COMEDK 2021 Question Paper with Solutions

The graph of stress versus strain is given below

From the graph it is clear that within the elastic limit, the corresponding stress refers to yield point.

Given, W = 50 J and q = 25 C

Potential difference, $\mathrm{V = {{Work\,done\,(W)} \over {Ch\arg e\,(q)}}}$

$\Rightarrow V = {{50} \over {25}} = 2V$

The interatomic forces decreases or becomes very weak at the boiling point of a liquid. Therefore, surface tension becomes zero.

The graph between stopping potential and frequency is a straight line, so stopping potential and hence maximum kinetic energy of the photoelectrons depends linearly on the frequency.

Total work done by the gas

${W_{Total}} = {W_{P \to Q}} + {W_{Q \to R}} + {W_{R \to S}} + {W_{S \to P}}$

Work done in isothermal process by ideal gas

${W_{P \to Q}} = \mu RT_1\log {{{V_2}} \over {{V_1}}}$ ..... (i)

Work done in adiabatic process by ideal gas

${W_{Q \to R}} = {{ \mu R({T_1} - {T_2})} \over {r - 1}}$ ..... (ii)

Work done in isothermal process by ideal gas,

${W_{R \to S}} = - \mu R{T_2}\log {{{V_3}} \over {{V_4}}}$ ...... (iii)

Work done in adiabatic process, by ideal gas

${W_{S \to P}} = {{ - \mu R({T_1} - {T_2})} \over {r - 1}}$ ..... (iv)

Adding Eqs. (i), (ii), (iii) and (iv) we get

${W_{Total}} = \mu R{T_1}\log {{{V_2}} \over {{V_1}}} - \mu R{T_2}\log {{{V_3}} \over {{V_4}}}$

Given, $y = 20\cos (\omega t + 4z)$

On comparing if with general equation,

$y = A\cos (kx+\omega t)$

we get, $A=20$

Therefore, amplitude of the wave is 20.

Given, $x = a{e^{ - pt}} + b{e^{qt}}$

$\Rightarrow v = {{dx} \over {dt}} = - pa{e^{ - pt}} + qb{e^{qt}}$

and $a = {{dv} \over {dt}} = {p^2}a{e^{ - pt}} + {q^2}b{e^{qt}}$

Acceleration is positive, so velocity goes on increasing with time.

If all the masses will remain same, then centre of mass will lie at the centre 'O. But as the mass at B is 4m, so the centre of mass of the system will shift towards B. So centre of mass will be on line OB.

The external torque changes the angular momentum, therefore the second law in rotational motion is given by,

${\tau _{ext}} = {{dL} \over {dt}}$

Lyman series of a hydrogen atom belongs to the ultraviolet region of electromagnetic spectrum.

From Kepler's second law of planetary motion,

$\frac{dL}{dt}=\frac{L}{2m}$

Thus, angular momentum remains conserved.

When an electron jumps from orbit $n_i$ to $n_f$, then the frequency of emitted photon is

$\nu = cR\left( {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right)$

Here, ${n_i} = 3,{n_f} = 2$

$\Rightarrow \nu = cR\left[ {{1 \over {{{(2)}^2}}} - {1 \over {{{(3)}^2}}}} \right] = {{5Rc} \over {36}}$

As we know that, Impulse = Force $\times$ Time

= [MLT$^{-2}$] $\times$ [T]

= [ML$^1$T$^{-1}$]

Momentum = Mass $\times$ Velocity

$p=mv$

$[p]=[m]\times[v]=[\mathrm{M}][\mathrm{LT^{-1}}]=[\mathrm{ML^1T^{-1}}]$

Thus, dimension of impulse represent dimensions of momentum.

If a wire is stretched to double its length, then

$\mathrm{Strain=\frac{Change\,in\,length}{Actual\,length}}=\frac{2L-L}{L}=1$

The centripetal force is balanced by the normal reaction exerted i.e. $m{v^2}/r \le \,{ \propto _s}N$

where, ${ \propto _s}$ is the coefficient of friction.

$\therefore$ ${{m{v^2}} \over r} = f \le \,{ \propto _s}N$

Given, $\lambda$ = 589 nm = 589 $\times$ 10$^{-9}$ m

d = 0.589 mm = 0.589 $\times$ 10$^{-3}$ m

In Young's double slit experiment, half angular width of central maxima is

$\sin \theta = {\lambda \over d} = {{589 \times {{10}^{ - 9}}} \over {0.589 \times {{10}^{ - 3}}}} = {10^{ - 3}} \Rightarrow \theta = {\sin ^{ - 1}}(0.001)$

The average energy associated with a particle of a monoatomic gas is ${3 \over 2}{K_B}T$.

The reactance of a capacitor of capacitance C connected to a AC source of frequency $f$ is given by

${X_C} = {1 \over {\omega C}} = {1 \over {2\pi fC}}$

Given, ${f_1} = 20$ cm, ${f_2} = 40$ cm

Since, power $P = {1 \over f} \Rightarrow {P_1} = {1 \over {{f_1}}} = {{100} \over {20}} = 5D$

and ${P_2} = {1 \over {{f_2}}} = {{100} \over {40}} = 2.5D$

Effective power, $P = {P_1} + {P_2} = 5 + 2.5 = 7.5D$

For an infinite plane sheet as shown below

As, the lines of force are parallel to the current surface of cylinder, the flux passing through the curved surface is zero. The flux through the plane-end faces of the cylinder is ${\phi _E} = EA + EA = 2EA$

By Gauss' law,

${\phi _E} = \int {E\,.\,ds = {q \over {{\varepsilon _0}}} \Rightarrow 2EA = {{\sigma A} \over {{\varepsilon _0}}} \Rightarrow E = {\sigma \over {2{\varepsilon _0}}}}$

The collision of the molecules of an ideal gas is taken as elastic.

The equation of motion of this system about mean position is given by

$F = {k_1}x + {k_2}x$

$\Rightarrow m{{{d^2}x} \over {d{t^2}}} = {k_1}x + {k_2}x \Rightarrow {{{d^2}x} \over {d{t^2}}} = x\left( {{{{k_1} + {k_2}} \over m}} \right)$

$\because$ $a = {{{d^2}x} \over {d{t^2}}} = {\omega ^2}x$

$\therefore$ ${\omega ^2} = {{{k_1} + {k_2}} \over m}$ or $\omega = \sqrt {{{{k_1} + {k_2}} \over m}}$

Given, n = 20 cm$^{-1}$ = 2000 m$^{-1}$

B = 20 mT = 20 $\times$ 10$^{-3}$ T

Using, $B = \,{ \mu _0}nI$

$20 \times {10^{ - 3}} = 4\pi \times {10^{ - 7}} \times 2000 \times I$

$\Rightarrow I = {{20 \times {{10}^{ - 3}}} \over {4\pi \times {{10}^{ - 7}} \times 2000}} = 8\,A$

As we know that, PE of satellite revolving at height $h$ from the surface of the earth.

$= - {{G{M_e}m} \over {{R_e} + h}}$

KE of satellite, $= {1 \over 2}{{G{M_e}m} \over {({R_e} + h)}}$

Therefore, PE = $-$2KE

$\therefore$ $r = - 2$

The increasing order of frequency is Microwave < UV rays < X-rays

The frequency is the inverse of wavelength. So, the order of increasing wavelength is

X-rays < UV rays < Microwave

According to Kirchhoff's first law, redrawing the circuit, we get

Current at $A=3-2=1A$

Current at $B=1+5=6A$

Current at $D=1+4=5A$

Current at $C=6+5=11A$

$\therefore\quad I=11-3=8A$

For an ideal gas coefficient of volume expansion can be deduced as

According to ideal gas equation,

$pV = nRT$ ..... (i)

at constant pressure, $p\Delta V = nR\Delta T$ ..... (ii)

Dividing Eq. (ii) by Eq. (i), we get

${{\Delta V} \over V} = {{\Delta T} \over T}$ ..... (iii)

But ${{\Delta V} \over V} = \alpha \Delta T$ .... (iv)

$\therefore$ From Eq. (iii) and (iv), we get

For ideal gas $\alpha = {1 \over T}$

So, for an ideal gas, coefficient of volume expansion is given by ${1 \over T}$.

Given, $R = 10k\Omega = {10^4}\Omega$

$C = 100pF = 100 \times {10^{ - 12}}F$

Time constant of RC circuit,

$\tau = RC = {10^4} \times 100 \times {10^{ - 12}} = {10^{ - 6}}\,s$

For demodulation ${1 \over {{f_C}}} < < RC \Rightarrow {f_C} > > {1 \over {RC}}$

or ${f_C} > > {1 \over {{{10}^{ - 6}}}} = {10^6}\,Hz$

$\therefore$ ${f_C} > > 1\,MHz$

The output of gate is given by

$Y = \overline {A\,.\,B}$

It is a NAND gate output and its truth table is given by

Since, frequency of spring oscillator is smallest among all the given options, hence small time interval cannot be measured with spring oscillator.

The statement given in the option (c) is incorrect, it can be corrected as

As hypothesis is a supposition without assuming that it is true.

Given, $y = ut + g{t^2}$

Comparing with equation of motion $y = ut + {1 \over 2}a{t^2}$

we get, $a = 2g$

$\therefore$ Force $= m \times a = m \times 2g = 2mg$

The magnetic field through the circle is constant as current through the straight wire is constant. So,

${{d{\phi _B}} \over {dt}} = 0$

As, $E = {{d{\phi _B}} \over {dt}} = 0$

$\therefore$ $I = {E \over R} = 0$

Here, no current is flowing in circle.

Given, ${ \propto _r} = 6000$

As we know, ${ \propto _r} = 1 + {X_m}$

$\Rightarrow {X_m} = { \propto _r} - 1 = 6000 - 1 = 5999$

Energy released in the nuclear process is

$E = m{c^2} - m{c^2}{\left( {{2 \over 3}} \right)^2} = \left( {1 - {4 \over 9}} \right)m{c^2} = {5 \over 9}m{c^2}$

Hence, energy released decrease by a factor of $\frac{5}{9}$.

The de-Broglie wavelength is given by

$\lambda = {h \over p} = {h \over {mv}}$

$\Rightarrow {{{\lambda _e}} \over {{\lambda _p}}} = {{{h \over {m{v_e}}}} \over {{h \over {m{v_p}}}}} = {{{v_p}} \over {{v_e}}} = {{{{{v_e}} \over 4}} \over {{v_e}}} = {1 \over 4}$ ($\therefore$ ${v_p} = {1 \over 4} \times {v_e}$)

Force on a charge particle moving in a uniform magnetic field at angle $\theta$ is given by

$F = qvB\sin \theta \Rightarrow F \propto q$

Hence, ${q_1} = q$ and ${q_2} = 2q$

$\therefore$ ${{{F_1}} \over {{F_2}}} = {{{q_1}} \over {{q_2}}} = {q \over {2q}} = {1 \over 2}$

For a freely falling body, $v^2-u^2=2gx$ ($\therefore u=0$)

Therefore, $v^2=2gx$ or $x=\frac{v^2}{2g}$

Thus, graph will be a parabola, symmetric along +ve X-axis.

Given, frequencies, $\nu_1=\nu-1$ and $\nu_2=\nu+1$

Beats = Difference in frequencies

$=(\nu+1)-(\nu-1)=2$

Green house gases are CO$_2$, CH$_4$, N$_2$O, CF$_x$Cl$_x$ and O$_3$. Thus, H$_2$O is not a green house gas.

Since, the incidence light beam is unpolarised, the intensity of light on being transmitted through first nicol polaroid, will become

$I=\frac{I_0}{2}$

So, according to law of Malus, intensity on passing through second nicol will be

$I' = I{\cos ^2}\theta = {{{I_0}} \over 2}{\cos ^2}60\Upsilon = {{{I_0}} \over 2} \times {\left( {{1 \over 2}} \right)^2} = {{{I_0}} \over 8}$

The error in measurement of length 1.02 cm is $\pm$0.01 cm and error in measurement of length 9.89 cm is $\pm$0.01 cm.

$\therefore$ The relative error is 1.02 cm

$= \left[ { \pm {{0.01} \over {1.02}} \times 100\% } \right] = \pm 0.98\% = \pm 1\%$

Similarly, the relative error in 9.89 cm

$= \left[ { \pm {{0.01} \over {9.89}} \times 100\% } \right] = \pm 0.1\%$

Therefore, relative errors in measurement of two masses are $\pm$1% and $\pm$0.1%.

$y = \log \omega t$ is a monotonically increasing function. If cannot repeat its value, therefore it is a non-periodic function.

LED is a light emitting diode which emits light in the spectral range of visible light is the range of wavelengths 0.4 $\propto$ m to 0.7 $\propto$ m. The energy corresponding to these wavelength is from 1.8 eV to 3 eV. Hence, for the fabrication of LED in the visible range, the minimum band gap energy should be 1.8 eV. It convert electrical energy into light energy.

Insulator can be charged by rubbing or friction with another body. For example, or rubbing a glass rod with the silk gets negatively charged while the glass rod gets positively charged.

When kinetic energy is conserved in a collision then the collision is said to be elastic.

In an adiabatic process,

$pV^\gamma$ = constant (k)

where $p$ is pressure, V is volume and $\gamma$ is the ratio of the specific heats.

Given, ${{{C_p}} \over {{C_V}}} = \gamma = {3 \over 2}$

Since, for adiabatic process $p{V^\gamma } = k$

$\Rightarrow p{V^{3/2}} = k \Rightarrow \log p + {3 \over 2}\log V = \log k$

$\therefore$ ${{\Delta p} \over p} + {3 \over 2}{{\Delta V} \over V} = 0$

$\Rightarrow {{\Delta V} \over V} \times 100 = - {2 \over 3}\left( {{{\Delta p} \over p} \times 100} \right) = - {2 \over 3} \times {2 \over 3} = - {4 \over 9}\%$

Negative sign implies that volume decreases by ${4 \over 9}\%$.

Given, $|e| = 220\,V,di = 0.9\,A,\nu = 50\,Hz$

As, $\nu = {1 \over t} \Rightarrow t = {1 \over \nu } = {1 \over {50}}$

Since, $|e| = L{{di} \over {dt}} \Rightarrow 220 = L \times {{0.9} \over {1/50}}$

$\Rightarrow L = {{220} \over {0.9 \times 50}} = 4.88H \simeq 5H$

The resonant frequency of a LC circuit is given by

$f = {1 \over {2\pi \sqrt {LC} }}$

when $L' = {L \over 3}$, as per question,

$f' = f \Rightarrow {1 \over {2\pi \sqrt {L'C'} }} = {1 \over {2\pi \sqrt {LC} }}$

$\Rightarrow LC = L'C' \Rightarrow C' = {{LC} \over {L'}} = {{LC} \over {L/3}} = 3C$

Pressure at a point inside a liquid is given by

$p = {F \over A} = \rho hg$

Therefore, at same height, pressure in a fluid at rest will be the same.

Volume of material remain same after stretching. As volume remains same,

${A_1}{l_1} = {A_2}{l_2}$

Here, ${l_2} = n{l_1}$

$\Rightarrow {A_2} = {{{A_1}{l_1}} \over {{l_2}}} = {{{A_1}} \over n}$

Resistance of wire after stretching,

${R_2} = {{\rho {l_2}} \over {{A_2}}} = \rho {{n{l_1}} \over {{A_1}/n}} = {n^2}\rho {{{l_1}} \over {{R_1}}} = {n^2}R$ ($\because$ $R = \rho {{{l_1}} \over {{A_1}}}$)

Time period of simple harmonic motion,

$T = 2\pi \sqrt {\left| {{\theta \over \alpha }} \right|} = 2\pi \sqrt {{1 \over {mH}}}$

$\Rightarrow T \propto \sqrt I \Rightarrow T \propto \sqrt m$ [$\therefore$ $I = {{m{l^2}} \over {12}}$]

${{{T_1}} \over {{T_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} = \sqrt {{m \over {4m}}} = {1 \over 2}$

$\Rightarrow {T_2} = 2{T_1} = 2T$ [$\therefore$ ${T_1} = T$]

so, motion remains simple harmonic with time period 2T.

For motion under central forces, the total mechanical energy i.e. the sum of kinetic energy and potential energy of a body is constant. The familiar example is the free fall of an object under gravity.

Infrared rays are used for the treatment of muscles ache. Due to their heating properties they are used in infrared therapy.

Given, $\mathrm{\lambda=1.3~\mu m=1.3\times10^{-6}~m}$

Bandwidth of channel, BW = 20 kHz = 20 $\times$ 10$^3$ Hz

Optical source frequency,

$f = {c \over \lambda } = {{3 \times {{10}^8}} \over {1.3 \times {{10}^{ - 6}}}}$

$= 2.3 \times {10^{14}}$ Hz

$\therefore$ Number of channels, $n = {f \over {BW}} = {{2.3 \times {{10}^{14}}} \over {20 \times {{10}^3}}}$

$= 1.15 \times {10^{10}}$

Given $M = {{{f_0}} \over {{f_e}}} = 9 \Rightarrow {f_0} = 9{f_e}$

Also, ${f_0} + {f_e} = 20$

$\Rightarrow 9{f_e} + {f_e} = 20 \Rightarrow {f_e} = 2$ cm

$\therefore$ ${f_0} = 9 \times 2 = 18$ cm

Using relation, $\tan \theta = {{{X_L} - {X_C}} \over R}$

$\Rightarrow \theta = {\tan ^{ - 1}}\left( {{{{X_L} - {X_C}} \over R}} \right) = {\tan ^{ - 1}}\left( {{Z \over R}} \right)$

Here, $Z = \sqrt 3 R$

$\Rightarrow \theta = {\tan ^{ - 1}}\left( {{{\sqrt 3 R} \over R}} \right) = {\tan ^{ - 1}}(\sqrt 3 ) = 60\Upsilon$

Given, n = 2 min to 10 min = 5 half-lives and $N_0=80$ g

We know, ${N \over {{N_0}}} = {\left( {{1 \over 2}} \right)^n}$

$\Rightarrow N = {{{N_0}} \over {{2^n}}} = \left( {{{80} \over {{2^5}}}} \right) = {{80} \over {32}} = 2.5\,g$