26 Aug 2021 Shift 1

The magnetic field intensity at a distance r from centre on the axis of a coil of radius a carrying current i is given by $\text{B}_{{\text{axis} }}=\frac{\\mu _{0}}{4 \pi} \frac{2 \pii a^{2}}{(a^{2}+r^{2})^{3 / 2}}$ ...(i) The magnetic field intensity at the centre of the coil is given by $\text{B}_{{\text{centre} }}=\frac{\\mu _{0}}{4 \pi} \frac{2 \pii}{a}$ ...(ii) According to question, the fractional change in the magnetic fieldintensity is given by $\frac{\Delta\text{B}}{\text{B}}=1-{\text{B}_{{\text{axis} }}}/{\text{B}_{{\text{centre} }}}$ ...(iii) Substituting the values of $\text{B}_{{\text{axis} }}$ and $\text{B}_{{\text{centre} }}$ from Eqs. (i) and (ii) respectively in Eq. (iii), we get $\frac{\Delta\text{B}}{\text{B}}=1-{\frac{\\mu _{0}}{4 \pi} \frac{2 \pii a^{2}}{(a^{2}+r^{2})^{3 / 2}}}/{\frac{\\mu _{0}}{4 \pi} \frac{2 \pii}{a}}$ $=1-\frac{a^{3}}{(a^{2}+r^{2})^{3 / 2}}$ $=1-{a^{3}}/{a^{3}(1+\frac{r^{2}}{a^{2}})^{3 / 2}}$ $=1-(1+\frac{r^{2}}{a^{2}})^{-3 / 2}$ Given that, r < a, applying Binomial expansion, $(1-x)^{-n}=1-x+\frac{2 x^{2}}{2 !}-\frac{3 \times 2 x^{3}}{3 !} ...(-1)^{n} x^{n},$ $=(1+n x)$ we get $\frac{\Delta\text{B}}{\text{B}}=1-(1-\frac{3}{2} \frac{r^{2}}{a^{2}})$ $\frac{\Delta\text{B}}{\text{B}}=1-1+\frac{3}{2} \frac{r^{2}}{a^{2}}$ $\frac{\Delta\text{B}}{\text{B}}=\frac{3}{2} \frac{r^{2}}{a^{2}}$, which is the required expression.

In the given figure, the magnitude of vectors OA, OB and OC areequal. Let it be m. From figure, we can write ${\text{OA}}=m(\cos 30° \hat{i}+\sin 30° \hat{j})$ $=m(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j})$ ${\text{OB}}=m(\cos 60° \hat{i}-\sin 60°\hat{j})$ $=m(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j})$ ${\text{OC}}=m(\cos 45°(-\hat{i})+\sin 45°\hat{j})$ $=m\{\frac{-1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}\}$ According to the question,${\text{OA}}+{\text{OB}}-{\text{OC}}=m(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j})+m(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j})$ $-m(-\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j})$ $=m \hat{i}(\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt{2}})+m \hat{j}(\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}})$ ∴ Angle with X - axis is given by $\tan \theta=\frac{y}{x}={m(\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}})}/{m(\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt{2}})}$ $\Rightarrow \tan \theta={\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}/{\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{\sqrt{2}}{2}}$ $\Rightarrow \tan \theta=\frac{1-\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\sqrt{2}}$ $\Rightarrow \theta=\tan ^{-1}(\frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}})$

According to the question, Velocity of car B w.r.t. mirror of car $\text{A}, v_{\text{BM}}=40 {\text{m}}/ {\text{s}}$ Distance of car B from mirror of car $\text{A}, u=1.9 {\text{m}}=190 {\text{cm}}$ Focal length of mirror, f = 10 cm We have to find the velocity of image of car B in the mirror, i.e. $\text{v}_{\text{IM}}$ We know that, $v_{\text{IM}}=-m^{2} v_{\text{BM}}$ ....(I) where, m is the magnification produced by mirror of car A, $\Rightarrow m=\frac{\text{f}}{\text{f}-u}=\frac{10}{10-(-190)}$ $=\frac{10}{200}=\frac{1}{20}$ Substituting the values in Eq. (i), we get $v_{\text{IM}}=-(\frac{1}{20})^{2} \times 40$ $=-\frac{1}{400} \times 40=-0.1 {\text{m}} / {\text{s}}$ Here, negative sign shows that, the speed of image w.r.t. mirror ofcar A is in opposite direction.

The uniform spherical shell is shown in the figure below Inside the shell, there is no mass, so by applying Gauss's law forgravitation $\∮{g} .{d} {a}=-4 \pi\text{GM}_{{\text{enclosed} }}$ where, g = gravitational field intensity. da = area enclosed G = gravitational constant and $\text{M}_{{\text{enclosed} }}=$ mass enclosed . As, $\text{M}_{{\text{enclosed} }}=0$ So, gravitational field intensity is zero inside the shell, i.e. g = 0 ...(i) We also know that, $g=-\frac{\text{dV}}{\text{dr}}$ where, dV is change in potential due to gravity ⇒ V = constant [using Eq. (i)] So, gravitational field is zero everywhere inside the shell and gravitational potential (V) is constant everywhere.

The given situation can be shown as below. Points A and B are at same horizontal level. So, the pressure at A and B will be equal i.e. $p_{\text{A}}=p_{\text{B}} ....$ (i) If $p_{0}$ is the atmospheric pressure, then from Bernoulli's theorem, $p_{\text{A}}=p_{0}-{2 \text{T}}/{r_{\text{A}}}+(h+y) \rho g$ $p_{\text{B}}=p_{0}-{2 \text{T}}/{r_{\text{B}}}+\rho g y$ Substituting the value of $p_{\text{A}}$ and $p_{\text{B}}$ in Eq. (i), we get $p_{0}-{2 \text{T}}/{r_{\text{A}}}+(h+y) \rho g=p_{0}-{2 \text{T}}/{r_{\text{B}}}+\rho g y$ $\Rightarrow 2 \text{T}({1}/{r_{\text{B}}}-{1}/{r_{\text{A}}})=\rho g y-(h+y) \rho g$ $\Rightarrow 2 \text{T}({1}/{r_{\text{B}}}-{1}/{r_{\text{A}}})=-\rho g h$ or$2 \text{T}({1}/{r_{\text{A}}}-{1}/{r_{\text{B}}})=\text{cg} h$ ...(ii) Here, $r_{\text{A}}=2.5 {\text{mm}}=2.5 \times 10^{-3} {\text{m}}$, and $\text{T}=7.3 \times 10^{-2} {\text{N}} / {\text{m}}$, $r_{\text{B}}=4 {\text{mm}}=4 \times 10^{-3} {\text{m}}, {g}=10 {\text{m}} / {\text{s}}^{2}$ and $\rho =1 \times 10^{3} {\text{kgm}}^{-3}$ Substituting the given value in Eq. (ii), we get $h=\frac{2 \times 7.3 \times 10^{-2}}{1 \times 10^{3} \times 10}(\frac{1}{2.5 \times 10^{-3}}-\frac{1}{4 \times 10^{-3}})$ $h=\frac{2 \times 7.3 \times 10^{-2}}{10^{-3} \times 10^{4}}(\frac{1}{2.5}-\frac{1}{4})$ $h=2.19 \times 10^{-3} {\text{m}}=2.19 {\text{mm}}$

Given, heat supplied to the system, $\frac{\Delta\text{Q}}{\Deltat}=6000 {\text{J}} / {\min}=\frac{6000}{60} {\text{J}} / {\text{s}}=100 {\text{J}} / {\text{s}}$ Power delivered, $\text{P}=\frac{\Delta\text{W}}{\text{t}}=90 {\text{W}}$ Increase in internal energy, $\Delta\text{U}=2.5 \times 10^{3} {\text{J}}$ From first law of thermodynamics, we have $\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$ or $\frac{\Delta\text{Q}}{\Deltat}=\frac{\Delta\text{U}}{\Deltat}+\frac{\Delta\text{W}}{\Deltat}$ ...(i) Substituting the given values in Eq. (i), we get $100=\frac{2.5 \times 10^{3}}{\Deltat}+90$ $\Rightarrow 10=\frac{2.5 \times 10^{3}}{\Deltat}$ $\Rightarrow \Deltat=\frac{2.5 \times 10^{3}}{10}$ ⇒ $\Deltat=2.5 \times 10^{2} {\text{s}}$

Magnetic energy is stored in an inductor, whereas energy isdissipated through resistor. Given, $\text{U}_{m}=64 {\text{J}}$ i = 8 A and P = 640 W As, magnetic energy, $\text{U}_{m}=\frac{1}{2} \text{Li}^{2}$ $\Rightarrow 64=\frac{1}{2} \times \text{L} \times 64$ ⇒ L = 2 H Energy dissipated, $\text{P}=\text{i}^{2} \text{R}$ \Rightarrow 640 64 = R R = 10 Ω Time constant of R-L circuit is given by $\tau =\frac{\text{L}}{\text{R}}$ $\Rightarrow \tau =\frac{2}{10}=\frac{1}{5}=0.2 {\text{s}}$

The circuit diagram can be drawn as, Average power of an L-C-R circuit is given by $\text{P}_{{\text{av}}}=\text{V}_{{\text{rms}}} \text{i}_{{\text{rms}}} \cos \phi$ For $\text{P}_{{\text{av} }}$ to be maximum, $\cos \phi=1$ ...(i) We know that, $\cos \phi=\frac{\text{R}}{\text{Z}}$, where, R = resistance and Z = impedance of L-C-R circuit. Now, $\text{Z}=\sqrt{(\text{X}_{\text{L}}-\text{X}_{\text{C}})^{2}+\text{R}^{2}}$ As, $\cos \phi=1=\frac{\text{R}}{\text{Z}}$ [using Eq, (i)] \Rightarrow R = Z $\Rightarrow \text{R}=\sqrt{(\text{X}_{\text{L}}-\text{X}_{\text{C}})^{2}+\text{R}^{2}}$ Squaring both sides, we get $\text{R}^{2}=(\text{X}_{\text{L}}-\text{X}_{\text{C}})^{2}+\text{R}^{2}$ $\Rightarrow (\text{X}_{\text{L}}-\text{X}_{\text{C}})^{2}=0$ $\Rightarrow \text{X}_{\text{L}}=\text{X}_{\text{C}}$ Since, $\text{X}_{\text{C}}=\frac{1}{\omega\text{C}}=\frac{1}{2 \pi\text{f} \text{C}}$ and $\text{X}_{\text{L}}=250 \pi\Omega$ (given) $\Rightarrow 250 \pi=\frac{1}{2 \pi\times 50 \times \text{C}}$ $\Rightarrow C=\frac{1}{2 \pi^{2} \times 50 \times 250}$ $\Rightarrow \text{C}=4 \times 10^{-6} {\text{F}}=4 \\mu {\text{F}}$

The given circuit can be drawn as The output, $\text{Z}={\ov \text{A}} .{\ov \text{B}}$ $=\ov {\text{A}+\text{B}}$ (Using de-Morgan theorem) This is the expression of a NOR gate. So, the logic operation carriedout by the given circuit is NOR. Alternate Method The given question can be solved using truth table, which is as given below A B X Y Z 0 0 1 1 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 0 This truth table is similar to that of a NOR gate.

The frequency of the emitted radiation by a particular hydrogen like ion when it makes transition from $n_{1}=n_{{\text{initial} }}$ to $n_{2}=n_{{\text{final} }}$ is given by $\text{f}=\text{R} \text{Z}^{2} \text{C}(\frac{1}{n_{2}}-\frac{1}{n_{1}})$ where, c = velocity of light and R = Rydberg's constant. For H-atom, Z = 1 So, for transition from n = 3 to n = 1 $\text{f}_{1}=\text{R}_{\text{C}}(1-\frac{1}{3^{2}})$ ...(i) Similarly for transition from n = 2 to n = 1, $\text{f}_{2}=\text{R}_{C}(1-\frac{1}{2^{2}})$ ...(ii) Dividing Eq. (i) by Eq. (ii), we get $\frac{\text{f}_{1}}{\text{f}_{2}}={{\text{Rc}}(1-\frac{1}{9})}/{{\text{Rc}}(1-\frac{1}{4})}=\frac{\frac{8}{9}}{{3}{4}}=\frac{32}{27}$ Here, $\text{f}_{1}=2.92 \times 10^{15} {\text{Hz}}$ $\Rightarrow \frac{2.92 \times 10^{15}}{\text{f}_{2}}=\frac{32}{27}$ $\Rightarrow \text{f}_{2}=\frac{2.92 \times 10^{15} \times 27}{32}$ $=2.46 \times 10^{15} {\text{Hz}}$

By Einstein's photoelectric equation, we have $\Rightarrow \text{hf}=\phi+({\text{KE}})_{\max }$ where, f is the frequency, phi is the work-function of metal and $({\text{KE}})_{\max }$ is the maximum kinetic energy of emitted photoelectron. $\Rightarrow \frac{\text{hc}}{\lambda}=\phi+{\text{eV}}_{0}$ $\Rightarrow {\text{eV}}_{0}=\frac{\text{hc}}{\lambda}-\phi$ Here, $\lambda=280 {\text{nm}}$ and $\phi=2.5 {\text{eV}}$ $\Rightarrow {e}(\text{V}_{0})_{1}=\frac{1240}{280}-2.5$ $\Rightarrow {\text{e}}(\text{V}_{0})_{1}=1.93 {\text{eV}}$ $\Rightarrow (\text{V}_{0})_{1}=1.93 {\text{V}}$ Similarly, stopping potential for the light of wavelength of 400 nm, ${\text{e}}(\text{V}_{0})_{2}=\frac{1240}{400}-2.5$ $\Rightarrow (\text{V}_{0})_{2}=0.6 {\text{V}}$ ∴ Change in stopping potential, $\Delta\text{V}=(\text{V}_{0})_{1}-(\text{V}_{0})_{2}$ $=1.93-0.6=1.33 ≃ 1.3 {\text{V}}$

The given circuit diagram can be drawn as Redrawing the above circuit diagram as Equivalent resistance of the circuit between point A and B is given as ${1}/{\text{R}_{{\text{eq}}}}=\frac{1}{\text{R}_{1}}+\frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}+\frac{1}{\text{R}_{4}}$ $=\frac{1}{4}+\frac{1}{12}+\frac{1}{6}+\frac{1}{8}$ $=\frac{6+2+4+3}{24}$ $=\frac{15}{24}$ $=\frac{24}{15} \Omega=1.6 \Omega$ $\text{R}_{{\text{total} }}=1.6+0.6=2.2 \Omega$ $\text{P}={\text{V}^{2}}/{\text{R}_{{\text{total} }}}=\frac{2.2 \times 2.2}{2.2}$ = 2.2 W

The given situation is shown below There can be two cases here Case 1 Considering the outer spherical shell as non-conducting.So, electric field for any point with r < R $\text{E}_{{\text{inside} }}=0$ and for $r > \text{R}, \text{E}=\frac{\text{kQ}}{r^{2}}$ In this case, option (b) is correct. Case 2 Considering the outer spherical shell as conducting. Now,charge will be induced on it.$r < \text{R}, \text{E}=0$ $\text{R} \le r < a, \text{E}=\frac{\text{kQ}}{r^{2}}$ $a \le r < b, E=0$ $r \ge b, \text{E}=\frac{\text{kQ}}{r^{2}}$ In this case, option (a) is correct.

The rms speed of the molecule of gas is given by $v_{{\text{rms}}}=\sqrt{\frac{3 \text{RT}}{\text{M}}}$ According to the, question T is constant. Moreover R is also a constant. So, we get $v_{{\text{rms}}} \propto{1}/{\sqrt{\text{M}}}$ $(\text{M}_{{\text{H}}}) \text{M}_{{\text{hydrogen} }}=1.00784 {\text{u}} \approx 1 {\text{u}}$ $(\text{M}_{{\text{O}}}) \text{M}_{{\text{oxygen} }}=15.999 {\text{u}} \approx 16 {\text{u}}$ $(\text{M}_{{\text{CO}}_{2}}) \text{M}_{{\text{carbon} \text{dioxide} }}=44 {\text{u}}$ $(v_{{\text{rms}}})_{{\text{hydrogen} }} > (v_{{\text{rms}}})_{{\text{oxygen} }} > (v_{{\text{rms}}})_{{\text{carbon} \text{dioxide} }}$ $\Rightarrow v_{{\text{H}}} > v_{{\text{O}}} > v_{{\text{CO}}_{2}}$

Least count of a screw gauge is given by Least count $(\text{LC})={{ \text{Pitch} }}/{{ \text{Number} \text{of} \text{circular} \text{scale} \text{divisions} }}$ $\Rightarrow {\text{LC}}=\frac{0.5}{50}$ ...(i) The actual reading (true reading) = Main scale reading + Least count × Circular division − Zero error $=5+(\frac{0.5}{50}) \times 20-\frac{0.5}{50} \times 5$ [∴ Zero error = Least count × Coinciding division] $=5+\frac{0.5}{50}(20-5)$ $=5+\frac{0.5}{50}(15)$ $=5+\frac{15}{100}$ $=\frac{515}{100} {\text{mm}}=5.15 {\text{mm}}$

Let the resistance of iron wire be $\text{R}_{1}$ and that of copper nickel alloy wire be $\text{R}_{2}$ $r_{1}=r_{2}=1 {\text{mm}}=10^{-3} {\text{m}}$ $\rho _{1}=12 \\mu \Omega{\text{cm}}$ $=12 \times 10^{-6} \Omega{\text{cm}}$ $=12 \times 10^{-8} \Omega{\text{m}}$ $\rho _{2}=51 \\mu \Omega{\text{cm}}$ $=51 \times 10^{-6} \Omega{\text{cm}}$ $=51 \times 10^{-8} \Omega{\text{m}}$ For parallel combination, $\text{R}_{{\text{eq}}}=\frac{\text{R}_{1} \text{R}_{2}}{\text{R}_{1}+\text{R}_{2}}$ $3={\frac{\rho _{1} \text{l}}{\pir_{1}^{2}} \frac{\rho _{2} \text{l}}{\pir_{2}^{2}}}/{\frac{\rho _{1} \text{l}}{\pir_{1}^{2}}+\frac{\rho _{2}}{\pir_{2}^{2}}}$ $\Rightarrow 3={\frac{(12 \times 10^{-8}) \text{l}}{\pi\times 10^{-6}}\times \frac{51 \times 10^{-8} \text{l}}{\pi\times 10^{-6}}}/{\frac{12 \times 10^{-8} \text{l}}{\pi\times 10^{-6}}+\frac{51 \times 10^{-8} \text{l}}{\pi\times 10^{-6}}}$ On solving, l = 97 m

Given, $\text{M}=1000 {\text{kg}}, a=20 {\text{m}} / {\text{s}}^{2}$ $v_{{\text{relative} }}=500 {\text{m}} / {\text{s}}, g=10 {\text{m}} / {\text{s}}^{2}$ The given situation is shown below $\text{F}_{{\text{thrust} }}=\frac{d}{d t}(\text{M} v_{{\text{relative} }})$ $\Rightarrow \text{F}_{{\text{thrust} }}=v_{{\text{relative} }}(\frac{d \text{M}}{d t})$ By Newton's second law of motion, $\Rightarrow \text{F}_{{\text{thrust} }}-\text{M} g=\text{M} a$ $\Rightarrow v_{{\text{relative} }}(\frac{d \text{M}}{d t})-\text{M} g=\text{M} a$ $\Rightarrow 500(\frac{d \text{M}}{d t})-1000 \times 10=1000 \times 20$ $\Rightarrow 500(\frac{d \text{M}}{d t})=1000(20+10)$ $\Rightarrow 500(\frac{d \text{M}}{d t})=1000 \times 30$ $\Rightarrow \frac{d \text{M}}{d t}=\frac{1000 \times 30}{500}=60$ $\Rightarrow \frac{d \text{M}}{d t}=60 {\text{kg}} / {\text{s}}$

Dimension of energy, $\text{E}=[{\text{ML}}^{2} {\text{T}}^{-2}]$ Angular momentum, $\text{L}=[{\text{ML}}^{2} {\text{T}}^{-1}]$ Mass, M = [M] and gravitational constant, $\text{G}=[{\text{M}}^{-1} {\text{L}}^{3} {\text{T}}^{-2}]$ The dimension of P in formula, $\text{P}=\text{E} \text{L}^{2} \text{M}^{-5} \text{G}^{-2}$ is given as follows [\text{P}]=\frac{[\text{E}]\[\text{L}^{2}]$}{[\text{M}^{5}]$$[\text{C}^{2}]$}$$={[{\text{ML}}^{2} {\text{T}}^{-2}]$ \times $[{\text{M}}^{2} {\text{L}}^{4} {\text{T}}^{-2}]$}/{[{\text{M}}^{5}]$$[{\text{M}}^{-2} {\text{L}}^{6} {\text{T}}^{-4}]$}$$=[{\text{M}}^{0} {\text{L}}^{0} {\text{T}}^{0}]$

The given situation is shown below Resistivity, ρ = 200 Ωm $\text{C}=2 {\text{pF}}=2 \times 10^{-12} {\text{F}}$ V = 40 V K or $\epsilon_r$ = 50 $\text{i}_{{\text{leakage} }}=?$ $\text{C}=\frac{\text{K} \epsilon_{0} \text{A}}{d}$(K is the dielectric constant or relative permittivity) and $\text{R}=\frac{\rho d}{A}$ Now, charge will be discharged through the resistance between theplates. Now, time constant (τ) of discharging, $\tau =\text{RC}=\frac{\rho d}{\text{A}} \times \frac{\text{K} \epsilon_{0} \text{A}}{d}$ $\tau = \rho \text{K}\epsilon_0$ For a given R-C circuit, the discharged current is given by $\text{i}=\frac{\text{Q}}{\text{R} \text{C}} {\text{e}}^{-\frac{t}{\text{RC}}}$ $\text{i}=\frac{\text{Q}}{\text{p} \text{K} \epsilon_{0}} e^{-\frac{t}{\text{p} \text{K} \epsilon_{0}}}$ The above discharge current is the leakage current, $\text{i}_{{\text{leakage} }}=\frac{\text{Q}}{\rho \text{K} \epsilon_{0}} e^{-\frac{t}{\rho \text{K} \epsilon_{0}}}$ Maximum leakage current, $(\text{i}_{0})_{{\text{leakage} }}=\frac{\text{Q}}{\rho \text{K} \epsilon_{0}}=\frac{\text{C} \text{V}}{\rho \text{K} \epsilon_{0}}$ $=\frac{2 \times 10^{-12} 40}{200 \times 50 \times 8.85 \times 10^{-12}}$ = 903 µA = 0.9 mA ∴ $(\text{i}_{0})_{{\text{leakage} }}=0.9 {\text{mA}}$

We know that, the pentavalent impurities have excess free electrons. So, when a silicon semiconductor is doped with pentavalent impurities, the electron density increases. But on the other hand, the whole semiconductor is electrically neutral, therefore net charge on n-type semiconductor is zero. Hence, statement I true but statement II is false.

The given situation is shown below as Applying law of conservation of energy, we get $\text{K}_{{\text{initial} }}+\text{U}_{{\text{initial} }}=\text{K}_{{\text{final} }}+\text{U}_{{\text{final} }}$ $\text{K}_{{\text{initial} }}=0$ $\Rightarrow \text{U}_{{\text{initial} }}=-1 \times 10 \times \frac{1}{2}$ ( $1 {\text{m}}$ lie below DATUM level, so negative) $\text{K}_{{\text{final} }}=?$ $\text{U}_{{\text{final} }}=-3 \times 10 \times \frac{3}{2}(3 {\text{m}}$ lie below DATUMlevel, so negative) $0-1 \times 10 \times \frac{1}{2}=\text{K}_{ {\text{final} }}-3 \times 10 \times \frac{3}{2}$ $\Rightarrow -5=\text{K}_{{\text{final} }}-45$ $\Rightarrow \text{K}_{{\text{final} }}=45-5$ = 40 J

The electric field in a plane electromagnetic wave is given by {\text{E}}=200 \cos \[\frac{(0.5 × 10^{3}) x}{m}-(1.5 × 10^{11} ;{{\text{rad}}}/{{s}} × t)]$ {{\text{V}}}/{{m}} \hat{j}$∴$\text{E}{0}=200 {\text{V}} / {\text{m}}$Intensity provided by electric field is given by$\text{l}=\frac{1}{2} \epsilon{0} \text{E}{0}^{2} \text{C}$The radiation pressure exerted by electromagnetic wave on surfaceis given by$\text{P}=\frac{2 \text{l}}{c}=\frac{2 \times \frac{1}{2} \epsilon{0} \text{E}{0}^{2} \text{C}}{c}=\epsilon{0} \text{E}{0}^{2}$$\text{P}=\epsilon{0} \text{E}_{0}^{2}$$=8.85 \times 10^{-12} \times (200)^{2}$$=\frac{354}{10^{9}}$According to question,$\frac{x}{10^{9}}=\frac{354}{10^{9}}$ ∴ x = 354

As source and detector move away from each other as shownbelow $\text{table} {←}_{\text{Source}},, {\to } _{\text{Detector}}$ $v_{s}=$ velocity of source with respect to ground = 20 m/s $v_{d}=$ velocity of detector with respect to ground = 20 m/s According to Doppler's effect, $\text{f}_{{\text{apparent} }}={\text{f}(v_{{\text{sound} }}-v_{d})}/{(v_{{\text{sound} }}+v_{s})}$ $\Rightarrow 1800=\frac{\text{f}(340-20)}{(340+20)}$ $1800=\text{f}(\frac{320}{360})$ $\text{f}=\frac{1800 \times 36}{32}$ $=\frac{1800 \times 9}{8}$ $=\frac{16200}{8}$ $\Rightarrow \text{f}=2025 {\text{Hz}}$

Given, $m_{1}=m_{2}=m$ $r_{1}=r_{2}=r=5 {\text{cm}}=5 \times 10^{-2} {\text{m}}$ $u_{1}=u_{2}=35 {\text{m}} / {\text{s}}$ $t_{1}=0$ and $t_{2}=3 {\text{s}}$ When both ball will collide, $(\text{S}_{1})_{{\text{vertical} }}=(\text{S}_{2})_{{\text{vertical} }}$ Let after time t both ball will collide. $(\text{S}_{1})_{{\text{vertical} }}=u_{1} t-\frac{1}{2} g t^{2}$ $(\text{S}_{2})_{{\text{vertical} }}=u_{2}(t-3)-\frac{1}{2} g(t-3)^{2}$ From Eq. (i), we have $\Rightarrow u_{1} t-\frac{1}{2} g t^{2}=u_{2}(t-3)-\frac{1}{2} g(t-3)^{2}$ $\Rightarrow 35 t-\frac{1}{2} \times 10 \times t^{2}$ $=35(t-3)-\frac{1}{2} \times 10(t-3)^{2}$ $\Rightarrow 35 t-35(t-3)=5 t^{2}-5(t-3)^{2}$ $\Rightarrow 35(t-t+3)=5[t^{2}-(t-3)^{2}]$ $35 \times 3=5(t+t-3)(t-t+3)$ $\Rightarrow \frac{105}{5}=(2 t-3)(3)$ $\Rightarrow \frac{21}{3}=2 t-3$ $\Rightarrow 7=2 t-3$ $\Rightarrow 2 t=10$ $\Rightarrow t=5 {\text{s}}$ So, the height at which ball will collide is $(\text{S}_{1})_{{\text{vertical} }}=\text{u}_{1} t-\frac{1}{2} g t^{2}$ $=35 \times 5-\frac{1}{2} \times 10 \times (5)^{2}$ $=175-125$ $=50 {\text{m}}$

The given situation is shown below $r_{\text{A}}=6 {\text{cm}}$ $r_{\text{B}}=3 {\text{cm}}$ Excess pressure inside the soap bubble is given by $(\Delta\text{p})_{{\text{excess} }}={4 \text{S}}/{r_{\text{A}}}+{4 \text{S}}/{r_{\text{B}}}$ According to the given question, ${4 \text{S}}/{r_{\text{A}}}+{4 \text{S}}/{r_{\text{B}}}={4 \text{S}}/{\text{R}_{{\text{eq}}}}$ $\Rightarrow {1}/{r_{\text{A}}}+{1}/{r_{\text{B}}}={1}/{\text{R}_{\text{eq}}}$ $\Rightarrow \frac{1}{6}+\frac{1}{3}={1}/{\text{R}_{\text{eq}}}$ $\Rightarrow \frac{1+2}{6}={1}/{\text{R}_{{\text{eq}}}}$ $\Rightarrow {1}/{\text{R}_{{\text{eq}}}}=\frac{1}{2}$ $\Rightarrow \text{R}_{{\text{eq}}}=2 {\text{cm}}$

The amplitude modulated wave is represented by $\text{C}_{m}(t)=10(1+0.2 \cos 12560 t)$ $\sin (111 \times 10^{4} t) {\text{V}}$ Comparing with standard equation, $\text{C}_{m}(t)=(\text{A}_{\text{C}}+\text{A}_{m} \sin \omega_{m} t) \sin \omega_{\text{C}} t$ $\omega_{m}=12560 {\text{rad}} / {\text{s}}$ $\Rightarrow 2 \pi\text{f}_{m}=12560$ $\text{f}_{m}=\frac{12560}{2 \pi}=\frac{12560 \times 7}{2 \times 22} \simeq 2000 {\text{Hz}}$ ≃ 2 kHz

In the given figure, The direction of magnetic moment is from South to North. When second magnet is kept in the magnetic field produced by firstmagnet, the torque acting on second magnet is given by $\tau ={\text{M}}_{2} \times {\text{B}}_{1}$ $|\tau |=\text{M}_{2} \text{B}_{1} \sin \theta$ (∴ theta = 90°) $|\tau |=\text{M}_{2} \text{B}_{1}$ Point P lie on equitorial position, so $\text{B}_{1}=\frac{\\mu _{0}}{4 \pi} \frac{\text{M}_{1}}{r^{3}}$ $|\tau |=(\frac{\\mu _{0}}{4 \pi} \frac{\text{M}_{1}}{r^{3}}) \text{M}_{2}$ $\text{M}_{1}=\text{M}_{2}=1 {\text{A}}-{\text{m}}^{2}, r=1 {\text{m}}$ (given) $|\tau ||=\frac{10^{-7} \times 1 \times 1}{(1)^{3}}$ $\Rightarrow |\tau |=10^{-7} {\text{Nm}}$ $\Rightarrow |\tau |=1 \times 10^{-7} {\text{Nm}}$

Equation of standing wave is given by $y=1.0 \cos (1.57) x \sin (78.5) t$ Comparing above equation with standard equation of wave given as $y=2 \text{A} \cos k x \sin \omegat$ $k=1.57,2 {\text{A}}=1$ \Rightarrow $\text{A}=0.5$ For node, $\cos k x=0$ $\Rightarrow k x=\frac{\pi}{2}$ $\Rightarrow (1.57) x=\frac{\pi}{2}$ $\Rightarrow x=\frac{\pi}{1.57 \times 2} {\text{cm}}$ $\Rightarrow x=\frac{\pi}{3.14} {\text{cm}}$ $\Rightarrow x=\frac{3.14}{3.14} {\text{cm}}$ $\Rightarrow x=1 {\text{cm}}$