COMEDK 2020 Question Paper with Solutions

We know that, angular velocity,

$\omega = {{2\pi } \over T}$

$\Rightarrow T = {{2\pi } \over \omega }$

$\Rightarrow {{{T_2}} \over {{T_1}}} = {{{\omega _1}} \over {{\omega _2}}}$

$= {{{\omega _1}} \over {2{\omega _1}}}$ [$\because$ ${\omega _2} = 2{\omega _1}$]

$\Rightarrow {{{T_2}} \over {{T_1}}} = {1 \over 2}$

$\Rightarrow {T_2} = {{{T_1}} \over 2}$ .... (i)

According to Kepler's law,

$$\begin{aligned} T^2 & \propto R^3 \\ \Rightarrow \quad & R\propto T^{2 / 3} \\ \Rightarrow \quad \frac{R_2}{R_1} & =\left(\frac{T_2}{T_1}\right)^{2 / 3} \\ & =\left(\frac{T_1 / 2}{T}\right)^{2 / 3} \\ & =\left(\frac{1}{2}\right)^{2 / 3} \\ R_2 & =\frac{R_1}{2^{2 / 3}} \\ & =\frac{R}{4^{1 / 3}} \end{aligned}$$

The given situation is shown in figure.

Moment of inertia of the systems about an axis perpendicular to the square and passing through the centre is given as

$I = m{\left( {{l \over {\sqrt 2 }}} \right)^2} + m{\left( {{l \over {\sqrt 2 }}} \right)^2} + m{\left( {{l \over {\sqrt 2 }}} \right)^2} + m{\left( {{l \over {\sqrt 2 }}} \right)^2}$

$= 4{{m{l^2}} \over 2} = 2m{l^2}$ ..... (i)

If $k$ be the radius of gyration, then

$I=4mk^2$ ..... (ii)

$\therefore$ From Eqs. (i) and (ii), we have

$2ml^2=4mk^2$

$\Rightarrow k=\frac{l}{\sqrt2}$

Let the frequency of the sound is $f$.

In first case,

$$\begin{aligned} f & =100 \pm 5 \\ & =105 \mathrm{~Hz} \text { or } 95 \mathrm{~Hz} \end{aligned}$$

The frequency of second harmonic $=2 f$

According to second case,

$$\begin{aligned} 2 f & =200 \pm 10 \\ & =210 \mathrm{~Hz} \text { or } 190 \mathrm{~Hz}\end{aligned}$$

If 5 beats are produced with frequency $105 \mathrm{~Hz}$, then frequency of second harmonic will be $210 \mathrm{~Hz}$.

i.e.

$$\begin{array}{r} 2 f=210 \\ \Rightarrow f=105 \mathrm{~Hz} \\ =105 \mathrm{~s}^{-1} \end{array}$$

Electric field due to thin spherical shell at a distance x outside from the centre is given as

$E = {1 \over {4\pi {\varepsilon _0}}}\,.\,{Q \over {{x^2}}}$ ..... (i)

where, $Q = s\,.\,A$

$=s \times 4 \pi R^2$

$\therefore$ From Eq. (i), we get

$E =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{s \times 4 \pi R^2}{x^2}=\frac{s R^2}{\varepsilon_0 x^2}$

$\therefore E \propto \frac{1}{x^2}$

Given, for steel wire,

Area, $A_1=3\times10^{-5}$ m$^2$

Length, $L_1=4.7$ m

For copper wire,

$A_2=4\times10^{-5}$ m$^2$, $L_2=3.5$ m

If $Y_1$ and $Y_2$ be the Young's modulus for the material steel wire and copper wire respectively, then

${{{Y_1}} \over {{Y_2}}} = {{{{{F_1}{L_1}} \over {{A_1}\Delta l}}} \over {{{{F_2}{L_2}} \over {{A_2}\Delta l}}}}$

$\Rightarrow {{{Y_1}} \over {{Y_2}}} = {{{L_1}{A_2}} \over {{L_2}{A_1}}}$ [$\because$ ${F_1} = {F_2}$]

$= {{4.7 \times 4 \times {{10}^{ - 5}}} \over {3.5 \times 3 \times {{10}^{ - 5}}}} = 1.8$

$\Rightarrow {Y_1}:{Y_2} = 1.8:1$

Given, number of turns in moving coil

galvanometer, $N_i=28$

Initial sensitivity, $S_i=S$

Final sensitivity, $S_f=S_i+25\%$ of $S_i$

$=S+25\%$ of $S$

$=S+\frac{S}{4}=\frac{5s}{4}$

Since, sensitivity of moving coil galvanometer is given as

$S = {{NAB} \over K}$

$\Rightarrow S \propto N$ [A, B and K are constant]

$\Rightarrow {{{S_f}} \over {{S_i}}} = {{{N_f}} \over {{N_i}}}$

$\Rightarrow {{{{5S} \over 4}} \over S} = {{{N_f}} \over {28}}$

$\Rightarrow {N_f} = {{28 \times 5} \over 4} = 35$

Given, ${K_{copper}}:{K_{brass}} = 4:1$

$\Rightarrow {{{K_{copper}}} \over {{K_{brass}}}} = {4 \over 1}$

Let the temperature of the free surface of brass = $\theta$

Rate of flow of heat will be equal in both the plates as connected in series.

i.e. $H = {K_{brass}}{{A(\theta - 20)} \over l} = {K_{copper}}{{A(20 - 0)} \over l}$

$\Rightarrow {{\theta - 20} \over {20}} = {{{K_{copper}}} \over {{K_{brass}}}} = {4 \over 1}$

$\Rightarrow \theta - 20 = 80$

$\theta = 100\Upsilon C$

Given,

$v = 2 \times {10^5}\widehat i$

$B = (\widehat i - 4\widehat j - 3\widehat k)\,T$

Charge on electron, $q = 1.6 \times {10^{ - 19}}\,C$

$\therefore$ Force, $F = q(v \times B)$

$= 1.6 \times {10^{ - 19}}[(2 \times {10^5}\widehat i) \times (\widehat i - 4\widehat j - 3\widehat k)]$

= 1.6 \times {10^{ - 19}}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {2 \times {{10}^5}} & 0 & 0 \cr 1 & { - 4} & { - 3} \cr } } \right|

$F = 1.6 \times {10^{ - 19}}(6 \times {10^5}\widehat j - 8 \times {10^5}\widehat k)$

$= 9.6 \times {10^{ - 14}}\widehat j - 12.8 \times {10^{ - 14}}\widehat k$

$\Rightarrow |F| = \sqrt {{{(9.6 \times {{10}^{ - 14}})}^2} + {{( - 12.8 \times {{10}^{ - 14}})}^2}}$

$= {10^{ - 14}}\sqrt {{{(9.6)}^2} + {{(12.8)}^2}}$

$= 16 \times {10^{ - 14}}\,N$

$= 1.6 \times {10^{ - 13}}\,N$

In step-up transformer, number of turns in secondary coil is greater than the number of turns in the primary coil.

i.e. N$_S$ > N$_P$

Let u be the initial velocity of vertical projection and t be the time taken by the body to reach at height h from the ground.

$\therefore$ using equation, $h = ut + {1 \over 2}a{t^2}$

Here, $u = u,a = - g$

$\therefore$ $h = ut - {1 \over 2}g{t^2}$

$\Rightarrow g{t^2} - 2ut + 2h = 0$

$\therefore$ $t = {{2u\, \pm \,\sqrt {4{u^2} - 4g \times 2h} } \over {2g}}$

$= {{u\, \pm \,\sqrt {{u^2} - 2gh} } \over g}$

It means t has two values, $t_1$ and $t_2$. where $t_1$ is time taken by the body in reaching maximum height and $t_2$ be the time taken by the body in reaching ground from maximum height.

$\therefore$ ${t_1} = {{u\, + \,\sqrt {{u^2} - 2gh} } \over g}$

and ${t_2} = {{u\, - \,\sqrt {{u^2} - 2gh} } \over g}$

$\therefore$ ${t_1} + {t_2} = {{2u} \over g}$ or $u = {{g({t_1} + {t_2})} \over 2}$

Given, temperature of sink,

$$\begin{aligned} T_2 & =-39^{\circ} \mathrm{C} \\ & =(273-39) \mathrm{K} \\ & =234 \mathrm{~K} \end{aligned}$$

Temperature of source,

$$\begin{aligned} T_1 & =0^{\circ} \mathrm{C} \\ & =(273+0) \mathrm{K} \\ & =273 \mathrm{~K} \end{aligned}$$

$\therefore$ Efficiency,

$$\begin{aligned} \eta & =1-\frac{T_2}{T_1} \\ & =1-\frac{234}{273} \\ & =\frac{39}{273} \\ & =0.142 \\ & =14.2 \% \end{aligned}$$

When a charged particle of charge $q$ is accelerated through a potential $V$, then de-Broglie wavelength is given as

$$\begin{aligned}& \lambda=\frac{h}{\sqrt{2 m q V}} \\ & \Rightarrow \quad \lambda \propto \frac{1}{\sqrt{m q}} \\ & \Rightarrow \quad \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_p}{m_\alpha}} \cdot \sqrt{\frac{q_p}{q_\alpha}} \\ & \Rightarrow \quad \frac{\lambda_\alpha}{\lambda}=\sqrt{\frac{m_p}{4 m_p}} \sqrt{\frac{q_p}{2 q_p}} \\ & {\left[\begin{array}{ll} \because m_\alpha=4 m_p \\ \text { and } q_\alpha=2 q_p \end{array}\right] } \\ & \Rightarrow \quad \frac{\lambda_\alpha}{\lambda}=\frac{1}{\sqrt{8}} \\ & \Rightarrow \quad \lambda_\alpha=\frac{\lambda}{2 \sqrt{2}} \end{aligned}$$

Refractive index, $\alpha = A + {B \over {{\lambda ^2}}}$

According to principle of homogeneity,

Dimension of ${B \over {{\lambda ^2}}} =$ Dimension of $\alpha$

$\Rightarrow [B] = [ \propto ][{\lambda ^2}]$

$= [{M^0}{L^0}{T^0}][{L^2}]$

$= [{L^2}]$

= Dimension of area

In L-C-R series resonance circuit,

${X_L} = {X_C}$

$\therefore$ Impedance, $Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} = \sqrt {{R^2} + {0^2}}$

$\Rightarrow Z = R$

$\therefore$ Power factor, $\cos \phi = {R \over Z} = {R \over R} = 1$

Which is not zero.

Since, circuit behaves likes a resistive circuit. So, power loss occurs always in this circuits.

i.e. wattless current is zero.

The given situation is shown below.

Resistance of potentiometer wire, $R_P=10~\Omega$

Emf, $E=2.5~V$

Total resistance, $R'=R_P+R=(10+R)~\Omega$

Current thruogh potentiometer wire,

$I=\frac{E}{R_P}=\frac{2.5}{10+R}$

Since, potential difference across length $AL=\frac{L}{2}$,

$V = I\left( {{{{R_P}} \over 2}} \right)$

$\Rightarrow 1 = {{2.5} \over {10 + R}} \times {{10} \over 2}$

$\Rightarrow 10 + R = 12.5$

$\Rightarrow R = 2.5~\Omega$

The given situation is shown below

$$\begin{aligned} r_1 & =1 \mathrm{~cm} \\ & =1 \times 10^{-2} \mathrm{~m} \\ r_2 & =5-1 \\ & =4 \mathrm{~cm} \\ & =4 \times 10^{-2} \mathrm{~m} \end{aligned}$$

$$\begin{aligned} \text { Required work done } & =\text { Changes in potential energy } \\ & =k q_A q_B\left[\frac{1}{r_2}-\frac{1}{r_1}\right] \\ & =9 \times 10^9 \times 3 \times 10^{-9} \times 10^{-9} \\ & {\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{1 \times 10^{-2}}\right] } \\ & =1.35 \times 10^{-7} \mathrm{~J} \end{aligned}$$

From the given diagram, we see that lenses $L_1$ and $L_3$ are identical but contribute opposite sense, hence their combined effect of making image is ineffective. Hence, only lens $L_2$ forms the image.

Given,

$$\begin{aligned} V_1 & =1 \mathrm{~mm}^3 \\ & =10^{-9} \mathrm{~m}^3 \\ p_1 & =1 \mathrm{~atm} \\ & =1 \times 10^5 \mathrm{~N} / \mathrm{m}^2 \\ T_1 & =27+273 \\ & =300 \mathrm{~K} \\ T_2 & =627+273 \\ & =900 \mathrm{~K} \end{aligned}$$

Under adiabatic condition,

$$\begin{aligned} p_1^{1-\gamma} T_1^\gamma & =p_2^{1-\gamma} T_2^\gamma \\ \Rightarrow \quad\left(\frac{p_2}{p_1}\right)^{1-\gamma} & =\left(\frac{T_1}{T_2}\right)^\gamma \\ \Rightarrow \quad\left(\frac{p_2}{p_1}\right)^{1-1.5} & =\left(\frac{300}{900}\right)^{1.5} \\ \Rightarrow \quad\left(\frac{p_2}{p_1}\right)^{-0.5} & =\left(\frac{1}{3}\right)^{1.5} \\ \Rightarrow \quad\left(\frac{p_1}{p_2}\right)^{0.5} & =\left(\frac{1}{3}\right)^{1.5} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad \frac{p_1}{p_2}=\left(\frac{1}{3}\right)^{\frac{1.5}{0.5}}=\left(\frac{1}{3}\right)^3=\frac{1}{27} \\ & \Rightarrow \quad p_2=27 p_1=27 \times 1 \times 10^5 \\ & =27 \times 10^5 \mathrm{~N} / \mathrm{m}^2 \\ & \end{aligned}$$

Frequency of incident light,

$v=10^{15} \mathrm{~Hz}$

Work function, $\phi_0=25 \mathrm{eV}$

Energy of incident photon,

$$\begin{aligned}E & =h v \\ & =6.62 \times 10^{-34} \times 10^{15} \\ & =6.62 \times 10^{-19} \mathrm{~J} \\ & =\frac{6.62 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV} \\ & =4.1 \mathrm{eV} \end{aligned}$$

According to Einstein's photoelectric equation,

$$\begin{array}{llrl} & K_{\max }=h v-\phi_0 \\ \Rightarrow & e V_0=h v-\phi_0 \\ \Rightarrow & e V_0=4.1 e V-2.5 \mathrm{eV} \\ \Rightarrow & e V_0=1.6 \mathrm{eV} \\ \Rightarrow & V_0=1.6 \mathrm{~V} \end{array}$$

Given, force $F=20 \mathrm{~N}$

$m=5 \mathrm{~kg}$

Net force on the body,

$F_{\text {net }}=F-f_s$

[where, $f_s=$ friction]

$\therefore$ Work done by displacing $2 \mathrm{~m}$ is given as

$$\begin{aligned}W & =F_{\text {net }} \times \text { displacement } \\ & =\left(F-f_s\right) 2 \\ & =\left(20-f_s\right) 2 \\ \Rightarrow \quad W & =40-2 f_s \end{aligned}$$

According to work-energy theorem,

Work done = gain in kinetic energy

$\Rightarrow 40-2 f_s =10$

$\Rightarrow 2 f_s=40-10 =30$

$\Rightarrow f_s =\frac{30}{2}$

$=15 \mathrm{~N}$

Wavelength of ground waves have frequency greater than the wavelength of sky waves and space waves both.

Given, wavelength of monochromatic light, $\lambda = 5867$ $\mathop A\limits^o$

Refractive index, $\alpha$ = $1.6$

Let t be the thickness of the plastic.

Since, the centre of the screen appears dark when plastic sheet is introduced, hence path difference,

$\alpha t = {\lambda \over 2}$ or ${{3\lambda } \over 2}$

when, $\alpha t = {{3\lambda } \over 2}$

$\Rightarrow t = {{3\lambda } \over {2 \alpha }} = {{3 \times 5867} \over {2 \times 1.6}} = 5500\mathop A\limits^o$

The given situation is shown below

Magnetic field at point P,

$B = {{{ \mu _0}} \over {2\pi }}\,.\,{I \over r}$

$= 2 \times {10^{ - 7}} \times {{90} \over {1.5}}$

$= 1.2 \times {10^{ - 5}}\,T$ (South)

When a $p$-$n$ junction diode is connected in forward bias, then its potential barrier decreases because more number of charge carriers (electrons in n-side and holes in p-side) move through the junction. Due to increase of drifting of charge carriers, electric current is increased.

Slope of distance-time graph gives the velocity of the particle. Since at the point C, slope of the curve is maximum, hence maximum instantaneous velocity of particle occurs at point C.

If $\alpha$ and $\beta$ be the longitudinal strain and lateral strain respectively, then

$\alpha =\frac{l_2-l_1}{l_1}=10^{-3}=0.001$

$\Rightarrow \frac{l_2}{l_1} =1.001$ ...... (i)

$\text { Poisson's ratio, } \sigma=\frac{\text { Lateral strain }}{\text { Longitudinal strain }}$

$$\begin{array}{rlrl} =\frac{\beta}{\alpha} \\ \Rightarrow & \beta=\sigma \alpha=0.1 \times 10^{-3}=10^{-4} \\ \text { But, } & \beta=\frac{r_1-r_2}{r_1} \end{array}$$

$\frac{r_1-r_2}{r_1}=10^{-4}=0.0001$

$\Rightarrow \frac{r_2}{r_1}=0.9999$ ..... (ii)

Hence, percentage increase in volume

$$\begin{aligned} & =\frac{V_2-V_1}{V_1} \times 100 \\ & =\frac{\pi r_2^2 l_2-\pi r_1^2 l_1}{\pi r_1^2 l_1} \times 100 \\ & =\left[\left(\frac{r_2}{r_1}\right)^2\left(\frac{l_2}{l_1}\right)^2-1\right] \times 100 \\ & =\left[(0.9999)^2 \times(1.0001)^2-1\right] \times 100 \\ & =0.0008 \times 100=0.08 \% \end{aligned}$$

For emission of energy, electron of H-atom must fall from higher energy state to lower energy state. Hence, options (a) and (c) are not possible.

In option (b), $n_1=2,n_2=1$

$\therefore$ Energy of emitted photon is given as

$E' = - 13.6{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

$\Rightarrow h\nu ' = - 13.6{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{1^2}}}} \right)$

$\nu ' = {{13.6{Z^2}} \over h} \times {3 \over 4}$ ...... (i)

In option (c), ${n_1} = 2,{n_2} = 6$, hence energy of emitted photon is given as

$E'' = - 13.6{Z^2}\left( {{1 \over {{6^2}}} - {1 \over {{2^2}}}} \right)$

$\Rightarrow h\nu '' = - 13.6{Z^2} \times {8 \over {36}}$

$\Rightarrow \nu '' = {{13.6{Z^2}} \over h} \times {8 \over {36}}$ ..... (ii)

Hence, from Eqs. (i) and (ii), we get

$\nu ' > \nu ''$

Given, equation of progressive wave,

$y=15 \sin (660 \pi t-0.02 \pi x)$

Comparing this equation with standard equation of progressive wave as $y=A \sin \left(\frac{2 \pi}{T} \cdot t-\frac{2 \pi}{\lambda} \cdot x\right)$

We get,

\begin{array} & & \frac{2 \pi}{T} =660 \pi \\ \Rightarrow & T =\frac{2 \pi}{660 \pi} \\ \Rightarrow & T =\frac{1}{330} \end{array}

$\therefore$ Frequency of the wave,

$$\begin{aligned} \nu & =\frac{1}{T} \\ & =\frac{1}{1 / 330}=330 \mathrm{~Hz} \end{aligned}$$

According to the equation of continuity,

$Av$ = constant

Therefore, velocity of the fluid is minimum at the widest part and maximum at the narrowest part of the horizontal pipe.

Again, according to Bernoulli's principle for horizontal pipe,

$p+\frac{1}{2}\rho v^2=$ constant

Hence, $p$ is more, $v$ is less and vice-versa.

Therefore, when a fluid is flowing through the horizontal pipe of variable area of cross-section, then velocity is maximum at the narrowest part of the pipe and pressure is maximum at the widest part of the pipe.

Given, $Q_1+Q_2=0.8$ C

$r=30$ cm = 0.3 m

$\therefore$ According to Coulomb's law

$F = k{{{Q_1}{Q_2}} \over {{r^2}}}$

$\Rightarrow F = k{{{Q_1}(0.8 - {Q_1})} \over {{r^2}}}$

The force on Q$_1$ will be maximum if

${{dF} \over {d{Q_1}}} = 0$

$\Rightarrow {d \over {dQ}}k{{{Q_1}(0.8 - {Q_1})} \over {{r^2}}} = 0$

$0.8 - 2{Q_1} = 0$

$\Rightarrow {Q_1} = 0.4$ C

$\therefore$ ${Q_2} = 0.8 - {Q_1}$

$= 0.8 - 0.4 = 0.4$ C

Given, $m_1=200$ g = 0.2 kg

$m_2=300$ g = 0.3 kg

${l_1} = 50 - 20 = 30$ cm

$\Rightarrow {l_1} = 0.3$ m

${l_2} = 70 - 50 = 20$ cm = 0.2 m

$\therefore$ Moment of inertia of the system about an axis passing through 50 cm mark (i.e. mid-point)

$I = {m_1}l_1^2 + {m_2}l_2^2$

$= 0.2 \times {(0.3)^2} + 0.3 \times {(0.2)^2}$

$= 0.036$ kg m$^2$

Given, initial surface area, A$_1$ = A

Final surface area, A$_2$ = A$_1$ $-$ 50% of A

A$_2$ = A $-$ 50% of A = $A-\frac{A}{2}=\frac{A}{2}$

Initial surface energy, $E_1=T.2A=2TA$ ..... (i)

Final surface energy, E$_2$ = Surface tension $\times$ Surface area

$\Rightarrow {E_2} = T \times 2({A_1} - {A_2})$