31 Aug 2021 Shift 2

Let inner and outer radii of hollow cylindrical column are r and R, respectively. Given that, r = 50 cm, R = 100 cm Mass supported on four columns, $\text{M}=50 \times 10^{3} {\text{kg}}$ Mass supported on each column, $m=\frac{\text{M}}{4}$ $\Rightarrow m=\frac{50 \times 10^{3}}{4}=12.5\times 10^{3} {\text{kg}}$ Now, weight, $w=m g=12.5 \times 9.8 \times 10^{3} {\text{N}}=1225 \times 10^{5} {\text{N}}$ Area of cross-section of each column $\text{A}=\pi(\text{R}^{2}-r^{2})$ $=3.14\{(100)^{2}-(50)^{2}\} \times 10^{-4} {\text{m}}^{2}=2.35 {\text{m}}^{2}$ Young's modulus, $\text{Y}=2.0 \times 10^{11} {\text{Pa}}$ By using Hooke's law, Stress = Y × Strain ∴ Compressive strain $={{ \text{Stress} }}/{\text{Y}}=\frac{\text{W}}{\text{AY}}$ Substituting the values, we get Compressive strain $=\frac{1.225 \times 10^{5}}{2.35 \times 2.0 \times 10^{11}}=2.60 \times 10^{-7}$

Given, electric current, I = 1.5 A Side of triangle, $l=9 {\text{cm}}=9 \times 10^{-2} {\text{m}}$ Let OD = OE = OF = r $\text{AD}=\frac{l}{2}$ ${\text{In}} \Delta\text{AOD}, \frac{\text{OD}}{\text{AD}}=\tan 30°$ $\text{OD}=\text{AD} \tan 30°$ $r=\text{OD}=\frac{l}{2} \times \frac{1}{\sqrt{3}}=\frac{l}{2 \sqrt{3}}$ Magnetic field due to AB at O, $\text{B}_{1}=\frac{\mu{0}^\text{I}}{4 \pir}[\sin 60°+\sin 60°]=\frac{\mu{0}^\text{I}}{4 \pir} \times \frac{2 \sqrt{3}}{2}$ $\Rightarrow \text{B}_{1}=\frac{\mu_{0} d \sqrt{3}}{4 \pir}$ (perpendicular inward) Now, total magnetic field at O due to all sides, $\text{B}=3 \text{B}_{1}=\frac{\mu_{0}}{4 \pir} 3 \sqrt{3}$ By substituting the values, we get $\text{B}=\frac{10^{-7} \times 1.5 \times 3 \sqrt{3} \times 2 \sqrt{3}}{9 \times 10^{-2}}=3 \times 10^{-5} {\text{T}}$ (inside the plane of triangle)

Given, mass of each sphere, M= 1.5 kg Radius of each sphere, $\text{R}=50 {\text{cm}}=50 \times 10^{-2} {\text{m}}$ Distance between centre of spheres, l = 5 m By using parallel axis theorem, moment of inertia of system is $\text{I}=2[\frac{2}{5} \text{MR}^{2}+\frac{\text{Ml}^{2}}{4}]$ $\text{I}=2 \text{M}[\frac{2}{5} \text{R}^{2}+\frac{\text{l}^{2}}{4}]$ Substituting the values, we get $\text{I}=2 \times 1.5[\frac{2}{5} \times (50 \times 10^{-2})^{2}+\frac{(5)^{2}}{4}]$ $=19.05 {\text{kg}}-{\text{m}}^{2}$

Given, force vectors ${\text{P}} \perp {\text{Q}}$, i.e. $\theta=90°$ Let resultant of P + Q = x and resultant of P − Q = y $\therefore x^{2}=\text{P}^{2}+\text{Q}^{2}+2 \text{PQ} \cos 90°=\text{P}^{2}+\text{Q}^{2}$ ...(i) $y^{2}=\text{P}^{2}+\text{Q}^{2}-2 \text{PQ} \cos 90°=\text{P}^{2}+\text{Q}^{2}$ ...(ii) When $\theta_1$ is the angle between (P + Q) and (P − Q), then, their resultant is given by $|{x}+{y}|=\sqrt{3(\text{P}^{2}+\text{Q}^{2})}$ $\therefore \sqrt{x^{2}+y^{2}+2 x y \cos \theta_{1}}=\sqrt{3 \text{P}^{2}+3 \text{Q}^{2}}$ ...(iii) Substituting the values of $x^{2}$ and $y^{2}$ in Eq. (iii), we get $\text{P}^{2}+\text{Q}^{2}+\text{P}^{2}+\text{Q}^{2}+2 \sqrt{\text{P}^{2}+\text{Q}^{2}} \sqrt{\text{P}^{2}+\text{Q}^{2}} \cos \theta_{1}=3 \text{P}^{2}+3 \text{Q}^{2}$ $2 \text{P}^{2}+2 \text{Q}^{2}+(2 \text{P}^{2}+2 \text{Q}^{2}) \cos \theta_{1}=3 \text{P}^{2}+3 \text{Q}^{2}$ $\Rightarrow \cos \theta_{1}=\frac{\text{P}^{2}+\text{Q}^{2}}{2(\text{P}^{2}+\text{Q}^{2})}=\frac{1}{2}$ $\therefore \theta_{1}=60°$ When $\theta_2$ is the angle between (P + Q) and (P − Q), then the magnitude of their resultant is given by $|{x}+{y}|=\sqrt{2(\text{P}^{2}+\text{Q}^{2})}$ $x^{2}+y^{2}+2 x y \cos \theta_{2}=2 \text{P}^{2}+2 \text{Q}^{2}$ ...(iv) Substituting the values, we get $\text{P}^{2}+\text{Q}^{2}+\text{P}^{2}+\text{Q}^{2}+2(\text{P}^{2}+\text{Q}^{2}) \cos \theta_{2}$ $=2 \text{P}^{2}+2 \text{Q}^{2}$ cos $\theta_2 = 0$ $\Rightarrow \theta_2 = 90°$ So, $\theta_{1}=60°$ and $\theta_{2}=90°$ and $\theta_{1}<\theta_{2}$ Hence, both statement I and statement II are true.

Given that, energy of free electron, $\text{E}_{1}=2.6 {\text{eV}}$ We know that, the energy of H-atom in its first excited state (n = 2), $\text{E}_{2}=\frac{-13.6}{2^{2}}=-\frac{13.6}{4}{\text{eV}}$ Now, the energy of emitted photons will be the difference ofthese two energies $\text{E}_{1}$ and $\text{E}_{2}$. $\Delta\text{E}=\text{E}_{1}-\text{E}_{2}$ $\therefore \text{hv}=[2.6-(-\frac{13.6}{4})] {\text{eV}}$ $\Rightarrow {\text{V}}=\frac{(10.4+13.6) \times 1.6 \times 10^{-19}}{4 \times 6.6 \times 10^{-34}}=1.45 \times 10^{9} {\text{MHz}}$

Given that, $r_{1}$ and $r_{2}$ be the radii of inner and outer shells. A material of thermal conductivity (K) is filled between region. Consider an elementary sphere of radius r and thickness dr. Now, thermal resistance of elementary sphere $\text{dR}=\frac{\text{dr}}{\text{K} 4 \pir^{2}}$ $[\because inner surface area,$\text{A}=4 \pir^{2}$]$ So, total thermal resistance, $R=\int\text{dR}=\int^{r_2}_{r_1} \frac{d r}{4 \piK r^{2}}=-\frac{1}{4 \pi\text{K}}[\frac{1}{r}]_{r_{1}}^{r_{2}}$ $\text{R}=\frac{r_{2}-r_{1}}{4 \pi\text{K} r_{1} r_{2}}$ Hence, heat current (or the rate of flow of heat) $\frac{\text{dQ}}{d t}=\frac{\Delta\theta}{\text{R}}={(\theta_{2}-\theta_{1})}/{\frac{(r_{2}-r_{1})}{4 \pi\text{K} r_{1} r_{2}}}=\frac{4 \pi\text{K} r_{1} r_{2}(\theta_{2}-\theta_{1})}{r_{2}-r_{1}}$

In figure $\text{A}, \text{V}_{\text{A}}$ and $\text{V}_{\text{B}}$ are two input voltages (either $5 {\text{V}}$ or $0 {\text{V}}$) When $\text{V}_{\text{A}}=0, \text{V}_{\text{B}}=0$ Both $\text{D}_{1}$ and $\text{D}_{2}$ are OFF, so $\text{V}_{0}=0$, When $\text{V}_{\text{A}}=1$ i.e. $5 {\text{V}}, \text{V}_{\text{B}}=0$ i.e. $0 {\text{V}}$ $\text{D}_{1}$ is ON but $\text{D}_{2}$ is OFF So, $\text{V}_{0}=1$ i.e. $5 {\text{V}}$ When $\text{V}_{\text{A}}=0$ i.e. $0 {\text{V}}, \text{V}_{\text{B}}=1$ i.e. $5 {\text{V}}$ $\text{D}_{1}$ is OFF and $\text{D}_{2}$ is ON So, $\text{V}_{0}=1$ i.e. $5 {\text{V}}.$ Similarly, when $\text{V}_{\text{A}}=\text{V}_{\text{B}}=1$ Both $\text{D}_{1}$ and $\text{D}_{2}$ are ON So, $\text{V}_{0}=1$ i.e. $5 {\text{V}}.$ We get the final truth table on the basis of inputs and their corresponding output. A B Y 0 0 0 0 1 1 1 0 1 1 1 1 ∴ The given circuit (A) is an OR gate. In figure B, it is a transistor in common emitter configuration which gives high output at low input and low output at high input. Hence, this circuit behaves as NOT Gate.

At same temperature kinetic energy of proton and electron are same ${\text{KE}}_{p}={\text{KE}}_{e}$ $\Rightarrow \frac{p_{p}^{2}}{2 m_{p}}=\frac{p_{e}^{2}}{2 m_{e}}$ $\frac{p_{p}^{2}}{p_{e}^{2}}=\frac{m_{p}}{m_{e}}$ ...(i) Using uncertainty in determining the position, $\Deltax \approx \frac{h}{\Deltap}$ Ratio,$\therefore \frac{\Deltax_{e}}{\Deltax_{p}}=\frac{\Deltap_{p}}{\Deltap_{e}}=\sqrt{\frac{m_{p}}{m_{e}}}$ [from Eq. (i)]

Given that, a bob of mass m suspended by a thread of length l undergoes SHM with time period, $\text{T}_{1}=\text{T}=2 \pi\sqrt{\frac{l}{g}}$ ...(i) In liquid, effective gravity, $g_{{\text{eff} }}=g(\frac{\sigma-\rho }{\sigma})$ where, ρ = density of liquid and sigma = density of body (bob). Given, $\rho =\frac{\sigma}{4}$ $\Rightarrow \sigma=4 \rho$ Substituting this value, we get $g_{{\text{eff}}}=g(\frac{4 \rho -\rho }{4 \rho })=\frac{3 g}{4}$ Now, the new time period of pendulum in liquid is $\text{T}'=2 \pi\sqrt{{\text{l}'}/{g_{{\text{eff}}}}}$ ...(ii) Here,$l'=l+\frac{l}{3}=\frac{4 \text{I}}{3}$ Substituting the values in Eq. (ii), we get $\text{T}'=2 \pi\sqrt{\frac{4 l}{3} \times \frac{4}{3 g}}$ $=(2 \pi\sqrt{\frac{l}{g}}) \frac{4}{3}$ $\Rightarrow \text{T}'=\frac{4 \text{T}}{3}$ [from Eq. (i)]

Three forces $\text{F}_{1}, \text{F}_{2}$ and $\text{F}_{3}$ are acting on a body and if this body is in equilibrium, then resultant of these three forces must be zero i.e. ${\text{F}}_{ {\text{net} }}={\text{F}}_{1}+{\text{F}}_{2}+{\text{F}}_{3}=0$ $\Rightarrow {\text{F}}_{1}+{\text{F}}_{2}=-{\text{F}}_{3}$ This situation can be shown graphically by three concurrent forces at 120° with each others. or, by three forces in the same order along three sides of a triangle. Hence, both statement I and statement II are true.

When the velocity (v), time (T) and force (F) are chosen as base quantities. Then, mass is given by $m \proptov^{x} \text{T}^{y} \text{F}^{z}$ ...(i) Using dimensional formula of all quantities, [{\text{ML}}^{0} {\text{T}}^{0}]=\[{\text{LT}}^{-1}]^{{x}}[{\text{T}}]^{{y}}$[{\text{MLT}}^{-2}]^{{z}}$$[\text{M}^{1} \text{L}^{0} \text{T}^{0}]=[\text{M}^{\text{Z}} \text{L}^{x+z} \text{T}^{-x+y-2 z}]$Comparing the powers of dimensions on both sides, we get z = 1$x+z=0$and$-x+y-2 z=0$$\Rightarrow x+1=0$$\Rightarrow x=-1$and$-(-1)+y-2(1)=0$$\Rightarrow 1+y-2=0$$\Rightarrow y=1$Substituting these values in Eq. (i), we get$m \propto v^{-1} \text{T}^{1} \text{F}^{1}$$\Rightarrow m=[\text{FTv}^{-1}]$

The equation of magnetic field vector of an electromagnetic field is ${\text{B}}=\frac{\text{B}_{0}}{\sqrt{2}}(\hat{i}+\hat{j}) \cos (k z-\omegat) {\text{T}}$ Electric charges $q_{1}$ at $\text{P}(0,0, \frac{\pi}{k})=4 \pi\text{C}$ $q_{2}$ at $\text{Q}(0,0, \frac{3 \pi}{k})=2 \pi\text{C}$ At $t=0$ and $\text{P}(0,0, \frac{\pi}{k})$, where $z=\frac{\pi}{k}$ Magnetic field vector, ${\text{B}}=\frac{\text{B}_{0}}{\sqrt{2}}(\hat{i}+\hat{j}) \cos (k \frac{\pi}{k}-0) {\text{T}}$ $=\frac{\text{B}_{0}}{\sqrt{2}}(\hat{i}+\hat{j}) \cos \pi{\text{T}}=-\frac{\text{B}_{0}}{\sqrt{2}}(\hat{i}+\hat{j}) {\text{T}}$ and at $t=0$ and $\text{Q}(0,0, \frac{3 \pi}{k})$, where $z=\frac{3 \pi}{k} .$ Magnetic field vector, ${\text{B}}=\frac{\text{B}_{0}}{\sqrt{2}}(\hat{i}+\hat{j}) \cos (k \frac{3 \pi}{k}-0)$ $=-\frac{\text{B}_{0}}{\sqrt{2}}(\hat{i}+\hat{j}) {\text{T}}$ [\because Using cos pi = cos 3pi = − 1] Velocity of charges $q_{1}$ and $q_{2}, {v}=0.5 c \hat{i}$ We know that, force on charge in magnetic field is given by ${\text{F}}=q({v} \times {\text{B}})$ ∴ For $q_{1}, {\text{F}}_{1}=4 \pi[0.5 c \hat{i} \times \{-\frac{\text{B}_{0}}{\sqrt{2}}(\hat{i}+\hat{j})\}] {\text{N}}$ ${\text{F}}_{1}=-\frac{2 \pic \text{B}_{0}}{\sqrt{2}} \hat{k} {\text{N}}$ Similarly, for $q_{2}$ $\Rightarrow {\text{F}}_{2}=2 \pi[0.5 c \hat{i} \times \{-\frac{\text{B}_{0}}{\sqrt{2}}(\hat{i}+\hat{j})\}] {\text{N}}$ $\Rightarrow {\text{F}}_{2}=-\frac{\pic \text{B}_{0}}{\sqrt{2}} \hat{k} {\text{N}}$ Hence, ratio of magnitudes of ${\text{F}}_{1}$ to ${\text{F}}_{2}$ $\frac{\text{F}_{1}}{\text{F}_{2}}={\frac{2 \pic \text{B}_{0}}{\sqrt{2}}}/{\frac{\pic \text{B}_{0}}{\sqrt{2}}}=\frac{2}{1}$ $\Rightarrow \text{F}_{1}: \text{F}_{2}=2: 1$

In the given circuit, 5 Ω resistance is shorted. So, it can be discarded. Now, we get a resolved circuit as shown below In parallel, $\text{R}_{1}=\frac{2 \times 2}{2+2}=1 \Omega$ In series, $\text{R}_{2}=\text{R}_{1}+2=1+2=3 \Omega$ Here, all the three resistances(3Ω) are parallel. ∴ Equivalent resistance across A and B, ${1}/{\text{R}_{{\text{eq}}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$ $\Rightarrow \text{R}_{{\text{eq} }}=1 \Omega$

In the given statements (a), (b) and (c) are correct. Only (d) is wrong. Since, if electric field lines of force (or electric field) is parallel to Gaussian surface, then angle between normal to surface and electric field is theta = 90°. ∴ By equation of electric flux phi = EA cos theta = EA cos 90° = 0 Hence, electric flux will be zero in this situation. Hence, option (c) is the correct.

Given, Volume, $\text{V}=500 {\text{cm}}^{3}=5 \times 10^{2} \times 10^{-6} {\text{m}}^{3}$ $=5 \times 10^{-4} {\text{m}}^{3}$ Pressure, $p=400 {\text{kPa}}=4 \times 10^{5} {\text{Pa}}$ Temperature, T = 300 K Total mass, m = 0.76 g We know that, Molar mass of hydrogen, $\text{M}_1$= 2 g Molar mass of oxygen, $\text{M}_2$ = 32 g Let $n_1$ and $n_2$ be the number of moles of hydrogen and oxygen, respectively. Total number of moles of mixture, $n=n_{1}+n_{2}$ By using ideal gas equation, pV = nRT Substituting the values, we get $4 \times 10^{5} \times 5 \times 10^{-4}=n \times 8.314 \times 300$ n= 0.08 mol Then,$n_{1}+n_{2}=0.08$ Let mass of hydrogen in mixture = $m_1$ and mass of oxygen in oxygen = $m_2$ $n_{1} \text{M}_{1}+n_{2} \text{M}_{2}=m$ $[\because n=\frac{m}{\text{M}} \Rightarrow m=n \text{M}]$ Substituting the values, we get $2 n_{1}+32 n_{2}=0.76$ $n_{1}+16 n_{2}=0.38$ ...(ii) Solving Eqs. (i) and (ii), we get $n_{1}=0.06$ and $n_{2}=0.02$ ∴ Ratio of masses, $\frac{m_{2}}{m_{1}}=\frac{n_{2} \text{M}_{2}}{n_{1} \text{M}_{1}}={0.02 \times 32}{0.06 \times 2}=\frac{16}{3}$ i.e. $m_{2}: m_{1}=16: 3$

Let a block of mass M splits into two masses $m_1$ and $m_2$ in ratio 1: 2. Then, mass of smaller part, $m_{1}=\frac{\text{M}}{1+2}=\frac{\text{M}}{3}$ Mass of bigger part, $m_{2}=2 \text{M} / 3$ Given, speed of mass M, v = 40 m/s Speed of mass $m_{1}, v_{1}=60 {m} / {s}$ Let speed of mass $m_{2}, v_{2}=v$ Using conservation of linear momentum, ${p}_{i}={p}_{\text{f}}$ $\text{M} {v}=m_{1} {v}_{1}+m_{2} {v}_{2}$ $\text{M} \times 40=\frac{\text{M}}{3} \times 60+\frac{2 \text{M}}{3} \times v$ \Rightarrow = 30 m/s ∴ The fractional change in kinetic energy, $\frac{\Delta\text{K}}{\text{K}}={\text{K}_{\text{f}}-\text{K}_{i}}/{\text{K}_{i}}={\text{K}_{\text{f}}}/{\text{K}_{i}}-1$ $=\frac{\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}}{\frac{1}{2} \text{M} v^{2}}-1$ ={\frac{1}{2}\[\frac{\text{M}}{3} × (60)^{2}+\frac{2 \text{M}}{3}× (30)^{2}]$}/{\frac{1}{2} \text{M} \times (40)^{2}}-1=\frac{9}{8}-1=\frac{1}{8}$

In the given figure, the magnetic field is outward, means N-pole is formed on outward face and S-pole is formed on inward (back) face. As the induced current is anti-clockwise which is also making N-pole on outward face. According to the Lenz's law, the induced current always opposes the nature by which it is produced. Hence, induced current is in the direction, so the strength of N-pole is maintained as it is decreasing.

As we know that, for a body executing SHM the average kinetic energy is equal to average potential energy over a complete time period i.e. ${\text{KE}}_{{\text{av}}}={\text{PE}}_{{\text{av}}}=\frac{1}{4} m \omega^{2} {a}^{2}$ ∴ Total mechanical energy at any time = Sum of kinetic and potential energy $=\frac{1}{2} m \omega^{2} a^{2}=$ constant. where, m = mass of body executing SHM, omega = angular frequency and a = amplitude of SHM. Hence, (c) and (d) are true.

According to statement I and II, full wave rectifier with capacitor and inductor are as shown in figure. In figure (A), the capacitor is connected in parallel to load resistance $(\text{R}_{\text{L}})$, as it can block DC current. ∴ DC current can pass only through $\text{R}_\text{L}$ which gives DC output. In figure (B), as the AC current is time varying current which can produce an opposition in current due to inductor that cancel the effect of AC in the load resistance (\text{R}\text{L}$). Hence, pure DC output is obtained across$\text{R}{\text{L}}$. Hence, statement I and statement II are true.

Given that, radius of Earth $=\text{R}_{\text{E}}$ Height above Earth's surface, h = r Depth below Earth surface, d = r We know that, acceleration due to gravity at height (r) and depth (r) is (for $r < \text{R}_{\text{E}}$) $g_{h}=g({\text{R}_{\text{E}}}/{\text{R}_{\text{E}}+h})^{2}=g({\text{R}_{\text{E}}}/{\text{R}_{\text{E}}+r})^{2}$ ...(i) and$g_{d}=g({\text{R}_{\text{E}}-d}/{\text{R}_{\text{E}}})=g({\text{R}_{\text{E}}-r}/{\text{R}_{\text{E}}})$ ...(ii) $\therefore \frac{g_{d}}{g_{h}}={g({\text{R}_{\text{E}}-r}/{\text{R}_{\text{E}}})}/{g({\text{R}_{\text{E}}}/{\text{R}_{\text{E}}+r})^{2}}$ [from Eqs. (i) and (ii)] $\Rightarrow \frac{g_{d}}{g_{h}}={(\text{R}_{\text{E}}-r)(\text{R}_{\text{E}}+r)^{2}}/{\text{R}^{3}}={\text{R}_{\text{E}}^{3}+\text{R}_{\text{E}}^{2} r-\text{R}_{\text{E}} r^{2}-r^{3}}/{\text{R}_{\text{E}}^{3}}$ $1+{r}/{\text{R}_{\text{E}}}-{r^{2}}/{\text{R}_{\text{E}}^{2}}-{r^{3}}/{\text{R}_{\text{E}}^{3}}$

Given, maximum frequency of audio signal = 6 kHz Available bandwidth = 6 MHz Bandwidth of each station =2 × Maximum frequency of audio signal $=2 \times 6 \times 10^{3} {\text{Hz}}=12 \times 10^{3} {\text{Hz}}$ Number of stations that can be broadcasted within the bandwidth of $6 {\text{MHz}}=\frac{6 \times 10^{6}}{12 \times 10^{3}}=500$ Thus, the correct answer is 500.

Given, capacity of capacitor, C = 200µF EMF of battery = Potential difference across capacitor, V = 200 V Dielectric constant of slab, K = 2 $\text{U}_{1}=\frac{1}{2} \text{CV}^{2}=\frac{1}{2} \times 200 \times (200)^{2} \mu\text{J}=4 \text{J}$ New capacity with dielectric slab, C′ = KC = 200 × 2 = 400µF Since, the battery remains connected. Then, V′ = V = 200 V Now, electrostatic energy in capacitor with dielectric $\text{U}_{2}=\frac{1}{2} \text{C} '\text{V}'=\frac{1}{2} \times 400 \times (200)^{2} \mu\text{J}$ = 8 J Hence, change in the electrostatic energy, $\Delta\text{U}=\text{U}_{2}-\text{U}_{1}=8-4=4 {\text{J}}$ Thus, the correct answer is 4.

Given, number of turns, $n=1000 {\text{m}}^{-1}$ Electric current, i = 0.75 A Relative permeability of 1st material $\mu_{1}=500$ Relative permeability of 2nd material $\mu_{2}=750$ we know that $\mu_{r}=1+\chi \Rightarrow \chi$ $=\mu_{r}-1$ For 1st and 2nd material, $\chi _{1}=\chi _{1}-1=500-1=499$ and $\chi _{2}=\mu_{2}-1=750-1=749$ Now, magnetic intensity, M = ni Let $\text{M}_{1}$ and $\text{M}_{2}$ be the magnetisation and $m_{1}$ and $m_{2}$ be the magnetic moments of material 1st and 2nd. Using, $\frac{\text{I}_{1}}{\text{M}}=\chi _{1}$ ...(i) and $\frac{\text{I}_{2}}{\text{M}}=\chi _{2}$ ...(ii) Dividing Eqs. (i) and (ii), $\frac{\text{I}_{1}}{\text{I}_{2}}=\frac{\chi _{1}}{\chi _{2}}$ ...(iii) We also know that, $\text{Magnetisation}={{ \text{Dipole} \text{moment} \in \text{core} \text{material} }}/{ { \text{Volume} }}$ i.e $\text{I}=\frac{m}{\text{V}}$ $\frac{\text{I}_{1}}{\text{I}_{2}}=\frac{m_{1}}{m_{2}}$ ...(iv) From Eqs. (iii) and (iv), we get $\frac{m_{1}}{m_{2}}=\frac{\chi _{1}}{\chi _{2}}=\frac{499}{749}$ Hence, fraction change in magnetic moment, $\frac{\Deltam}{m_{1}}=\frac{m_{2}-m_{1}}{m_{1}}=(\frac{m_{2}}{m_{1}}-1)$ $=(\frac{749}{499}-1)=\frac{250}{499}$ Thus, value of χ is 250.

$\text{Av}^2$ versus x graph is shown below. Using equation of motion, $v^{2}=u^{2}+2 a x$ ...(i) We know that, the equation of straight line. y = mx + c ...(ii) Comparing Eqs. (i) and (ii), Slope,$m=2 a=\frac{80-20}{30-0}=\frac{60}{30}=2$ Now, the acceleration, $a=\frac{2}{2}=1 {\text{m}} / {\text{s}}^{2}$

Given, slit separation, $d=0.3 {\text{mm}}=0.3 \times 10^{-3} {\text{m}}$ Distance between slit and screen, D = 1.5 m Distance between 4th bright fringes $=2 \text{Y}_{4}=2.4 {\text{cm}}$ Position of 4th fringe, $\text{Y}_{4}=\frac{2.4}{2}=1.2 {\text{cm}}=1.2 \times 10^{-2} {\text{m}}$ Using equation, $\text{Y}_{n}=\frac{n \lambda\text{D}}{d}$ For n = 4 $1.2 \times 10^{-2}=\frac{4 \times \lambda\times 1.5}{0.3 \times 10^{-3}}$ ∴ Wavelength, $\lambda=6 \times 10^{-7} {\text{m}}$ Using c = lambdaf $\Rightarrow \text{f}=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6 \times 10^{-7}}=5 \times 10^{14} {\text{Hz}}$ Hence, the frequency of light is $5 \times 10^{14} {\text{Hz}}$. So, correct answer is 5.

Given, 9 divisions of main scale are equal to 10 divisions of Vernier scale. i.e. 9 MSD = 10 VSD $\Rightarrow {\text{VSD}}=\frac{9}{10} {\text{MSD}}$ ...(i) Size of 1 main scale division, 1 MSD = 1 mm Now, least count, LC = 1MSD − 1VSD ...(ii) Using Eqs. (i) and (ii), we get ${\text{LC}}=1 {\text{MSD}}-\frac{9}{10} {\text{MSD}}=\frac{1}{10} {\text{MSD}}=\frac{1}{10} {\text{mm}}$ While measuring the diameter of bob. Main Scale Reading, MSR = 10 mm Vernier Scale Reading, VSR = 8 Zero error, e = 0.04 cm Now, diameter, $d=[{\text{MSR}}+{\text{LC}} \times {\text{VSR}}]-{e}$ $=(10 {\text{mm}}+\frac{1}{10} \times 8 {\text{mm}})-0.04 {\text{cm}}$ $=(10.8) {\text{mm}}-0.04 {\text{cm}}$ $=1.08 {\text{cm}}-0.04 {\text{cm}}=1.04 {\text{cm}}$ Radius, $r=\frac{d}{2}=\frac{1.04}{2} {\text{cm}}=0.52 {\text{cm}}=52 \times 10^{-2} {\text{cm}}$ ∴ Correct answer is 52.

Given, sample of an ideal gas with (gamma = 1.5) is taken through adiabatic process, where Initial volume, $\text{V}_{1}=1200 {\text{cm}}^{3}=12 \times 10^{-4} {\text{m}}^{3}$ Final volume, $\text{V}_{2}=300 {\text{cm}}^{3}=3 \times 10^{-4} {\text{m}}$ Initial pressure, $p_{1}=200 {\text{kPa}}=2 \times 10^{5} \times {\text{Pa}}$ Let final pressure = $p_2$ Using equation of adiabatic process, $p_{1} \text{V}_{1}^{\gamma}=\rho _{2} \text{V}_{2}^{\gamma}$ $p_{2}=p_{1}(\frac{\text{V}_{1}}{\text{V}_{2}})^{\gamma}=2 \times 10^{5}(\frac{12 \times 10^{-4}}{3 \times 10^{-4}})^{1.5}$ $=16 \times 10^{5} {\text{Pa}}$ Work done in adiabatic process, $\text{W}=\frac{p_{1} \text{V}_{1}-p_{2} \text{V}_{2}}{\gamma-1}$ Substituting the values, we get $\text{W}=\frac{2 \times 10^{5} \times 12 \times 10^{-4}-16 \times 10^{5} \times 3 \times 10^{-4}}{1.5-1}$ $=-480 {\text{J}}$ Hence, work done in adiabatic compression is 480 J.

We know that, at very high frequency capacitive reactance becomes negligible i.e. short circuit (SC) and inductive reactance becomes very large i.e. open circuit (OC). i.e. $\text{X}_{\text{C}} \to 0$ and $\text{X}_{\text{L}} \to \\infty$. Now, the circuit can be rearranged as shown in figure. Hence, equivalent resistance, $\text{R}_{{\text{eq}}}=1+\frac{2 \times 2}{2+2}=2 \Omega$ Thus, correct answer is 2.

Given, the base of a prism is equilateral triangle of side 10 cm as shown in figure. At minimum deviation, Angle of incident = Angle of prism i.e. i = A = 60° Angle of refraction, $r=\frac{\text{A}}{2}={60°}{2}=30°$ Let µ be the refractive index. Then, by Snell's law of refraction, $\mu=\frac{\sin i}{\sin r}=\frac{\sin 60°}{\sin 30°}= {\frac{\sqrt{3}}{2}}/{\frac{1}{2}}=\sqrt{3}$ Now, in \Delta ABP, AP = AB cos 30° $=\frac{10 \sqrt{3}}{2}=5 \sqrt{3} {\text{cm}}$ $=5 \sqrt{3} \times 10^{-2} {\text{m}}$ i.e. optical distance travelled by light along AP, $d=\mu\times \text{AP}=\sqrt{3} \times 5 \sqrt{3} \times 10^{-2}=15 \times 10^{-2} {\text{m}}$ Now, the time taken by light $(c=3 \times 10^{8} {\text{m}} / {\text{s}})$ to travel from P to A, $t=\frac{d}{\text{C}}=\frac{15 \times 10^{-2}}{3 \times 10^{8}} {\text{s}}=5 \times 10^{-10} {\text{s}}$ Thus, correct answer is 5.

Given that, initial current, $\text{I}_{1}=4 {\text{A}}$ Final current, $\text{I}_{2}=2 \text{I}_{1}=8 {\text{A}}$ Initial heat dissipated, $\text{H}_{1}=192 {\text{J}}$ Initial time, $t_{1}=1 {\text{s}}$ Final time, $t_{2}=5 {\text{s}}$ Let final heat dissipated = $\text{H}_2$ By Joule's law of heating, $\text{H} \propto \text{I}^2\text{RT}$ Since resistance remains same at initial and final condition, $\therefore \frac{\text{H}_{2}}{\text{H}_{1}}=\frac{\text{I}_{2}^{2} \text{R} t_{2}}{\text{I}_{1}^{2} \text{R} t_{1}}=\frac{\text{I}_{2}^{2} t_{2}}{\text{I}_{1}^{2} t_{1}}$ Substituting the given values, we get $\frac{\text{H}_{2}}{192}=(\frac{8}{4})^{2} \times \frac{5}{1}$ $\Rightarrow \text{H}_2$ = 3840 J

The order of potential energy of above conformations is The fully eclipsed form is least stable due to repulsion between bulky $(-{\text{CH}}_{3})$ methyl group at front and rear carbon atom. ∴ It has maximum potential energy. While the repulsion in anti form is minimum. ∴ It has minimum potential energy.