COMEDK 2022 Question Paper with Solutions

As Young's modulus Y for two wires is different, hence strain is different but stress is same as equal force F on each wire having same diameter and area of cross-section act on both wires.

If fringe widths of air and water are $\beta_{\mathrm{air}}$ and $\mathrm{{\beta _{water}}}$ respectively, then fringe width, $\mathrm{{\beta _{water}} = {{{\beta _{air}}} \over \mu }}$

Given, $\mu = {4 \over 3},{\beta _{air}} = 0.4$

Hence, ${\beta _{water}} = {{0.4} \over {4/3}} = 0.3$ mm

Given, collector resistance = $\mathrm{R_{out}=4~k\Omega}$

$\mathrm{\beta=100}$

$\mathrm{R_{in}=2~k\Omega}$

$\mathrm{V_{out}=5~V}$

$\therefore$ Voltage amplification,

$\mathrm{{A_V} = \beta {{{R_{out}}} \over {{R_{in}}}} = {{{V_{out}}} \over {{V_{in}}}}}$

$\because$ $\mathrm{{V_{in}} = {{{V_{out}} \times {R_{in}}} \over {\beta \times {R_{out}}}}}$

$= {{5 \times 2 \times {{10}^3}} \over {100 \times 4 \times {{10}^3}}}$

$= {1 \over {40}} = 25$ mV

As work done = area under p-V curve, work done is maximum in case I.

Change in internal energy is independent of the path from A to B. Therefore, in all cases change in internal energy is same.

The most commonly used material in LEDs is gallium arsenide.

From the diagram we get,

$\tan 30^\circ = {1 \over {\sqrt 3 }} = {x \over {10}}$

$\Rightarrow x = {{10} \over {\sqrt 3 }}$

If a man slides down with some acceleration, then its apparent weight decreases. For critical condition, rope can bear only $\frac{2}{3}$ of his weight. If a is the minimum acceleration, then tension in the rope

$=m(g-a)=$ breaking strength

$\Rightarrow m(g-a)=\frac{2}{3}mg$

$\Rightarrow a=g-\frac{2g}{3}=\frac{g}{3}$

As, acceleration due to gravity at height $h$,

$$\begin{aligned} & g^{\prime}=\frac{G M}{(R+h)^2}, \\ & \Rightarrow \quad \frac{g}{16}=\frac{G M}{R^2}\left(\frac{R^2}{(R+h)^2}\right)=g \frac{R^2}{(R+h)^2}\left(\because g^{\prime}=\frac{g}{16}\right) \\ & \Rightarrow \quad \frac{1}{4}=\frac{R}{R+h} \Rightarrow h=3 R \\ \end{aligned}$$

Lyman series lies in the ultraviolet region whereas Paschen, Brackett and Pfund series lie in the infrared region.

The potential energy is given by

$$\begin{aligned} U & =\int F \cdot d r=\int \frac{K}{r^2} d r=K \int r^{-2} d r \\ & =K\left[\frac{r^{-1}}{-1}\right]=\frac{-K}{r} \end{aligned}$$

The kinetic energy is given by

$K=\frac{1}{2} m v^2=\frac{1}{2}\left(\frac{K}{r}\right) \quad\left[\because \frac{m v^2}{r}=\frac{K}{r^2}\right]$

Total energy, $E=U+K$

$=\frac{-K}{r}+\frac{K}{2r}=\frac{-K}{2r}$

Radius of the circular disc $=R$

Suppose, centre of mass of remaining portion is at a distance $C_2$ from the centre $C$.

Now, $A_1\left(C C_1\right)=A_2\left(C C_2\right)$

$\therefore CC_2=\frac{A_1}{A_2}(CC_2)$

For square, $(\text { side })^2+(\text { side })^2=(\text { radius })^2$

$$\begin{array}{r} 2(\text { side })^2=R^2 \\ (\text { side })^2=\frac{R^2}{2} \\ \text { side }=\frac{R}{\sqrt{2}} \end{array}$$

Now, $A_1\left(C C_1\right)=A_2\left(C C_2\right)$

$A_1=$ area of the square $=\left(\frac{R}{\sqrt{2}}\right)^2$

$A_2=$ area of the remaining part of the circle

$=\pi R^2-\left(\frac{R}{\sqrt{2}}\right)^2$

$$\begin{aligned} C C_2 & =\frac{\left(\frac{R}{\sqrt{2}}\right)^2}{\pi R^2-\left(\frac{R}{\sqrt{2}}\right)^2}\left(\frac{R}{2}\right) \\ & =\frac{R^2}{2\left(\pi R^2-\frac{R^2}{2}\right)} \cdot \frac{R}{2} \\ & =\frac{R^2}{2 \pi R^2-R^2} \cdot \frac{R}{2} \\ & =\frac{R \cdot R \cdot R}{R^2(2 \pi-1) 2}=\frac{R}{4 \pi-2}\end{aligned}$$

Given, speed of car, $u=54 \mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}$

Since, car stops in $3 \mathrm{~s}$ on applying brakes, so acceleration can be calculated from first equation of motion,

$$\begin{aligned} & \quad v=u+a t \Rightarrow 0=15+a(3) \\ & \Rightarrow \quad a=-\frac{15}{3}=-5 \mathrm{~ms}^{-2} \end{aligned}$$

Using second equation of motion,

$$\begin{aligned} s & =u t+\frac{1}{2} a t^2 \\ & =15 \times 3+\frac{1}{2} \times(-5) \times(3)^2 \\ & =4 5-22.5=22.5 \mathrm{~m} \end{aligned}$$

Resultant current is superposition of two currents, i.e. $I$ (instantaneous total current)

$=6+I_0 \sin \omega t$

$\mathrm{DC}$ ammeter will read average value

$$\begin{aligned} & =\overline{6+I_0 \sin \omega t}\\ & =6 \mathrm{~A} \end{aligned}$$

$\left(\because \overline{I_0 \sin \omega t}=0\right)$

$\mathrm{AC}$ ammeter will read average value

$$\begin{aligned} & =\sqrt{\overline{\left(6+I_0 \sin ^2 \omega t\right)^2}} \\ & =\sqrt{36+12 I_0 \sin \omega t+I_0^2 \sin ^2 \omega t}\end{aligned}$$

$\left(\because \overline{I_0 \sin \omega t}=0\right)$

Since, $\overline{\sin ^2 \omega t}=\frac{1}{2}$ and $I_{\mathrm{rms}}=8=\frac{I_0}{\sqrt{2}}$

$$\begin{aligned} \therefore \mathrm{AC} \text { reading }=\sqrt{36+\frac{I_0^2}{2}} & =\sqrt{36+64} \\ & =10 \mathrm{~A}\end{aligned}$$

If we differentiate the equation of motion, $x=1-t-t^2$ w.r.t. time, we get velocity.

$v=\frac{dx}{dt}=-1-2t$

Comparing with $v=u+at$, we have $u=-1~\mathrm{ms^{-1}}$ and $a=-2~\mathrm{ms^{-2}}$

At $t=0,x=1$ m. Then, $u$ and $a$ both are negative.

Hence, $x$-coordinate of particle will go on decreasing.

The given network can be simplified as shown in following figure

The equivalent capacitance between P and Q is given by

${1 \over {{C_{eq}}}} = {1 \over 3} + {1 \over 3} + {1 \over 3} = 1\,\mu F$

${C_{eq}} = 1\,\mu F$

Given, emf of cell, $E=2V$

Load resistance, $R_L=1.5~\Omega$

Current flowing the circuit is

$I=\frac{E}{R+r}$

Power transferred to the load, R,

$P = {I^2}R = {\left( {{E \over {R + r}}} \right)^2}\,.\,R = {{{E^2}R} \over {{{(R + r)}^2}}}$

Power P will be maximum, if

$dP/dR = 0$

$\Rightarrow {d \over {dR}}{{{E^2}R} \over {{{(R + r)}^2}}} = 0$

$$\begin{array}{ll} \Rightarrow & \frac{E^2(R+r)^2-E^2 \cdot R \cdot 2(R+r)}{(R+r)^3}=0 \\ \Rightarrow & E^2(R+r)[R+r-2 R]=0 \\ \Rightarrow & E^2(R+r)(r-R)=0 \\ \Rightarrow & r-R=0 \\ \Rightarrow & r=R \end{array}$$

$\therefore$ Maximum power, $P=\frac{E^2 R}{(R+r)^2}=\frac{E^2 R}{(R+R)^2}=\frac{E^2}{4 R}$

$\Rightarrow \quad P=\frac{2^2}{4 \times 1.5}=\frac{1}{3}=0.33 \mathrm{~W}$

According to fluid dynamics, for maximum level of water in tank,

$A_1v_1=A_2v_2$

Given, $A_1=A$,

velocity, $v_1=4$ ms$^{-1}$

area of hole $A_2=\frac{A}{2}$

$\Rightarrow A\times4=\frac{A}{2}\times\sqrt{2gh}$

$\Rightarrow 16=\frac{2gh}{4}$

$\Rightarrow h=\frac{16\times4}{2\times10}=3.2$ m

Let number of turns of the regular hexagon $=n$

Now, $n \times 6 m=24 m$

$\therefore \quad n=4$

Magnetic moment, $M=n I A=4 I A$

$\therefore$ Area of hexagon

$$\begin{aligned} & =\frac{1}{2} m^2 \sin ^2 60^{\circ}+m \cdot 2 m \sin 60^{\circ}+\frac{1}{2} m^2 \sin 120^{\circ} \\ & =\frac{\sqrt{3}}{4} m^2+\sqrt{3} m^2+\frac{\sqrt{3} m^2}{4} \\ & =\frac{6 \sqrt{3} m^2}{4}=\frac{3 \sqrt{3} m^2}{2} \end{aligned}$$

$$\begin{aligned} \text { Hence, magnetic moment } & =4 I\left(\frac{3 \sqrt{3} m^2}{2}\right) \\ & =6 \sqrt{3} I m^2 \end{aligned}$$

As,

$K=\frac{1}{2} m v^2$

$\Rightarrow \quad K=\frac{1}{2 m} p^2$

$\Rightarrow \quad p^2=2 m K$

$\Rightarrow \quad p=\sqrt{2 m K}$

Suppose, $p_1$ is the momentum of the body and $K_1$ = kinetic energy of the body initially. As the kinetic enery of the body is increased by $300 \%$, new kinetic energy of the body $=K_1+3 K_1$

Hence, percentage change in momentum

$$\begin{aligned} & =\frac{p_2-p_1}{p_1} \times 100 \\ & =\frac{\sqrt{2 m K_2}-\sqrt{2 m K_1}}{\sqrt{2 m K_1}} \times 100 \\ & =\frac{\sqrt{2 m} \sqrt{K_2}-\sqrt{K_1}}{\sqrt{2 m} \sqrt{K_1}} \times 100 \\ & =\frac{\sqrt{K_1+3 K_1}-\sqrt{K_1}}{\sqrt{K_1}} \times 100 \\ & =\frac{\sqrt{4 K_1}-\sqrt{K_1}}{\sqrt{K_1}} \times 100=100 \% \end{aligned}$$

The variation of g with angular velocity $\omega$ is given by

$g'=g-R\omega^2$

If earth were to spin faster, that is angular velocity increases, then except at poles, the weight of bodies will decrease at all places.

Frequency received by bat after reflection from wall-1,

${f_1} = f\left[ {{{v + u} \over {v - u}}} \right] = 4.5 \times {10^4}\left[ {{{330 + 6} \over {330 - 6}}} \right]$

Given, $f = 4.5 \times {10^4}$ Hz

$\Rightarrow {f_1} = 4.67 \times {10^4}$ Hz

Frequency received by bat after reflection from wall-2,

${f_2} = f\left[ {{{v - u} \over {v + u}}} \right]$

$= 4.5 \times {10^4}\left[ {{{330 - 6} \over {300 + 6}}} \right]$

$= 4.34 \times {10^4}$ Hz

From the given circuit diagram, output $Y$ is given as

$$\begin{aligned} & Y=(\overline{\bar{A}}+\overline{\bar{B}}) A B \\ & =(\overline{\bar{A}} \cdot \overline{\bar{B}}) A B \quad \text { [By De morgan's law, } \overline{A+B}=\bar{A} \cdot \bar{B}] \\ & =(A B)(A B) \quad[\because A \cdot A=A] \\ & =A B \\ & =\text { output of AND gate. } \end{aligned}$$

Given, $A_m=20$ cm

$A_c=40$ cm

$\therefore$ Modulation index,

$\mu=\frac{A_m}{A_c}=\frac{20}{40}=0.5$

Given, $d=0.2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m}$

$D=1 \mathrm{~m}, \lambda=500 \mathrm{~nm}=5 \times 10^{-7} \mathrm{~m}$

The distance between 6th maxima and 10th minima is given as

$$\begin{aligned}\Delta x & =\left(x_{10}\right)_{\text {dark }}-\left(x_6\right)_{\text {bright }} \\ & =\frac{(2 \times 10-1) D \lambda}{2 d}-\frac{6 D \lambda}{d} \\ & =\frac{D \lambda}{d}\left[\frac{19}{2}-6\right) \\ & =\frac{1 \times 5 \times 10^{-7}}{2 \times 10^{-4}}\left[\frac{7}{2}\right]=\frac{35}{4} \times 10^{-3} \mathrm{~m} \\ & =8.75 \times 10^{-3} \mathrm{~m} \\ & =8.75 \mathrm{~mm} \end{aligned}$$

Hence, $8 \mathrm{~mm}$ is closest to $8 \mathrm{~mm}$.

Force on a moving charged particle in uniform magnetic field,

$F=Bqv\sin\theta$ ..... (i)

Since, charge particle moves along the axis of circular current carrying loop, therefore, $\theta=0^\circ$ or 180$^\circ$.

When $\theta=0^\circ,F=Bqv\sin0^\circ$ [from Eq. (i)]

$F=0$

When $\theta=180^\circ,F=Bqv\sin180^\circ$

$F=0$

The energy emitted by a body, in the form of radiation on account of its temperature, is called thermal radiation. These radiations are heat radiations and travel along straight lines with speed of light.

According to the question,

$x=5 \sin \left(4 t-\frac{\pi}{6}\right)$

When $x=3$,

$$\begin{aligned} & \Rightarrow \quad \frac{3}{5}=\sin \left(4 t-\frac{\pi}{6}\right) \\ & \Rightarrow \quad 4 t-\frac{\pi}{6}=37^{\circ} \\ & \text { Hence, velocity }=\frac{d x}{d t}=5 \times 4 \cos \left(4 t-\frac{\pi}{6}\right) \\ & =5 \times 4 \times \cos 37^{\circ} \\ & =16 \mathrm{~ms}^{-1} \\ & \left(\because \cos 37^{\circ}=\frac{4}{5}\right) \\ &\end{aligned}$$

At neutral point

$B = {B_H}$

$\Rightarrow {{{\mu _0}} \over {4\pi }} \times {M \over {{{({r^2} + {l^2})}^{3/2}}}} = {B_H}$

$\Rightarrow {10^{ - 7}} \Rightarrow {{m \times 2l} \over {{{({r^2} + {l^2})}^{3/2}}}} = 1.2 \times {10^{ - 5}}$

$\Rightarrow {10^{ - 7}} \times {{m \times 2 \times 3 \times {{10}^{ - 2}}} \over {{{[{{(4 \times {{10}^{ - 2}})}^2} + {{(3 \times {{10}^{ - 2}})}^2}]}^{3/2}}}} = 1.2 \times {10^{ - 5}}$

$\Rightarrow {{6m \times {{10}^{ - 4}}} \over {{{[{{(5 \times {{10}^{ - 2}})}^2}]}^{3/2}}}} = 1.2$

$\Rightarrow {{6m \times {{10}^{ - 4}}} \over {125 \times {{10}^{ - 6}}}} = 1.2$

$\Rightarrow m = {{1.2 \times 125 \times {{10}^{ - 2}}} \over 6} = 0.25$ A-m$^2$

As we know,

$$\begin{aligned} & \text { power factor }=\frac{R}{Z} \\ & \qquad \begin{aligned} & =\frac{R}{\sqrt{R^2+\omega L^2}} \\ \text { Given } \omega & =100, L=400 \mathrm{mH} \\ & =\frac{30}{\sqrt{(30)^2+\left(100 \times 400 \times 10^{-3}\right)^2}} \\ & =\frac{30}{\sqrt{900+160000 \times 10^{-3 \times 2}}} \\ & =\frac{30}{50}=0.6 \end{aligned} \end{aligned}$$

Magnetic Lorenz force acting on each electron,

$F=evB$ ..... (i)

As, current, $i=neAv$

$ev=\frac{i}{nA}$

Therefore, substituting value of $ev$ into Eq. (i),

$F=\frac{i}{nA}B$

Given, $i=5~A,n=20$ turns, $B = 0.1$ T

$F = eVB = {{iB} \over {nA}} = {{5 \times 0.1} \over {{{10}^{29}} \times {{10}^{ - 5}}}} = 5 \times {10^{ - 25}}N$

Given that, moment of inertia of disc,

$I_1=4$ kg - m$^2$

$I_2=8$ kg - m$^2$

Angular velocities of discs,

$\omega_1=16$ rad/s

$\omega_2=4$ rad/s

From angular momentum conservation principle,

$I_1\omega_1+I_2\omega_2=(I_1+I_2)\omega$

$\Rightarrow 4\times16+8\times4=(4+8)\omega$

$\Rightarrow 64+32=12\omega$

$\Rightarrow \omega=\frac{96}{12}=8$ rad/s

During $\alpha$-decay, atomic mass of parent nuclei is decreased by 4 units while atomic number is decreased by 2 units.

From two similar triangles,

$\frac{f}{\frac{d}{2}}=\frac{f}{2f}$

$h=\frac{d}{4}$

Speed of EM wave in medium is given as

$v = {c \over {\sqrt {{\mu _r}{\varepsilon _r}} }}$

Volume of $\mathrm{wood=\frac{Mass}{Density}}$

$=\frac{120}{600}=0.2~\mathrm{m^3}$

Weight of displaced water,

$vdg=0.2\times1000\times10=2000~\mathrm{N}$

$Mg=vdg$

$\Rightarrow (120+m)g=2000$

$\Rightarrow m=80~\mathrm{kg}$

In general, nth mode of a string fixed at ends has frequency,

$\nu=\frac{nv}{2l}$

where, $n=1,2,3,...$

where, $v$ is the velocity of wave and $l$ is the length of string.

In fourth normal mode, $n=4$

$v=\frac{4\nu}{2l}$

Given, $\nu=500$ Hz, $l=2$ m

Hence, $500=\frac{4\nu}{2\times2}$

or, $v=\frac{500\times4}{4}=500$ m/s

Given, $q_1=-20 \mu \mathrm{C}=-2 \times 10^{-5} \mathrm{C}$

$q_2=60 \mu \mathrm{C}=6 \times 10^{-5} \mathrm{C}$

Force of attraction between the both charges, kept at a certain distance $r$,

$$\begin{aligned} F_1 & =\frac{K q_1 q_2}{r^2} \\ & =\frac{K \times 2 \times 10^{-5} \times 6 \times 10^{-5}}{r^2} \\ & =\frac{12 \times 10^{-10} K}{r^2} \end{aligned}$$

When both charges are touched, then new charge on both spheres,

$$\begin{aligned} q_1^{\prime} & =q_2^{\prime}=\frac{q_1+q_2}{2}=\frac{-20+60}{2}=20 \mu \mathrm{C} \\ & =2 \times 10^{-5} \mathrm{C} \end{aligned}$$

$\therefore$ Force between the two spheres,

$F_2=\frac{K q_1^{\prime} \cdot q_2^{\prime}}{r^2}$

$=\frac{K \cdot 2 \times 10^{-5} \times 2 \times 10^{-5}}{r^2}$

$=\frac{4 \times 10^{-10}}{r^2} K$

$\therefore$ ${{{F_1}} \over {{F_2}}} = {{{{12 \times {{10}^{ - 10}}K} \over {{r^2}}}} \over {{{4 \times {{10}^{ - 10}}} \over {{r^2}}}}} = {{12} \over 4} = {3 \over 1}$

$\therefore$ ${F_1}:{F_2} = 3:1$

According to Doppler's effect, the apparent frequency or sound when both source and observer are moving towards each other

$\Rightarrow f = {{(v + {v_0}){f_0}} \over {(v - {v_0})}}$

$\Rightarrow f = \left( {{v \over {v - {v_s}}}} \right){f_0} + \left( {{{{v_0}} \over {v - {v_s}}}} \right){f_0}$

$\Rightarrow f = \left( {{{{f_0}} \over {v - {v_s}}}} \right){v_0} + \left( {{{{f_0}} \over {v - {v_s}}}} \right)v$