27 Aug 2021 Shift 1

A circular disc is placed in xy-plane with its centre at the origin as shown below Consider an elemental ring of thickness dr and radius r. Now, the area of the elemental ring can be given by $\text{dA}=2 \pi\text{rdr}$ The charge stored in this elemental ring, $\text{dq}=\sigma\text{dA}$ Now, the electric field at the point on Z-axis at a distance of Z from origin can be given by $\text{dE}={\text{kdqZ}}/{(\text{r}^{2}+\text{Z}^{2})^{\frac{3}{2}}}$ Substituting the value of dq and dA in above equation, we get $\text{dE}=\frac{\text{kZ} \sigma(2 \pi\text{rdr})}{(\text{r}^{2}+\text{Z}^{2})^{3 / 2}}$ $=\frac{\sigma\text{Z}}{2 \epsilon_{0}}(\frac{\text{rdr}}{(\text{r}^{2}+\text{Z}^{2})^{3 / 2}})$ Calculating the total electric field by integrating the above expression from r = 0 to r = R, we get $\text{E}={\sigma\text{Z}}{2 \epsilon_{0}} \int_{0}^{\text{R}} \frac{\text{rdr}}{(\text{r}^{2}+\text{Z}^{2})^{3 / 2}}$ Put $r^{2}+\text{Z}^{2}=u^{2}$ $\Rightarrow 2 \text{rdr}=2 \text{udu}$ $\Rightarrow \text{rdr}=\text{ud}$ For lower limit, r = 0 ⇒ u = Z Upper limit, r = R $\Rightarrow \text{u}=\sqrt{\text{R}^{2}+\text{Z}^{2}}$ $\text{E}=\frac{\sigma\text{Z}}{2 \epsilon_{0}} \int_{\text{Z}}^{\sqrt{\text{R}^{2}+\text{Z}^{2}}} \frac{\text{udu}}{\text{u}^{3}}$ $\text{E}=\frac{\sigma\text{Z}}{2 \epsilon_{0}} \int_{Z}^{\sqrt{\text{R}^{2}+\text{Z}^{2}}}\frac{d u}{u^{2}}$ $=\frac{\sigma\text{Z}}{2 \epsilon_{0}} \int_{\text{Z}}^{\sqrt{\text{R}^{2}+\text{Z}^{2}}} u^{-2} d u$ =\frac{\sigma\text{Z}}{2 \epsilon_{0}}\[\frac{u^{-2+1}}{(-2+1)}]{\text{Z}}^{\sqrt{\text{Z}^{2}+\text{R}^{2}}}$$=\frac{\sigma\text{Z}}{2 \epsilon{0}}[-\frac{1}{u}]{\text{Z}}^{\sqrt{\text{Z}^{2}+\text{R}^{2}}}$$=\frac{\sigma\text{Z}}{2 \epsilon{0}}[-{1}/{\sqrt{(\text{Z}^{2}+\text{R}^{2})}}+\frac{1}{\text{Z}}]$$=\frac{\sigma\text{Z}}{2 \epsilon_{0}}[\frac{1}{\text{Z}}-{1}/{\sqrt{(\text{Z}^{2}+\text{R}^{2})}}]$$=\frac{\sigma}{2 \epsilon_{0}}[1-{\text{Z}}/{\sqrt{(\text{Z}^{2}+\text{R}^{2})}}]$Thus, the electric field at the point on Z-axis at a distance of Z from origin is$\frac{\sigma}{2 \epsilon_{0}}[1-{\text{Z}}/{\sqrt{(\text{Z}^{2}+\text{R}^{2})}}]$

Initial number of radioactive nuclei, $\text{N}_{0}=10^{10}$ Half-life, $t_{1 / 2}=1 {\min}=60 {\text{s}}$ Time, $t=30 {\text{s}}$ We know that, number of nuclei remaining in radioactive decay after time t is given by $\frac{\text{N}}{\text{N}_{0}}=(\frac{1}{2})^{n}=(\frac{1}{2})^{\frac{\text{t}}{\text{t}_{1 / 2}}}$ Substituting the values in above relation, we get $\frac{\text{N}}{10^{10}}=(\frac{1}{2})^{30 / 60}=(\frac{1}{2})^{1 / 2}=\frac{1}{\sqrt{2}}$ $\Rightarrow \text{N}=\frac{1}{\sqrt{2}} \times 10^{10}=0.707 \times 10^{10}$ $\Rightarrow \text{N} \approx 7 \times 10^{9}$ Thus, the radioactive nuclei remaining after $30 {\text{s}}$ is $7 \times 10^{9}$.

Relative magnetic permeability $(\\mu _{r})$ It is the ratio of permeability of medium to the permeability of free space i.e., $\\mu _{r}=\frac{\\mu _{m}}{\\mu _{0}}$ As, it is a ratio of permeabilities. So, it is unitless and dimensionless quantity. Power factor $(\cos \phi)$ It is cosine of phase difference between alternating current and alternating voltage. Hence, it has only a numerical value, so it is also a unitless and dimensionless quantity. Permeability of free space $(\\mu _{0})$ It is the ratio of magnetising field induction (B) to magnetising field intensity (H) i.e. $\\mu _{0}=\frac{\text{B}}{\text{H}}$ $\Rightarrow [\\mu _{0}]=\frac{[\text{B}]}{[\text{H}]}$ ...(i) Since, $\text{H}=\text{nl}$ where, n = number of turns per unit length and I = current $\therefore [\text{H}]=[\text{n}][\text{I}]$ =\[{\text{L}}^{-1}]$[{\text{A}}]=[{\text{AL}}^{-1}]$We know that, force on a current carrying conductor in magnetic field B is given as F = IBl where, l = length of the conductor, B = magnetic field and I = current.$\therefore \text{B}=\frac{\text{F}}{\text{II}}$$\Rightarrow [\text{B}]=\frac{[\text{F}]}{[\text{I}][\text{I}]}={[{\text{MLT}}^{-2}]$}/{[{\text{A}}][{\text{L}}]}=[{\text{MT}}^{-2} {\text{A}}^{-1}]$\therefore From Eq. (i), we have$[\μ_{0}]={[{\text{MT}}^{-2}{\text{A}}^{-1}]$}/{[{\text{AL}}^{-1}]$}=[{\text{MLT}}^{-2} {\text{A}}^{-2}]$Hence, permeability of free space is not a dimensionless quantity. Quality factor It is the ratio of energy stored to the energy dissipated per cycle. Quality factor$={{ \text{Energy} \text{stored} }}/{{ \text{Energy} \text{dissipated} \text{per} c\text{ycle} }}$ As, it is ratio of energies so, it will be unitless and dimensionless quantity.

Unit of intensity of electric field E is ${\text{Vm}}^{-1}$. Unit of intensity of magnetising field H is ${\text{Am}}^{-1}$. Unit of E / H can be calculated as ${{ \text{Unit} \text{of} } \text{E}}/{{ \text{Unit} \text{of} } \text{H}}={{\text{Vm}}^{-1}}/{{\text{Am}}^{-1}}={{\text{V}}}{{\text{A}}}={\text{ohm}}$. Thus, the unit of E / H will be ohm.

From given figure, we can write all the forces in vector notation as ${\text{OP}}=(20 \sin 30° \hat{i}+20 \cos 30° \hat{j}) {\text{N}}$ $=(10 \hat{i}+17.32 \hat{j}) {\text{N}}$ ${\text{OQ}}=(10 \cos 30° \hat{i}+10 \sin 30° \hat{j}) {\text{N}}$ $=(8.66 \hat{i}+5 \hat{j}) {\text{N}}$ ${\text{OR}}=(20 \cos 45° \hat{i}-20 \sin 45° \hat{j}) {\text{N}}$ $=(14.14 \hat{i}-14.14 \hat{j}) {\text{N}}$ ${\text{OS}}=(-15 \cos 45° \hat{i}-15 \sin 45° \hat{j}) {\text{N}}$ $=(-10.60 \hat{i}-10.60 \hat{j}) {\text{N}}$ ${\text{OT}}=(-15 \sin 60°\hat{i}+15 \cos 60°\hat{j}) {\text{N}}$ $=(-12.99 \hat{i}+7.5 \hat{j}) {\text{N}}$ The resultant of these forces is given as F = OP + OQ + OR + OS + OT ${\text{F}}=[(10 \hat{i}+17.32 \hat{j})+(8.66 \hat{i}+5 \hat{j}) +(14.14 \hat{i}-14.14 \hat{i})+(-10.60 \hat{i}-10.60 \hat{j}) +(-12.99 \hat{i}+7.5 \hat{j})]$ ${\text{F}}=9.21 \hat{i}+5.08 \hat{j} {\text{N}}$ ${\text{F}} \approx 9.25 \hat{i}+5 \hat{j} {\text{N}}$ Thus, the resultant force is ${\text{F}}\approx 9.25 \hat{i}+5 \hat{j} {\text{N}}$

Given, initial load, $\text{M}_{1}=185 {\text{kg}}$ Initial pressure, $\text{p}_{1}=76 {\text{cm}}$ of Hg Initial temperature, $\text{T}_{1}=27° {\text{C}}=(273+27) {\text{K}}=300 {\text{K}}$ Final pressure at height, $\text{p}_{2}=45 {\text{cm}}$ of Hg Final temperature, $\text{T}_{2}=-7° {\text{C}}=[273+(-7)] {\text{K}}=266 {\text{K}}$ Volume of balloon is constant, $\text{V}_{1}=\text{V}_{2}=\text{V}$ We know that, for air, $\text{p}=\rho \text{RT}$ Now, for initial condition, we can write $\rho _{1}=\rho _{1} \text{RT}_{1}$ ...(i) Similarly, for final condition, we can write $\text{P}_{2}=\rho _{2} \text{RT}_{2}$ ...(ii) Dividing Eq. (i) by Eq. (ii), we get $\frac{\text{p}_{1}}{\text{p}_{2}}=\frac{\rho _{1} \text{T}_{1}}{\rho _{2} \text{T}_{2}}$ Substituting the given values in above expression, we get ${76 {\text{cm}} { \text{of} } {\text{Hg}}}/{45 {\text{cm}} { \text{of} } {\text{Hg}}}=\frac{\rho _{1}}{\rho _{2}}({300 {\text{K}}}/{266 {\text{K}}})$ $\Rightarrow \frac{\rho _{1}}{\rho _{2}}=\frac{76 \times 266}{45 \times 300}$ We know that, density of balloon can be written as $\rho \frac{\text{M}}{\text{V}}$ As, volume is constant ρ ∝ M We can write, $\frac{\rho _{1}}{\rho _{2}}=\frac{\text{M}_{1}}{\text{M}_{2}}=\frac{76 \times 266}{45 \times 300}$ $\Rightarrow \frac{\text{M}_{1}}{\text{M}_{2}}=\frac{20216}{13500}$ Substituting the value of $\text{M}_{1}$ in above expression, we get $\text{M}_{2}=185 \times \frac{13500}{20216}=123.54 {\text{kg}}$ Thus, the load carried by balloon will be 123.54 kg.

The given situation is shown in the following ray diagram. In concave mirror, when object is placed beyond C the image isformed between C and focus of mirror. Consider radius of curvature of mirror is R. Object distance, $\text{u}=-(\text{R}+\text{d}_{1})$ Image distance, $\text{v}=-(\text{R}-\text{d}_{2})$ Focal length of mirror, $\text{f}=\frac{-\text{R}}{2}$ Now, from mirror formula, we have $\frac{1}{\text{f}}=\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{\text{u}+\text{v}}{\text{uv}}$ $\Rightarrow \text{uv}=\text{f}(\text{u}+\text{v})$ Substituting the values in above expression, we get $[-(\text{R}+\text{d}_{1})][-(\text{R}-\text{d}_{2})]$ $=-\frac{\text{R}}{2}(\text{R}+\text{d}_{1}+\text{R}-\text{d}_{2})$ $\Rightarrow 2(\text{R}^{2}-\text{d}_{2} \text{R}+\text{d}_{1} \text{R}-\text{d}_{1} \text{d}_{2})$ $=2 \text{R}^{2}+\text{d}_{1} \text{R}-\text{d}_{2} \text{R}$ $\Rightarrow 2 \text{R}^{2}-2 \text{d}_{2} \text{R}+2 \text{d}_{1} \text{R}-2 \text{R}^{2}-\text{d}_{1} \text{R}+\text{d}_{2} \text{R}$ $=2 \text{d}_{1} \text{d}_{2}$ $\Rightarrow \text{d}_{1} \text{R}-\text{d}_{2} \text{R}=2 \text{d}_{1} \text{d}_{2}$ $\Rightarrow \text{R}=\frac{2 \text{d}_{1} \text{d}_{2}}{\text{d}_{1}-\text{d}_{2}}$ Thus, radius of curvature of mirror is $\frac{2 \text{d}_{1} \text{d}_{2}}{\text{d}_{1}-\text{d}_{2}}$.

Length of ${\text{arc}}, \text{I}=4.4 {\text{ly}}=4.4 \times 9.46 \times 10^{15}$ $=4.1624 \times 10^{16} {\text{m}}$ $[\because 1 \text{Iy}=9.46 \times 10^{15} {\text{m}}]$ Angle subtended by arc, theta = 4 s Number of revolution, n = 4 Speed of body, $v=8 {\text{AUs}}^{-1}=1.5 \times 10^{11} \times 8=1.2 \times 10^{12} {\text{ms}}^{-1} [\because 1 {\text{AU}}=1.5 \times 10^{11} {\text{m}}]$ We know that, length of arc subtended by an angle, $\text{I}=\text{R} \theta$ Here, R = radius of arc. Substituting the values in above expression, we get $4.1624 \times 10^{16}=\text{R} \times 4 \times (\frac{1°}{3600}) \times (\frac{\pi}{180°})$ $(\because 1 {\text{s}}=\frac{1}{3600} {\text{degree}})$ $\text{R}=2.146 \times 10^{21} {\text{m}}$ We know that, time = distance/speed $t=\frac{n \times 2 \pi\text{R}}{\text{V}}$ (\because Distance = n × length of circular path ) Substituting the values in above expression, we get $t=\frac{4 \times 2 \pi\times 2.146 \times 10^{21}}{1.2 \times 10^{12}}$ $=4.5 \times 10^{10} {\text{s}}$ Thus, the time required is $4.5 \times 10^{10} {\text{s}}$.

The given circuit diagram is shown below The battery connected in the circuit is a DC source and capacitor does not allow direct current to pass through it. Therefore, there will no current pass through upper branch of circuit. The equivalent resistance of remaining circuit will be $\text{R}_{{\text{eq}}}=\text{R}_{{\in}}+4 \Omega$ Here, $\text{R}_{{\in}}$ is internal resistance of battery. $\text{R}_{{\text{eq}}}=1+5=5 \Omega$ From Ohm's law, we know that $\text{I}={\text{V}}/{\text{R}_{{\text{eq}}}}=\frac{5}{5}=1 {\text{A}}$ Now, voltage drop across branch AB is calculated as $\text{V}_{\text{AB}}=1 \times 4=4 {\text{V}}$ Voltage across upper branch will also be 4V as the connection is parallel. Now, charge flowing from upper branch is given as $\text{Q}=\text{C}_{{\text{eq}}} \text{V}_{\text{AB}}$ ...(i) Calculating the equivalent capacitance, ${1}/{\text{C}_{{\text{eq}}}}=\frac{1}{4}+\frac{1}{2+2}=\frac{1}{4}+\frac{1}{4}$ ${\text{C}_{{\text{eq}}}}=2$ Substituting the value of equivalent capacitance and voltage drop in Eq. (i), we get $\text{Q}=2 \times 4 {\text{V}}=8 \\mu {\text{C}}$ Therefore, the charge stored in 4µF will be 8µC.

Let the length of side of square plate is l. The diagram of square plate having an axis passing from centre as $\text{X}-\text{X}'$ and an axis passing through one of the corner and perpendicular to plane of square plate $\text{Z}-\text{Z}'$ is shown below. The distance between the centre of square plate and the corner of square plate is $d=\sqrt{(\frac{1}{2})^{2}+(\frac{1}{2})^{2}}=\frac{1}{\sqrt{2}}$ We know that, moment of inertia of square plate about centre, $\text{I}_{\text{X} \text{X}'}=\text{MI}^{2} / 6$ By parallel axis theorem, moment of inertia of square plate about axis $\text{Z}-\text{Z}'$ will be $\text{I}_{\text{ZZ}'}=\text{I}_{\text{XX}'}+\text{M}(\frac{1}{\sqrt{2}})^{2}$ $\text{I}_{\text{ZZ}'}=\frac{\text{MI}^{2}}{6}+\frac{\text{MI}^{2}}{2}$ $=\text{MI}^{2}[\frac{1}{6}+\frac{1}{2}]$ $\text{I}_{\text{ZZ}'}=\frac{2 \text{MI}^{2}}{3}$ Thus, the moment of inertia about an axis passing through corner of square plate and perpendicular to plane of plate is $2 {\text{MI}}^{2} / 3 .$

The common emitter amplifier circuit is shown below An amplifier is a device which is used to amplify the amplitude of input signals and produce an amplified version as output. In CE-configuration base is input terminal, collector is output terminal and emitter is common (grounded). For a proper operation of amplifier, the emitter base junction is forward biased and collector base junction is reversed biased. As amplifier amplifies the input signal, therefore the current gain and voltage gain of circuit should be high. To get higher current gain and voltage gains, the circuit should operate in active region as in active region for CE-circuit, $\text{I}_{\text{C}}=\beta\text{I}_{\text{B}}$ The output current at collector is beta times the input current at base. Thus, the transistor must operate only in active region.

Given, $\text{pT}^{3}= \text{constant}$ ...(i) From ideal gas equation, we have $\text{pV}=\text{nRT}$ $\Rightarrow \text{p}=\frac{\text{nRT}}{\text{V}}$ From Eq. (i), we get $\therefore (\frac{\text{nRT}}{\text{V}}) \text{T}^{3}= \text{constant}$ $\Rightarrow \text{nRT}^{4} \text{V}^{-1}= \text{constant}$ Differentiate above expression, we get \text{nR}\[4 \text{T}^{3} \text{V}^{-1} \text{dT}-\frac{\text{T}^{4}}{\text{V}^{2}} \text{dV}]$=0$$\Rightarrow 4 \text{dT}=\frac{\text{T}}{\text{V}} \text{dV}$...(ii) We know that,$\text{dV}=\gamma\text{VdT}$Here, gamma = coefficient of volume expansion.$\Rightarrow \gamma=\frac{\text{dV}}{\text{VdT}}$...(iii) From Eq. (ii),$\frac{4}{\text{T}}=\frac{\text{dV}}{\text{VdT}}$...(iv) On comparing Eqs. (iii) and (iv), we get$\gamma=\frac{4}{\text{T}}$Thus, the coefficient of volume expansion is$4 /\text{T}$.

As per Einstein's theory of photoelectric effect when a photon strikes the metal surface it ejects an electron. If we increase the intensity of light, the number of photons incident on surface will increase and thus, the number of electrons ejected willincrease proportionally and the photocurrent will also increase. Kinetic energy of ejected electrons is given as $\frac{1}{2} \text{m}_{\text{e}} \text{v}_{\text{e}}^{2}={\text{eV}}_{0}$ Here, $\text{V}_{0}$ is stopping potential. The stopping potential of surface is constant and it will not increase with increase in intensity of light, so the kinetic energy of ejected electrons will not increase. Thus, increasing the intensity of light will increase the number of photons incident and the kinetic energy will remain unchanged.

Given, radius of loop is R and velocity of the bar magnet is v. According to Faraday's law of electromagnetic induction, the emf induced in loop is given by ${\text{e}}=-\frac{\text{Nd} \phi}{\text{dt}}$ ...(i) As initially the bar magnet is moving towards the coil, so the rate of change of magnetic flux linked with the coil will increase. Thus, initially the emf induced in coil will decrease as per Eq. (i) (due to -ve sign). When magnet reaches the middle point in the coil, for that moment the emf induced will be equal to zero because in this situation, change in magnetic flux associated with bar magnet becomes zero. Now, after that when magnet moves forward, the magnetic flux again starts changing but the polarity of magnetic flux change will be reversed. So, the emf will now increase in opposite direction. The graph for change in emf of coil is shown below.

Given, mass of 1 st ion, $m_{1}=4 {\text{amu}}$ Mass of 2 nd ion, $m_{2}=16 {\text{amu}}$ Charge on 1 st ion, $q_{1}=+2 {e}$ Charge on 2 nd ion, $q_{2}=+3 {e}$ Kinetic energy of both ion is same, ${\text{KE}}_{1}={\text{KE}}_{2}$ The radius of path traced by a charged particle in magnetic field is given as $r={\sqrt{2 \text{mKE}}}/{\text{qB}}$ $\Rightarrow r \propto\frac{\sqrt{m}}{q}$ $\Rightarrow \frac{r_{1}}{r_{2}}=\sqrt{\frac{m_{1}}{m_{2}}} \times (\frac{q_{2}}{q_{1}})$ $\Rightarrow \frac{r_{1}}{r_{2}}=\sqrt{\frac{4}{16}} \times (\frac{+3 e}{2 e})$ $=\frac{1}{2} \times \frac{3}{2}=\frac{3}{4}$ $\Rightarrow r_{1}=\frac{3}{4} r_{2}$ $\Rightarrow r_{1} < r_{2}$ Now, the deflection of ion in region with constant magnetic field as shown in figure is denoted by theta. From diagram, $\sin \theta=\frac{d}{r}$ $\theta\propto\frac{1}{r}$ [\because d = constant] As, $r_{1} < r_{2}$ from above expression $\theta'_{1} > \theta_{2}$. Thus, deflection of lighter ion will be more than the deflection of heavier ion.

Given, focal length of first lens, $\text{f}_{1}=+10 {\text{cm}}$ Focal length of second lens, $\text{f}_{2}=-10 {\text{cm}}$ Focal length of third lens, $\text{f}_{3}=+30 {\text{cm}}$ Object distance from first lens, $\text{u}_{1}=-30 {\text{cm}}$ Using lens formula for first lens, we get $\frac{1}{\text{f}_{1}}=\frac{1}{\text{v}_{1}}-\frac{1}{\text{u}_{1}}$ $\Rightarrow \frac{1}{-10}=\frac{1}{\text{v}_{1}}-\frac{1}{-30}$ $\Rightarrow \frac{1}{\text{v}_{1}}=\frac{1}{10}-\frac{1}{30}=\frac{2}{30}$ $\text{v}_{1}=15 {\text{cm}}$ Thus, the image formed by first lens is 15 cm to the right of first lens. The object distance for second lens will be $u_{2}=15 {\text{cm}}-5 {\text{cm}}=10 {\text{cm}}$ Using lens formula for first lens, we get $\frac{1}{\text{f}_{2}}=\frac{1}{v_{2}}-\frac{1}{u_{2}}$ $\Rightarrow \frac{1}{-10}=\frac{1}{v_{2}}-\frac{1}{10}$ $\Rightarrow \frac{1}{v_{2}}=\frac{1}{-10}+\frac{1}{10}=0$ $\Rightarrow v_{2}=\\infty$ Thus, the object distance for third lens, $u_3$ = ∞ . Using lens formula for third lens, we get $\frac{1}{\text{f}_{3}}=\frac{1}{v_{3}}-\frac{1}{u_{3}}$ $\Rightarrow \frac{1}{30}=\frac{1}{v_{3}}-\frac{1}{\\infty }$ $\frac{1}{v_{3}}=\frac{1}{30}$ $v_{3}=30 {\text{cm}}$ The final image will form 30 cm from the third lens. The distance of final image from object will be $30+10+5+30=75 {\text{cm}}$ Thus, the distance of final image formed from object O is 75 cm.

Given, radius of oil drop, $r=2.0 \times 10^{-5} {\text{m}}$ Density of oil drop, $\rho =1.2 \times 10^{3} {\text{kgm}}^{-3}$ Viscosity of liquid, $\eta =1.8 \times 10^{-5} {\text{Nsm}}^{-2}$ To neglect buoyancy, consider the density of air, ρa = 0 Viscous force acting on drop can be given as $F=6 \pi\eta \text{rV}$ ...(i) Here, v is terminal velocity. The terminal velocity of drop can be given as $v=\frac{2 r^{2}(\rho -\rho _{a}) g}{9 \eta }$ $\Rightarrow v=\frac{2 \times (2.0 \times 10^{-5})^{2}(1.2 \times 10^{3}-0) \times 9.8}{9 \times 1.8 \times 10^{-5}}$ $=5.807 \times 10^{-2} {\text{ms}}^{-1}$ Substituting all values in Eq. (i), we get $\text{F}=6 \pi\times 1.8 \times 10^{-5} \times 2.0 \times 10^{-5} \times 5.807 \times 10^{-2}$ $=3.94 \times 10^{-10} {\text{N}}$ $\simeq 3.9 \times 10^{-10} {\text{N}}$

Given, equation of electromagnetic wave, $\text{E}=50 \sin (500 x-10 \times 10^{10} \text{t}) {\text{V}} / {\text{m}}$ The standard equation of electromagnetic wave, $\text{E}=\text{E}_{0} \sin (k x-\omega\text{t}) {\text{V}} / {\text{m}}$ Comparing with standard equation, we get $\text{E}_{0}=50 {\text{V}} / {\text{m}}, \omega=10^{10} {\text{rad}} {\text{s}}^{-1}, k$ $=500 {\text{rad}} {\text{m}}^{-1}$ Speed of wave in medium will be $v=\frac{\omega}{k}$ $=\frac{10 \times 10^{10}}{500}$ $=2 \times 10^{8} {\text{ms}}^{-1}$ $=\frac{2}{3} \times 3 \times 10^{8} {\text{m}} / {\text{s}}$ $=\frac{2}{3} \text{c}$ $[\because c=3 \times 10^{8} {\text{m}} / {\text{s}}= \text{speed} \text{of} \text{light} ]$

Given, number of cells, n = 5 Internal resistance of each cell, r = 1Ω Emf of each cell, e = 5 V When all cells are connected in parallel as shown below. Potential will remain same as, $\text{V}_{\text{P}}=5 {\text{V}}$ Net resistance in parallel combination will be given as $\text{R}_{\text{p}}=\text{R}+[\frac{1}{\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\frac{1}{r}}]=\text{R}+\frac{r}{5}$ $\text{R}_{\text{p}}=\text{R}+\frac{1}{5}$ (\because r = 1 Ω) ...(i) When all cells are connected in series as shown below The net potential will increase as cells are connected in series, $\text{V}_{\text{S}}=5+5+5+5+5=25 {\text{V}}$ The net resistance of circuit will be equivalent of sum of all resistances as all are connected in series. $\text{R}_{\text{s}}=r+r+r+r+r+\text{R}=5 r+\text{R}$ $\text{R}_{\text{s}}=5+\text{R}$ ( \because r = 1Ω) ...(ii) By Ohm's law, current flowing in circuit is given as $\text{I}=\text{V} / \text{R}$ As current in both series and parallel combination is same, ${\text{V}_{\text{p}}}/{\text{R}_{\text{p}}}={\text{V}_{\text{s}}}/{\text{R}_{\text{s}}}$ $\Rightarrow \frac{5}{\text{R}+\frac{1}{5}}=\frac{25}{5+\text{R}}$ [From Eqs. (i) and (ii)] $\Rightarrow 25+5 \text{R}=25 \text{R}+5$ ⇒ R = 1 Ω

It is given that motion is free simple harmonic motion. Consider a spring mass system performing free simple harmonic motion. Displacement of mass is given as $x=x_{0} \sin \omega t$ The potential energy stored in spring for displacement x can be given as ${\text{PE}}=\frac{1}{2} k x^{2}$ Substituting the value of x, we get ${\text{PE}}=\frac{1}{2} k x_{0}^{2} \sin ^{2} \omegat$ Here, k is spring constant and $x_0$ is also constant. $\text{PE}=\text{C} \sin ^{2} \omegat$ $(\therefore \text{C}=\frac{1}{2} k x_{0}^{2})$ Simplify the value of $\sin^2$ omegat, we get $\text{PE}=\text{C}[\frac{1-\cos2\omega\text{t}}{2}]$ The value of function cos2omegat will always lies between −1 and 1. Thus, the value of potential energy will be greater than or equal to zero. 0 ≤ PE This condition is only satisfies by graph (d).

Given, total mass of body is 2M. Potential energy is maximum at $\text{M} / m=x / 1$ The arrangement of masses to form a square is shown in diagram. The gravitational potential energy of a body is given $\text{U}=-\frac{\text{GMm}}{\text{r}}$ For the given system, the potential energy will be $\text{U}_{\text{T}}=-{{\text{Gm}}(\text{M}-\text{m})}/{d}-{{\text{Gm}}(\text{M}-\text{m})}/{d}-{{\text{Gm}}(\text{M}-\text{m})}/{d} -\frac{\text{Gm}(\text{M}-\text{m})}{d}-\frac{\text{Gm}^{2}}{(\sqrt{2} d)^{2}}-\frac{\text{G}(\text{M}-\text{m})^{2}}{(\sqrt{2} d)^{2}}$ $\text{U}_{\text{T}}=-\frac{4 \text{Gm}(\text{M}-\text{m})}{d}-\frac{\text{Gm}^{2}}{\sqrt{2} d}-\frac{\text{G}(\text{M}-\text{m})^{2}}{\sqrt{2} d}$ For maximum potential energy, ${\text{dU}_{\text{T}}}/{\text{dm}}=0$ $-\frac{4 \text{G}}{d}[\text{M}-2 \text{m}]-\frac{\text{G}}{\sqrt{2} d}[2 m]$ $-\frac{\text{G}}{\sqrt{2} d}[2(\text{M}-\text{m}) \times -1]=0$ $\Rightarrow 4 \text{M}-8 \text{m}+\sqrt{2} \text{m}=\sqrt{2}(\text{M}-\text{m})$ $(4-\sqrt{2}) \text{M}=(8-2 \sqrt{2}) \text{m}$ $\frac{\text{M}}{\text{m}}=\frac{2(4-\sqrt{2})}{4-\sqrt{2}}=2$ Comparing it with the given value, we get x = 2 Thus, potential energy will be maximum when x = 2.

Given, equation of alternating current, $i=[\sqrt{42} \sin (\frac{2 \pi}{\text{T}} t)+10] {\text{A}}$ From given equation, we get $i=i_{1}+i_{2}$ where, $i_{1}=\sqrt{42} \sin (\frac{2 \pi}{\text{T}} t) {\text{A}}$ and $i_{2}=10 {\text{A}}$ Now, $i_{1}$ is oscillating current, whereas $i_{2}$ is direct current and its value does not change with time. $(i_{1})_{{\text{rms}}}=\frac{\sqrt{42}}{\sqrt{2}}=\sqrt{21} {\text{A}}$ $(i_{2})_{{\text{rms}}}=10 {\text{A}}$ We know that, $i_{{\text{rms}}}^{2}=(i_{1})_{{\text{rms}}}^{2}+(i_{2})_{{\text{rms}}}^{2}$ Substituting the values, we get $i_{{\text{rms}}}=\sqrt{(\sqrt{21})^{2}+10^{2}}=\sqrt{121}$ $\Rightarrow i_{{\text{rms}}}=11 {\text{A}}$ Thus, RMS value of given equation of current is 11 A.

Length of conducting wire, l = 24 a Resistance of wire is R. Ratio of magnetic moment of coil for equilateral triangle to square is $1: \text{sqrty}$. Side of equilateral triangle is a and also side of square is a. Number of turns in equilateral triangle will be $\text{n}_{\text{T}}=\frac{24 a}{3 a}=8$ [ \because Length of one turn is 3a] Number of turns in square will be $\text{n}_{\text{S}}=\frac{24 a}{4 a}=6$ [ \because Length of one turn is 4a] Magnetic moment of coil is given as M = nIA The ratio of magnetic moment for triangular coil to square coil is given as ${\text{M}_{\text{T}}}/{\text{M}_{\text{S}}}={\text{n}_{\text{T}} \text{IA}_{\text{T}}}/{\text{n}_{\text{S}} \text{IA}_{\text{S}}}$ (\because I is same in both coils) Substituting the values in above expression, we get $\frac{1}{\sqrt{y}}={(8 \times \frac{\sqrt{3}}{4} a^{2})}/{6 \times a^{2}}=\frac{8 \sqrt{3}}{24}=\frac{1}{\sqrt{3}}$ $\Rightarrow \sqrt{y}=\sqrt{3}$ ⇒ y = 3 Hence, the value of y is 3.

In the given circuit diagram shown in question, we can observe that diode $\text{D}_{1}$ and diode $\text{D}_{2}$ are in forward biasing. So, in forward biasing the diodes will offer zero resistance. Hence, given circuit diagram is redrawn as The input voltage will become 0V and thus, the input current will be 0A. The output current will also be zero. Thus, the output voltage should be equal to 5 V.

Given, value of each resistance, R = 10 Ω Emf of battery, e = 20V Internal resistance of battery, r = 10 Ω Current in parallel connection is 20 times current in series combination, $i_{p}=20 i_{s}$. Net resistance in parallel combination will be given as $\text{R}_{\text{p}}=r+[{1}/{\frac{1}{\text{R}}+\frac{1}{\text{R}}+\frac{1}{\text{R}} ....+n}]$ $=r+\frac{\text{R}}{n}$ ${\text{R}}_{\text{p}}=10+\frac{10}{n}$ (\because r = 10 Ω and R = 10 Ω )...(i) In series combination, The net resistance of circuit will be equivalent to sum of all resistances as all are connected in series. $\text{R}_{\text{s}}=[\text{R}+\text{R}+.....+\text{n}]+\text{r}=\text{n} \text{R}+\text{r}$ $\text{R}_{\text{s}}=10 \text{n}+10$ (\because r = 10 Ω and R = 10 Ω)...(ii) By Ohm's law, current flowing in circuit is given as $i=\text{V} / \text{R}$ As, $\text{i}_{\text{p}}=20\text{i}_{\text{s}}$ ${\text{V}_{\text{p}}}/{\text{R}_{\text{p}}}=20 \frac{\text{V}_{s}}{\text{R}_{s}}$ $\Rightarrow \frac{20}{10+\frac{10}{n}}=\frac{20 \times 20}{10 n+10}$ $[\because\text{V}_{\text{p}}=\text{V}_{\text{s}}={\text{e}}=20 {\text{V}}]$ $\Rightarrow \frac{20 n}{10 n+10}=\frac{400}{10 n+10}$ ⇒ 20n = 400 n = 20 Thus, the value of number of resistances n is 20.

Given, velocity of car X, $\text{v}_{\text{x}}=36 {\text{kmh}}^{-1}=36 \times \frac{5}{18} {\text{ms}}^{-1}=10 {\text{ms}}^{-1}$ Velocity of car $\text{Y}, \text{v}_{\text{y}}=72 {\text{kmh}}^{-1}$ $=72 \times \frac{5}{18} {\text{ms}}^{-1}=20 {\text{ms}}^{-1}$ Frequency of sound heard by passenger in car Y emitted by passenger in car X, f′ =1320 Hz Velocity of sound, $v_s$ = 340 m/s. By Doppler's effect, $\text{f}'={\text{f}(\text{v}_{\text{s}} \pm v_{\text{O}})}/{(v_{s} \∓v_{s})}$ Velocity of observer is equal to velocity of car, $\text{v}_{\text{o}}=20 {\text{ms}}^{-1}$ Velocity of source is equal to velocity of car X, $\text{v}_{\text{s}}=10 {\text{ms}}^{-1}$ Substituting the values in above expression, we get $1320=\frac{\text{f}(340+20)}{(340-10)}$ $\Rightarrow 1320=\frac{360 \text{f}}{330}$ $\Rightarrow \text{f}=1320 \times \frac{11}{12}=1210 {\text{Hz}}$ Thus, the actual frequency is 1210 Hz.

Velocity of body is $v=\sqrt{5000+24 x} {\text{m}} / {\text{s}}$ Acceleration of body can be calculated as $a=\frac{d v}{d t}=\frac{d v}{d x} \times \frac{d x}{d t}$ $\Rightarrow a=v \frac{d v}{d x}$ ...(i) $\frac{d v}{d x}=\frac{d}{d x}(\sqrt{5000+24 x})$ $\Rightarrow \frac{d v}{d x}=\frac{1}{2(\sqrt{5000+24 x})} \times 24$ $=\frac{12}{\sqrt{5000+24 x}}$ Substituting the value of $\frac{d v}{d x}$ in Eq. (i), we get $a=\sqrt{5000+24 x} \times \frac{12}{\sqrt{5000+24 x}}$ $=12 {\text{m}} / {\text{s}}^{2}$ Thus, the acceleration of body is $12 {\text{m}} / {\text{s}}^{2}$.

Given, thermal resistance of rod CD, $\text{R}_{\text{CD}}=10.0 {\text{kW}}^{-1}$ Temperature of end $\text{A}, \text{T}_{\text{A}}=200° {\text{C}}$ Temperature of end $\text{B}, \text{T}_{\text{B}}=100° {\text{C}}$ Temperature of end $\text{D}, \text{T}_{\text{D}}=125° {\text{C}}$ As rods are given as identical, then $\text{R}_{\text{CD}}=\text{R}_{\text{CB}}=10 {\text{kW}}^{-1}$ As, C is mid-point of AB, and as we know R ∝ length $\text{R}_{\text{AC}}=\text{R}_{\text{BC}}=5 {\text{kW}}^{-1}$ Now, we know that heat current is given as $\frac{\text{Q}}{t}=\frac{\Delta\text{T}}{\text{R}}$ Here, \Delta T is temperature difference and R is thermal resistance. Now, for point C, consider temperature of point C as T. Consider at point C heat is coming in from A and going out at point B and D. Therefore, we can write, ${\text{T}_{\text{A}}-\text{T}}/{\text{R}_{\text{AC}}}={\text{T}-\text{T}_{\text{D}}}/{\text{R}_{\text{C} \text{D}}}={\text{T}-\text{T}_{\text{B}}}/{\text{R}_{\text{BC}}}$ $\Rightarrow {200-\text{T}}/{\text{R}_{\text{AC}}}={\text{T}-125}/{\text{R}_{\text{CD}}}+{\text{T}-100}/{\text{R}_{\text{BC}}}$ $\Rightarrow \frac{200-\text{T}}{5}=\frac{\text{T}-125}{10}, \frac{\text{T}-100}{5}$ $\Rightarrow 400-2 \text{T}=\text{T}-125+2 \text{T}-200$ $5 \text{T}=725° {\text{C}}$ $\text{T}=145° {\text{C}}$ The heat current in CD is $\text{I}_{\text{CD}}={\text{T}-\text{T}_{\text{D}}}/{\text{R}_{\text{CD}}}$ $=\frac{145-125}{10}=\frac{20}{10}=2 {\text{W}}$ Thus, the value of P is 2.