17 Mar 2021 Shift 2

Velocity of rubber ball when it strikes to the ground, $v_{0}=\sqrt{2 {gh}_{0}}$ Using the formula of coefficient of restitution, ${e}=\;\frac{\;\text{ velocity after collision }\;}{\;\text{ velocity before collision }\;}$ $\Rightarrow \;\; e=\;\frac{v_{1}-0}{v_{0}-0} \Rightarrow e=\;{v_{1}}/{\sqrt{2 g h_{0}}}$ $\Rightarrow \;\; v_{1}=e \sqrt{2 g h_{0}}$...(i) As, initial height of the ball $=h_{0}$. $\therefore$ The first height of the rebound, $h_{1}=\;\frac{v_{1}^{2}}{2 g}$ $\Rightarrow \;\; h_{1}=e^{2} h_{0}$ [Using Eq. (i)] The nth height of the ball to the rebound, $h_{n}=\;\frac{v_{n}^{2}}{2 g} \Rightarrow h_{n}=e^{2 n} h_{0}$ The velocity of the ball after $n$th rebound, $v_{n}=e^{n} v_{0}$ Now, the total distance travelled by the ball after nth rebound is $H=h_{0}+2 h_{1}+2 h_{2}+2 h_{3} . . .$ $H=h_{0}+2 e^{2} h_{0}+2 e^{4} h_{0}+2 e^{6} h_{0} . . .$ $H=h_{0}[1+2 e^{2}(1+e^{2}+e^{4}+e^{6} . . .)]$ Using the formula, $1+e^{2}+e^{4}+. . .=\;\frac{1}{1-e^{2}}$ $\Rightarrow \;\; H=h_{0}[1+2 e^{2}(\;\frac{1}{1-e^{2}})]$ $\Rightarrow \;\; H=h_{0}(\;\frac{1+ {e}^{2}}{1- {e}^{2}})$ $\Rightarrow \;\; H=5(\;\frac{1+(0.81)}{1-(0.81)}) \;\; ( \because e^{2}=\;\frac{h_{1}}{h_{0}}=\;\frac{81}{100})$ $H=47.6 {m}$ Now, the total time taken by the ball to come to rest $T=t_{0}+2 t_{1}+2 t_{2}+. . .$ $T=\sqrt{\;\frac{2 h_{0}}{g}}+2 \sqrt{\;\frac{2 h_{1}}{g}}+2 \sqrt{\;\frac{2 h_{2}}{g}}+. . .$ $T=\sqrt{\;\frac{2 h_{0}}{g}}[1+2 e+2 e^{2}+. . .]$ $T=\sqrt{\;\frac{2 h_{0}}{g}}[1+2 e(1+e+e^{2}+. . .)]$ $T=\sqrt{\;\frac{2 h_{0}}{g}}(\;\frac{1+e}{1-e}) \Rightarrow T=\sqrt{\;\frac{2(5)}{10}}(\;\frac{1+\sqrt{0.81}}{1-\sqrt{0.81}})$ $T=19 {s}$ The average velocity, $v_{\;\text{avg }\;}=\;\frac{H}{T}=\;\frac{47.6}{19}$ $=2.5 {m} / {s}$ Hence, the average velocity is $2.5 \frac{m}{s}$.

For polyatomic gas molecule has 3 rotational degrees of freedom, 3 translational degrees of freedom, and 2 vibrational modes. So, number of vibrational degrees of freedom $=2(2)=4$ $\therefore$ Total number of degrees of freedom, $f=3+3+4=10$ Here, ratio of molar specific heats is given as $\beta=\;\frac{C_{p}}{C_{v}}$ In terms of degrees of freedom, $\beta=1+\;\frac{2}{f}$ Substituting the value of $f$ in above equation, we get $\beta=1+\;\frac{2}{10} \Rightarrow \beta=1.2$

Given, mass block, $m=1 {kg}$ Initial amplitude, $A_{0}=12 {cm}$ Final amplitude, $A=6 {cm}$ The time taken to reduce the amplitude, $t=2 {min}=120 {s}$ Using the expression of damped oscillation, $A=A_{0} e^{-\;\frac{b}{2 m} t}$ Substituting the values in the above equation, we get $6=12 e^{-\;\frac{b(120)}{2(1)}}$ $\Rightarrow \;\; e^{60 b}=2$ or $60 b=10 \log 2$ $\Rightarrow \;\; b=\;\frac{0.693}{60}=1.16 \times 10^{-2} \frac{kg}{s}$

The given circuit, with the output of the respective gates is as given below.The Boolean expression of the output is $Y=(\overline{A \cdot B}) \cdot (A+B)$ $\Rightarrow \;\; Y=(\overline{A}+\overline{B}) \cdot (A+B)$ (using de-Morgan's theorem, $\overline{x \cdot y}=\overline{x}+\overline{y}$ ) $\Rightarrow \;\; Y=\overline{A} \cdot B+\overline{B} \cdot A$ This represents the Boolean expression for XOR gate. Alternate solution This question can also be verified from the following truth table.AB$P=\overline{A \cdot B}$ $Q=A+B$ $Y=P\cdot Q$00100011111011111010Hence, the output of this circuit represents the output of a XOR gate.

Given, The volumetric strain is $\;\frac{\Delta V}{V}=1.36 \%$ The depth beneath the water surface, $h=2 {km}$ The pressure inside the water surface up to $2 {km}$, $p=\rho g h$ Substituting the values in the above equation, we get $p=1000 \times 9.81 \times 2000$ $p=19.62 \times 10^{6} {Pa}$ The bulk modulus of the object, $\beta=\;\frac{p}{\frac{\Delta V}{V}}$ Substituting the values in the above equation, we get $\beta=\;\frac{19.62 \times 10^{6}}{\frac{1.36}{100}}$ $\beta=1.44 \times 10^{9} {N} / {m}^{2}$ Hence, the ratio of the hydraulic stress to the corresponding hydraulic strain will be $1.44 \times 10^{9} {N} / {m}^{2}$.

Given Height of satellite from planet's surface, $h=11 R$ So, the total distance from the centre of planet $=11 R+R=12 R$ Time period of planet $=24 {h}$ Similarly, the total distance of second satellite from the centre of planet $=2 R+R=3 R$ Using the Kepler's law $T^{2} \propto R^{3}$ $\Rightarrow \;\; (\;\frac{T_{1}}{T_{2}})=(\;\frac{R_{1}}{R_{2}})^{3 / 2}$ Substituting the values in the above equation, we get $(\;\frac{24}{T})=(\;\frac{12 R}{3 R})^{3 / 2}$ $\therefore$ The time period of another satellite is $3 {h}$.

Given, The frequency of the sound wave, $f=245 {Hz}$ The speed of the travelling wave, $v=300 {m} / {s}$ As, total distance of to and fro motion is $6 {cm}$. Hence, the amplitude of the wave, $A=6 / 2=3 {cm}=0.03 {m}$ As we know, Angular frequency is given as $\omega=2 \pi f$ Substituting the value of $f$, we get $\omega=2 \pi(245)$ $\omega=1.54 \times 10^{3} {rad} / {s}$ Propagation constant is given as $k=\;\frac{\omega}{v}$ Substituting the value of $v$ and $\omega$, we get $k=\;\frac{1.54 \times 10^{3}}{300} \Rightarrow k=5.1 {m}^{-1}$ General mathematical expression for a travelling wave is given as $y=A \sin (k x-\omega t)$ Substituting the values in the above equation, we get $y=0.03 \sin (5.1 x-1.5 \times 10^{3} t)$

In graph (A) vertical line is isochoric process, so the graph (A) is incorrect. In graph (B) horizontal line is isobaric process, so the graph (B) is incorrect. In graph (C) and (D) vertical line is isothermal process and the curve line is adiabatic process. So, the correct representation of thermodynamic process graph is (C) and (D). So, the correct option is (b)

Given, equation of velocity, $v \;\; =v_{0}+g t+F t^{2}$ $\;\text{ At }\; t \;\; =0, x=0$ $\;\text{ At }\; t \;\; =1 s x=?$ We know that, $v=\;\frac{d x}{d t}$ $d x=v d t$ Integrating the above equation with proper limits, we get $int_{0}^{x} d x=int_{0}^{t-1} v d t$ $x-0=int_{0}^{t-1}(v_{0}+g t+F t^{2}) d t$ \Rightarrow \;\; x=\[v_{0} t+;\frac{g t^{2}}{2}+;\frac{F t^{3}}{3}]{0}^{1} ⇒ x=$[v{0}(1)+;\frac{g(1)^{2}}{2}+;\frac{F(1)^{3}}{3}]$-0$$\Rightarrow ;; x=[v_{0}+;\frac{g}{2}+;\frac{F}{3}]$

Given, carrier signal, $C(t)=25 \sin (2.512 \times 10^{10} t)$ Message signal, $m(t)=5 \sin (1.57 \times 10^{8} t)$ Angular frequency of the message signal, $\omega_{m}=1.57 \times 10^{8} {rads}^{-1}$ Frequency of the message signal, of the message $f_{m}=\;\frac{\omega_{m}}{2 \pi}$ Substituting the values in the above equation, we get $f_{m}=\;\frac{1.57 \times 10^{8}}{2 \pi} {Hz}$ Bandwidth of the modulated signal is $\beta=2 f_{m}$ Substituting the values in the above equation, we get $\beta=2(\;{1.57 \times 10^{ {B}}}/{2 \pi})$ $\beta=50 {MHz}$ Hence, the bandwidth of the modulated signal is $50 {MHz}$.

The given circuit can be drawn as Since in series combination, the current through each resistance remains same. So, equivalent resistance of the circuit is given as $R_{\;\text{equivalent }\;}=R+r_{1}+r_{2}$ and equivalent emf, $E_{\;\text{equivalent }\;}=2 E+E=3 E$ From Ohm's law, I = {E_{\;\text{equivalent}}/{R_{\;\text{equivalent}\; }}\Rightarrow I=\frac{3 E}{R+r_{1}+r_{2}} When potential difference is zero across the first cell, then potential positive terminal is equal to the potential at negative terminal. $V_{P}=V_{N}$ $2 E=I r_{1}$ Substituting the values in the above equation, we get $2 E=\;\frac{3 E}{R+r_{1}+r_{2}} r_{1}$ $2 R+2 r_{1}+2 r_{2}=3 r_{1} \Rightarrow R=\;\frac{r_{1}-2 r_{2}}{2}$ $\Rightarrow \;\; R=\;\frac{r_{1}}{2}-r_{2}$

For the straight wire, the expression of the magnetic field, $B_{1}=B_{3}=\;\frac{\mu _{0} I}{4 \pi r}$ For the semicircular wire, the expression of the magnetic field, $B_{2}=\;\frac{\mu _{0} I}{4 r}$ Hence, the total magnetic field at the point $P$, $B=B_{1}+B_{2}+B_{3}$ Substituting the values in the above equation, we get $B=\;\frac{\mu _{0} I}{4 \pi r}+\;\frac{\mu _{0} I}{4 r}+\;\frac{\mu _{0} I}{4 \pi r} \Rightarrow B=\;\frac{\mu _{0} I}{4 \pi r}(1+1+\pi)$ $B=\;\frac{\mu _{0} I}{4 \pi r}(2+\pi)$

As, A is directly connected to the positive terminal of the battery, $V_{A}=10 V$ and $V_{C}=0$ By nodal analysis at $B$, $\;\frac{V_{B}-10}{100}+\;\frac{V_{B}-V_{D}}{15}+\;\frac{V_{B}-0}{10}=0$ $53 V_{B}-20 V_{D}=30$...(i) By nodal analysis at $D$, $\;\frac{V_{D}-10}{60}+\;\frac{V_{D}-V_{B}}{15}+\;\frac{V_{D}-0}{5}=0$ $-4 V_{B}+17 V_{D}=10$....(ii) Solving Eqs. (i) and (ii) by substitution method, we get $V_{D}=0.792 {V} \Rightarrow V_{B}=0.865 {V}$ The current through the galvanometer, $I=\;\frac{V_{B}-V_{D}}{R}$ Substituting the values in the above equation, we get $I=\;\frac{0.865-0.792}{15} \Rightarrow I=4.87 {mA}$

We know that, the expression of maximum velocity during oscillation, $V_{\max }=A \omega$ Given, $V_{\max }(A)=V_{\max }(B) \Rightarrow A_{1} \omega_{1}=A_{2} \omega_{2}$ Now, the ratio of the amplitude during oscillation, $\;\frac{A_{1}}{A_{2}}=\;\frac{\omega_{2}}{\omega_{1}}$ We know that, $\omega=\sqrt{k / m}$ where, $k$ is the spring constant, $m$ is the mass of the object. Substituting the value of $\omega$ in Eq. (i), we get $\;\frac{A_{1}}{A_{2}}=\sqrt{\;\frac{k_{2}}{k_{1}}}$

A. In a purely resistive $A C$ circuit, the phase difference between the current and voltage is zero. B. In a purely inductive ${AC}$ circuit, the current lags the voltage, so the phase difference between the current and voltage is $\pi / 2$. C. In a purely capacitive AC circuit the current leads the voltage, so the phase difference between the current and voltage is $\pi / 2$. D. The phase difference between current & voltage in an $L-C-R$ series circuit is $\varphi =\tan ^{-1} \;\frac{(X_{C}-X_{L})}{R}$ $\therefore$ The correct match is A-(2), B-(3), C-(1), D-(4).

Let $v$ is the speed of the third block $C$. The velocity of centre of mass of $A$ and $B$ is $v_{C M}=\;\frac{v}{2}$ The spring is compressed maximum by $x$ distance. Using the law of conservation of energy. $\;\frac{1}{2} m v^{2}=\;\frac{1}{2} m(\;\frac{v}{2})^{2}+\;\frac{1}{2} m(\;\frac{v}{2})^{2}+\;\frac{1}{2} k x^{2}$ $\Rightarrow \;\; \;\frac{1}{4} m v^{2}=\;\frac{1}{2} k x^{2} \Rightarrow x=\sqrt{\;\frac{m v^{2}}{2 k}}$ $\Rightarrow \;\; x=v \sqrt{\;\frac{m}{2 k}}$

When an electron jumps from the higher energy level to $n=2$ orbit, Balmer series of the line spectrum is obtained. The Balmer series of the hydrogen atom lies in the visible region. However, Brackett and Paschen series of hydrogen atom lies in the infrared region and Lyman series of hydrogen atom lies in the ultraviolet region.

We know that, the photoelectric effect equation, ${KE}=h f-\varphi$ Here, $K E=$ kinetic energy of the electrons $h=$ Planck's constant $f=$ frequency of the light and $\varphi =$ work-function of the photocathodes. For two identical photocathodes, $\;\frac{1}{2} m v_{1}^{2}=h f_{1}-\varphi$....(i) $\;\frac{1}{2} m v_{2}^{2}=h f_{2}-\varphi$.....(ii) Since, the material of photocathodes is same, so the value of the work-function $\varphi$ is same. Subtracting Eq. (ii) from Eq. (i) we get $\;\frac{1}{2} m v_{1}^{2}-\;\frac{1}{2} m v_{2}^{2}=h f_{1}-h f_{2}$ $v_{1}^{2}-v_{2}^{2}=\;\frac{2 h}{m}(f_{1}-f_{2})$

As we know, the inductive reactance is directly proportional to the frequency of the $A C$ circuit $\;\text{ i.e., }\;\;X_{L}=\omega L$ $\Rightarrow \;X_{L}=2 \pi f L$ $( \because \omega=2 \pi f)$ Here, $f$ is the frequency of the $A C$ circuit, $L$ is the inductive resistance and $X_{L}$ is the inductive reactance. When the frequency of an AC circuit is halved, then the inductive reactance of the circuit is also halved. i.e. $X_{L}^{'}=\;\frac{X_{L}}{2}$ Using Ohm's law, $\;\; I=\;\frac{V}{X_{L}}$ When the frequency is halved, then the current $I^{'}=\;\frac{V}{X_{L}^{'}} \Rightarrow I^{'}=\;\frac{V}{X_{L} / 2}$ $I^{'}=2 I$ The current becomes doubled.

As we know the moment of inertia of the solid sphere, $I=\;\frac{2}{5} m R^{2}$Acceleration of sphere on inclined plane, $a=\;{g \sin \theta}/{1+\;\frac{I}{m R^{2}}}$ Substituting the values in the above equation, we get $a=\;{g \sin 30^{\circ}}/{1+\;\frac{\;\frac{2}{5} m R^{2}}{m R^{2}}}=\;\frac{10+1{/}2}{(1+2{/}5)}=\;\frac{5 \times 5}{7}$ $a=3.5 \frac{m}{s}^{2}$ At the top of incline, $v=u+a t \Rightarrow 0=1-3.5 t$ $t=1 {/} 3.5 {s}$ $\therefore$ Total time $=$ time of ascent $+$ time of decent As, time of ascent $=$ time of decent $\Rightarrow$ Total time $=2$ (time of ascent) Total time $=2(1{/}3.5) {s}$ Total time $=0.57 {s}$ Hence, the total time taken to sphere to return the starting point $A$ is $0.57 {s}$.

Intensity of the electromagnetic radiation is given as $I=\;\frac{\;\text{ Energy }\;}{\;\text{ Area }\; \times \;\text{ Time }\;}=\;\frac{\;\text{ Power }\;}{\;\text{ Area }\;}=U_{\;\text{avg }\;} C=\;\frac{1}{2} C \epsilon_{0} E_{0}^{2}$ Here, $U_{\;\text{avg }\;}=$ average energy density, $C=$ speed of radiation in air (or vacuum), $\epsilon_{0}=$ permittivity of the free space and $E_{0}=$ peak value of the electric field. $\Rightarrow \;\; P \propto E_{0}^{2} \Rightarrow \;\frac{P_{1}}{P_{2}}=\;\frac{E_{1}^{2}}{E_{2}^{2}}$ $\Rightarrow \;\; \;\frac{E_{1}}{E_{2}}=\sqrt{\;\frac{P_{1}}{P_{2}}}$ Substituting the values in the above equation, we get $\Rightarrow \;\; \;\frac{E_{1}}{E_{2}}=\sqrt{\;\frac{60}{100}}$ $\Rightarrow \;\; E_{1}=\sqrt{\;\frac{3}{5}} E_{2}=\sqrt{\;\frac{3}{5}}=E$ $( \because E_{2}=E)$ Comparing with $\sqrt{\;\frac{x}{5}} E$, we get $x=3$

Given, mass of the body, $m=1 {kg}$ Coefficient of static friction, $\mu =1 {/} \sqrt{3}$ Let's draw the free body diagram of the block Using the condition of the equilibrium In $x$-direction, $F \cos \theta=f=\varphi =\mu N$...(i) In $y$-direction, $F \sin \theta+N=m g$ $\Rightarrow \;\; N=m g-F \sin \theta$....(ii) Substituting the value of $N$ in Eq. (i), we get $F=\;\frac{\mu m g}{\cos \theta+\mu \sin \theta} \Rightarrow F=\;{\mu m g}/{\sqrt{1+\mu ^{2}}}$ Substituting the values in the above equation, we get $F=\;{\;\frac{1}{\sqrt{3}} \times 10}/{\sqrt{1+(\;\frac{1}{\sqrt{3}})^{2}}} \Rightarrow F=5 {N}$ Hence, the body move by applying minimum possible force of $5 {N}$. So, the value of $F$ will be 5 .

The free body diagram for the wooden block is shown belowUsing the condition of the equilibrium, In the $x$-direction, the summation of all the forces is to be zero. $f-T \;\; =0$ $\mu R-T \;\; =0$ $T \;\; =0.5 R$...(i) In the $y$-direction, , the summation of all the forces is to be zero. $T+R-90=0$ $\Rightarrow \;\; 0.5 R+R-90=0$ $R=60 {N}$ Hence, the normal force on the wooden block is $60 {N}$ Using the Eq. (i), $T=0.5(60)=30 {N}$ Hence, the maximum value of the tension in the rope, so that wooden block will not move is $30 {N}$.

Given, the radius of the spherical drop $r=(\;\frac{3}{40 \pi})^{1 {/} 3} \times 10^{-3} {cm}$ The volume of 100 spherical drops, $V=100 \times \;\frac{4 \pi r^{3}}{3}$ Substituting the value of the radius in the above equation, we get $V=100 \times \;\frac{4 \pi}{3} \times \;\frac{3}{40 \pi} \times 10^{-9}$ $V=10^{-8} {cm}^{3}$ Now, Volume of the film $=$ Area of the film $\times$ Thickness of the film $\Rightarrow$ Volume of the film $=4 \times t$ The film is made up of 100 spherical drops. $\therefore$ The volume of the film $=$ Volume of 100 spherical drops $4 t=10^{-8} \Rightarrow t=25 \times 10^{-12} {m}$ The thickness of the oleic layer, $t_{0}=0.01 \times t=25 \times 10^{-14} {m}$ Comparing with $x \times 10^{-14}$, we get $x=25$

As we know, the force on the particle is given as $F=-\;\frac{d U(r)}{d r}$ Given, $U(r)=U_{0} r^{4}$ $\Rightarrow \;\; F=-\;\frac{d(U_{0} r^{4})}{d r} \Rightarrow F=-4 U_{0} r^{3}$ As we know, the force on the particle moving in a circular orbit of radius $r$ will be centripetal force $|\;\frac{m v^{2}}{r}|=4 U_{0} r^{3}$ $\Rightarrow v^{2} \propto r^{4}$ $v \propto r^{2}$ .....(i) Hence, the velocity of the particle is square of the radius of the orbit. Using the Bohr's quantisation condition, which states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\;\frac{h}{2 \pi}$, here, $h$ is Planck's constant. i.e, $n \;\frac{h}{2 \pi}=m v r$ [From Eq. (i)] $\Rightarrow \;\; r^{3} \propto n \Rightarrow r \propto n^{1 / 3}$ Comparing the above relation with $n^{1 {/} \alpha}$, we get $\alpha=3$

Given, The electric field in the region, $E=\;\frac{2}{5} E_{0} \hat{i}+\;\frac{3}{5} E_{0} \hat{j}$ Here, $E_{0}=4 \times 10^{3} \;\frac{ {N}}{ {C}}$ Area of the rectangular surface, $A=0.4 {m}^{2}$ The direction of electric field vector and area vector is same, so the angle between the electric field vector and area vector is 0 . As we know the expression of electric flux, $\varphi =E \cdot A \cos \theta$...(i) Here, $E$ is the electric field vector, and $A$ is the surface area of the surfaces. Consider the surface parallel to the $Y-Z$ plane, so the area vector, A=0.4 i {m}^{2} Substituting the values in Eq. (i), we get $\varphi =E \cdot A \cos 0^{\circ} \Rightarrow \varphi =\;\frac{2}{5} E_{0}(0.4)$ $\Rightarrow \;\; \varphi =\;\frac{2}{5}(4 \times 10^{3})(0.4)=640 {Nm}^{2} {C}^{-1}$ Hence, the electric flux of the surface parallel to the $Y-Z$ plane is $640 {Nm}^{2} {C}^{-1}$.

As we know the centre of mass of the quarter disc, $=(\;\frac{4 R}{3 \pi}, \;\frac{4 R}{3 \pi})$ Since $R=a$ $\therefore$ The centre of mass of the quarter disc is $(\;\frac{4 a}{3 \pi}, \;\frac{4 a}{3 \pi})$. Hence, the value of $x$ is 4 .

Given, Image distance, $v=10 {m}$ Object distance, $u=(\;\frac{3}{2}) \times 10-15 {m}$ and $n_{2}=\;\frac{3}{2} n_{1}$ $( \because n \propto \;\frac{1}{\lambda})$ Using the lens Maker's formula, $\;\frac{n_{2}}{v}-\;\frac{n_{1}}{u}=(\;\frac{n_{2}-n_{1}}{R})$ Substituting the values in the above equation, we get $\;\frac{\;\frac{3}{2} n_{1}}{10}-\;\frac{n_{1}}{-15}=(\;\frac{\;\frac{3}{2} n_{1}-n_{1}}{R}) \Rightarrow R=\;\frac{30}{13} {m}$ The radius of the curved surface is $30 / 13 {m}$. Comparing with $x / 13$, we get $x=30$