31 Aug 2021 Shift 1

Given, horizontal velocity of helicopter, $u_{x}=v$ Height of the helicopter from ground = h The given condition can be drawn in the figure given below Here, M is the position of man. Let, time taken by food packet to reach ground from helicopter = t Acceleration due to gravity, $g=10 {\text{ms}}^{-2}$ Initial velocity of food packet along Y-axis, $u_{y}=0$ Since, $h=u_{y} t+\frac{1}{2} g t^{2}$ $\therefore t=\sqrt{\frac{2 h}{g}}$ and distance travelled by helicopter in time t along horizontal direction $\Rightarrow x=u_{x} t$ $=v \sqrt{\frac{2 h}{g}}$ Hence, distance of helicopter from man, $\text{H}=\sqrt{h^{2}+x^{2}}=\sqrt{h^{2}+(v \sqrt{\frac{2 h}{g}})^{2}}$ $=\sqrt{h^{2}+\frac{v^{2} 2 h}{g}}$

Given, logic inputs are given as $(\text{A}, \text{B})=(0,0),(0,1),(1,0)$ and $(1,1)$ And the logic circuit is drawn as Output of the given logic circuit is given as $\text{Y}=\ov {\text{AB} .(\text{A}+\text{B})}$ $=\overline{\text{AB}}+(\overline{\text{A}+\text{B}})$ [using de-Morgan's theorem] $=(\overline{\text{A}}+\overline{\text{B}})+\overline{\text{A}} .\overline{\text{B}}$ $=\overline{\text{A}}+\overline{\text{B}}+\overline{\text{A}} \overline{\text{B}}$ Now, by using truth table for above logic circuit A B $\ov \text{A}$ $\ov \text{B}$ $\ov \text{A} \ov \text{B}$ P Q Y 0 0 1 1 1 0 0 1 0 1 1 0 0 0 1 1 1 0 0 1 0 0 1 1 1 1 0 0 0 1 1 0

Charge of particle $\text{A}, \text{q}_{\text{A}}=20 \\mu {\text{C}}$ Charge of particle $\text{B}, \text{q}_{\text{B}}=-5 \\mu \text{C}$ Separation between A and $\text{B}, \text{R}=5 {\text{cm}}=5 \times 10^{-2}$ The given situation is shown below. Here, $\text{q}_{\text{A}}=20 \\mu \text{C}, \text{q}_{\text{B}}=-5 \\mu \text{C}, r=5 {\text{cm}}$ Let, third charge particle of charge $\text{q}_{\text{C}}$ is placed at point C to the right of charge B, so that net electric force would be equal to zero. This is because of following reason. As the net electric force on particle C should be equal to zero, so the forces due to charged particles A and B must be in opposite direction. Hence, the particle should be placed on the line AB. As particles A and B having charges of opposite signs, so charged particle C cannot be between A and B. Also, A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. Now, let position of C is at distance x from right of B. ∴ Force on C due to $\text{B}(\text{F}_{\text{CB}})=$ Force on C due to $\text{A}(\text{F}_{\text{CA}})$ $\Rightarrow {\text{k} \text{q}_{\text{B}} \text{q}_{\text{C}}}/{x^{2}}={\text{k} \text{q}_{\text{A}} \text{q}_{\text{C}}}/{(x+5 / 100)^{2}}$ $\Rightarrow \frac{(x+5 / 100)^{2}}{x^{2}}={\text{q}_{\text{A}}}/{\text{q}_{\text{B}}}$ $\Rightarrow \frac{20}{5}=(\frac{x+5 / 100}{x})^{2}$ $\Rightarrow 2=\frac{x+5 / 100}{x}$ $\Rightarrow 2 x=x+5 / 100$ ⇒ x = 5 cm

Given, initial efficiency of engine, $\eta _{1}=1 / 4$ Let, initial temperature of $\sin k=\text{T}_{2}$ Final temperature of $\sin k=\text{T}_{2}-58° {\text{C}}$ Final efficiency of engine, $\eta _{2}=2 \eta _{1}=2 / 4=1 / 2$ Temperature of source $=\text{T}_{1}$ Since, $\eta =1-\frac{\text{T}_{2}}{\text{T}_{1}}$ $\therefore \eta _{1}=\frac{1}{4}=1-\frac{\text{T}_{2}}{\text{T}_{1}}$ $\Rightarrow \frac{\text{T}_{2}}{\text{T}_{1}}=1-\frac{1}{4}$ $\Rightarrow \text{T}_{2} / \text{T}_{1}=3 / 4$ $\Rightarrow \text{T}_{1}=\frac{4}{3} \text{T}_{2}$ ...(i) and$\eta _{2}=\frac{1}{2}=1-\frac{\text{T}_{2}-58}{\text{T}_{1}}$ $\Rightarrow \frac{\text{T}_{2}-58}{\text{T}_{1}}=1-\frac{1}{2}$ $\Rightarrow \frac{\text{T}_{2}-58}{4 / 3 \text{T}_{2}}=\frac{1}{2}$[From Eq. (i)] $\Rightarrow 2 \text{T}_{2}-116=\frac{4}{3} \text{T}_{2}$ $\Rightarrow \frac{2 \text{T}_{2}}{3}=116$ $\Rightarrow \text{T}_{2}=\frac{116 \times 3}{2}=58 \times 3=174° {\text{C}}$

Given, focal length of concave lens, f' = -f Object distance from lens, u = -f Image distance from lens = v Magnification = m By using lens formula, $\frac{1}{\text{f}'}=\frac{1}{v}-\frac{1}{u}$ $\Rightarrow \frac{1}{v}=\frac{1}{\text{f}'}+\frac{1}{u}$ $=-\frac{1}{\text{f}}-\frac{1}{\text{f}}=-2 / \text{f}$ $\Rightarrow v=-\text{f} / 2$ $\therefore |v|=\frac{\text{f}}{2}$ $\therefore m=\frac{v}{u}=\frac{-\text{f} / 2}{-\text{f}}=\frac{1}{2}$

According to given question, the decay is as shown below $\text{A} { \to }_{\lambda\text{A}}\text{B} {\to }_{\lambda\text{B}}\text{C} (\text{Stable})$ Let $\text{N}_{0}$ be the original number of atoms in the radioactive nucleus A. Initially at t = 0, Number of atoms of B in the sample is zero. According to radioactive decay law, i.e. $\text{N}=\text{N}_{0} {e}^{-\lambdat}$ ...(i) With the increase in time t , the number of atoms of B will start increasing and reaches a maximum value. After reaching the maximum value, B will start decaying into C. Since, number of atoms varies with time following an exponential law i.e. Eq. (i). So, the graph between number of atoms and t will be exponential in nature. Thus, the correct option is (b).

Given, number of turns = N Inner radius = a Outer radius = b Current = I Consider an element of thickness dr at a distance r from the centre.Now, number of turns in dr element, $\text{dN}=\frac{\text{N}}{(b-a)} \text{dr}$ Magnetic field due to this element at the centre, $\text{dB}=\frac{\\mu _{0}}{2} \frac{\text{dN} .\text{I}}{r}$ $\therefore \text{dB}=\frac{1}{2} \frac{\\mu _{0} \text{I}}{r} \frac{\text{N}}{(b-a)} \text{dr}$ ∴ Total magnetic field at the centre, $\text{B}=\int_{a}^{b} \text{dB}=\int_{a}^{b} \frac{\\mu _{0} \text{IN}}{2(b-a)} \frac{dr}{r}$ $=\frac{\text{N}}{2} \frac{\\mu _{0} \text{I}}{(b-a)} \ln r|_{a} ^{b}$ $=\frac{\text{N}}{2} \frac{\\mu _{0} \text{l}}{(b-a)} \ln (\frac{b}{a})$

Given, mass of body 1 = M Mass of body 2 = m Initial speed of body $1, u_{1}=v_{0}$ Initial speed of body $2, u_{1}=0$ Final speed of body 1 and $2=v_{1}$ and $v_{2}$ Angle made by body 1 and 2 after collision with respect to initial direction $=\theta_{1}, \theta_{2}$ By using law of conservation of momentum, Along X-axis $\text{Mv}_{0}+m .0=\text{M} v_{1} \cos \theta_{1}+\text{mv}_{2} \cos \theta_{2}$ If $\theta_{1}=\theta_{2}=\theta$ $\therefore \text{Mv}_{0}=\text{Mv}_{1} \cos \theta+m v_{2} \cos \theta$ ...(i) Along Y-axis, $\text{Mv}_{1} \sin \theta_{1}=m v_{2} \sin \theta_{2}$ [since, $\theta_{1}=\theta_{2}=\theta$] $v_{2}=\frac{\text{Mv}_{1}}{m}$...(ii) Substituting the value of $v_{2}$ in Eq. (i), we get $\text{Mv}_{0}=\text{Mv}_{1} \cos \theta+m(\frac{\text{Mv}_{1}}{m}) \cos \theta$ $=2 \text{Mv}_{1} \cos \theta$ $v_{1}=\frac{v_{0}}{2 \cos \theta}$ ...(iii) By using law of conservation of energy along X-axis $\frac{1}{2} \text{Mv}_{0}^{2}=\frac{1}{2} \text{Mv}_{1}^{2}+\frac{1}{2} mv_{2}^{2}$ $\Rightarrow \text{Mv}_{0}^{2}=\text{Mv}_{1}^{2}+m(\frac{\text{M}}{m} v_{1})^{2}$ [from Eq. (ii)] $\Rightarrow \text{Mv}_{0}^{2}=\text{Mv}_{1}^{2}+\frac{\text{M}^{2} v_{1}^{2}}{m}={\text{Mv}_{1}^{2}}{m}(m+\text{M})$ $\Rightarrow v_{0}^{2}=(\frac{v_{0}}{2 \cos \theta})^{2}(\frac{m+\text{M}}{m})$ [From Eq. (iii)] $\Rightarrow v_{0}^{2}=\frac{v_{0}^{2}}{4 \cos ^{2} \theta}(1+\frac{\text{M}}{m})$ $\Rightarrow 4 \cos ^{2} \theta=1+\frac{\text{M}}{m}$ $\Rightarrow \frac{\text{M}}{m}=4 \cos ^{2} \theta-1$ [ For largest possible value of $\frac{\text{M}}{m}, \theta=0$ ] $=4 \cos ^{2} 0°-1=4-1=3$

Given, masses and radii of Earth and Moon are $(m_{1}, \text{R}_{1}),(m_{2}, \text{R}_{2})$ and separation between their centre of mass = r If v be the minimum escape velocity of particle of mass m, then by using law of conservation of energy, Initial energy = Final energy = 0 $\frac{1}{2} mv^{2}-\frac{\text{G} m_{1} m}{r / 2}-\frac{\text{G} m_{2} m}{r / 2}=0$ $\Rightarrow \frac{1}{2} m v^{2}=\frac{2 \text{G} m}{r}(m_{1}+m_{2})$ $\Rightarrow v=\sqrt{\frac{4 \text{G}}{r}(m_{1}+m_{2})}$

Given, side of one square loop = a Side of 2nd square loop = b $b>>a$ Current = i Since, phi = Mi $\text{M}=\frac{\phi}{i}=\frac{\text{B} .\text{A}}{i}$ where, M = mutual inductance, phi = magnetic flux, B = magnetic field and A = area. Mutual inductance by 4 straight conductor, $\text{M}=\frac{4 \text{B} .\text{A}}{i}$ $\Rightarrow \text{M}={4[\frac{\\mu _{0}}{4 \pi} \frac{i}{b / 2}(\sin \pi/ 4+\sin \pi/ 4)] a^{2}}/{i}$ $\Rightarrow \text{M}=\frac{4}{i}(\frac{\\mu _{0}}{4 \pi} \frac{2 i}{b}(1 / \sqrt{2}+1 / \sqrt{2}) a^{2})$ $=(\frac{\\mu _{0}}{4 \pi} \frac{1}{b} 8 \sqrt{2} a^{2})=(\frac{\\mu _{0}}{4 \pi} 8 \sqrt{2} \frac{a^{2}}{b})$

According to given circuit diagram, Let I current is flowing through the circuit. ∴ Kirchhoff's voltage loop equation will be $10 \sin \omegat-\text{IR}-3=0$ $\text{IR}=\text{V}=10 \sin \omegat-3$ ∴ At time, $t=0, \text{V}=0-3=-3 \text{V}$ ...(i) $t=\text{T} / 4, \text{V}=10 \sin (\frac{2 \pi}{\text{T}}).\frac{\text{T}}{4}-3$ $=10-3=7 {\text{V}}$ ...(ii) $t=\text{T} / 2$ to T, diode is in reverse bias. ∴ V = 0 ...(iii) We see that at $t=0, \text{V}=-3$ and $\frac{\text{T}}{2} \\le t \\le \text{T}, \text{V}=0$ Therefore, no waveforms exist in option (a), (b) (c) and (d), which satisfies above condition. Hence, no option is correct.

Given, weight of rod, $w=10 {\text{kg}} {\text{ms}}^{-2}$ Area of cross- section, $\text{A}=100 {\text{cm}}^{2}=100 \times 10^{-4} {\text{m}}^{2}$ Length of rod, $\text{I}=20 {\text{cm}}=20 \times 10^{-2} {\text{m}}$ Young's modulus $(\text{Y})=2 \times 10^{11} {\text{Nm}}^{-2}$ Let, elongation = Δl Since, Young's modulus, $\text{Y}=\frac{\text{dmgx}}{\text{A} \Delta\text{I}}$ Mass of element dx at distance $x, d m=\frac{m}{\text{I}} d x$ $\Rightarrow \Delta\text{I}={(\frac{m g}{\text{I}})}/{\text{AY}} x d x$ $\Rightarrow \Delta\text{I}=\frac{m g}{\text{AIY}} \frac{x^{2}}{2}|_{0} ^{\text{I}}$ $\Rightarrow \Delta\text{I}=\frac{m g}{\text{AIY}} \frac{\text{I}^{2}}{2}$ $=\frac{m g l}{2 \text{AY}}=\frac{w l}{2 \text{AY}}$ $=\frac{10 \times 20 \times 10^{-2}}{2 \times 100 \times 10^{-4} \times 2 \times 10^{11}}$ $=5 \times 10^{-10} {\text{m}}$

According to the given figure, we have two plane mirror, placed at 90º to each other. Since, image formed by plane mirror are virtual, erect, same size and at same distance behind mirror. Let $\text{I}_{1}$ is the image of P due to mirror $\text{M}_{1}$ and $\text{I}_{2}$ is the image of P due to mirror $\text{M}_{2}$ Therefore, the shortest distance between images i.e., between $\text{I}_{1}$ and $\text{I}_{2}$ $\text{I}_{1} \text{I}_{2}=\sqrt{(4 a)^{2}+(2 a)^{2}}$ $=\sqrt{16 a^{2}+4 a^{2}}$ $=a \sqrt{20}=a 2 \sqrt{5}$ $=a \times 2 \times 2.3=4.6 a$

As we know that, The dimensional formula of following terms are (a) [ Torque ] = [ Radius ] × [ Force ] =\[{\text{M}}^{0} {\text{LT}}^{0}]$$[{\text{M}} {\text{L}} {\text{T}}^{-2}]$=[{\text{ML}}^{2} {\text{T}}^{-2}]$(b) [Impulse] = [Force] [Time]$=$[{\text{MLT}}^{-2}]$[{\text{T}}]=[{\text{MLT}}^{-1}]$(c) [ Tension ] = [ Force ]$= [{\text{M}} {\text{L}} {\text{T}}^{-2}]$(d) [Surface tension] =${[{ \text{Force} }]}/{[{ \text{Length} }]}= {[{\text{ML}} {\text{L}}^{-2}]$}/{[{\text{L}}]}=[{\text{MT}}^{-2}]$ Hence, option (a) is the correct.

Given, Change in pressure with volume $\frac{\text{dp}}{\text{dV}}=-\text{ap}$ At $\text{V}=0, \text{p}=\text{p}_{0}$ Number of mole n = 1 Gas constant, $\text{R}=8.314 {\text{J}} {\text{K}}^{-1}$ From given equation, $\frac{\text{dp}}{\text{p}}=-\text{adV}$ $\int^{\text{p}}_{\text{p}_0} \frac{\text{dp}}{\text{p}}=-a \int^{v}_{0} \text{dV}$ $\Rightarrow [\ln \text{p}]_{\text{p}_{0}}^{\text{p}}=-a[\text{V}]_{0}^{\text{V}}$ $\Rightarrow \ln (\text{p} / \text{p}_{0})=-a(\text{V}-0)$ Taking anti log on both side, $\Rightarrow \text{p} / \text{p}_{0}={e}^{-{\text{aV}}}$ $\Rightarrow \text{p}=\text{p}_{0} {e}^{-{\text{aV}}}$ By using ideal gas law, pV = nRT $\Rightarrow (\text{p}_{0} {e}^{-{\text{aV}}}) \text{V}=\text{nRT}$ $\Rightarrow \frac{1}{\text{nR}}(\text{p}_{0} {e}^{-{\text{aV}}}) \text{V}=\text{T}$ $\Rightarrow \frac{1}{\text{R}}(\text{p}_{0} {e}^{-{\text{aV}}}) \text{V}=\text{T}$ [\because n = 1]...(i) For maximum temperature $(\text{T}_{\max })$, On differentiating both side with respect to V, we get \Rightarrow \frac{1}{n \text{R}}\[\text{p}{0} .{e}^{-a \text{V}} .(-a \text{V}) \text{V}+\text{p}{0} {e}^{-a \text{V}}]$=\frac{d \text{T}}{d \text{V}}=0$$\Rightarrow \text{p}{0} e^{-a \text{V}}(1-a \text{V}^{2})=0$$\Rightarrow 1=a \text{V}^{2}$$\Rightarrow \text{V}^{2}=1 / a$$\Rightarrow \text{V}=1 / a$Substituting the value in Eq. (i), we get$\text{T}=\frac{1}{\text{R}}(\text{p}{0} {e}^{-a .1 / a}) 1 / {a}=\frac{\text{p}{0} {e}^{-1}}{\text{R}}(1 / {a})=\frac{\text{p}{0}}{{e} \text{R} {a}}$

As we know that, Dimensional formula of volume $=[{\text{M}}^{0} {\text{L}}^{3} {\text{T}}^{0}]$ ...(i) Since, $\text{F}=6 \pi\eta r v$ $\therefore \eta \propto\frac{\text{F}}{r v}$ where, η is viscosity and F is force. \therefore [\eta ]={[{\text{ML}} {\text{T}}^{-2}]\}/{[{\text{L}} .{\text{L}} {\text{T}}^{-1}]$}=[{\text{M}} {\text{L}}^{-1} {\text{T}}^{-1}]$So,$[\frac{\pi\text{pa}^{4}}{8 \eta \text{L}}]=\frac{[\text{ML}^{-1} \text{T}^{-2}]$$[\text{L}^{4}]$}{[\text{ML}^{-1} \text{T}^{-1}]$$[\text{L}^{1}]$}$$=[\text{M}^{0} \text{L}^{3} \text{T}^{-1}]$...(ii) Since, Eq. (i) is not equal to Eq (ii), so option (a) is wrong. Now, since formula of capillary rise in tube,$h=\frac{2 s \cos \theta}{\rho g r}$Dimensional formula of LHS part, ∴ [h] = [L] Dimensional formula of RHS part$=\frac{[{s}]}{[\rho ][{g}][r]}={[{\text{MT}}^{-2}]$}/{[{\text{ML}}^{-3}]$$[{\text{LT}}^{-2}]$[{\text{L}}]}=[{\text{L}}]$Hence,$h=\frac{2 s \cos \theta}{\rho r g}$is dimensionally correct. So, option (b) will also be dimensionally correct. In option (c),$\text{J}=\epsilon\frac{\text{dE}}{\∂ t}$...(iii)$\Rightarrow \text{J}=\epsilon\frac{\text{E}}{t}$Dimension of current density J is calculated as Since,$\text{J}=\frac{\text{l}}{\text{A}}$$\therefore [\text{J}]=\frac{[\text{l}]}{[\text{A}]}={[{\text{A}}]}/{[\text{L}^{2}]$}$$\Rightarrow [\text{J}]=[{\text{AL}}^{-2}]$...(iv) Again, we know that$\text{E}=\frac{1}{4 \pi\epsilon} .\frac{q}{r^{2}}$$\Rightarrow \epsilon \text{E}=\frac{1}{4 \pi} .\frac{q}{r^{2}}$$\frac{\Rightarrow \epsilon\text{E}}{t}=\frac{1}{4 \pi} .\frac{q}{t r^{2}}$$\Rightarrow [\frac{\epsilon\text{E}}{t}]=\frac{[q]}{[t]$[r^{2}]$}={[{\text{AT}}]}/{[{\text{T}}]$[{\text{L}}^{2}]$}$$[\frac{\epsilon\text{E}}{t}]=[{\text{AL}}^{-2}]$...(v) Form Eqs. (iv) and (v), we see that Eq. (iii) is dimensionally correct. In option (d) W = \tau \theta and$\tau =r × \text{F}$So, dimensional formula of$[\tau ][\theta]=[r][\text{F}]=[{\text{L}}]$[{\text{MLT}}^{-2}]$=[{\text{ML}}^{2} {\text{T}}^{-2}]$$=[{\text{W}}]$

As we know that, Angular momentum, $\text{L}=r \times p=m v r \sin \theta$ where, m = mass,v = velocity,r = radius, $\theta=$ angle between v and $r=90°$ (for circular motion) Since, $m, v, r$ ∴ L will always remain same in magnitude and direction.

Given, inductor, capacitor and resistors are connected in series and have same reactance i.e., $\text{X}_{\text{L}}=\text{X}_{\text{C}}=\text{R}$ Since, impedance $(\text{Z})=\sqrt{\text{R}^{2}+(\text{X}_{\text{L}}-\text{X}_{\text{C}})^{2}}$ $\therefore \text{Z}=\sqrt{\text{R}^{2}+(\text{R}-\text{R})^{2}}=\text{R}$

Let de-Broglie wavelength of proton and electron are $\lambda_{p}$ and $\lambda_{e}$. Kinetic energy of electron and proton are $\text{K}_{\text{e}}$ and $\text{K}_{\text{p}}$. Momentum of electron and proton are $\text{p}_{\text{e}}$ and $\text{p}_{\text{P}}$. According to question, $\lambda_{\text{p}}=\lambda_{\text{e}}=\lambda$ By using de-Broglie wavelength, $\lambda=\frac{h}{p}={h}/{\sqrt{2 \text{mK}}}$ where, m is mass and h is Planck's constant. $\therefore p=\frac{h}{\lambda}$ ...(i) From Eq. (i) $\text{p}_{\text{e}}=\text{p}_{\text{p}}$ as $\lambda_{e}=\lambda_{p}$ Again, $\text{K}=\frac{h^{2}}{2 m \lambda}$ ...(ii) From Eq. (ii) $\therefore {\text{K}_{\text{e}}}/{\text{K}_{\text{p}}}={\text{m}_{\text{p}}}/{\text{m}_{\text{e}}} \times {\lambda_{\text{p}}}/{\lambda_{\text{e}}}$ $\Rightarrow {\text{K}_{\text{e}}}/{\text{K}_{\text{p}}}={\text{m}_{\text{p}}}/{\text{m}_{\text{e}}}$ $[\because \lambda_{\text{e}}=\lambda_{\text{p}}]$ $\because m_{p}>m_{e}$ $\text{K}_{e}>\text{K}_{p}$

Given, Shunt resistance, $\text{R}_{{\text{sn}}}=5 \Omega$ Let galvanometer resistance = G Total, current = I Voltage across galvanometer $(\text{V}_{\text{G}})=$ Voltage across shunt resistance $\text{R}_{{\text{sn} }}$ $\text{I}_{\text{G}} \times \text{G}=\text{I}_{{\text{sn}}} \times \text{R}_{{\text{sn}}}$ $0.021 \times \text{G}=0.981 \times 5$ $\Rightarrow \text{G}=\frac{98 \times 5}{2}=49 \times 5=245 \Omega$

Given, decrease in volume $(\Delta\text{V} / \text{V})=-\frac{0.5}{100}$ Bulk modulus of rubber, $\text{B}=9.8 \times 10^{8} {\text{Nm}}^{-2}$ Density of sea water, $\rho =10^{3} {\text{kgm}}^{-3}$ Acceleration due to gravity, $g=9.8 {\text{ms}}^{-2}$ Let h be the depth at which ball is dipped. Since, $\text{B}=\frac{\Delta\text{p}}{-\Delta\text{V} / \text{V}}$ where, $\Delta\text{p}$ is change in pressure $=\rho \text{gh}$ $\therefore \rho \text{gh}=-\text{B} \frac{\Delta\text{V}}{\text{V}}$ $\Rightarrow h=-\frac{1}{\rho g} \text{B} \frac{\Delta\text{V}}{\text{V}}$ $\Rightarrow h=-\frac{1}{10^{3} \times 9.8} \times 9.8 \times 10^{8} \times (-\frac{0.5}{100})$ $=0.5 \times 10^{8-3-2}=0.5 \times 10^{3}=500 {\text{m}}$

Given, mass of particle, m = 1kg Spring constant, $k=100 {\text{Nm}}^{-1}$ Let time period is T. Kinetic energy (KE) = Potential energy (PE) = E Amplitude = A Angular velocity = omega Wave displacement = x KE in SHM = PE in SHM $\Rightarrow \frac{1}{2} m \omega^{2}(\text{A}^{2}-x^{2})=\frac{1}{2} k x^{2}$ $\Rightarrow m \omega^{2}(\text{A}^{2}-x^{2})=k x^{2}$ ...(i) Since, force, $\text{F}=m \omega^{2} x=k x$ $\Rightarrow k=m \omega^{2}$ Substituting the value in Eq. (i), we get $\Rightarrow m \omega^{2}(\text{A}^{2}-x^{2})=m \omega^{2} x^{2}$ $\Rightarrow \text{A}^{2}-x^{2}=x^{2}$ $\Rightarrow \text{A}^{2}=2 x^{2}$ $\Rightarrow x=\text{A} / \sqrt{2}$ Since, wave displacement, $x=\text{A} \sin \omegat$ $\Rightarrow \frac{\text{A}}{\sqrt{2}}=\text{A} \sin \omegat$ $\Rightarrow \sin (\pi/ 4)=\sin \omegat$ $\Rightarrow \omegat=\pi/ 4$ $\Rightarrow \frac{2 \pi}{\text{T}} .t=\frac{\pi}{4}$ $\Rightarrow t=\frac{\text{T}}{8}$ Comparing the given value in the question i.e. $\frac{\text{T}}{x}$, the value of x = 8

Given, Height of transmitting anteena $(\text{H}_{\text{T}})+$ height of receiving anteena $(\text{H}_{\text{R}})=160 {\text{m}}$ Radius of earth, $\text{R}_{\text{e}}=6400 {\text{km}}$ Let, range = d If height of transmitting antenna be x then, height of receiving anteena = 160 - x We know that, $d=\sqrt{2 \text{Rh}_{\text{T}}}+\sqrt{2 \text{Rh}_{\text{R}}}=\sqrt{2 \text{R}}[\sqrt{h_{\text{T}}}+\sqrt{h_{\text{R}}}]$ $=\sqrt{2 \times \text{R}}[\sqrt{x}+\sqrt{160-x}]$ $=\sqrt{2 \text{R}}[\sqrt{x}+\sqrt{160-x}]$ For maximum range $\frac{d}{d x}(d)=\frac{d}{d x} \sqrt{2 \text{R}}[\sqrt{x}+\sqrt{160-x}]=0$ $\Rightarrow \frac{d}{d x}[\sqrt{x}+\sqrt{160-x}]=0$ $\Rightarrow \frac{1}{2} x^{-1 / 2}-\frac{1}{2}(160-x)^{-1 / 2}=0$ $\Rightarrow \frac{1}{2} \frac{1}{\sqrt{x}}=\frac{1}{2} \frac{1}{\sqrt{160-x}}$ $\Rightarrow x=160-x$ $\Rightarrow 2 x=160$ ⇒ x = 80 ∴ Maximum range $(d_{\max })=\sqrt{2} \text{R}[\sqrt{x}+\sqrt{160}-x]$ $=\sqrt{2 \times 6400}[\sqrt{80}+\sqrt{160-80}]$ $=80 \sqrt{2}[\sqrt{\frac{80}{1000}}+\sqrt{\frac{80}{1000}}]$ $=80 \sqrt{2} \times 2 \sqrt{\frac{80}{1000}}$ $=80 \sqrt{2} \times \frac{2 \sqrt{80}}{10 \sqrt{10}}=16 \sqrt{\frac{2 \times 80}{10}}$ $=16 \times 4=64 {\text{km}}$

Let the sides of square be a. ∴ Total length or perìmeter of square $=4 a$ If the radius of shape of circle be r, then $\therefore 2 \pir=4 a$ $\Rightarrow r=\frac{4 a}{2 \pi}$ $\Rightarrow r=\frac{2 a}{\pi}$ Since, resistance of each side of square = 3 Ω ∴ Total resistance of square $=4 \times 3=12 \Omega$ i.e. resistance of length $2 \pi r=12 \Omega$ \Rightarrow Resistance of $\pir=6 \Omega$ Now, equivalent resistance of circle diametrically opposite $(\text{R}_{{\text{eq}}})=\frac{\text{R}_{1} \text{R}_{2}}{\text{R}_{1}+\text{R}_{2}}=\frac{6 \times 6}{6+6}=3 \Omega$

Given, linear mass density, $\lambda=9 \times 10^{-4} {\text{kg}} / {\text{m}}$ Tension, $\text{T}'=900 {\text{N}}$ Resonance frequency, $\text{f}_{1}=500 {\text{Hz}}$ and $\text{f}_{2}=550 {\text{Hz}}$ Let, length of wire be l. As we know that,$\text{f}_{\text{n}}={\text{n}}{2 \text{I}} \sqrt{\frac{\text{T}}{\lambda}}$ $\Rightarrow 500=\frac{n}{2 l} \sqrt{\frac{900}{9 \times 10^{-4}}}=\frac{n}{2 \text{l}} \sqrt{10^{6}}$ $\Rightarrow 500=\frac{n}{2 l} \times 10^{3}$ $\Rightarrow \frac{n}{2 \text{l}}=\frac{500}{1000}=\frac{1}{2}$ $\Rightarrow \frac{n}{\text{l}}=1$ ⇒ n = l ...(i) and$550=\frac{n+1}{2 I} \sqrt{\frac{9 \times 10^{6}}{9}}$ $\Rightarrow 550=\frac{n+1}{2 l} \times 10^{3}$ $\Rightarrow 550={\text{I}+1}{2 \text{I}} \times 10^{3}$ [From Eq, (i)] $\Rightarrow \frac{55}{100}=\frac{\text{I}+1}{2 \text{I}}$ $\Rightarrow \frac{11}{20}=\frac{\text{I}+1}{2 \text{I}}$ $\Rightarrow 22 l=20 l+20$ $\Rightarrow 2 \text{l}=20$ ⇒ l = 10 m

According to given circuit diagram, As we know that, parallel equivalent resistance, ${1}/{\text{R}_{{\text{eq}}}}={1}{\text{R}_{1}}+\frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}+...$ and series equivalent resistance, $\text{R}_{{\text{eq}}}=\text{R}_{1}+\text{R}_{2}+\text{R}_{3}+...$ Let the net resistance across a and b be R' $\therefore \frac{1}{\text{R}'}=\frac{1}{2+2+6}+\frac{1}{4+6}$ ${a}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}$ R' = 5 Ω Hence, total resistance, $\text{R}=\text{R}'+1=5+1=6 \Omega$ According to current division rule, current in upper branch, $\text{I}_{1}=\text{I} .\frac{4+6}{(2+2+6)+(4+6)}$ $=1 .\frac{10}{20}=\frac{1}{2}=\frac{1}{2} .\frac{\text{V}}{\text{R}}=\frac{1}{2} \times \frac{12}{6}=1 {\text{A}}$ Again, according to current division rule, current in 15Ω resistor $\text{I}_{15}=\text{I}_{1} .\frac{10}{10+15}=1 \times \frac{2}{5}=0.4 {\text{A}}$ ∴ Voltage drop across15Ω resistor, $\text{V}_{15}=\text{I}_{15} \times 15=0.4 \times 15=6 {\text{V}}$

Given, initial speed of block, $u=40 {\text{ms}}^{-1}$ Let total mass of block = m ∴ Broken masses, $m_{1}=m_{2}=\frac{m}{2}$ Final speed of $m_{1}, v_{1}=60 {\text{ms}}^{-1}$ As we know that, kinetic energy, ${\text{KE}}=1 / 2 {mv}^{2}$ ∴ Initial kinetic energy ${\text{KE}}_{i}=\frac{1}{2} m u^{2}$ $=\frac{1}{2} m(40)^{2}=800 m$ ...(i) By using law of conservation of momentum, $m u=m_{1} v_{1}+m_{2} v_{2}$ $\therefore v_{2}=\frac{m \times 40-m / 2 \times 60}{m / 2}$ $=\frac{10 m}{m / 2}=20 m s^{-1}$ So, final kinetic energy ${\text{KE}}_{\text{f}}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}$ $=\frac{1}{2} \frac{m}{2}(v_{1}^{2}+v_{2}^{2})=\frac{m}{4}(60^{2}+20^{2})$ $=m / 4(3600+400)$ $=1000 m$ ...(ii) ∴ Divide Eq (ii) by Eq. (i), we get ${{\text{KE}}_{\text{f}}}/{{\text{KE}}_{i}}={1000 {\text{m}}}/{800}=\frac{5}{4} {\text{m}}$ So, x = 5