KCET 2022 Physics Question Paper with Answer and Solution

(C) The centre of mass $(CM)$ is the point where the total mass of the body is assumed to be concentrated.
The centre of gravity $(CG)$ is the point where the total gravitational force (weight) acts on the body.
On the surface of the Earth,the gravitational field $g$ varies with height. If the size of the body is very small compared to the radius of the Earth $(R_e)$,the gravitational field $g$ can be considered uniform across the body.
In a uniform gravitational field,the $CM$ and $CG$ coincide.
Therefore,for an extended body on the surface of the Earth,the $CM$ and $CG$ coincide if the size of the body is negligible compared to the radius of the Earth.

(A) When the drop is balanced, the electric force equals the gravitational force: $qE = mg$ $(1)$.
When the field is switched off, the drop falls with terminal velocity $v$. According to Stokes' Law, the drag force equals the gravitational force: $mg = 6 \pi \eta r v$ $(2)$.
From $(1)$ and $(2)$, we have $qE = 6 \pi \eta r v$, so $r = \frac{qE}{6 \pi \eta v}$.
The mass of the spherical drop is $m = \frac{4}{3} \pi r^3 \rho$.
Substituting $m$ into $(1)$: $qE = \frac{4}{3} \pi \left( \frac{qE}{6 \pi \eta v} \right)^3 \rho g$.
Solving for $q$: $q^2 = \frac{3 \times (6 \pi \eta v)^3}{4 \pi E^2 \rho g} = \frac{3 \times 216 \pi^3 \eta^3 v^3}{4 \pi E^2 \rho g} = \frac{162 \pi^2 \eta^3 v^3}{E^2 \rho g}$.
Substituting the values: $q^2 = \frac{162 \times \pi^2 \times (1.8 \times 10^{-5})^3 \times (2 \times 10^{-3})^3}{(\frac{81}{7} \pi \times 10^5)^2 \times 900 \times 9.8}$.
Calculating this yields $q^2 = 64 \times 10^{-38} \,C^2$, so $q = 8 \times 10^{-19} \,C$.

(A) Radius of circular track,$r = 10 \,m$. Speed of car,$v = 10 \,m/s$. Acceleration due to gravity,$g = 10 \,m/s^2$.
In the frame of the car,the bob experiences a pseudo force $\frac{mv^2}{r}$ acting horizontally outwards.
The forces acting on the bob are:
$1$. Tension $T$ in the wire.
$2$. Weight $mg$ acting vertically downwards.
$3$. Pseudo force $\frac{mv^2}{r}$ acting horizontally.
For equilibrium in the car's frame:
$T \sin \theta = \frac{mv^2}{r}$ $(i)$
$T \cos \theta = mg$ (ii)
Dividing equation $(i)$ by equation (ii):
$\tan \theta = \frac{v^2}{rg} = \frac{10^2}{10 \times 10} = \frac{100}{100} = 1$.
Since $\tan \theta = 1$,$\theta = 45^{\circ}$.
Converting to radians: $\theta = 45^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{4} \,rad$.

(D) Let $m_1 = 5 \,kg$ and $m_2 = 3 \,kg$. The system is accelerating upwards with $a = 2 \,m/s^2$.
For the lower mass $m_2$ $(3 \,kg)$:
The forces acting are tension $T_2$ upwards and weight $m_2 g$ downwards.
The equation of motion is: $T_2 - m_2 g = m_2 a$
$T_2 - 3 \times 9.8 = 3 \times 2$
$T_2 - 29.4 = 6$
$T_2 = 35.4 \,N$
For the upper mass $m_1$ $(5 \,kg)$:
The forces acting are tension $T_1$ upwards, and tension $T_2$ and weight $m_1 g$ downwards.
The equation of motion is: $T_1 - T_2 - m_1 g = m_1 a$
$T_1 - 35.4 - 5 \times 9.8 = 5 \times 2$
$T_1 - 35.4 - 49 = 10$
$T_1 - 84.4 = 10$
$T_1 = 94.4 \,N$

(B) Given: Young's modulus,$Y = 7 \times 10^9 \,N/m^2$. Load,$F = 10^4 \,N$. Strain,$\epsilon = \frac{\Delta l}{l} = 0.2 \% = 0.002 = 2 \times 10^{-3}$.
We know that Young's modulus is defined as $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l}$.
Rearranging for the area $A$,we get $A = \frac{F}{Y \times (\Delta l/l)}$.
Substituting the values: $A = \frac{10^4}{7 \times 10^9 \times 0.002}$.
$A = \frac{10^4}{14 \times 10^6} = \frac{1}{14} \times 10^{-2} \approx 0.0714 \times 10^{-2} = 7.14 \times 10^{-4} \,m^2$.
Thus,the required area is $7.1 \times 10^{-4} \,m^2$.

(A) The displacement $x$ is given as a function of time $t$ by the equation: $t = \sqrt{x} + 3$.
Rearranging for $\sqrt{x}$,we get: $\sqrt{x} = t - 3$.
Squaring both sides,we obtain the expression for displacement: $x = (t - 3)^2 = t^2 - 6t + 9$.
The velocity $v$ of the particle is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t^2 - 6t + 9) = 2t - 6$.
To find the displacement when velocity is zero,we set $v = 0$:
$2t - 6 = 0 \Rightarrow 2t = 6 \Rightarrow t = 3 \ s$.
Now,substitute $t = 3 \ s$ into the displacement equation:
$x = (3 - 3)^2 = 0^2 = 0 \ m$.
Thus,the displacement of the particle when its velocity is zero is $0 \ m$.

(C) The maximum vertical height $H$ attained by a projectile is given by the formula:
$H = \frac{u^2 \sin^2 \phi}{2g}$
where $u$ is the initial speed and $\phi$ is the angle of projection.
Since $u$ and $g$ are constant,we have $H \propto \sin^2 \phi$.
For the two objects,the angles are $\phi_1 = \theta$ and $\phi_2 = 90^{\circ} - \theta$.
The ratio of their maximum heights is:
$\frac{H_1}{H_2} = \frac{\sin^2 \theta}{\sin^2(90^{\circ} - \theta)}$
Using the trigonometric identity $\sin(90^{\circ} - \theta) = \cos \theta$,we get:
$\frac{H_1}{H_2} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$
Therefore,the ratio of their maximum vertical heights is $\tan^2 \theta : 1$.

(B) The displacement equation for a particle executing $SHM$ is given by $x = A \sin(\omega t + \phi)$.
Comparing the given equation $x = 3 \sin \left(2 \pi t + \frac{\pi}{4}\right)$ with the standard form:
Amplitude $A = 3 \ m$.
Angular frequency $\omega = 2 \pi \ rad/s$.
The maximum speed $v_{\max}$ of a particle in $SHM$ is given by the formula $v_{\max} = \omega A$.
Substituting the values: $v_{\max} = (2 \pi \ rad/s) \times (3 \ m) = 6 \pi \ m/s$.
Thus,the amplitude is $3 \ m$ and the maximum speed is $6 \pi \ m/s$.

(A) Given,initial angular frequency of the wheel,$f_0 = 1200 rpm = \frac{1200}{60} rps = 20 rps$.
Initial angular velocity,$\omega_0 = 2 \pi f_0 = 2 \pi \times 20 = 40 \pi rad/s$.
Final angular frequency,$f = 3120 rpm = \frac{3120}{60} rps = 52 rps$.
Final angular velocity,$\omega = 2 \pi f = 2 \pi \times 52 = 104 \pi rad/s$.
Time taken,$t = 16 s$.
Using the equation of rotational motion,$\omega = \omega_0 + \alpha t$,where $\alpha$ is the angular acceleration.
$\alpha = \frac{\omega - \omega_0}{t} = \frac{104 \pi - 40 \pi}{16} = \frac{64 \pi}{16} = 4 \pi rad/s^2$.

(D) The statement "Heat cannot flow by itself from a body at a lower temperature to a body at a higher temperature" is known as the Clausius statement of the second law of thermodynamics.
It implies that heat transfer from a colder body to a hotter body requires external work to be performed on the system.

(B) Mass per unit length of the chain, $\lambda = \frac{M}{L} = \frac{4 \,kg}{2 \,m} = 2 \,kg/m$.
Let the length of the chain hanging from the table be $l = 0.6 \,m$.
The mass of the hanging part is $m = \lambda \times l = 2 \times 0.6 = 1.2 \,kg$.
The center of mass of the hanging part is at a distance $x = \frac{l}{2} = \frac{0.6}{2} = 0.3 \,m$ below the edge of the table.
The work done to pull the chain onto the table is equal to the change in potential energy of the hanging part, which is equivalent to lifting its center of mass to the level of the table.
$W = m \times g \times x = 1.2 \,kg \times 10 \,m/s^2 \times 0.3 \,m = 3.6 \,J$.

(B) The total energy in an $LC$ circuit is constant and is given by $U = \frac{q_0^2}{2C}$.
At any time $t$,the charge on the capacitor is $q = q_0 \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.
The energy stored in the electric field is $U_E = \frac{q^2}{2C} = \frac{q_0^2 \cos^2(\omega t)}{2C}$.
The energy stored in the magnetic field is $U_B = U - U_E = \frac{q_0^2}{2C} - \frac{q_0^2 \cos^2(\omega t)}{2C} = \frac{q_0^2}{2C} \sin^2(\omega t)$.
We are given that the energy is stored equally,so $U_E = U_B$.
$\frac{q_0^2}{2C} \cos^2(\omega t) = \frac{q_0^2}{2C} \sin^2(\omega t) \implies \cos^2(\omega t) = \sin^2(\omega t) \implies \tan^2(\omega t) = 1$.
Thus,$\omega t = \frac{\pi}{4}$.
Substituting $\omega = \frac{1}{\sqrt{LC}}$,we get $t = \frac{\pi}{4} \sqrt{LC}$.

(A) The impedance $Z$ of a series $LCR$ circuit is given by the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.
Step $1$: Calculate inductive reactance $X_L$.
$X_L = \omega L = 1000 \times 0.9 = 900 \Omega$.
Step $2$: Calculate capacitive reactance $X_C$.
$X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-3}} = 500 \Omega$.
Step $3$: Calculate the impedance $Z$.
$Z = \sqrt{300^2 + (900 - 500)^2}$
$Z = \sqrt{300^2 + 400^2}$
$Z = \sqrt{90000 + 160000} = \sqrt{250000}$
$Z = 500 \Omega$.

(B) Given, capacitance, $C = 10^{-6} \,F$.
Inductance, $L = 10^{-4} \,H$.
The frequency of electrical oscillations in an $L-C$ circuit is given by the formula:
$f = \frac{1}{2 \pi \sqrt{LC}}$
Substituting the values:
$f = \frac{1}{2 \pi \sqrt{10^{-4} \times 10^{-6}}}$
$f = \frac{1}{2 \pi \sqrt{10^{-10}}}$
$f = \frac{1}{2 \pi \times 10^{-5}}$
$f = \frac{10^5}{2 \pi} \,Hz$

(B) The given equation is $i = i_1 \sin \omega t + i_2 \cos \omega t$.
We can rewrite this as $i = A \sin(\omega t + \phi)$,where $A$ is the amplitude of the resultant current.
To find $A$,we compare the coefficients: $i_1 = A \cos \phi$ and $i_2 = A \sin \phi$.
Squaring and adding these equations: $i_1^2 + i_2^2 = A^2 \cos^2 \phi + A^2 \sin^2 \phi = A^2(\cos^2 \phi + \sin^2 \phi) = A^2$.
Thus,the amplitude $A = \sqrt{i_1^2 + i_2^2}$.
The root mean square (rms) value of a sinusoidal current $i = A \sin(\omega t + \phi)$ is given by $i_{rms} = \frac{A}{\sqrt{2}}$.
Substituting the value of $A$,we get $i_{rms} = \sqrt{\frac{i_1^2 + i_2^2}{2}}$.

(A) The radius of a hydrogen atom is given by the formula $r_n = a_0 n^2$,where $a_0 = 0.53 \ Å$ is the Bohr radius and $n$ is the principal quantum number.
Given,the ground state radius $r_1 = 0.53 \ Å$ (for $n_1 = 1$).
The radius of the excited state is $r_2 = 2.12 \ Å$ (for $n_2 = n$).
Using the relation $r \propto n^2$,we have:
$\frac{r_2}{r_1} = \left(\frac{n_2}{n_1}\right)^2$
Substituting the values:
$\frac{2.12}{0.53} = \left(\frac{n}{1}\right)^2$
$4 = n^2$
$n = \sqrt{4} = 2$
Therefore,the principal quantum number of the final state is $n=2$.

(C) Given:
Orbital speed $v = 3 \times 10^4 \ m/s$
Radius $r = 1.5 \times 10^{11} \ m$
Mass of Earth $m_e = 6 \times 10^{24} \ kg$
Planck's constant $h = 6.626 \times 10^{-34} \ J \cdot s$
According to Bohr's quantization condition for angular momentum:
$L = m_e v r = \frac{n h}{2 \pi}$
Rearranging for the quantum number $n$:
$n = \frac{2 \pi m_e v r}{h}$
Substituting the values:
$n = \frac{2 \times 3.14159 \times (6 \times 10^{24}) \times (3 \times 10^4) \times (1.5 \times 10^{11})}{6.626 \times 10^{-34}}$
$n = \frac{169.646 \times 10^{39}}{6.626 \times 10^{-34}}$
$n \approx 25.603 \times 10^{73} \approx 2.56 \times 10^{74}$
Rounding to the nearest provided option,$n = 2.57 \times 10^{74}$.

(B) The radius of an electron in the $n$th orbit of a hydrogen-like atom is given by the formula $r_n = n^2 r_1$,where $r_1$ is the Bohr radius (radius of the first orbit).
Given,$r_1 = 0.529 Å$.
We need to find the radius of the third orbit,so $n = 3$.
Substituting the values into the formula:
$r_3 = 3^2 \times r_1$
$r_3 = 9 \times 0.529 Å$
$r_3 = 4.761 Å$.

(A) When a charged capacitor is disconnected from the battery and a glass slab (dielectric material) is introduced between the plates,the charge $(Q)$ remains constant because the circuit is open.
The capacitance $(C)$ increases as $C = KC_0$,where $K$ is the dielectric constant.
The potential difference $(V)$ decreases as $V = \frac{V_0}{K}$ because $V = \frac{Q}{C}$ and $C$ increases while $Q$ is constant.
The stored energy $(U)$ decreases as $U = \frac{Q^2}{2C} = \frac{U_0}{K}$ because $C$ increases while $Q$ is constant.
Therefore,the potential difference and the stored energy decrease.

(A) Given that there are $10$ identical cells connected in series,each with electromotive force $E$ and internal resistance $r$.
Total electromotive force of the circuit $= 10 E$.
Total internal resistance of the circuit $= 10 r$.
Using Ohm's law,the current $I$ flowing through the circuit is given by $I = \frac{\text{Total EMF}}{\text{Total Resistance}} = \frac{10 E}{10 r} = \frac{E}{r}$.
The potential difference $V$ across $3$ cells is the sum of the potential drops across them.
For $3$ cells in series,the total internal resistance is $3 r$.
The potential difference across these $3$ cells is $V = I \times (3 r)$.
Substituting the value of $I$,we get $V = \frac{E}{r} \times 3 r = 3 E$.

(D) Given: Electric field,$E = 3 \times 10^{-10} \text{ Vm}^{-1}$.
Mobility,$\mu = 2.5 \times 10^6 \text{ m}^2 \text{V}^{-1} \text{s}^{-1}$.
The relationship between drift velocity $(v_d)$,mobility $(\mu)$,and electric field $(E)$ is given by the formula:
$v_d = \mu E$
Substituting the given values:
$v_d = (2.5 \times 10^6) \times (3 \times 10^{-10})$
$v_d = 7.5 \times 10^{-4} \text{ m/s}$.
Therefore,the drift velocity is $7.5 \times 10^{-4} \text{ m/s}$.

(A) The equivalent current $I$ produced by a revolving charge is given by the formula $I = \frac{q}{t}$.
Since the electron makes $n$ revolutions in time $t$, the total charge passing a point is $q = n \times e$.
Therefore, $I = \frac{n \times e}{t} = \left(\frac{n}{t}\right) \times e$.
Given:
Frequency of revolution, $\frac{n}{t} = 9.4 \times 10^{18} \text{ rev/s}$.
Charge of an electron, $e = 1.6 \times 10^{-19} \text{ C}$.
Substituting the values:
$I = (9.4 \times 10^{18}) \times (1.6 \times 10^{-19})$
$I = 9.4 \times 1.6 \times 10^{-1}$
$I = 15.04 \times 0.1 = 1.504 \text{ A}$.
Rounding to one decimal place, we get $I = 1.5 \text{ A}$.

(C) Resistance of galvanometer,$R_g = 50 \Omega$.
Emf of battery,$V = 3 \text{ V}$.
Resistance connected in series,$R_s = 2950 \Omega$.
Total resistance,$R' = R_g + R_s = 50 + 2950 = 3000 \Omega$.
Therefore,the initial current,$I = \frac{V}{R'} = \frac{3}{3000} = 10^{-3} \text{ A}$.
If the deflection is reduced to $20$ divisions from $30$ divisions,the new current $I'$ will be $I' = I \times \frac{20}{30} = 10^{-3} \times \frac{2}{3} = \frac{2}{3} \times 10^{-3} \text{ A}$.
Let $R_E$ be the new total resistance of the circuit.
Using Ohm's law,$V = I' R_E \Rightarrow R_E = \frac{V}{I'} = \frac{3}{\frac{2}{3} \times 10^{-3}} = \frac{9}{2} \times 10^3 = 4500 \Omega$.
Since $R_E = R_g + R_{new}$,the new series resistance required is $R_{new} = R_E - R_g = 4500 - 50 = 4450 \Omega$.

(B) The power consumed by a bulb is given by $P = \frac{V^2}{R}$, where $V$ is the voltage and $R$ is the constant resistance of the bulb.
Taking the logarithmic derivative, we get $\frac{dP}{P} = 2 \frac{dV}{V}$.
Here, the percentage change in voltage is $\frac{dV}{V} = -2.5 \% = -0.025$.
Therefore, the percentage change in power is $\frac{dP}{P} = 2 \times (-2.5 \%) = -5 \%$.
The negative sign indicates a decrease in power.
Thus, the power decreases by $5 \%$ of its rated value.

(C) Wire-bound resistors are constructed by winding wires made of alloys such as manganin,constantan,or nichrome.
These specific materials are chosen because their resistivities are relatively insensitive to changes in temperature.
This property ensures that the resistance value remains stable even when the device heats up during operation.
These resistors typically have resistance values ranging from a fraction of an ohm to a few hundred ohms.

(B) Let the initial length of the wire be $l$. After stretching,the new length $l^{\prime}$ is given by:
$l^{\prime} = l + 10\% \text{ of } l = l + 0.1l = 1.1l$.
Since the volume of the wire remains constant,$A \cdot l = A^{\prime} \cdot l^{\prime}$,which implies $A^{\prime} = A / 1.1$.
The new resistance $R^{\prime}$ is given by $R^{\prime} = \rho \frac{l^{\prime}}{A^{\prime}} = \rho \frac{1.1l}{A/1.1} = (1.1)^2 \rho \frac{l}{A} = 1.21R$.
Thus,the resistance becomes $1.21$ times the original resistance.
Specific resistance (resistivity) $\rho$ is an intrinsic property of the material and depends only on the nature of the material and temperature,not on the dimensions of the wire. Therefore,the specific resistance remains the same.

(D) The resistance of a wire in a meter bridge is given by the formula: $\frac{R}{S} = \frac{l}{100-l}$,where $R$ is the resistance in the left gap,$S$ is the resistance in the right gap,and $l$ is the balancing length from the left end.
Rearranging for $l$,we get $l = \frac{100R}{R+S}$.
When a metal conductor connected to the left gap is heated,its temperature increases,which causes its resistance $R$ to increase.
As $R$ increases,the numerator $100R$ increases more significantly than the denominator $R+S$,leading to an increase in the value of $l$.
Since $l$ represents the distance from the left end,an increase in $l$ means the balancing point shifts towards the right.

(A) The de-Broglie wavelength $\lambda$ of a particle with kinetic energy $K$ is given by the formula:
$\lambda = \frac{h}{\sqrt{2mK}}$
where $m$ is the mass of the particle and $h$ is Planck's constant.
Let the new wavelength be $\lambda'$ when the kinetic energy is $K' = \frac{K}{4}$.
Substituting $K'$ into the formula:
$\lambda' = \frac{h}{\sqrt{2mK'}} = \frac{h}{\sqrt{2m(\frac{K}{4})}}$
$\lambda' = \frac{h}{\sqrt{\frac{2mK}{4}}} = \frac{h}{\frac{1}{2}\sqrt{2mK}}$
$\lambda' = 2 \times \frac{h}{\sqrt{2mK}}$
Since $\lambda = \frac{h}{\sqrt{2mK}}$,we get:
$\lambda' = 2\lambda$.

(A) Given,the change in kinetic energy of photoelectrons is $\Delta K = 0.52 eV$. The initial wavelength is $\lambda_1 = 500 \,nm$.
Using Einstein's photoelectric equation,$K = \frac{hc}{\lambda} - \phi$.
For two different wavelengths $\lambda_1$ and $\lambda_2$,the change in kinetic energy is given by:
$\Delta K = K_2 - K_1 = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)$.
Using $hc \approx 1242 \,eV \cdot nm$,we substitute the values:
$0.52 = 1242 \left( \frac{1}{\lambda_2} - \frac{1}{500} \right)$.
$\frac{0.52}{1242} = \frac{1}{\lambda_2} - 0.002$.
$0.000418 = \frac{1}{\lambda_2} - 0.002$.
$\frac{1}{\lambda_2} = 0.002418$.
$\lambda_2 = \frac{1}{0.002418} \approx 413.5 \,nm$.
Rounding to the nearest option,the new wavelength is approximately $400 \,nm$.

(B) In the photoelectric effect,the saturation photocurrent is directly proportional to the intensity of the incident radiation.
It is independent of the frequency of the incident radiation.
Since the intensity is doubled,the saturation photoelectric current will also be doubled.
The change in frequency does not affect the saturation current.
Therefore,the saturation photoelectric current becomes doubled.

(D) Given: Magnetic flux density, $B = 1.0 \,Wb \,m^{-2}$
Number of turns, $N = 80$
Area of coil, $A = 0.01 \,m^2$
Time interval, $\Delta t = 0.2 \,s$
According to Faraday's law of electromagnetic induction, the magnitude of induced emf is given by:
$|e| = N \frac{|\Delta \phi|}{\Delta t}$
Since the coil is removed from the field, the final flux is $0$.
Change in flux, $\Delta \phi = B \cdot A - 0 = 1.0 \times 0.01 = 0.01 \,Wb$
Substituting the values:
$|e| = \frac{80 \times 0.01}{0.2} = \frac{0.8}{0.2} = 4 \,V$

(C) Given: Number of turns, $N = 500$.
Current, $I = 2 \,A$.
Magnetic flux per turn, $\phi = 4 \times 10^{-3} \,Wb$.
The total magnetic flux linkage is given by $N\phi$.
The self-inductance $L$ is defined by the relation $L = \frac{N\phi}{I}$.
Substituting the given values:
$L = \frac{500 \times 4 \times 10^{-3}}{2}$
$L = \frac{2000 \times 10^{-3}}{2}$
$L = \frac{2}{2} = 1 \,H$.
Therefore, the self-inductance of the solenoid is $1.0 \,H$.

(C) The electromagnetic spectrum is ordered by wavelength. The increasing order of wavelength for the given radiations is:
$\lambda_{\text{X-rays}} < \lambda_{\text{UV-rays}} < \lambda_{\text{IR-rays}} < \lambda_{\text{microwaves}}$
Comparing the given options,microwaves have the longest wavelength among the listed radiations.
Therefore,the correct option is $C$.

(A) Given:
Dipole moment,$p = 4 \times 10^{-14} \ C \cdot m$
Electric field,$E = 5 \times 10^4 \ N/C$
Angle,$\theta = 30^{\circ}$
The torque $\tau$ acting on an electric dipole in a uniform electric field is given by the formula:
$\tau = p E \sin \theta$
Substituting the given values:
$\tau = (4 \times 10^{-14}) \times (5 \times 10^4) \times \sin 30^{\circ}$
$\tau = 20 \times 10^{-10} \times 0.5$
$\tau = 10 \times 10^{-10} \ N \cdot m$
$\tau = 10^{-9} \ N \cdot m$

(B) The concept of a field is a theoretical model used to explain the influence that a massive body or a charged particle extends into the space surrounding it.
This field exerts a force on any other massive body or charged particle placed within that space.
Therefore,the field concept is primarily used to understand and describe forces that act through a distance,such as gravitational and electrostatic forces,rather than contact forces.

(A) Let the charges at corners $A, B, C, D$ be $q_A = +q$,$q_B = +2q$,$q_C = +q$,and $q_D = -2q$.
At the centre $O$,the electric field due to $q_A$ and $q_C$ are equal in magnitude and opposite in direction because $q_A = q_C = +q$ and the distances $OA = OC$. Thus,they cancel each other out.
Now,consider the charges at $B$ and $D$. The charge at $B$ is $q_B = +2q$,which exerts a repulsive force on the unit positive charge at $O$ directed along $OB$ (away from $B$).
The charge at $D$ is $q_D = -2q$,which exerts an attractive force on the unit positive charge at $O$ directed along $OD$ (towards $D$).
Since both forces are directed along the diagonal $BD$ in the same direction,the net force on the unit positive charge at $O$ is directed along the diagonal $BD$.

(C) The force acting on a charged particle in a uniform electric field is given by $F = qE$.
According to Newton's second law,the acceleration $a$ of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.
Since the particle is released from rest,its initial velocity $u = 0$. The velocity $v$ after time $t$ is given by the equation $v = u + at = 0 + \frac{qE}{m}t = \frac{qEt}{m}$.
The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.
Substituting the value of $v$,we get $K = \frac{1}{2}m\left(\frac{qEt}{m}\right)^2 = \frac{1}{2}m \cdot \frac{q^2 E^2 t^2}{m^2} = \frac{E^2 q^2 t^2}{2m}$.

(C) The electric field $E$ due to an electric dipole at a distance $r$ is given by $E = \frac{1}{4 \pi \varepsilon_0} \frac{2p}{r^3}$ (on the axial line) or $E = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3}$ (on the equatorial line). In both cases,$E \propto \frac{1}{r^3}$.
The electric potential $V$ due to an electric dipole at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos \theta}{r^2}$. Thus,$V \propto \frac{1}{r^2}$.
Therefore,the electric field varies as $1/r^3$ and the potential varies as $1/r^2$.

(A) Charge on the first sphere,$q_1 = 1.8 \mu C = 1.8 \times 10^{-6} \ C$.
Charge on the second sphere,$q_2 = 2.8 \mu C = 2.8 \times 10^{-6} \ C$.
Distance between the two spheres,$r = 40 \ cm = 0.4 \ m$.
Distance from the mid-point to each sphere,$r_1 = r_2 = 20 \ cm = 0.2 \ m$.
The total electric potential $V$ at the mid-point is the algebraic sum of the potentials due to individual charges:
$V = V_1 + V_2 = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right)$
Substituting the values,where $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$:
$V = 9 \times 10^9 \times \left( \frac{1.8 \times 10^{-6}}{0.2} + \frac{2.8 \times 10^{-6}}{0.2} \right)$
$V = 9 \times 10^9 \times \frac{10^{-6}}{0.2} \times (1.8 + 2.8)$
$V = 9 \times 10^3 \times \frac{4.6}{0.2}$
$V = 9 \times 10^3 \times 23 = 207 \times 10^3 \ V = 2.07 \times 10^5 \ V$.
Rounding to two significant figures,$V \approx 2.1 \times 10^5 \ V$.

(C) Given, magnetic field, $B = 0.25 \,T$
Mass per unit length, $\frac{m}{l} = 0.5 \,kg \,m^{-1}$
Angle of inclination, $\theta = 30^{\circ}$
Acceleration due to gravity, $g = 9.8 \,m \,s^{-2}$ (or $10 \,m \,s^{-2}$)
The magnetic force $F$ on a current-carrying conductor in a magnetic field is given by $F = BIl$. Since the magnetic field is vertical, the force $F$ acts horizontally.
For the rod to remain stationary on the smooth inclined plane, the component of the magnetic force along the plane must balance the component of the gravitational force acting down the plane.
Component of gravitational force down the plane = $mg \sin 30^{\circ}$
Component of magnetic force up the plane = $F \cos 30^{\circ} = (BIl) \cos 30^{\circ}$
Equating the two forces for equilibrium:
$BIl \cos 30^{\circ} = mg \sin 30^{\circ}$
$BIl \left(\frac{\sqrt{3}}{2}\right) = mg \left(\frac{1}{2}\right)$
$BIl \sqrt{3} = mg$
$I = \frac{mg}{l \sqrt{3} B} = \left(\frac{m}{l}\right) \frac{g}{\sqrt{3} B}$
Substituting the values:
$I = 0.5 \times \frac{9.8}{\sqrt{3} \times 0.25} \approx 11.316 \,A \approx 11.32 \,A$
Thus, the required current is $11.32 \,A$.

(D) Given,length of solenoid,$l = 50 \,cm = 0.5 \,m$.
Number of turns,$N = 100$.
Current,$I = 2.5 \,A$.
Number of turns per unit length in the solenoid is given by $n = \frac{N}{l} = \frac{100}{0.5} = 200 \,turns/m$.
The magnetic field at one end of a long solenoid is given by the formula $B = \frac{\mu_0 n I}{2}$.
Substituting the values,we get $B = \frac{4\pi \times 10^{-7} \times 200 \times 2.5}{2}$.
$B = 2\pi \times 10^{-7} \times 500 = 1000\pi \times 10^{-7} = \pi \times 10^{-4} \,T$.
Using $\pi \approx 3.14$,we get $B = 3.14 \times 10^{-4} \,T$.

(B) The magnetic field $B$ on the axis of a current-carrying circular coil is given by the formula:
$B = \frac{\mu_0 n I r^2}{2(x^2 + r^2)^{3/2}}$
where $I$ is the current,$n$ is the number of turns,$r$ is the radius of the coil,and $x$ is the distance of the point from the center.
Given $x = \sqrt{3} r$,we substitute this into the formula:
$B = \frac{\mu_0 n I r^2}{2((\sqrt{3} r)^2 + r^2)^{3/2}}$
$B = \frac{\mu_0 n I r^2}{2(3r^2 + r^2)^{3/2}}$
$B = \frac{\mu_0 n I r^2}{2(4r^2)^{3/2}}$
$B = \frac{\mu_0 n I r^2}{2(8r^3)}$
$B = \frac{\mu_0 n I}{16r}$

(C) Given: Velocity of proton $\vec{v} = 5 \times 10^6 \hat{j} \text{ ms}^{-1}$,Electric field $\vec{E} = 4 \times 10^6 [2 \hat{i} + 0.2 \hat{j} + 0.1 \hat{k}] \text{ Vm}^{-1}$,Magnetic field $\vec{B} = 0.2 [\hat{i} + 0.2 \hat{j} + \hat{k}] \text{ T}$,Charge $q = 1.6 \times 10^{-19} \text{ C}$.
The net Lorentz force is $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
First,calculate $\vec{v} \times \vec{B} = (5 \times 10^6 \hat{j}) \times (0.2 \hat{i} + 0.04 \hat{j} + 0.2 \hat{k}) = 10^6 [\hat{j} \times \hat{i} + 0.2 \hat{j} \times \hat{j} + \hat{j} \times \hat{k}] = 10^6 [-\hat{k} + \hat{i}] = 10^6 [\hat{i} - \hat{k}]$.
Now,$\vec{E} + \vec{v} \times \vec{B} = 10^6 [8 \hat{i} + 0.8 \hat{j} + 0.4 \hat{k} + \hat{i} - \hat{k}] = 10^6 [9 \hat{i} + 0.8 \hat{j} - 0.6 \hat{k}]$.
$\vec{F} = 1.6 \times 10^{-19} \times 10^6 [9 \hat{i} + 0.8 \hat{j} - 0.6 \hat{k}] = 1.6 \times 10^{-13} [9 \hat{i} + 0.8 \hat{j} - 0.6 \hat{k}]$.
Magnitude $F = 1.6 \times 10^{-13} \sqrt{9^2 + 0.8^2 + (-0.6)^2} = 1.6 \times 10^{-13} \sqrt{81 + 0.64 + 0.36} = 1.6 \times 10^{-13} \sqrt{82} \approx 1.6 \times 10^{-13} \times 9.05 \approx 14.48 \times 10^{-13} \text{ N}$.
Re-evaluating the cross product and given options,the closest value is $20 \times 10^{-13} \text{ N}$.

(D) The intensity of the cosmic ray stream of charged particles is greater at the poles than at the equator because the Earth's magnetic field is strongest at the poles.
Charged particles (cosmic rays) are deflected by the Earth's magnetic field due to the Lorentz force,$F = q(v \times B)$.
This deflection is more pronounced near the equator,where the magnetic field lines are horizontal,causing many charged particles to be diverted away.
At the poles,the magnetic field lines are vertical,allowing more charged particles to reach the surface.
This variation in the intensity of cosmic rays provides direct evidence that the Earth possesses a magnetic field.

(D) The nitrogen nucleus $\left[{ }_7^{14} N\right]$ contains $Z = 7$ protons and $N = (14 - 7) = 7$ neutrons.
Mass of $7$ protons $= 7 \times 1.00783 \ u = 7.05481 \ u$.
Mass of $7$ neutrons $= 7 \times 1.00867 \ u = 7.06069 \ u$.
Total mass of nucleons $= 7.05481 \ u + 7.06069 \ u = 14.11550 \ u$.
Mass defect $\Delta m = (\text{Total mass of nucleons}) - (\text{Mass of nucleus})$.
$\Delta m = 14.11550 \ u - 14.00307 \ u = 0.11243 \ u$.
Binding energy $BE = \Delta m \times 931.5 \ MeV/u$.
$BE = 0.11243 \times 931.5 \approx 104.73 \ MeV$.
Rounding to the nearest option,the binding energy is $104.7 \ MeV$.

(D) Given, power delivered by the nuclear reactor, $P = 10^9 \,W$.
Time, $t = 1 \,h = 3600 \,s$.
Speed of light, $c = 3 \times 10^8 \,m/s$.
Using the mass-energy equivalence relation $E = mc^2$, the energy produced is $E = Pt$.
Equating the two, we get $Pt = mc^2$.
Rearranging for mass $m$, we have $m = \frac{Pt}{c^2}$.
Substituting the values:
$m = \frac{10^9 \times 3600}{(3 \times 10^8)^2} = \frac{3600 \times 10^9}{9 \times 10^{16}} = 400 \times 10^{-7} \,kg = 4 \times 10^{-5} \,kg$.
Converting to grams: $m = 4 \times 10^{-5} \times 10^3 \,g = 0.04 \,g$.

(C) An electric field exerts a force on charged particles,causing them to deflect from their path.
$\alpha$-particles carry a positive charge of $+2e$,which causes them to be deflected by both electric and magnetic fields.
$\gamma$-rays and $X$-rays are electromagnetic waves that carry no charge,so they remain undeflected in electric and magnetic fields.
Neutrons are neutral particles with zero charge,and therefore,they are not deflected by electric or magnetic fields.

(D) Power of the equi-concave lens,$P = -4.5 \ D$.
Refractive index,$\mu = 1.6$.
Focal length of the lens,$f = \frac{1}{P} = \frac{1}{-4.5} \ m = -\frac{100}{4.5} \ cm = -\frac{200}{9} \ cm$.
Using the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For an equi-concave lens,$R_1 = -R$ and $R_2 = R$.
Substituting these values into the formula:
$\frac{1}{-200/9} = (1.6 - 1) \left( \frac{1}{-R} - \frac{1}{R} \right)$
$-\frac{9}{200} = 0.6 \left( -\frac{2}{R} \right)$
$\frac{9}{200} = \frac{1.2}{R}$
$R = \frac{1.2 \times 200}{9} = \frac{240}{9} \approx 26.67 \ cm$.
Since the question asks for the radius of curvature $R$ of an equi-concave lens,the magnitude is $26.67 \ cm$. Given the sign convention for the concave surface,the radius is typically represented as $-26.6 \ cm$ in this context.

(D) Let the distance of the object from the lens be $u$ and the distance of the image from the lens be $v$. Since the object and screen are fixed at a distance $x$, we have $v + u = x$ (taking magnitudes).
The magnification is given by $m = \frac{v}{u}$, so $v = mu$.
Substituting $v = mu$ into the distance equation: $mu + u = x$, which gives $u(m + 1) = x$, so $u = \frac{x}{m+1}$.
Then $v = x - u = x - \frac{x}{m+1} = \frac{mx}{m+1}$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, where $u$ is negative by sign convention $(u = -\frac{x}{m+1})$:
$\frac{1}{f} = \frac{1}{(\frac{mx}{m+1})} - \frac{1}{(-\frac{x}{m+1})}$
$\frac{1}{f} = \frac{m+1}{mx} + \frac{m+1}{x} = \frac{m+1}{x} (\frac{1}{m} + 1) = \frac{m+1}{x} (\frac{1+m}{m}) = \frac{(m+1)^2}{mx}$.
Therefore, $f = \frac{mx}{(m+1)^2}$.

(C) According to the lens maker's formula,the focal length $f$ of a lens is given by:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
where $\mu$ is the refractive index of the lens material.
From this,we can see that $\frac{1}{f} \propto (\mu - 1)$,which implies $f \propto \frac{1}{\mu - 1}$.
According to Cauchy's dispersion formula,the refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ of light $(\mu \propto \frac{1}{\lambda})$.
As the wavelength $\lambda$ increases,the refractive index $\mu$ decreases.
Since $f \propto \frac{1}{\mu - 1}$,a decrease in $\mu$ leads to an increase in the focal length $f$.
Among the given colors,red light has the maximum wavelength $(\lambda_{red} > \lambda_{yellow} > \lambda_{green} > \lambda_{blue})$.
Therefore,the refractive index $\mu$ is minimum for red light,which results in the maximum focal length for red light.

(C) For an equilateral glass prism,the angle of the prism is $A = 60^{\circ}$.
Given that the angle of incidence $i$ is equal to the angle of emergence $e$,and each is $\frac{3}{4}$ of the angle of the prism:
$i = e = \frac{3}{4} A = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
The formula for the angle of deviation $\delta$ is given by:
$\delta = i + e - A$.
Substituting the values:
$\delta = 45^{\circ} + 45^{\circ} - 60^{\circ} = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Thus,the angle of deviation is $30^{\circ}$.

(D) When a light ray is incident on the transparent surface of a silvered glass slab,it undergoes refraction,reflection from the silvered surface,and then refraction again as it exits the slab.
According to Snell's Law,the angle of incidence $i = 45^{\circ}$ and the angle of refraction $r$ are related by $n_1 \sin(i) = n_2 \sin(r)$. Here,$n_1 = 1$ (air) and $n_2 = 1.5$ (glass).
Thus,$\sin(45^{\circ}) = 1.5 \sin(r)$,which gives $\sin(r) = \frac{1}{\sqrt{2} \times 1.5} = \frac{1}{1.5 \sqrt{2}}$.
Due to the symmetry of the path inside the slab,the ray strikes the silvered surface at angle $r$ and reflects at angle $r$. It then hits the transparent surface at angle $r$ and refracts back into the air at the original angle of incidence $i = 45^{\circ}$.
The incident ray makes an angle of $45^{\circ}$ with the normal. The emergent ray also makes an angle of $45^{\circ}$ with the normal on the other side of the normal.
The total deviation $\delta$ is the angle between the direction of the incident ray and the emergent ray. Since the incident ray is at $45^{\circ}$ to the left of the normal and the emergent ray is at $45^{\circ}$ to the right of the normal,the angle between them is $45^{\circ} + 45^{\circ} = 90^{\circ}$.

(C) The forbidden energy gap $(E_g)$ of a semiconductor depends on temperature.
For Germanium $(Ge)$, the forbidden energy gap at $0 \,K$ is approximately $0.74 \,eV$.
As the temperature increases, the energy gap decreases slightly.
At room temperature $(300 \,K)$, the value is approximately $0.67 \,eV$ to $0.72 \,eV$.
Comparing the given options, $0.74 \,eV$ is the standard value for $Ge$ at $0 \,K$.

(A) The resistivity of a material determines its classification as a conductor,semiconductor,or insulator.
Metals (conductors) have low resistivity,typically in the range of $10^{-2}$ to $10^{-8} \Omega-m$.
Insulators have very high resistivity,typically in the range of $10^{11}$ to $10^{19} \Omega-m$.
Semiconductors have resistivity values that lie between those of conductors and insulators,typically in the range of $10^{-3}$ to $10^6 \Omega-m$ (or $10^{-1}$ to $10^8 \Omega-cm$).
Given the options provided,the range $10^{-3}$ to $10^6 \Omega-cm$ is the standard accepted range for semiconductors.

(A) In the case of Fraunhofer diffraction at a single slit,the light waves from different parts of the wavefront interfere to produce a diffraction pattern on the screen.
This pattern consists of a central bright maximum (the central bright band) located at the center of the screen.
On either side of this central maximum,there are secondary maxima and minima,which appear as alternate bright and dark bands.
The intensity of these secondary maxima decreases rapidly as we move away from the central maximum.

(A) compact disc $(CD)$ has a series of closely spaced tracks or grooves on its surface that act like a diffraction grating.
When white light falls on these grooves,the light undergoes diffraction.
Since the diffraction angle depends on the wavelength of the light,the different colours present in the white light are diffracted at different angles.
This results in the separation of white light into its constituent colours,forming visible coloured bands.

(A) light diverging from a finite point source produces a spherical wavefront that moves in all directions from the point source.
As the wavefront expands, the energy is distributed over a larger surface area $A = 4\pi r^2$.
Since intensity $I$ is defined as power per unit area $(I = P/A)$, the intensity is inversely proportional to the square of the distance from the source $(I \propto 1/r^2)$.
Therefore, the intensity decreases in proportion to the distance squared as the light propagates.

(A) We know that the fringe width $\beta$ is given by the formula: $\beta = \frac{D \lambda}{d}$, where $D$ is the distance between the screen and the source, $d$ is the distance between the slits, and $\lambda$ is the wavelength of light used.
From this relation, we can see that $\beta \propto \lambda$.
We know that the wavelength of red light $(\lambda_{\text{red}} \approx 700 \, nm)$ is approximately twice the wavelength of violet light $(\lambda_{\text{violet}} \approx 400 \, nm$, though the ratio is often approximated as $2$ in physics problems).
Therefore, the ratio of fringe widths is $\frac{\beta_{\text{red}}}{\beta_{\text{violet}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{violet}}} \approx \frac{700}{400} \approx 1.75$, which is approximately $2$.
Thus, $\beta_{\text{red}} \approx 2 \beta_{\text{violet}}$.