 2 \pi \sqrt{\frac{R_E}{g}}$
Given values:
$R_E = 6400 \ km = 6.4 \times 10^6 \ m$
$g = 10 \ ms^{-2}$
$\pi = 3.14$
Substituting these values into the formula:
$T = 2 \times 3.14 \times \sqrt{\frac{6.4 \times 10^6}{10}}$
$T = 6.28 \times \sqrt{6.4 \times 10^5} = 6.28 \times \sqrt{64 \times 10^4}$
$T = 6.28 \times 800 = 5024 \ s$
To convert the time into minutes:
$T = \frac{5024}{60} \ min \approx 83.73 \ min$
Therefore, the period of revolution is $83.73$ minutes.

(A) geostationary satellite orbits around the Earth above the equator such that it remains stationary as seen from the Earth.
This means the satellite completes one revolution around the Earth in the same time it takes for the Earth to complete one rotation about its axis.
The rotational period of the Earth is $24 \text{ h}$.
Therefore,the time period of a geostationary satellite is $24 \text{ h}$.

(A) Given: Length of wire,$l = 1 \ m$; Mass of wire,$m = 10 \times 10^{-3} \ kg$; Tension,$T = 100 \ N$.
The speed of a transverse wave in a stretched string is given by the formula: $v = \sqrt{\frac{T}{\mu}}$,where $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{m}{l} = \frac{10 \times 10^{-3} \ kg}{1 \ m} = 10 \times 10^{-3} \ kg/m$.
Substituting the values into the formula:
$v = \sqrt{\frac{100}{10 \times 10^{-3}}} = \sqrt{\frac{100}{0.01}} = \sqrt{10000} = 100 \ ms^{-1}$.
Therefore,the speed of the transverse wave is $100 \ ms^{-1}$.

(B) The force between two protons is the same as the force between a proton and a neutron. This is a characteristic property of the strong nuclear force,which is charge-independent. The strong nuclear force acts between nucleons (protons and neutrons) and is responsible for holding the nucleus together. It is extremely strong at short distances and becomes negligible at larger distances.

(C) Given: Radius of the coil $r = 10 \text{ cm} = 0.1 \text{ m}$.
Number of turns $N = 100$.
Current $I = 1 \text{ A}$.
The magnetic moment $M$ of a current-carrying coil is given by the formula $M = N I A$,where $A$ is the area of the coil.
The area $A = \pi r^{2} = \pi \times (0.1 \text{ m})^{2} = 0.01 \pi \text{ m}^{2}$.
Substituting the values into the formula:
$M = 100 \times 1 \times (0.01 \times 3.142) \text{ A m}^{2}$.
$M = 100 \times 0.03142 \text{ A m}^{2} = 3.142 \text{ A m}^{2}$.
Thus,the magnetic moment of the coil is $3.142 \text{ A m}^{2}$.

(A) Given: Resistance of galvanometer $ G = 100 \Omega $; Range of ammeter $ I = 1 \text{ A} $; Full scale deflection current $ I_{g} = 5 \text{ mA} = 5 \times 10^{-3} \text{ A} $.
To convert a galvanometer into an ammeter,a shunt resistance $ S $ is connected in parallel with it.
The formula for shunt resistance is $ S = \frac{I_{g} G}{I - I_{g}} $.
Substituting the given values:
$ S = \frac{5 \times 10^{-3} \times 100}{1 - 5 \times 10^{-3}} $
$ S = \frac{0.5}{1 - 0.005} $
$ S = \frac{0.5}{0.995} \Omega $
$ S = \frac{500}{995} \Omega = \frac{100}{199} \Omega $.
Wait,calculating $ \frac{0.5}{0.995} $ gives $ \frac{500}{995} = \frac{100}{199} \approx 0.5025 \Omega $.
Checking the options provided,$ \frac{5}{9.95} = \frac{500}{995} = \frac{100}{199} $. Thus,option $ A $ is correct.

(A) The magnetic energy $E$ stored in a solenoid with inductance $L$ carrying a current $I$ is given by the formula:
$E = \frac{1}{2} L I^2$
Given:
Inductance $L = 2 \ H$
Current $I = 1 \ A$
Substituting the values into the formula:
$E = \frac{1}{2} \times 2 \times (1)^2$
$E = 1 \times 1 = 1 \ J$
Therefore,the magnetic energy stored in the solenoid is $1 \ J$.

(C) Magnetic susceptibility $\chi$ is a dimensionless quantity that indicates the degree of magnetization of a material in response to an applied magnetic field.
For diamagnetic materials, $\chi$ is small and negative.
For paramagnetic materials, $\chi$ is small and positive.
For ferromagnetic materials, $\chi$ is large and positive.
Given that the susceptibility is $ 400 $, which is a large positive value, the material belongs to the class of ferromagnetic materials.

(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.
At half-life $t = T$,the number of undecayed nuclei $N(T) = \frac{N_0}{2}$.
Substituting these values into the decay law:
$\frac{N_0}{2} = N_0 e^{-\lambda T}$
$\frac{1}{2} = e^{-\lambda T}$
Taking the natural logarithm on both sides:
$\ln(1/2) = -\lambda T$
$-\ln(2) = -\lambda T$
$\lambda T = \ln(2) = \log_{e} 2$.

(D) The resolving power of a telescope is given by the formula:
$RP = \frac{D}{1.22 \lambda}$
Where $D$ is the diameter of the objective and $\lambda$ is the wavelength of light.
Given:
$D = 200 \text{ cm} = 2 \text{ m}$
$\lambda = 5000 \text{ \AA} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}$
Substituting the values:
$RP = \frac{2}{1.22 \times 5 \times 10^{-7}}$
$RP = \frac{2}{6.1 \times 10^{-7}}$
$RP = \frac{2}{6.1} \times 10^{7}$
$RP \approx 0.32786 \times 10^{7}$
$RP \approx 3.28 \times 10^{6}$
Therefore, the resolving power of the telescope is $3.28 \times 10^{6}$.

(B) In a potentiometer experiment, the emf $E$ of a cell is directly proportional to its balancing length $L$, given by the relation $E \propto L$.
Given for the first cell: $E_{1} = 1.25 \,V$ and $L_{1} = 30 \,cm$.
For the second cell: $E_{2} = ?$ and $L_{2} = 40 \,cm$.
Using the ratio formula: $\frac{E_{1}}{E_{2}} = \frac{L_{1}}{L_{2}}$.
Substituting the values: $\frac{1.25}{E_{2}} = \frac{30}{40}$.
Simplifying the equation: $\frac{1.25}{E_{2}} = \frac{3}{4}$.
Solving for $E_{2}$: $E_{2} = 1.25 \times \frac{4}{3} = \frac{5}{3} \approx 1.666 \,V$.
Rounding to two decimal places, we get $E_{2} \simeq 1.67 \,V$.

(D) According to Malus's Law,the intensity of the emergent light is given by:
$I = I_{0} \cos^{2} \theta$
where $I$ is the intensity of the emergent polarized light,$I_{0}$ is the intensity of the incident polarized light,and $\theta$ is the angle between the pass axes of the two polarizers.
Given that $\theta = 60^{\circ}$,we substitute this value into the formula:
$I = I_{0} \cos^{2}(60^{\circ})$
Since $\cos(60^{\circ}) = 1/2$,we have:
$I = I_{0} \times (1/2)^{2}$
$I = I_{0} \times (1/4)$
$I = I_{0} / 4$
Thus,the intensity of the emergent polarized light from the second polarizer is $I_{0} / 4$.

(B) The quantization of charge is given by the formula $q = n e$,where $q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \ C)$.
To find the number of electrons $n$,we rearrange the formula: $n = \frac{q}{e}$.
Given the charge $q = 1 \ nC = 1 \times 10^{-9} \ C$.
Substituting the values: $n = \frac{1 \times 10^{-9} \ C}{1.6 \times 10^{-19} \ C}$.
$n = \frac{1}{1.6} \times 10^{10} = 0.625 \times 10^{10} = 6.25 \times 10^{9}$.
Therefore,the number of electrons present on the body is $6.25 \times 10^{9}$.

(D) Maximum power is dissipated in an $LCR$ circuit at the resonant frequency.
Given: Inductance $L = 5 \ mH = 5 \times 10^{-3} \ H$,Capacitance $C = 2 \ \mu F = 2 \times 10^{-6} \ F$.
The resonant frequency $f_R$ is given by the formula:
$f_R = \frac{1}{2 \pi \sqrt{LC}}$
Substituting the values:
$f_R = \frac{1}{2 \pi \sqrt{5 \times 10^{-3} \times 2 \times 10^{-6}}}$
$f_R = \frac{1}{2 \pi \sqrt{10 \times 10^{-9}}} = \frac{1}{2 \pi \sqrt{10^{-8}}}$
$f_R = \frac{1}{2 \pi \times 10^{-4}}$
$f_R = \frac{10^4}{2 \pi} = \frac{10 \times 10^3}{2 \pi} = \frac{5 \times 10^3}{\pi} \ Hz$.

(D) In Gauss's law,the Gaussian surface is a closed,imaginary surface in three-dimensional space through which the flux of a vector field is calculated.
By definition,any surface element $d\vec{S}$ on a Gaussian surface is treated as a vector quantity,where the magnitude is the area of the element and the direction is normal to the surface.
Since the flux $\Phi_E = \oint \vec{E} \cdot d\vec{S}$ involves the dot product of the electric field vector and the area vector,the surface element itself is fundamentally a vector.
Therefore,the nature of the surface element used in the calculation is vector.

(C) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by the formula:
$V = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r}$
Given values:
Charge $q = 3 \text{ nC} = 3 \times 10^{-9} \text{ C}$
Distance $r = 9 \text{ cm} = 9 \times 10^{-2} \text{ m}$
Coulomb's constant $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \text{ N m}^2/\text{C}^2$
Substituting these values into the formula:
$V = (9 \times 10^{9}) \times \frac{3 \times 10^{-9}}{9 \times 10^{-2}}$
$V = \frac{9 \times 3 \times 10^{0}}{9 \times 10^{-2}}$
$V = 3 \times 10^{2} \text{ V} = 300 \text{ V}$
Therefore, the electric potential is $300 \text{ V}$.

(B) We know that the charge $Q$ on the capacitor remains constant if the battery is disconnected,or the potential difference changes based on the capacitance. Given $Q = CV$,we have $V = Q/C$,which implies $V \propto 1/C$.
The capacitance with air is $C_0 = \epsilon_0 A / d$.
The capacitance with a dielectric slab is $C = K C_0$,where $K$ is the dielectric constant.
Therefore,the ratio of the potentials is given by $V_{dielectric} / V_0 = C_0 / C = C_0 / (K C_0) = 1/K$.
Given $V_0 = 4 \ V$ and $V_{dielectric} = 2 \ V$.
Substituting the values: $2 / 4 = 1/K$.
This gives $1/2 = 1/K$,so $K = 2$.

(C) The electric field $E$ at a point outside a spherical conductor at a distance $r$ from its center is given by the formula:
$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$
Given values:
Charge $q = 3 \text{ nC} = 3 \times 10^{-9} \text{ C}$
Distance $r = 3 \text{ cm} = 3 \times 10^{-2} \text{ m}$
Constant $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \text{ Nm}^{2}\text{C}^{-2}$
Substituting these values into the formula:
$E = (9 \times 10^{9}) \times \frac{3 \times 10^{-9}}{(3 \times 10^{-2})^{2}}$
$E = \frac{9 \times 10^{9} \times 3 \times 10^{-9}}{9 \times 10^{-4}}$
$E = \frac{27}{9 \times 10^{-4}} = 3 \times 10^{4} \text{ Vm}^{-1}$
Thus,the electric field at a distance of $3 \text{ cm}$ from the center of the sphere is $3 \times 10^{4} \text{ Vm}^{-1}$.

(A) The magnetic force on a charged particle is given by the formula $\vec{F} = q(\vec{v} \times \vec{B})$.
For the force to be non-zero,the particle must be moving (i.e.,velocity $\vec{v} \neq 0$) and the velocity must not be parallel to the magnetic field $\vec{B}$.
If the particle is stationary,$\vec{v} = 0$,so $\vec{F} = 0$.
If the velocity is parallel or anti-parallel to the magnetic field,the cross product $\vec{v} \times \vec{B} = 0$,so $\vec{F} = 0$.
Therefore,for a magnetic force to exist,the particle must be moving,and the velocity vector must have a component perpendicular to the magnetic field vector.

(B) Given: Focal length of objective,$f_{o} = 1 \ cm$; Focal length of eyepiece,$f_{e} = 6 \ cm$; Tube length,$L = 30 \ cm$; Least distance of distinct vision,$D = 25 \ cm$.
For a compound microscope,the magnifying power $M$ when the final image is formed at the near point $(D)$ is given by:
$M = \frac{L}{f_{o}} \left(1 + \frac{D}{f_{e}}\right)$
Substituting the given values:
$M = \frac{30}{1} \left(1 + \frac{25}{6}\right)$
$M = 30 \left(\frac{6 + 25}{6}\right)$
$M = 30 \left(\frac{31}{6}\right)$
$M = 5 \times 31 = 155$.
Rounding to the nearest provided option,the magnification is approximately $150$.

(D) The fringe width is given as $\beta = 0.002 \text{ cm}$.
For a Young's double-slit interference pattern,the distance of the $n^{\text{th}}$ dark fringe from the central bright fringe is given by the formula $x_n = (n - 0.5) \beta$,where $n$ is the order of the dark fringe $(n = 1, 2, 3, \dots)$.
For the $5^{\text{th}}$ dark fringe,$n = 5$.
Substituting the values: $x_5 = (5 - 0.5) \times 0.002 \text{ cm}$.
$x_5 = 4.5 \times 0.002 \text{ cm} = 0.009 \text{ cm}$.
Converting to scientific notation: $0.009 \text{ cm} = 0.9 \times 10^{-2} \text{ cm}$.
Comparing this result with the given options,none of the options match the calculated value.
Therefore,the correct choice is $D$.

(C) The root mean square $(RMS)$ value of voltage,$ V_{rms} $,and the peak voltage,$ V_{0} $,of an $A$.$C$. source are related by the formula:
$ V_{rms} = \frac{V_{0}}{\sqrt{2}} $
Given that the multimeter reads the $RMS$ value,$ V_{rms} = 100 \,V $.
To find the peak voltage $ V_{0} $,we rearrange the formula:
$ V_{0} = V_{rms} \times \sqrt{2} $
Substituting the given value:
$ V_{0} = 100 \times 1.414 = 141.4 \,V $
Thus,the peak value of the voltage of the $A$.$C$. source is $ 141.4 \,V $.

(D) In an $n$-type semiconductor,although electrons are the majority charge carriers,the material as a whole remains electrically neutral.
This is because the semiconductor crystal is composed of atoms that are electrically neutral,containing an equal number of protons in the nucleus and orbiting electrons.
When a pentavalent impurity is added to create an $n$-type semiconductor,the added impurity atom itself is also electrically neutral.
Therefore,the total negative charge of the free electrons is exactly balanced by the total positive charge of the ionized donor atoms and the protons in the crystal lattice.

(D) The given circuit consists of two $NOR$ gates acting as $NOT$ gates, followed by a $NOR$ gate. Let the inputs be $A$ and $B$.
The first two gates are $NOR$ gates with both inputs tied together. The output of the first gate is $\overline{A+A} = \bar{A}$.
The output of the second gate is $\overline{B+B} = \bar{B}$.
These outputs are fed into the final $NOR$ gate. The output $Y$ is given by:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's theorem, $\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
Thus, the circuit represents an $AND$ gate.

$A$	$B$	$Y = A \cdot B$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

(C) Since the charges are equal in magnitude and placed in a uniform electric field $E$,the magnitude of the force acting on each charge is $F = qE$.
Since the magnitude of the force $F$ is the same for both charges,we have $F_{1} = F_{2}$.
Using Newton's second law,$F = ma$,we can write $m_{1}a_{1} = m_{2}a_{2}$.
Rearranging this gives the ratio of accelerations as $\frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}}$.
Given that $\frac{m_{1}}{m_{2}} = 0.5$,we have $\frac{m_{2}}{m_{1}} = \frac{1}{0.5} = 2$.
Therefore,the ratio of their accelerations is $\frac{a_{1}}{a_{2}} = 2$.

(B) For a transistor,the $ \beta $-current gain and $ \alpha $-current gain are related by the formula:
$\beta = \frac{\alpha}{1 - \alpha}$
Given that the $ \alpha $-current gain is $ 0.98 $.
Substituting the value into the formula:
$\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$
Thus,the $ \beta $-current gain of the transistor is $ 49 $.

(D) The resonant frequency $ f $ of a tuned amplifier circuit ($LC$ circuit) is given by the formula: $ f = \frac{1}{2 \pi \sqrt{LC}} $.
Rearranging the formula to solve for $ \sqrt{LC} $,we get: $ \sqrt{LC} = \frac{1}{2 \pi f} $.
Given the carrier frequency $ f = 2 \text{ MHz} = 2 \times 10^{6} \text{ Hz} $.
Substituting the value of $ f $ into the equation: $ \sqrt{LC} = \frac{1}{2 \pi \times (2 \times 10^{6})} $.
Calculating the denominator: $ \sqrt{LC} = \frac{1}{4 \pi \times 10^{6}} $.

(C) Let the two resistances be $R_{1}$ and $R_{2}$.
When they are connected in series,the equivalent resistance is $R_{s} = R_{1} + R_{2} = 6 \ \Omega$ $(1)$.
When they are connected in parallel,the equivalent resistance is $R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{4}{3} \ \Omega$ $(2)$.
Substituting $R_{1} + R_{2} = 6$ into equation $(2)$:
$\frac{R_{1} R_{2}}{6} = \frac{4}{3} \Rightarrow R_{1} R_{2} = 6 \times \frac{4}{3} = 8 \ \Omega^{2}$ $(3)$.
From $(1)$,$R_{2} = 6 - R_{1}$. Substituting this into $(3)$:
$R_{1}(6 - R_{1}) = 8 \Rightarrow 6R_{1} - R_{1}^{2} = 8 \Rightarrow R_{1}^{2} - 6R_{1} + 8 = 0$.
Solving the quadratic equation:
$(R_{1} - 4)(R_{1} - 2) = 0$.
Thus,$R_{1} = 4 \ \Omega$ or $R_{1} = 2 \ \Omega$.
If $R_{1} = 4 \ \Omega$,then $R_{2} = 2 \ \Omega$. If $R_{1} = 2 \ \Omega$,then $R_{2} = 4 \ \Omega$.
Therefore,the two resistances are $4 \ \Omega$ and $2 \ \Omega$.

(D) The gyromagnetic ratio is defined as the ratio of the magnetic dipole moment to the angular momentum of the electron,which is given by the formula: $\gamma = \frac{e}{2m_e}$.
Given,gyromagnetic ratio $\gamma = 8.8 \times 10^{10} \ C \ kg^{-1}$.
Charge of the electron $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula to solve for the mass of the electron $m_e$:
$m_e = \frac{e}{2\gamma}$.
Substituting the given values:
$m_e = \frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}$.
$m_e = \frac{1.6 \times 10^{-19}}{17.6 \times 10^{10}}$.
$m_e = \frac{16}{176} \times 10^{-29} \ kg$.
$m_e = \frac{1}{11} \times 10^{-29} \ kg$.
Thus,the mass of the electron is $\frac{1}{11} \times 10^{-29} \ kg$.

(D) When a charged particle moves in a uniform magnetic field $\vec{B}$ with a velocity $\vec{v}$ that is not parallel or perpendicular to the field,it can be resolved into two components:
$1$. The component $v_{\perp} = v \sin \theta$ perpendicular to the magnetic field,which provides the necessary centripetal force for circular motion.
$2$. The component $v_{\parallel} = v \cos \theta$ parallel to the magnetic field,which remains unaffected by the magnetic force and causes the particle to move linearly along the direction of the field.
The combination of these two motions—circular motion in the plane perpendicular to the field and linear motion along the field—results in a helical path.

(C) The colour code for the resistor is Green,Black,Violet,Gold.
According to the standard resistor colour code:
Green corresponds to the digit $5$.
Black corresponds to the digit $0$.
Violet corresponds to the multiplier $10^7$.
Gold corresponds to the tolerance $\pm 5\%$.
The resistance value is calculated as: $50 \times 10^7 \pm 5\% \ \Omega$.
Converting this to Megaohms $(M\Omega)$: $50 \times 10^7 \ \Omega = 500 \times 10^6 \ \Omega = 500 \ M\Omega$.
Including the tolerance,the final value is $500 \pm 5\% \ M\Omega$.

(A) The current $I$ flowing through the battery is given by the formula $I = \frac{E}{R_{eq} + r}$,where $E$ is the emf,$R_{eq}$ is the equivalent external resistance,and $r$ is the internal resistance.
Since the two resistors $R_1 = 2 \Omega$ and $R_2 = 6 \Omega$ are connected in parallel,their equivalent resistance $R_{eq}$ is calculated as:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{6} = \frac{3 + 1}{6} = \frac{4}{6} = \frac{2}{3} \Omega^{-1}$.
Therefore,$R_{eq} = \frac{3}{2} = 1.5 \Omega$.
Now,substituting the values into the current formula:
$I = \frac{2}{1.5 + 0.5} = \frac{2}{2.0} = 1 \text{ A}$.
Thus,the current flowing through the battery is $1 \text{ A}$.

(A) Given: Length of solenoid $l = 0.4 \,m$, number of turns $N = 400$, current $I = 5 \,A$.
The number of turns per unit length $n = \frac{N}{l} = \frac{400}{0.4} = 1000 \,m^{-1}$.
The magnetic field inside a long solenoid is given by $B = \mu_0 n I$.
Substituting the values: $B = (4\pi \times 10^{-7}) \times 1000 \times 5$.
$B = 4 \times 3.14159 \times 10^{-7} \times 5000$.
$B = 20 \times 3.14159 \times 10^{-4} = 62.83 \times 10^{-4} = 6.28 \times 10^{-3} \,T$.

(B) Given: Number of turns on secondary,$N_{s} = 50$; Number of turns on primary,$N_{p} = 1000$; Primary voltage,$V_{p} = 220 \ V$; Primary current,$I_{p} = 1 \ A$.
We know the transformer ratio formula: $\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$.
Using the relation $\frac{N_{s}}{N_{p}} = \frac{I_{p}}{I_{s}}$,we can find the output current $I_{s}$ as: $I_{s} = \frac{N_{p}}{N_{s}} \times I_{p}$.
Substituting the given values: $I_{s} = \frac{1000}{50} \times 1$.
$I_{s} = 20 \times 1 = 20 \ A$.
Thus,the output current of the transformer is $20 \ A$.

(A) The relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ in an electromagnetic wave is given by the equation:
$E = cB$,where $c$ is the speed of light in vacuum.
Given values are $E = 6 \text{ V m}^{-1}$ and $c = 3 \times 10^{8} \text{ m s}^{-1}$.
Rearranging the formula to solve for $B$:
$B = \frac{E}{c}$
Substituting the values:
$B = \frac{6}{3 \times 10^{8}}$
$B = 2 \times 10^{-8} \text{ T}$.
Therefore,the magnitude of the magnetic field vector at that point is $2 \times 10^{-8} \text{ T}$.

(A) The common potential $V$ of two capacitors connected in parallel is given by the formula: $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.
Given: $C_1 = 10 \ pF$,$V_1 = 200 \ V$,$C_2 = 20 \ pF$,$V_2 = 100 \ V$.
Substituting the values:
$V = \frac{(10 \times 10^{-12} \times 200) + (20 \times 10^{-12} \times 100)}{10 \times 10^{-12} + 20 \times 10^{-12}}$
$V = \frac{2000 \times 10^{-12} + 2000 \times 10^{-12}}{30 \times 10^{-12}}$
$V = \frac{4000 \times 10^{-12}}{30 \times 10^{-12}} = \frac{400}{3} \approx 133.33 \ V$.
Thus,the common potential is $133.33 \ V$.

(B) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:
$\lambda = \frac{12.27}{\sqrt{V}} \ \text{Å}$
Given that the potential difference $V = 100 \ V$,we substitute this value into the equation:
$\lambda = \frac{12.27}{\sqrt{100}}$
$\lambda = \frac{12.27}{10}$
$\lambda = 1.227 \ \text{Å}$
Thus,the de Broglie wavelength of the electron is $1.227 \ \text{Å}$.

(A) Given: Object distance $u = -20 \ cm$ (by sign convention).
Magnification $m = -3$ (since the image is real and magnified).
Using the magnification formula $m = -\frac{v}{u}$,we have:
$-3 = -\frac{v}{-20} \Rightarrow v = -60 \ cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,we get:
$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = \frac{-4}{60} = -\frac{1}{15}$.
Therefore,the focal length $f = -15 \ cm$.
The magnitude of the focal length is $15 \ cm$.

(A) The energy of an electron in the $ n^{\text{th}} $ orbit of a hydrogen atom is given by the formula: $ E_n = \frac{-13.6 \ eV}{n^2} $.
For the third orbit,we substitute $ n = 3 $ into the equation:
$ E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} $.
Calculating this,we get $ E_3 \approx -1.51 \ eV $.
Thus,the energy of the electron in the third orbit is $ -1.51 \ eV $.

(D) The hydrogen spectrum consists of five main spectral series: Lyman,Balmer,Paschen,Brackett,and Pfund.
$1$. The Lyman series lies in the ultraviolet region.
$2$. The Balmer series lies in the visible region.
$3$. The Paschen,Brackett,and Pfund series lie in the infrared region.
Therefore,the Balmer series is the correct answer.

(B) When light of suitable frequency falls on a metal plate,electrons are emitted. This phenomenon is known as the photoelectric effect.
According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of the emitted photo-electrons is given by:
$K_{\max} = hf - \phi_0$
where $h$ is Planck's constant,$f$ is the frequency of the incident light,and $\phi_0$ is the work function of the metal.
Since $h$ and $\phi_0$ are constants for a given metal,the maximum kinetic energy depends only on the frequency $(f)$ of the incident light.

(A) The average power dissipated in an $A.C.$ circuit is given by the formula: $P = I_{rms}^2 Z \cos \phi$,where $P$ is the power,$I_{rms}$ is the root mean square current,$Z$ is the impedance,and $\cos \phi$ is the power factor.
Given values are: $P = 2 \ W$,$I = 2 \ A$,and $Z = 1 \ \Omega$.
Substituting these values into the formula: $2 = (2)^2 \times 1 \times \cos \phi$.
$2 = 4 \times \cos \phi$.
$\cos \phi = \frac{2}{4} = 0.5$.
Thus,the power factor of the $A.C.$ circuit is $0.5$.

(B) Given,focal length $f = 10 \ cm = 0.1 \ m$.
Power of a lens $P$ is defined as the reciprocal of the focal length in meters.
$P = \frac{1}{f(m)} = \frac{1}{0.1 \ m} = 10 \ D$.
Thus,the power of the lens is $10 \ D$.