26 Aug 2021 Shift 2

Given masses of three liquids are equal, i.e. $m_{x}=m_{y}=m_{z}=m$ Temperature of mixture of liquids x and $y, \text{T}_{1}=16° {\text{C}}$ Temperature of mixture of liquids y and $z, \text{T}_{2}=26° {\text{C}}$ While mixing the liquids x and y, the heat energy lost by liquid y will be equal to the heat energy absorbed by liquid x. Consider the specific heat capacity of liquid x is $s_{x}$, of liquid y is $s_{y}$ and of liquid z is $s_{z}$. Now, for liquid x and y, using principle of calorimetry, Heat gained by liquid x = Heat lost by liquid y $m s_{x}(\text{T}_{1}-10)=m s_{y}(20-\text{T}_{1})$ $\Rightarrow s_{x}(16-10)=s_{y}(20-16)$ $\Rightarrow \frac{s_{x}}{s_{y}}=\frac{2}{3}$ ...(i) Similarly, for liquid y and z, using principle of calorimetry, Heat gained by liquid y = Heat lost by liquid z $m s_{y}(\text{T}_{2}-20)=m s_{z}(30-\text{T}_{2})$ $\Rightarrow s_{y}(26-20)=s_{z}(30-26)$ ...(ii) $\Rightarrow \frac{s_{y}}{s_{z}}=\frac{2}{3}$...(i) Multiply Eq. (i) by Eq. (ii), we get $\frac{s_{x}}{s_{z}}=\frac{4}{9}$ Consider the mixing of liquids x and z, let the temperature of mixture be $\text{T}_{3}$. Using principle of calorimetry for liquids x and z. Heat gained by liquid x = Heat lost by liquid z $m s_{x}(\text{T}_{3}-10)=m s_{z}(30-\text{T}_{3})$ $\Rightarrow s_{x}(\text{T}_{3}-10)=s_{z}(30-\text{T}_{3})$ $\Rightarrow \frac{s_{x}}{s_{z}}=\frac{30-\text{T}_{3}}{\text{T}_{3}-10}$ $\Rightarrow \frac{4}{9}=\frac{30-\text{T}_{3}}{\text{T}_{3}-10}$ $\Rightarrow 4 \text{T}_{3}-40=270-9 \text{T}_{3}$ $\Rightarrow \text{T}_{3}=23.84° {\text{C}}$

Given, initial kinetic energy, $\text{E}_{1}=\text{E}$ Initial de-Broglie wavelength, $\lambda_{1}=\lambda$ Consider the wavelength of particle changes to $75 \%$ of lambda after providing energy \Delta E to the particle. Hence, final wavelength of particle, $\lambda_{2}=0.75 \lambda$ Final energy, $\text{E}_{2}=\text{E}+\Delta\text{E}$ Relationship between de-Broglie wavelength and energy of particleis given as $\lambda={h}/{\sqrt{2 m \text{E}}}$ Here, h is Planck's constant and m is mass of particle which is alsoconstant term. Therefore, we get $\lambda\propto{1}/{\sqrt{\text{E}}}$ Thus, we can write the relationship as $\frac{\lambda_{1}}{\lambda_{2}}={\sqrt{\text{E}_{2}}}/{\sqrt{\text{E}_{1}}}$ $\Rightarrow \frac{\lambda}{0.75 \lambda}={\sqrt{\text{E}+\Delta\text{E}}}/{\sqrt{\text{E}}}$ $\Rightarrow \frac{4}{3}={\sqrt{\text{E}+\Delta\text{E}}}/{\sqrt{\text{E}}}$ Squaring both sides of above equation, we get $\Rightarrow \frac{16}{9}=\frac{\text{E}+\Delta\text{E}}{\text{E}}$ $\Rightarrow 16 \text{E}=9 \text{E}+9 \Delta\text{E}$ $\Delta\text{E}=\frac{7}{9} \text{E}$

Given, radius of horizontal circle, $r=\text{L} /\sqrt{2}$ Figure illustrating the particle of mass m moving in a horizontal circle, while suspended from a ceiling is shown In equilibrium condition at point A, $\text{T} \cos \theta=m g$ ...(i) $\text{T} \sin \theta=\frac{m v^{2}}{r}$ ...(ii) Divide Eq. (ii) by Eq. (i), $\tan \theta=v^{2} / r g$ $\Rightarrow v=\sqrt{r g \tan \theta}$ ...(iii) Now, from figure, we can write $\sin \theta=\frac{\text{L} / \sqrt{2}}{L}=\frac{1}{\sqrt{2}}$ \Rightarrow theta = 45° Substituting the value of theta in Eq. (iii), we get $v=\sqrt{r g \tan 45°}=\sqrt{r g}$ Thus, the value of speed of particle is $v=\sqrt{r g}$.

Given, volume of container, $\text{V}=4.0 \times 10^{-3} {\text{m}}^{3}$ Gas constant, $\text{R}=8.3 {\text{J}} \;{\text{mol}}^{-1} {\text{K}}^{-1}$ Number of moles of hydrogen, $n_{1}=1$ Number of moles of carbon dioxide, $n_{2}=2$ Temperature of mixture, T = 400 K Using Dalton's law to calculate the pressure of mixture as follows $p=p_{1}+p_{2}$ $=\frac{n_{1} \text{R} \text{T}}{\text{V}}+\frac{n_{2} \text{R} \text{T}}{\text{V}}$ $=\frac{\text{R} \text{T}}{\text{V}}(n_{1}+n_{2})$ $=\frac{\text{R} \text{T}}{\text{V}}(1+2)=\frac{3 \text{R} \text{T}}{\text{V}}$ $=\frac{3 \times 8.3 \times 400}{4.0 \times 10^{-3}}=24.9 \times 10^{5} {\text{Pa}}$ Thus, the pressure of mixture of gases is $24.9 \times 10^{5} {\text{Pa}}$. Thus, the pressure of mixture of gases is $24.9 \times 10^{5} {\text{Pa}}$.

Given, angle between A and B, theta = 120° The angle between A and - B, alpha = 180° − theta = 180° − 120° = 60° Now, from diagram the resultant vector of A and −B will be A − B and the angle between A − B and A is denoted by beta. So, angle between A and −B is calculated as, phi = 180°− alpha = 180°− 60° = 120° Now, from parallelogram law of vector addition angle beta can becalculated as follows $\tan \beta=\frac{-\text{B} \sin \phi}{-\text{A}+(-\text{B} \cos \phi)}$ $=\frac{-\text{B} \sin 120°}{-\text{A}-\text{B} \cos 120°}={-\text{B} \frac{\sqrt{3}}{2}}/{-\text{A}+\frac{\text{B}}{2}}$ $=\frac{\sqrt{3} \text{B}}{2 \text{A}\-\text{B}}$ $\beta=\tan ^{-1}(\frac{\sqrt{3} \text{B}}{2 \text{A} - \text{B}})$ Hence, the angle between vector A and (A - B) is $\tan ^{-1}(\frac{\sqrt{3} \text{B}}{2 \text{A} - \text{B}})$.

Equation for light beam is given as $\text{E}=800 \sin (\omegat-\frac{x}{\text{C}})$ Here, c is speed of light. Comparing with standard equation, we get $\text{E}_{0}=800 {\text{Vm}}^{-1}$ As we know, the magnitude of magnetic field is given as $\text{E}_{0}=\text{B}_{0} \text{C}$ $\Rightarrow 800=\text{B}_{0} \times 3 \times 10^{8}$ $\Rightarrow \text{B}_{0}=\frac{800}{3 \times 10^{8}} {\text{T}}$ Maximum magnetic force exerted on electron will be $\text{F}_{\max }={\text{ev}} \text{B}_{0}$ $=(1.6 \times 10^{-19}) \times (3 \times 10^{7}) \times \frac{800}{3 \times 10^{8}}$ (Given, $v=3 \times 10^{7} {\text{ms}}^{-1}$ ) $\text{F}_{\max }=12.8 \times 10^{-18} \;{\text{N}}$ or $1.28 \times 10^{-17}$

Given, charges on two rings are +Q and -Q, respectively as shown below. Potential at centre of ring A due to charge on ring A and charge onring B is given as $\text{V}_{\text{A}}=\frac{k \text{Q}}{a}-{k \text{Q}}/{\sqrt{s^{2}+a^{2}}}$ Similarly, potential at centre of ring B due to charge on ring B and charge on ring A is given as $\text{V}_{\text{B}}=-\frac{k \text{Q}}{a}+{k \text{Q}}/{\sqrt{s^{2}+a^{2}}}$ Potential difference between centre of both the rings is $\Delta \text{V}=\text{V}_{\text{A}}-\text{V}_{\text{B}}$ $=\frac{k \text{Q}}{a}-{k \text{Q}}/{\sqrt{s^{2}+a^{2}}}-(-\frac{k \text{Q}}{a}+{k \text{Q}}/{\sqrt{s^{2}+a^{2}}})$ $=\frac{2 k \text{Q}}{a}-{2 k \text{Q}}/{\sqrt{s^{2}+a^{2}}}$ $=\frac{\text{Q}}{2 \pi\epsilon_{0}}(\frac{1}{a}-{1}/{\sqrt{s^{2}+a^{2}}})$ $(\because k=\frac{1}{4 \pi\epsilon_{0}} {\text{Nm}}^{2} {\text{C}}^{-2})$ Thus, potential difference between centre of both the rings is $\frac{\text{Q}}{2 \pi\epsilon_{0}}[\frac{1}{a}-{1}/{\sqrt{s^{2}+a^{2}}}]$

The given value of resistances are $2 \Omega, 4 \Omega, 6 \Omega$ and $8 \Omega$. The required value of combination is $46 / 3 \Omega$. In order to achieve the above mentioned values of resistance from given resistances, we will connect 2Ω and 4Ω resistance in parallel, then join 6 Ω and 8 Ω resistance in series with the combination. The circuit diagram for connection is shown below. $\therefore \text{R}_{{\text{eq}}}=(2 \| 4)+6+8=\frac{2 \times 4}{2+4}+14=\frac{46}{3} \Omega$ Thus, resistance of 2Ω and 4Ω are in parallel with 6Ω and 8Ω in series combination.

Given, the length of cylinder, L = 80 cm = 0.8 m The radius of cylinder, r = 20 cm = 0.2 m The moment of inertia about $\text{C} \text{D}, \text{I}_{\text{CD}}=2.7 {\text{kg}}-{\text{m}}^{2}$ Now, the moment of inertia of cylinder about its axis AB is $\text{I}_{\text{AB}}=\frac{1}{2} \text{M} r^{2}$ By parallel axis theorem, the moment of inertia of cylinder about CD will be calculated as $\text{I}_{\text{CD}}=\text{I}_{\text{AB}}+\text{M}(\frac{\text{L}}{2})^{2}$ $\Rightarrow 2.7=\frac{\text{M} r^{2}}{2}+{\text{M} \text{L}^{2}}{4}$ $\Rightarrow 2.7=\text{M}[\frac{(0.2)^{2}}{2}+{(0.8)^{2}}{4}]$ $\Rightarrow \text{M}=15 {\text{kg}}$ Density of cylinder is given as $\rho =\frac{\text{M}}{\text{V}}=\frac{\text{M}}{\pir^{2} \text{L}}$ $=\frac{15}{\pi(0.2)^{2}(0.8)}=149.2 {\text{kgm}}^{-3}$ $=1.49 \times 10^{2} {\text{kg}} {\text{m}}^{-3}$ Thus, the density of cylinder is $1.49 \times 10^{2} {\text{kgm}}^{-3}$.

Given, area of dielectrics with dielectric constant $\text{K}_{1}$ and $\text{K}_{2}$, $\text{A}_{1}=\text{A}_{2}=\text{A} / 2$ Thickness of dielectrics with dielectric constant $\text{K}_{1}$ and $\text{K}_{2}$, $d_{1}=d_{2}=d / 2$ Other half of the capacitor is still filled with air, so distance betweenplates in other half of capacitor will be d but the area of plates for airsection will be $\text{A}/2$. Calculating the capacitance of section of capacitor filled with air. $\text{C}_{a}={\epsilon_{0}(\frac{\text{A}}{2})}{d}=\frac{\epsilon_{0} \text{A}}{2 d}$ Now, the capacitance of section of capacitor filled with dielectricconstant $\text{K}_1$ can be calculated as $\text{C}_{1}=\frac{\text{K}_{1} \epsilon_{0} \text{A}_{1}}{d_{1}}={\text{K}_{1} \epsilon_{0}(\frac{\text{A}}{2})}/{\frac{d}{2}}=\frac{\text{K}_{1} \epsilon_{0} \text{A}}{d}$ Similarly, the capacitance of section of capacitor filled withdielectric constant $\text{K}_2$ can be calculated as $\text{C}_{2}=\frac{\text{K}_{2} \epsilon_{0} \text{A}_{2}}{d_{2}}={\text{K}_{2} \epsilon_{0}(\frac{\text{A}}{2})}/{\frac{d}{2}}=\frac{\text{K}_{2} \epsilon_{0} \text{A}}{d}$ Now, the dielectric sections are in series with each other, so equivalent capacitance of combination can be calculated as ${1}/{\text{C}_{{\text{eq}}}}=\frac{1}{\text{C}_{1}}+\frac{1}{\text{C}_{2}}=\frac{d}{\text{K}_{1} \epsilon_{0} \text{A}}+\frac{d}{\text{K}_{2} \epsilon_{0} \text{A}}$ $\Rightarrow \text{C}_{{\text{eq} }}=\frac{\epsilon_{0} \text{A}}{d}[\frac{\text{K}_{1}\text{K}_{2}}{\text{K}_{1}+\text{K}_{2}}]$ Now, $\text{C}_{{\text{eq} }}$ is in parallel with the section of capacitor filled with air. Total capacitance of combination can be calculated as $\text{C}_{\text{T}}=\text{C}_{a}+\text{C}_{e q}=\frac{\epsilon_{0} \text{A}}{2 d}+\frac{\epsilon_{0} \text{A}}{d}[\frac{\text{K}_{1} \text{K}_{2}}{\text{K}_{1}+\text{K}_{2}}]$ $\Rightarrow \text{C}_{\text{T}}=\frac{\epsilon_{0} \text{A}}{d}[\frac{1}{2}+\frac{\text{K}_{1} \text{K}_{2}}{\text{K}_{1}+\text{K}_{2}}]$

The bomb is dropped from a fighter plane flying in horizontaldirection only. Let us assume the velocity of fighter plane, ${v}_{{a}}=v_{x} \hat{i}$. As fighter plane is moving in horizontal direction only. Hence, it has only one component. When fighter plane drops the bomb, the bomb will free fall from the fighter plane and have a horizontal velocity in the direction of plane and vertical velocity due to gravity. Velocity of bomb can be given as, {v}_{b}=v_{x} {{i}}^{\^ }-v_{y} {{j}}^{\^ }. Velocity in vertical direction will be negative as bomb is moving downwards. Now, for the person sitting inside the plane, hence the relative velocity of bomb with respect to plane can be calculated as ${v}_{b a}={v}_{b}-{v}_{a}=v_{x} \hat{i}-v_{y} \hat{j}-v_{x} \hat{i}=-v_{y} \hat{j}$ The velocity of bomb with respect to person sitting in plane is only in vertical direction. Thus, the trajectory of bomb with respect to the person will be a straight line vertically down to the plane.

Given, at time $t=0, \text{N}_{\text{A}}(0)=2 \text{N}_{\text{B}}(0)$ Decay constant is same for both radioactive atoms as lambda. For A → B , ${d \text{N}_{\text{B}}(t)}/{d t}=\lambda\text{N}_{\text{A}}(t)-\lambda\text{N}_{\text{B}}(t)$ Substituting $\text{N}_{\text{A}}(0) {e}^{-\lambdat}$ for $\text{N}_{\text{A}}(t)$ in above expression, we get ${d \text{N}_{\text{B}}(t)}/{d t}=\lambda\text{N}_{\text{A}}(0) e^{-\lambdat}-\lambda\text{N}_{\text{B}}(t)$ $=2 \lambda\text{N}_{\text{B}}(0) {e}^{-\lambdat}-\lambda\text{N}_{\text{B}}(t)$ $\Rightarrow {d \text{N}_{\text{B}}(t)}/{d t}+\lambda\text{N}_{\text{B}}(t)=2 \lambda\text{N}_{\text{B}}(0) e^{-\lambdat}$ Multiplying both sides by ${e}^{\lambdat}$, we get ${e}^{\lambdat}[{d \text{N}_{\text{B}}(t)}/{d t}+\lambda\text{N}_{\text{B}}(t)]=2 \lambda\text{N}_{\text{B}}(0) {e}^{-\lambdat} \times {e}^{\lambdat}$ \Rightarrow \frac{d}{d t}\[\text{N}{\text{B}}(t) e^{\lambdat}]$=2 \lambda\text{N}{\text{B}}(0)$Integrating both sides,$[\text{N}{\text{B}}(t) e^{\lambdat}]=2 \lambda\text{N}{\text{B}}(0) t+\text{C}$...(i) Putting t = 0 in above expression,$[\text{N}{\text{B}}(0) e^{\lambda\times 0}]=2 \lambda\text{N}{\text{B}}(0) \times 0+\text{C}$$\Rightarrow \text{C}=\text{N}{\text{B}}(0)$Putting value of C in Eq. (i)$[\text{N}{\text{B}}(t) e^{\lambdat}]=2 \lambda \text{N}{\text{B}}(0) t+\text{N}{\text{B}}(0)$...(ii)$\Rightarrow \text{N}{\text{B}}(t)=\text{N}{\text{B}}(0)[1+2 \lambdat] {e}^{-\lambdat}$$\Rightarrow {\text{N}{\text{B}}(t)}/{\text{N}{\text{B}}(0)}=[1+2 \lambda t] {e}^{-\lambda t}$As,$\text{N}{\text{B}}(t)=\text{N}{\text{B}}(0)[1+2 \lambdat) {e}^{-\lambda t}$$\Rightarrow \text{N}{\text{B}}(t)=\text{C}[1+2 \lambdat] {e}^{-\lambdat}$...(iii) The maximum value of$\text{N}{\text{B}}(\text{t})$is obtained at${d \text{N}{\text{B}}(t)}/{d t}=0$From Eq. (ii),${d \text{N}{\text{B}}(t)}/{d t}=-\lambda\text{C}[1+2 \lambdat] e^{-\lambdat}+2 \text{C} \lambdae^{-\lambdat}=0$Solving the above expression,$t=\frac{1}{2 \lambda} {s}$Thus, maximum value of function will be at$t=\frac{1}{2 \lambda} {s}$. Hence, graph (c) is correct option.

Given, height of transmitting antenna, $\text{H}_{\text{T}}=50 {\text{m}}$ Height of receiving antenna, $\text{H}_{\text{R}}=80 {\text{m}}$ Radius of earth, $\text{R}=6400 {\text{km}}=6.4 \times 10^{6} {\text{m}}$ For maximum range of communication for LOS, $\text{d}_{\max }=\sqrt{2 \text{H}_{\text{T}} \text{R}}+\sqrt{2 \text{H}_{\text{R}} \text{R}}$ $\Rightarrow \text{d}_{\max }=\sqrt{2 \times 50 \times 6.4 \times 10^{6}} +\sqrt{2 \times 80 \times 6.4 \times 10^{6}}$ $\Rightarrow \text{d}_{\max }=10^{3}[\sqrt{640}+\sqrt{1024}]$ $=57.29 \times 10^{3} {\text{m}}$ $\text{d}_{\max }=57.28 {\text{km}}$ Hence, maximum range of communication for LOS is 57.28 km.

Given, power consumed by refrigerator $\text{P}=35 {\text{W}}=35 {\text{Js}}^{-1}$ ∴ Energy consumed by refrigerator per second, $\text{W}=35 {\text{J}}$ Lower temperature limit, $\text{T}_{\text{L}}=-10° {\text{C}}=263 {\text{K}}$ Higher temperature limit, $\text{T}_{\text{H}}=25° {\text{C}}=298 {\text{K}}$ The coefficient of performance of refrigerator is given as ${\text{COP}}={\text{T}_{\text{L}}}/{\text{T}_{\text{H}}-\text{T}_{\text{L}}}=\frac{\text{Q}}{\text{W}}$ Here, Q is the heat transferred by the refrigerator. Substituting the values, we get $\frac{263}{298-263}=\frac{\text{Q}}{35}$ $\Rightarrow \text{Q}=\frac{263}{35} \times 35=263 {\text{J}}/ {\text{s}}$ Thus, the average heat transfer by refrigerator per second is $263 {\text{Js}}^{-1}$.

Given, power rating of bulb, $\text{P}_{\text{B}}=500 {\text{W}}$ Voltage across bulb, $\text{V}_{\text{B}}=100 {\text{V}}$ Supply voltage, $\text{V}_{\text{S}}=200 {\text{V}}$ If a resistance R is attached in series with the bulb, then the voltage across resistance will be 100 V. Now, current flowing in circuit when bulb delivers power of 500 W is given as $\text{P}_{\text{B}}=\text{V}_{\text{B}} \text{I}$ $\Rightarrow 500=100 \times \text{I}$ $\Rightarrow \text{I}=5 {\text{A}}$ Same amount of current will flow from the resistance as it is connected in series. Using Ohm's law, V = IR ⇒ 100 = 5 × R ⇒ R = 20Ω Thus, the resistance connected in series is 20Ω.

Given, logic circuit can be drawn as shown below. The expression for output is $\text{Y}= \ov {\ov { \text{A}+ {\ov {(\text{A}+\text{B})}} ^{}}^{} + \overline{ \text{B} + {\overline{(\text{A}+\text{B})}}^{}} }$ $= { \ov {\ov { \text{A}+ {\ov {(\text{A}+\text{B})}} ^{}}^{}} . \overline{\overline{ \text{B} + {\overline{(\text{A}+\text{B})}}^{}} ^{}}}$ (Using de-Morgan's theorem, $\overline{x+y}=\ov x.\ov y$ $\text{Y}=(\text{A}+\ov {\text{A}+\text{B}}).(\text{B}+\overline{\text{A}+\text{B}})$ Draw the truth table for the logic circuit is shown below A B A + B $\ov {\text{A} + \text{B}}$ $\text{A} + \overline{\text{A} + \text{B}}$ $\text{B} + \ov {\text{A} + \text{B}}$ Y 0 0 0 1 1 1 1 0 1 1 0 0 1 0 1 0 1 0 1 0 0 1 1 1 0 1 1 1

Magnetic induction Magnetic force of induction can be given as $\text{F}=\text{q} \text{v} \text{B}$ \Rightarrow $\text{B}=\frac{\text{F}}{q v}$ [\text{B}]=\frac{[\text{F}]}{[q][v]}={[\text{M} \text{L} \text{T}^{-2}]\}/{[{\text{AT}}]$[{\text{LT}}^{-1}]$}=[{\text{M}}^{1} {\text{T}}^{-2} {\text{A}}^{-1}]$Magnetic Flux Magnetic flux is the product of magnetic inductionand area. \phi = B \cdot A$[\phi]=[\text{B}][\text{A}]=$[\text{M}^{1} \text{T}^{-2} \text{A}^{-1}]$[\text{L}^{2}]$$=[\text{M} \text{L}^{2} \text{T}^{-2} \text{A}^{-1}]$Magnetic permeability We know that, magnetic field inside the solenoid.$\text{B}=\μ_{0} \text{n} \text{I}$\Rightarrow$\μ_{0}=\frac{\text{B}}{n l}$(where, n is number of turns per unit length)$\Rightarrow [\μ_{0}]=\frac{[\text{B}]}{[n][l]}=\frac{[\text{M} \text{T}^{-2} \text{A}^{-1}]$}{[\text{L}^{-1}]$[\text{A}]}=[\text{M} \text{L} \text{T}^{-2} \text{A}^{-2}]$Magnetisation The magnetic dipole moment acquired per unitvolume is known as magnetisation.${\text{I}}={{\text{M}}}/{{\text{V}}}$$[\text{l}]={[{\text{L}}^{2} {\text{A}}]$}/{[{\text{L}}^{3}]$}=[{\text{L}}^{-1} {\text{A}}]$ Hence, no option is correct.

Given, angular frequency, $\omega=100 {\text{rad}} / {\text{s}}$ Capacitance of capacitor, $\text{C}=100 \\mu {\text{F}}=100 \times 10^{-6} {\text{F}}$ Inductance of inductor coil, L = 0.5 H Resistance in upper branch, $\text{R}_{1}=100 \Omega$ Resistance in lower branch, $\text{R}_{2}=50 \Omega$, In the given circuit consider current in upper branch be $i_{1}$ and current flowing in lower branch be $i_{2}$. The net current flowing in circuit will be I. Impedance of upper branch can be calculated as $\text{Z}_{1}=\sqrt{\text{X}_{\text{C}}^{2}+\text{R}_{1}^{2}}=\sqrt{(\frac{1}{\omega\text{C}})^{2}+\text{R}_{1}^{2}}$ $=\sqrt{(\frac{1}{100 \times 100 \times 10^{-6}})^{2}+100^{2}}$ $=\sqrt{100^{2}+100^{2}}=100 \sqrt{2} \Omega$ Impedance of lower branch can be calculated as $\text{Z}_{2}=\sqrt{\text{X}_{\text{L}}^{2}+\text{R}_{2}^{2}}$ $=\sqrt{(\omega\text{L})^{2}+\text{R}_{2}^{2}}$ $=\sqrt{(100 \times 0.5)^{2}+50^{2}}=\sqrt{50^{2}+50^{2}}$ $=50 \sqrt{2} \Omega$ Current flowing in upper branch, $i_{1}=\frac{\text{V}}{\text{Z}_{1}}=\frac{200}{100 \sqrt{2}}=\sqrt{2} {\text{A}}$ Phase of current in upper branch, $\cos \phi_{1}=\frac{\text{R}_{1}}{\text{Z}_{1}}=\frac{100}{100 \sqrt{2}}=\frac{1}{\sqrt{2}}$ $\Rightarrow \phi_{1}=45°$ Thus, in upper branch, current leads voltage by 45° as capacitor ispresent. Current flowing in lower branch ${i}_{2}$ is $i_{2}=\frac{\text{V}}{\text{Z}_{2}}=\frac{200}{50 \sqrt{2}}=2 \sqrt{2} {\text{A}}$ Phase of current in lower branch is $\cos \phi_{2}=\frac{\text{R}_{2}}{\text{Z}_{2}}=\frac{50}{50 \sqrt{2}}$ $\Rightarrow \phi_{2}=45°$ Thus, in lower branch current lags voltage by 45° as inductor ispresent. Thus, the net current, $\text{I}=\sqrt{i_{1}^{2}+i_{2}^{2}}$ $\text{I}=\sqrt{(\sqrt{2})^{2}+(2 \sqrt{2})^{2}}$ $\text{I}=\sqrt{10}=3.16 {\text{A}}$ Thus, no option in the given question is correct. If $\text{I}=i_{1}+i_{2}$ is taken then $\text{I}=\sqrt{2}+2 \sqrt{2}=4.24 {\text{A}}$ is obtained which is incorrect method of solution.

Increase in length of pendulum is 0.1%. i.e. $\frac{\Delta\text{L}}{\text{L}} \times 100=0.1$ Time period of pendulum is given by $\text{T}=2 \pi\sqrt{\frac{\text{L}}{g}}$ Here, 2 pi and g are constant. $\frac{\Delta\text{T}}{\text{T}} \times 100=\frac{1}{2} \frac{\Delta\text{L}}{\text{L}} \times 100$ $\Rightarrow \frac{\Delta\text{T}}{\text{T}} \times 100=\frac{1}{2} \times 0.1$ $\Delta\text{T}=0.05 \times \frac{\text{T}}{100}$ In one single day, the time in seconds is $\text{T}=24 \times 60 \times 60=86400 {\text{s}}$ $\therefore \Delta\text{T}=0.05 \times \frac{86400}{100}=43.2 {\text{s}}$ Thus, the error in time per day is 43.2 s.

Given, breaking stress of wire, $\sigma=\frac{24}{\pi} \times 10^{2} {\text{Nm}}^{-2}$ Free body diagram of 5kg block is given as where, a is common acceleration. Value of acceleration due to gravity, $g=10 {\text{ms}}^{-2}$ From free body diagram of block of mass 5 kg 5g − T = 5a \Rightarrow 5 × 10 − T = 5a ⇒ 50 − T = 5a ...(i) Free body diagram of 3 kg block is given as From free body diagram of block of mass 3 kg, T − 3g = 3a ⇒ T - 3 × 10 = 3a ⇒ T − 30 = 3a ...(ii) Add Eqs. (i) and (ii), we get 50 − T + T − 30 = 5a + 3a ⇒ 20 = 8a ⇒ $a=2.5 {\text{ms}}^{-2}$ Substituting the value of a in Eq. (i), we get 50 − T = 5 × 2.5 \Rightarrow T = 37.5 N Let us assume the minimum radius of wire is r. The breaking stress is expressed as $\sigma=\frac{\text{T}}{\pir^{2}}$ $\frac{24}{\pi} \times 10^{2}=\frac{37.5}{\pir^{2}}$ \Rightarrow $r^{2}=\frac{37.5}{24 \times 10^{2}}=\frac{1}{64}$ \Rightarrow $r=\frac{1}{8} {\text{m}}=\frac{100}{8} {\text{cm}}=12.5 {\text{cm}}$ Thus, the minimum radius of wire should be 12.5 cm.

Given, equation of wave of first string, $y_{1}=\text{A}_{1} \sin k(x-v t)$ Equation of wave of second string, $y_{2}=\text{A}_{2} \sin k(x-v t+x_{0})$ $\text{A}_{1}=12 {\text{mm}}, \text{A}_{2}=5 {\text{mm}}, x_{0}=3.5 {\text{cm}}$ and $k=6.28 {\text{cm}}^{-1}$ The equation of waves after substitution of values can beexpressed as $y_{1}=12 \sin 6.28(x-v t)$ and $y_{2}=5 \sin 6.28(x-v t+3.5)$ The phase difference between two waves can be calculated as $\Delta\phi=\frac{2 \pi}{\lambda} \Deltax$ From equation of waves, the value of $\Deltax=(x-v t+3.5)-(x-v t)=3.5 {\text{cm}}$ We know that, wave number is k = 2 pi / lambda. Now, the phase difference is \Delta phi = k\Delta x= 6.28 × 3.5 = 2 pi × 3.5 = 7 pi The amplitude of resulting wave can be calculated by using thefollowing relation, $\text{A}_{{\text{net} }}=\sqrt{\text{A}_{1}^{2}+\text{A}_{2}^{2}+2 \text{A}_{1} \text{A}_{2} \cos \Delta\phi}$ Substituting the values in above expression, $\text{A}_{{\text{net} }}=\sqrt{12^{2}+5^{2}+2 \times 12 \times 5 \times \cos 7 \pi}$ $=\sqrt{144+25+120 \times (-1)}$ $\text{A}_{{\text{net} }}=\sqrt{49}=7 {\text{mm}}$ Thus, the amplitude of resultant wave is 7 mm.

Given, intensity of light falling on screen is I. Intensity of light when $\text{P}_{1}$ and $\text{P}_{2}$ placed between screen and source is I/2. As per Malus's law, the intensity of transmitted light when passing through a pair of polariser and analyser having angle phi between their axis is given as $\text{I}=\text{I}_{0} \cos ^{2} \phi$ Initially let us consider the angle between analyser and polariser is 0°. Then, the intensity of transmitted light is I / 2. Thus, intensity of light become I / 2 after transmitting through 1st polariser as shown in the figure. If we rotate the polarise $\text{P}_{2}$ by angle phi, then the resulting intensity of light on screen will be $3 / / 8$. Now, for second condition, apply Malus's law, $\frac{3 \text{l}}{8}=\frac{\text{l}}{2} \cos ^{2} \phi$ $\cos ^{2} \phi=\frac{3}{4}$ \Rightarrow $\cos \phi=\frac{\sqrt{3}}{2}$ phi = 30°. Thus, $\text{P}_{2}$ should be rotated by $30°$ to get intensity of light $31/ / 8$ on screen.

Given, maximum value of accelerating potential, V = 12 kV = 12000 V, Speed achieved by a proton cyclotron, $\text{v} = \text{c} / 6$ where, c is the speed of light $=3 \times 10^{8} {\text{ms}}^{-1}$ Mass of proton, $m_{p}=1.67 \times 10^{-27} {\text{kg}}$ Charge of proton, $q_{p}=1.6 \times 10^{-19} {\text{C}}$ According to the question, assume a proton makes n revolutions in the cyclotron to achieve the speed v. The energy absorbed by proton due to the accelerating potential provided by the radio oscillator $=n \times 2 \times q_{p} \times \text{V}$ This energy absorbed by proton will be equal to the kinetic energy gain by the proton i.e., $2 n q_{p} \text{V}=\frac{m_{p} v^{2}}{2}$ Substituting the given values in above expression, we get $n \times 2 \times 1.6 \times 10^{-19} \times 12000$ $=\frac{1}{2} \times 1.67 \times 10^{-27} \times (\frac{3 \times 10^{8}}{6})^{2}$ $n \times 3.84 \times 10^{-15}=2.0875 \times 10^{-12}$ n = 543.62 As, n is number of revolutions, so it should be an integer. ∴ n = 543

Given, accuracy of acceleration due to gravity = 4% Accuracy of time period = 3% Energy stored in pendulum at any instant is given as $\text{E}=\frac{\text{mgL} \theta^{2}}{2}$ ...(i) Time period of pendulum is given by expression $\text{T}=2 \pi\sqrt{\frac{\text{L}}{g}}$ Rearrange the above expression for L, we get $\text{L}=g(\frac{\text{T}}{2 \pi})^{2}$ ...(ii) Substituting the value of L in Eq. (i), we get $\text{E}=\frac{m g^{2} \theta^{2}}{2}({\text{T}}{2 \pi})^{2}=\frac{m g^{2} \theta^{2} \text{T}^{2}}{8 \pi^{2}}$ Now, the accuracy in measurement of energy can be calculated as $\frac{\Delta\text{E}}{\text{E}} \times 100=2 \frac{\Deltag}{g} \times 100+2 \frac{\Delta\text{T}}{\text{T}} \times 100$ (\because m,theta and pi are constant) \Rightarrow $\frac{\Delta\text{E}}{\text{E}} \times 100=2 \times 4 \%+2 \times 3 \%=14 \%$ Thus, the accuracy to which E is known is 14%.

Given, radius of circular coil, r = 8.0 cm = 0.08 m Number of turns, N = 20 Angular speed of coil, $\omega=50 {\text{rad}} {\text{s}}^{-1}$ Magnetic field, $B=3.0 \times 10^{-2} {\text{T}}$ The maximum emf induced in a coil can be expressed as $\epsilon =\omega\text{NBA}$ Here, A is the area of coil. $\epsilon=\text{NB} \omega(\pir^{2})$ Substituting the given values, we get $\epsilon=20 \times 3.0 \times 10^{-2} \times 50 \times \pi\times (0.08)^{2}$ $=60.3 \times 10^{-2} {\text{V}}$ $≃ 60 \times 10^{-2} {\text{V}}$