26 Feb 2021 Shift 1

Given, distance between centre of ring and sphere, $d=\sqrt{8} R$ Since, gravitational field at the axis of ring, $E=\;\frac{G m d}{(d^{2}+R^{2})^{3 / 2}}$ Here, $G$ is the gravitational constant. $\Rightarrow \;\; E=\;\frac{G m R \sqrt{8}}{(8 R^{2}+R^{2})^{3 / 2}}=\;\frac{G m R \sqrt{8}}{(3 R)^{3}}$ $\Rightarrow \;\; E=\;\frac{G m R \sqrt{8}}{27 R^{3}}=\;\frac{G m \sqrt{8}}{27 R^{2}}$ Force between ring and sphere, $F=M E$ . . . (i) Substituting the value of $E$ in Eq. (i), we get $F=\;\frac{\sqrt{8}}{27} \;\frac{G m M}{R^{2}}$

$({*})$ Let, equivalent capacitance of two capacitors $C_{1}$ and $C_{2}$ connected in parallel be $C_{a}$ and equivalent capacitance of same, when connected in series be $C_{b}$. According to given data, $C_{ {a}}=\;\frac{15}{4} C_{b}$ . . . (i) Since, equivalent capacitance in parallel combination, $C_{ {eq}}=C_{1}+C_{2}$ $C_{ {a}}=C_{1}+C_{2}$ . . . (ii) and equivalent capacitance in series combination, $\;{1}/{C_{ {eq}}^{'}} \;\; =\;\frac{1}{C_{1}}+\;\frac{1}{C_{2}}$ $\;\frac{1}{C_{b}} \;\; =\;\frac{1}{C_{1}}+\;\frac{1}{C_{2}}=\;\frac{C_{2}+C_{1}}{C_{1} C_{2}}$ $C_{b} \;\; =\;\frac{C_{1} C_{2}}{C_{1}+C_{2}}$. . . (iii) Substituting Eqs. (ii) and (iii) in Eq. (i), we get $C_{1}+C_{2}=\;\frac{15}{4} \;\frac{C_{1} C_{2}}{C_{1}}+C_{2}$ $4(C_{1}+C_{2})^{2}=15 C_{1} C_{2}$ $\Rightarrow \;\; 4 C_{1}^{2}+4 C_{2}^{2}+8 C_{1} C_{2}=15 C_{1} C_{2}$ $\Rightarrow \;\; 4 C_{1}^{2}+4 C_{2}^{2}-7 C_{1} C_{2}=0$ On dividing both sides by $C_{1}^{2}$, we get $4+4(\;\frac{C_{2}}{C_{1}})^{2}-7 \;\frac{C_{2}}{C_{1}}=0$ $\;\text{ or }\; \;\; 4(\;\frac{C_{2}}{C_{1}})^{2}-7(\;\frac{C_{2}}{C_{1}})+4=0$ $\;\text{ If }\; \;\frac{C_{2}}{C_{1}}=x, \;\text{ then }\; 4 x^{2}-7 x+4=0$ By using the concept of quadratic equation, $\;x=\;{-b \pm \sqrt{b^{2}-4 a c}}/{2 a} \Rightarrow x=\;\frac{7 \pm \sqrt{49-64}}{8}$ $\Rightarrow \;\; x=\;\frac{C_{2}}{C_{1}}=\;\frac{7 \pm \sqrt{-15}}{8}=\;\frac{7 \pm \sqrt{15} i}{8}$

Given, work done by gas molecule, $W=\alpha^{2} \beta {e}^{-\beta x^{2} / k T}$ Here, $x$ is displacement, $k$ is Boltzmann constant, $\alpha$ and $\beta$ are constants and $T$ is temperature. Dimensional formula of $[W]=[ {ML}^{2} {T}^{-2}]$ $\therefore$ Dimensions of $[\alpha^{2} \beta]=[ {ML}^{2} {T}^{-2}]$ \Rightarrow \;\; \alpha=\[;\frac{M L^{2} T^{-2}}{beta}]^{1 / 2}$. . . (i) The term$[ {e}^{-beta x^{2} / k T}]$should be dimensionless, i.e.$[ {M}^{0} {L}^{0} {T}^{0}]$.$\Rightarrow ;; $[;\frac{\beta x^{2}}{k T}]$=[ {M}^{0} {L}^{0} {T}^{0}]$$\Rightarrow ;; [beta]=;\frac{[k][T]}{[x^{2}]$}$. . . (ii)$;\text{ Energy of gaseous molecule };(E)=;\frac{7}{2} k T$$[k]=[E] /[T]=[ {ML}^{2} {T}^{-2} {K}^{-1}]$Substituting the value of$k$in Eq. (ii), we get$[\beta]=;\frac{[M L^{2} T^{-2} K^{-1}]$[K]}{[L^{2}]$}=[M T^{-2}]$Substituting the value of$\beta$in Eq. (i), we get$[\alpha]={;\frac{[ {ML}^{2} {T}^{-2}]$}{[ {MT}^{-2}]$}}^{1 / 2}=[ {M} {L}^{1} {T}^{0}]$

By using Rydberg's formula, $\;\frac{1}{\lambda}=R[\;\frac{1}{n_{2}^{2}}-\;\frac{1}{n_{1}^{2}}]$ where, $R$ is Rydberg constant. For wavelength of third member of Lyman series $n_{2}=1$ and $n_{1}=4$. \therefore \;\; \;\frac{1}{\lambda_{1}}=R\[;\frac{1}{1^{2}}-;\frac{1}{4^{2}}]$=R[;\frac{1}{1}-;\frac{1}{16}]$. . . (i) For wavelength of first member of Paschen series,$n_{2}=3$and$n_{1}=4$$\therefore ;; ;\frac{1}{\lambda_{2}}=R$[;\frac{1}{3^{2}}-;\frac{1}{4^{2}}]$=R[;\frac{1}{9}-;\frac{1}{16}]$. . . (ii) On dividing Eq. (ii) by Eq. (i), we get$;\frac{\lambda_{1}}{\lambda_{2}}=;\frac{[\frac{1}{9}-\frac{1}{16}]}{[;\frac{1}{1}-;\frac{1}{16}]}=;\frac{7}{9 \times 15}=;\frac{7}{135}$

Given, height of object, $h_{o}=100 {cm}$ Focal length of mirror, $f=40 {cm}$ Height of image, $h_{i}=25 {cm}$ Nature of image is erect means virtual. As, $h_{i}$ is less than $h_{o}$, so mirror used is convex mirror. Hence, (d) is correct option, i.e. image is virtual, opposite and is convex.

Given, radius of earth $=R$ Distance of chord from centre of earth $=\;\frac{R}{2}$ Let $x_{1}$ be the radius of inner circle and $M$ be the mass of earth. $\therefore m^{'}(\;\text{ effective mass of earth }\;) \;\; =\;\frac{M}{\frac{4}{3} \pi R^{3}} \cdot \;\frac{4}{3} \pi x_{1}^{3}$ $\Rightarrow \;\; m^{'} \;\; =\;\frac{M}{R^{3}} x_{1}^{3}$ If $F$ is the gravitational force exerted by earth on particle at position $x$ and $\omega$ be the angular velocity in time period $T$, then $F \;\; =\;\frac{G m^{'} m}{x_{1}^{2}}=\;\frac{G m}{x_{1}^{2}} \cdot \;\frac{M}{R^{3}} x_{1}^{3}$ $\Rightarrow \;\; m \omega^{2} x_{1} \;\; =\;\frac{G M m x_{1}}{R^{3}}$ $\Rightarrow \;\; \omega \;\; =\sqrt{\;\frac{G M}{R^{3}}}$ $F=\;\frac{G m^{'} m}{x_{1}^{2}}=\;\frac{G m}{x_{1}^{2}} \cdot R^{3} x_{1}^{3}$ $\Rightarrow \;\; m \omega^{2} x_{1}=\;\frac{G M m x_{1}}{R^{3}}$ $\Rightarrow \;\; \omega=\sqrt{\;\frac{G M}{R^{3}}}$ Since, $\omega=\;\frac{2 \pi}{T}$ and $G M=g R^{2}$ Substituting the above values in Eq. (i), we get 1$\Rightarrow \;\; \;\frac{2 \pi}{T}=\sqrt{\;\frac{g R^{2}}{R^{3}}} \Rightarrow T=2 \pi \sqrt{\;\frac{R}{g}}$

Given, $i=i_{1} \sin \omega t+i_{2} \cos \omega t$ Let $I_{\;\text{rms }\;}$ be the rms current. $\therefore \;\; I_{ {rms}}=(\;\frac{i_{1}^{2}+i_{2}^{2}}{2})^{1 / 2}$ $\Rightarrow \;\; I_{ {rms}}=\;\frac{1}{\sqrt{2}}(i_{1}^{2}+i_{2}^{2})^{1 / 2}$

Given, density of material $=\rho$ Bulk modulus of elasticity $=K$ and applied pressure $=p$ Let change in volume and density be $\Delta V$ and $\Delta \rho$ respectively and initial volume and density be $V$ and $\rho$. Since, $K=\;\frac{p}{-\;\frac{\Delta V}{V}}$ . . . (i) and density $(\rho )=\;\frac{mass(m)}{\;\text{ volume }\;(V)}$ $\therefore \;\; \;\frac{\Delta \rho }{\rho }=-\;\frac{\Delta V}{V}$ Substituting it in Eq. (i), we get $\;\frac{-\Delta V}{V}=\;\frac{p}{K}=\;\frac{\Delta \rho }{\rho }$ $\therefore \;\; \Delta \rho =\;\frac{p \rho }{K}$

Given, radius of circle $=R$ Central fictitious force is, $F \propto \;\frac{1}{R^{3}}$. Let $T$ be the time period of revolution, $m, \omega$ be the mass and angular velocity of Earth. $\because \;\; F=m \omega^{2} R \propto \;\frac{1}{R^{3}}$ $\Rightarrow \;\; \omega^{2} \propto \;\frac{1}{R^{4}} \Rightarrow \omega=\;\frac{2 \pi}{T} \propto \;\frac{1}{R^{2}}$ $\Rightarrow \;\; T \propto R^{2}$

According to Kepler's second law of planetary motion, areal velocity of every planet moving around the sun should remain constant in elliptical orbit.

Let $v_{1}$ and $v_{2}$ are the speed of $P$ and $Q$ after collision. By using law of conservation of mementum, $m_{1} u_{1}+m_{2} u_{2} \;\; =m_{1} v_{1}+m_{2} v_{2}$ $M u+m \cdot 0 \;\; =M v_{1}+m v_{2}$ $\Rightarrow \;\; \;\; \;\frac{M(u-v_{1})}{m} \;\; =v_{2}$ and by using law of conservation of energy, $\;\;\frac{1}{2} m_{1} u_{1}^{2}+\;\frac{1}{2} m_{2} u_{2}^{2} \;\; =\;\frac{1}{2} m_{1} v_{1}^{2}+\;\frac{1}{2} m_{2} v_{2}^{2}$ $\;\Rightarrow \;M u^{2}+0 \;\; =M v_{1}^{2}+m v_{2}^{2}$ $\Rightarrow \;M(u^{2}-v_{1}^{2}) \;\; =m v_{2}^{2}$ $\Rightarrow \;M(u-v_{1})(u+v_{1}) / m \;\; =v_{2}^{2}$ $\;\frac{1}{2} m_{1} u_{1}^{2}+\;\frac{1}{2} m_{2} u_{2}^{2} \;\; =\;\frac{1}{2} m_{1} v_{1}^{2}$ $M u^{2}+0 \;\; =M v_{1}^{2}+m v^{2}$ $\Rightarrow \;\; M(u^{2}-v_{1}^{2}) \;\; =m v_{2}^{2}$ $\Rightarrow \;\; M(u-v_{1})(u+v_{1}) / m \;\; =v_{2}^{2}$ $\;\text{ Substituting the value of }\; M \;\frac{(u-v_{1})}{m} \;\text{ from Eq. (i) \in }\;\;$ $\;\text{ Eq. (ii), we get }\;$ $v_{2}(u+v_{1}) \;\; =v_{2}^{2}$ $u+v_{1} \;\; =v_{2}$ $\Rightarrow \;\; M\;≫ m$ $v_{1} \;\; =u$ $v_{2} \;\; =2 u$ Eq. (ii), we get $v_{2}(u+v_{1}) \;\; =v_{2}^{2}$ $u+v_{1} \;\; =v_{2}$ $M > \; > m$ $v_{1} \;\; =u$ $v_{2} \;\; =2 u$ Hence, option (c) is the correct.

Given, mass of solid sphere $=m$ Radius of solid sphere $=a$ Side of square $=b$ Let $I_{\;\text{net }\;}$ be the net moment of inertia of system. Moment of inertia of sphere, $I_{s}=\;\frac{2}{5} {ma}^{2}$ Axis of rotation is $B C$. $\therefore$ Moment of inertia of any body at distance, $( \because d=b)$ $d \;\; =m d^{2}=m b^{2}$ $\therefore \;\; I_{\;\text{net }\;} \;\; =4(\;\frac{2}{5} m a^{2})+2(m b^{2})$ ${ \therefore \;\; }_{\;\text{net }\;} \;\; =4(\;\frac{2}{5} m a^{2})+2(m b^{2})$ $\Rightarrow \;\; I_{\;\text{net }\;} \;\; =\;\frac{8}{5} m a^{2}+2 m b^{2}$

Given, in YDSE, the separation between two slits, $d=2 {mm}$ $=2 \times 10^{-3} {m}$ Distance between slit and screen, $D=1 {m}$ Wavelength of light, $\lambda=500 {nm}$ $=500 \times 10^{-9} {m}$ Let $B$ will be the fringe width $\therefore \;\; B \;\; =\;\frac{\lambda D}{d}=\;\frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}}=250 \times 10^{-6}$ $\;\; =0.25 {mm}$

Given, length of conductor $=L$ Charge on conductor $=Q$ According to figure, $O P=a=\;\frac{\sqrt{3}}{2} L, O Q=\;\frac{L}{2}$Let $\;\; P Q=r=\sqrt{O P^{2}+O Q^{2}}$ $\Rightarrow \;\; P Q=\sqrt{(\;\frac{\sqrt{3}}{2} L)^{2}+(\;\frac{L}{2})^{2}}=\sqrt{\;\frac{3}{4} L^{2}+\;\frac{L}{4}}=L$ and $E$ be the electric field at point $P$. Since, $E$ (due to finite wire) $=\;\frac{k \lambda}{a}(\sin \varphi _{1}+\sin \varphi _{2})$. . . (i) where, $k=$ Coulomb's constant $=\;\frac{1}{4 \pi \epsilon_{0}}$ $\lambda=$ linear charge density $=\;\frac{Q}{L}$ and $\sin \varphi =\sin \varphi _{2}=\;\frac{L / 2}{L}=\;\frac{1}{2}$ Substituting the above value in Eq. (i), we get $E=\;\frac{k \lambda}{a}=\;\frac{1}{4 \pi \epsilon_{0}} \;{Q}/{L \cdot \;\frac{\sqrt{3} L}{2}}=\;\frac{1}{2 \sqrt{3} \pi \epsilon_{0} L^{2}}$

Let series equivalent of spring constant $=k_{ {eq}}$ and $T$ be the time period. In series arrangement, $\;\frac{1}{k_{e q}}=\;\frac{1}{k_{1}}+\;\frac{1}{k_{2}}$ $\Rightarrow \;\; \;{1}/{k_{ {eq}}}=\;\frac{1}{k_{1}}+\;\frac{1}{k_{1}}=\;\frac{2}{k_{1}} \Rightarrow k_{ {eq}}=\;\frac{k_{1}}{2}$ As, $\;\; T=2 \pi \sqrt{\;\frac{m}{k_{1}}}$ where, $m$ is mass of body connected with spring. $\Rightarrow \;\; T \propto \sqrt{\;\frac{1}{k_{1}}}$ and $\;\; T^{'} \propto \sqrt{\;\frac{2}{k_{1}}} \Rightarrow T^{'}=\sqrt{2} T$

Let, $Q=$ heat current, $k=$ thermal conductivity, A = area, $I=$ length of capacitor and $\Delta \theta=$ change in temperature . $\therefore \;\; Q=\;\frac{k A \Delta \theta}{l}=\;\frac{\Delta \theta}{R} \;\; ( \because R=\;\frac{l}{k A})$ $\Rightarrow \;\; \;\frac{\theta_{2}-\theta}{R_{2}}=\;\frac{\theta-\theta_{1}}{R_{1}}$ $\Rightarrow \;R_{1} \theta_{2}-R_{1} \theta \;\; =R_{2} \theta-R_{2} \theta_{1}$ $\Rightarrow \;\theta \;\; =\;\frac{R_{1} \theta_{2}+R_{2} \theta_{1}}{R_{1}+R_{2}}$

As we know that, Resolving power of microscope $=\;\frac{1}{\Delta \theta}=\;\frac{2 u \sin \theta}{\lambda}$ i.e. resolving power $\propto \;\frac{1}{\lambda}$ and since, wavelength of electron emitted $(\lambda_{e}) <$ wavelength of visible light $(\lambda_{v})$ $\therefore$ Resolving power of electron microscope $>$ Resolving power of optical microscope. Hence, option (b) is the correct.

Given, energy gap of LED, $E=1.9 {eV}$ Speed of light in free space, $C=3 \times 10^{8} {ms}^{-1}$ Planck's constant, $h=6.63 \times 10^{-34} {J}- {s}$ As we know that, $E=\;\frac{h c}{\lambda}$ $\Rightarrow \;\; \lambda=\;\frac{h c}{E}$ $\Rightarrow \;\; \lambda \;\; =\;\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{19 \times 1.6 \times 10^{-19}}$ $\;\; =654 \times 10^{-9} {m}$ $\;\; =654 {nm}$ As, wavelength of red light is $600 {nm}$. $\therefore$ Required wavelength will be of red colour.

Given, radius of small drop $=r$ Radius of big drop $=R$ Surface tension $=T$ and mechanical equivalent of heat $=J$ As, small drops combine to form big drop. $\therefore$ Volume of big drop $(V_{B})=n \times$ Volume of small drop $(V_{S})$ $\Rightarrow \;\; \;\frac{4}{3} \pi R^{3}=n \cdot \;\frac{4}{3} \pi r^{3}$ $\Rightarrow \;\; n r^{3}=R^{3}$ $\Rightarrow \;\; r=\;\frac{R}{n^{1 / 3}}$. . . (i) Surface energy of small drop $(E_{S})=$ Surface tension $(T) \times$ Area $(A)$ $\Rightarrow \;\; E_{S}=n \times 4 \pi r^{2} T$ and $E_{B}=4 \pi R^{2} T$ Now, change in energy will be $\Delta E=E_{B}-E_{S}=4 \pi T(n r^{2}-R^{2})$ $\therefore$ Heat energy per unit volume $=\;\frac{\Delta E}{V}=\;\frac{4 \pi T(n r^{2}-R^{2})}{J \times \;\frac{4}{3} \pi R^{3}}$ $=\;\frac{3 T}{J}(\;\frac{n r^{2}}{R^{3}}-\;\frac{1}{R})=\;\frac{3 T}{J}(n \;\frac{R^{2}}{n^{2 / 3} R^{3}}-\;\frac{1}{R})$ =\;\frac{3 T}{J}\[;\frac{n^{1 / 3}}{R}-;\frac{1}{R}]$;;$[from Eq. (i)]$=;\frac{3 T}{J}[;\frac{1}{r}-;\frac{1}{R}]$

Given all resistances have same resistance $R$. Now, we can redraw the circuit as belowLet resistances be $R_{1}, R_{2}, R_{3}$ and $R_{4}$. $\because$ $\;\frac{R_{1}}{R_{3}}=\;\frac{R_{2}}{R_{4}}$ So, circuit will behave as a Wheatstone bridge and no current will flow through midthle resistor. $\therefore \;\; R_{ {eq}} \;\; =\;\frac{(R_{1}+R_{2})(R_{3}+R_{4})}{(R_{1}+R_{2})+(R_{3}+R_{4})}$ $\;\; =\;\frac{(R+R)(R+R)}{(R+R)+(R+R)}$ $\;\; =R$

Given, mass of $man(m)=60 {kg}$ Downward acceleration of lift, $a=1.8 {ms}^{-2}$ Let $T$ be the tension in the rope connected with lift, $g$ be the acceleration due to gravity $(10 {ms}^{-2})$. As, lift is moving in downward direction $\therefore \;\; m g-T \;\; =m a$ $\Rightarrow \;\; T \;\; =m g-m a=m(g-a)$ $\;\; =60(10-1.8)=60 \times 8.2$ $\;\; =492 {N}$ Hence, the weight of the man during downward acceleration is $492 {N}$.

Given, charge passing through circuit, $q=20 {C}$ Potential difference between two plates, $V=15 {V}$ Let $W$ be the amount of work done by battery. $\therefore \;\; W=q V=20 \times 15=300 {J}$

Given, forward resistance, $R_{1}=50 Ω$ Reverse resistance, $R_{2}=$ infinity Battery voltage $=6 {V}$ According to circuit diagram, In this case, diode $D_{1}$ is forward biased, whereas diode $D_{2}$ is reverse biased. So, $D_{2}$ will act as open circuit. By using Kirchhoff's voltage law, $\Rightarrow \;\; 6-50 I-130 I-120 I \;\; =0$ $\Rightarrow \;\; 6 \;\; =300 I$ $\Rightarrow \;\; I \;\; =\;\frac{6}{300}=\;\frac{1}{50}=\;\frac{2}{100}=0.02 {A}=20 {mA}$ Hence, current through $120 Ω=20 {mA}$

Given, power of bulb, $P=1000 {W}$Distance $d=2 {m}$Efficiency of bulb $=1.25 \%$,$\epsilon_{0}=8.854 \times 10^{-12} C^{2} {N}^{-1} {m}^{-1}$,$c=3 \times 10^{8} {ms}^{-1}$ $\therefore$ Intensity, $(I)=\;\frac{Power(P)}{Area(A)}=\;\frac{1}{2} \epsilon_{0} E^{2} c$ $\Rightarrow \;\; \;\frac{1.25}{100} \times \;\frac{1000}{4 \pi(2)^{2}}=\;\frac{1}{2} \times 8.854 \times 10^{-12}$ $\Rightarrow \;\; E=\sqrt{\;\frac{12.5}{16 \pi} \times \;\frac{2}{8.854 \times 10^{-4} \times 3} \times 10^{8}}$ $=13.7 {NC}^{-1}=137 \times 10^{-1} {NC}^{-1}$ or ${Vm}^{-1}$ $\therefore \;\; x=137$

Given, mass of box, $m=2 {kg}$ Force, $F=20 \hat{i}+10 \hat{j} {N}$ Initial speed of box, $u=0 {ms}^{-1}$ Time, $t=10 {s}$ Let acceleration of box is a and displacement along $X$-axis after $10 {s}$ is $S_{x}$. As, $\;\; F=m a$ \Rightarrow \;\; a=F / m=\;{20 \hat{i}+10 \frac\hat{j}}{2}=(10 \hat{i}+5 \hat{j}) {ms}^{-2} By second equation of motion along $X$-axis, $\;s_{x}=u_{x} t+\;\frac{1}{2} a_{x} t^{2}$ $\;s_{x}=0+\;\frac{1}{2} \times 10 \times (10)^{2}=500 m$ Hence, displacement along $X$-axis after $10 {s}$ is $500 {m}$.