COMEDK 2023 Morning Shift Question Paper with Solutions

Let total current through the 1 parallel combination be $I$, the current through the galv anometer be $I_g$ and the current through the shunt be $I-I_g$.

The shunted galvanometer is shown in the figure.

The potential difference, $V_{a b}=\left(V_a-V_b\right)$ is the same for both paths, so $I_g G=\left(I-I_g\right) S$

$\Rightarrow I_g(G+S)=I S \Rightarrow \frac{I_g}{I}=\frac{S}{S+G}$

The fraction of current passing through shunt

$$\begin{aligned} & =\frac{I-I_g}{I}=1-\frac{I_g}{I}=1-\frac{S}{S+G}=\frac{G}{S+G} \\ & =\frac{8}{2+8}=\frac{8}{10}=0.8 \mathrm{~A} \end{aligned}$$

Given, Escape velocity on the surface of the earth is given by

i.e:

$$\begin{aligned} & v_e=\sqrt{2 g R_e} \\ & v_e=\sqrt{\frac{2 G M_e}{R_e}} \quad \text{.... (i)} \end{aligned}$$

Mass of the moon, $M_m=\frac{M_e}{81}$

Radius of the moon, $R_m=\frac{R_e}{4}$

$\therefore$ Escape velocity on the surface of the moon

$v_m=\sqrt{\frac{2 G M_m}{R_m}}=\sqrt{\frac{2 G\left(\frac{M e}{81}\right)}{\frac{R_e}{4}}}=\frac{2 \sqrt{2}}{9} \sqrt{\frac{G M_e}{R_e}}$

From Eq. (i),

$=\frac{2}{9} \sqrt{\frac{2 G M_e}{R_e}}=\frac{2}{9} v_e$

Escape velocity $v_e=11.1 \mathrm{~km} / \mathrm{h}$

$=\frac{2}{9} \times 11.1=2.46 \mathrm{~km} \mathrm{~h}^{-1}$

According to conservation of angular momentum,

$I \omega=\text { constant }$

i.e. we can write,

$$\begin{aligned} I_1 \omega_1 & =I_2 \omega_2 \\ \text { or } \quad & M R^2 \omega =(M+4 m) R^2 \omega_2 \quad (\because \omega_1=\omega)\\ \text { or } \quad & \omega_2=\left(\frac{M}{M+4 m}\right) \omega\end{aligned}$$

$B=4\left[\frac{\mu_0}{4 \pi} \cdot \frac{I}{a}\left(\sin \phi_1+\sin \phi_2\right)\right]$

Here, $a=\frac{l}{2}$ and $\phi_1=\phi_2=45^{\circ}$

$$\begin{aligned} B & =4 \times \frac{\mu_0}{4 \pi} \frac{I}{(l / 2)}\left[\frac{1}{\sqrt{2}}+\frac{I}{\sqrt{2}}\right] \\ \therefore \quad B & =\frac{16}{\sqrt{2}}\left[\frac{\mu_0}{4 \pi} \frac{I}{l}\right]=8 \sqrt{2}\left(\frac{\mu_0}{4 \pi} \cdot \frac{I}{l}\right) \end{aligned}$$

Acceleration due to gravity at a depth $d$ below the surface of the earth is given by

$$\begin{aligned} g_{\text {depth }} & =g_{\text {surface }}\left(1-\frac{d}{R}\right) \\ & =g_{\text {surface }}\left[\frac{R-d}{R}\right]=d_{\text {surface }}\left(\frac{r}{R}\right) \end{aligned}$$

Also, for a point at height $h$ above surface,

$g_{\text {height }}=g_{\text {surface }}\left[\frac{R^2}{(R+h)^2}\right]$

Therefore, we can say that value of $g$ increases from centre to maximum at the surface and then decreases as depicted in graph (b).

We know that,

$e V_0=h \nu-\phi_0$ where, $V_0=$ stopping potential

and $\quad \nu=$ frequency of light.

Case I $e V_0=h \nu-\phi_0$ .... (i)

Case II $\frac{e V_0}{4}=\frac{h \nu}{2}-\phi_0$

$\Rightarrow \quad e V_0=2 h \nu-4 \phi_0$ .... (ii)

From Eqs. (i) an i (ii), we get

$h \nu-\phi_0 =2 h \nu-4 \phi_0$

$\Rightarrow h\nu = 3{\phi _0} \Rightarrow h\nu = 3h{\nu _0}\quad$ ($\because$ ${\phi _0} = h{\nu _0}$)

$\Rightarrow {\nu _0} = {{h\nu } \over {3h}} \Rightarrow {\nu _0} = {\nu \over 3}$

Given, ratio of radius of two wires,

$\frac{r_1}{r_2}=\frac{2}{1}$

and ratio in their lengths,

$\frac{l_1}{l_2}=\frac{1}{2}$

When same force is applied on them, then ratio change in their lengths,

$\frac{\Delta l_1}{\Delta l_2}=\text { ? }$

We known that, Young's modulus,

$$\begin{aligned} & Y=\frac{F l}{A \Delta l} \Rightarrow \Delta l=\frac{F l}{A Y} \\ &\quad \Delta l \propto \frac{l}{A} \\ \therefore \quad & \frac{\Delta l}{\Delta l_1} \frac{l}{A}=\frac{l_1 / A_1}{l_2 / A_2}=\frac{l_1 A_2}{l_2 A_2} \\ \Rightarrow \quad & \frac{\Delta l_1}{\Delta l_2}=\frac{l_1 / A_1}{r_2^2 / r_1^2}=\frac{1}{2} \cdot\left(\frac{1}{2}\right)^2=\frac{1}{8} \end{aligned}$$

Consider the string of length $l$ connected to a particle as shown in the figure

Speed of the particle is $v$. As, the particle is in uniform circular motion, the net force on the particle must be equal to centripetal force which is provided by the tension $(T)$.

$$\begin{aligned} \therefore \text { Net force } & =\text { Centripetal force } \\ & =\text { Tension in string } \end{aligned}$$

$\Rightarrow \quad \frac{m v^2}{l}=T$

Applying Archimedes' principle, submerged part $=$ replaced water

$\Rightarrow \quad \frac{4}{5} h \rho_{w o} g=h \times \rho_l \times g$

where, $\rho_w=$ density of water and $\rho_l=$ density of the liquid.

$$\begin{array}{ll} \therefore & \rho_l=\frac{4}{5} \times \rho_w=\frac{4}{5} \times 1000 \\ \Rightarrow & \rho_l=800 \mathrm{~kgm}^{-3} \end{array}$$

When object is in rarer medium and observer is in denser medium, then normal shift,

$d=(n-1) h$

where, $h=$ real depth $=y$

Now, apparent depth or the apparent height of the bird from the surface of the water $=y+(n-1) y=n. y$. The total distance of the bird as estimated by fish is $x+n y$.

From figure,

$C_1=C_2=C_3=C_4=3 \mu \mathrm{F} \quad$ (given)

$\because C_4$ and $C_3$ are in parallel, i.e.

$C_{\text {eq }}=C_4+C_3=(3+3) \mu \mathrm{F}=6 \mu \mathrm{F}$

Now arrangement of capacitors will be as follows

$\because C_1$ is in parallel with the series combination of $C_2$ and $C_{\text {eq }}^{\prime}$.

$\therefore C_{\text {eq }}^{\prime \prime}$ between $A$ and $B$

$=\frac{\left(C_2 \times C_{\text {eq }}^{\prime}\right)}{C_2+C_{\text {eq }}^{\prime}}+C_1 \Rightarrow\left(\frac{3 \times 6}{3+6}\right)+3=5 \mu \mathrm{F}$

$$\begin{array}{ll} \text {By using } v^{\prime}=v\left[\frac{v-v_0}{v-v_s}\right] \\ \Rightarrow \quad 2 v=v\left[\frac{v-v_0}{v-0}\right] \\ \Rightarrow \quad v_0=-v \end{array}$$

Negative sign indicates that observer is moving opposite to the direction of velocity of sound.

Charge, $q=n e$

$\Rightarrow \quad q=10^{19} \times 1.6 \times 10^{-19}=1.6 \mathrm{~C}$

As, the mass density of a nucleus is constant

$\therefore$ Mass density, $=\frac{3 m}{4 \pi R_0^3}$

Lyman series of $\mathrm{H}$-atom, we can write

$\frac{h c}{\lambda}=R h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

where, symbols have their usual meaning and for second line of Balmer series of $\mathrm{H}$-like ion

$\frac{h c}{\lambda}=Z^2 R h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$

Therefore, $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=Z^2\left(\frac{1}{4}-\frac{1}{16}\right)$

$$\begin{aligned} \left(1-\frac{1}{4}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\ \Rightarrow \quad Z & =2.19 \simeq 2 \end{aligned}$$

The approximately value of $Z$ will be 2.

For first slab,

Heat current, $H_1=\frac{K_1\left(\theta_1-\theta\right) A}{d_1}$

For second slab,

Heat current, $H_2=\frac{K_2\left(\theta-\theta_2\right) A}{d_2}$

As slabs are in series, $H_1=H_2$

$$\begin{aligned} \therefore \quad & \frac{K_1\left(\theta_1-\theta\right) A}{d_1}=\frac{K_2\left(\theta-\theta_2\right) A}{d_2} \\ \Rightarrow \quad & \theta=\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_2 d_1+K_1 d_2} \end{aligned}$$

The frequency of a vibrating string can be expressed by the formula:

Where:

• $f$ is the frequency of the string,
• $L$ is the length of the string,
• $T$ is the tension in the string, and
• $\mu$ is the linear mass density of the string.

Initially, the frequency of the string is $200 \mathrm{~Hz}$ when the length is $L$ and the tension is $T_1$. The formula becomes:

When the length of the string is doubled to $2L$ and the tension is altered to $T_2$, the frequency becomes $300 \mathrm{~Hz}$. The formula then changes to:

We can rearrange each equation to express the square root of tension over linear mass density:

Squaring both sides of each equation gives:

Take the ratio $\frac{T_2}{T_1}$:

Thus, the ratio of the new tension $T_2$ to the original tension $T_1$ is 9:1, indicating that the new tension is 9 times greater than the original tension.

The correct answer is Option A: 9 : 1.

Assume, Young's double slit experiment, amplitudes corresponding to the sources $S_1$ and $S_2$ be $a_1$ and $a_2$ and intensities be $I_1$ and $I_2$ respectively.

So, $I_{\max } \propto a_{\text {max }}^2$ and $I_{\text {min }} \propto a_{\text {min }}^2$

$\Rightarrow \frac{I_{\max }}{I_{\min }}=\left(\frac{a_{\max }}{a_{\min }}\right)^2=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2$

According to given question,

$$\begin{aligned} & \frac{I_{\max }}{I_{\min }}=\frac{9}{1} \\ & \Rightarrow \quad 9=\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}\right)^2 \\ & \Rightarrow \quad(3)^2=\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}\right)^2 \quad \text{(given)}\\ & \Rightarrow \quad 3=\frac{a_1 / a_2+1}{a_1 / a_2-1} \Rightarrow 3=\frac{\left(\frac{a_1+a_2}{a_2}\right)}{\left(\frac{a_1-a_2}{a_2}\right)} \Rightarrow 3=\frac{a_1+a_2}{a_1-a_2}\\ & \Rightarrow \quad 3\left(a_1-a_2\right)=a_1+a_2 \Rightarrow 3 a_1-3 a_2=a_1+a_2 \\ &\Rightarrow \quad 2 a_1=4 a_2 \Rightarrow \frac{a_1}{a_2}=\frac{4}{2} \Rightarrow a_1: a_2=2: 1 \end{aligned}$$

We know that, $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

$\Rightarrow \quad \frac{1}{\lambda}=R\left(\frac{1}{1_1^2}-\frac{1}{2^2}\right)=\frac{3 R}{4} \Rightarrow \lambda=\frac{4}{3 R} \mathrm{~cm}$

$$\begin{aligned} & \text { Here, } \frac{I_{\text {bright }}}{I_{\text {dark }}}=\frac{I_{\max }}{I_{\min }}=3 \quad \text{(given)}\\ & \Rightarrow \quad \frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{3}{1} \Rightarrow \frac{a_1+a_2}{a_1-a_2}=\sqrt{3} \\ & \Rightarrow \quad a_1+a_2=\sqrt{3}\left(a_1-a_2\right) \Rightarrow \frac{a_1}{a_2}=\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) \end{aligned}$$

For an ideal gas, mean kinetic energy per molecule

$=\frac{f}{2} k_B T$

For diatomic gas, degree of freedom,

$\begin{gathered} f=5 \\ \therefore \text { K.E per molecule }=\frac{5 k_B T}{2} \end{gathered}$

Net work done by the gas = Area enclosed in $p$-$V$ curve, i.e. area of $\triangle A B C$.

$W_{\text {net }}=\frac{1}{2} \times 5 \times 10^{-3} \times 4 \times 10^5 \mathrm{~J}=10^3 \mathrm{~J}=1000 \mathrm{~J}$

Here, phase difference

$\tan \phi=\frac{X_L-X_C}{R} \Rightarrow \tan \frac{\pi}{3}=\frac{X_L-X_C}{R}$

When $L$ is removed, $\tan \frac{\pi}{3}=\frac{X_C}{R}=\sqrt{3}$

$\therefore \quad X_C=\sqrt{3} R$ .... (i)

Similarly, when $C$ is removed

$$\begin{aligned} & \tan \frac{\pi}{3}=\frac{X_L}{R}=\sqrt{3} \\ \Rightarrow \quad X_L & =\sqrt{3} R \\ \Rightarrow \quad X_C & =X_L \quad \text{...(ii) [from Eq. (i)]} \end{aligned}$$

Now, $\tan \phi=0$

$\Rightarrow \quad \phi=0^{\circ}$

$\therefore$ Power factor, $\cos \phi=\cos 0^{\circ}=1$

When two capacitors are connected then their common potential difference is

$V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}$

Given, $C_1=2 \mu \mathrm{F}, V_1=14 \mathrm{~V}, C_2=5 \mu \mathrm{F}, V_2=0~V$

$\therefore \quad V=\frac{2 \times 14+5 \times 0}{2+5}=4 V$

When the balloon is descending down with an acceleration $a$,

$m g-B=m a\quad$ .... (i)

where, $B=$ buoyant force.

Here, we should assume that while removing some mass the volume of balloon and hence buoyant force will not change.

Let the new mass of the balloon is $m^{\prime}$, so

$B-m^{\prime} g=m^{\prime} a\quad$ ... (ii)

On solving Eqs. (i) and (ii), we get

$$\begin{aligned} & m g-m^{\prime} g=m a+m^{\prime} a \\ \Rightarrow & m(g-a)=m^{\prime}(g+a) \Rightarrow m^{\prime}=\frac{m(g-a)}{g+a} \end{aligned}$$

So, mass removed, $\Delta m=m-m^{\prime}$

$=m\left[1-\frac{(g-a)}{(g+a)}\right]=m\left[\frac{g+a-g+a}{g+a}\right]=\frac{2 m a}{g+a}$

$$\begin{aligned} \text { Here, } L & =1 \mathrm{H}, C=20 \mu \mathrm{F}, R=300 \Omega \\ X_L & =2 \pi f L=2 \pi\left(\frac{50}{\pi}\right) \times 1=100 \Omega \\ X_C & =\frac{1}{2 \pi f C}=\frac{1}{2 \pi\left(\frac{50}{\pi}\right) 20 \times 10^{-6}}=500 \Omega \\ \end{aligned}$$

$$\begin{aligned} \text { Impedance, } Z & =\sqrt{R^2+\left(X_C-X_L\right)^2} \\ & =\sqrt{(300)^2+(400)^2}=500 \Omega \end{aligned}$$

According to the question,

For wire 1 : Area of cross-section $=A_1$,

Force applied $=F_1$

and increase in length $=\Delta l$.

From the relation of Young's modulus of elasticity,

$Y=\frac{F l}{A \Delta l}$

Substituting the values for wire 1 in the above relation, we get

$\Rightarrow \quad Y_1=\frac{F_1 l_1}{A_1 \Delta l}\quad$ .... (i)

For wire 2: Area of cross-section $=A_2$

Force applied $=F_2$

increase in length $=\Delta l$

Similarly, $Y_2=\frac{F_2 l_2}{A_2 \Delta l}\quad$ .... (ii)

$\because$ Volume, $V=A l$ or $l=\frac{V}{A}$

Substituting the value of $l$ in Eqs. (i) and (ii), we get

$Y_1=\frac{F_1 V}{A_1^2 \Delta l} \text { and } Y_2=\frac{F_2 V}{A_2^2 \Delta l}$

As it is given that the wires are made up of same material,

i.e. $\quad Y_1=Y_2$

$\Rightarrow \quad \frac{F_1 V}{A_1^2 \Delta l}=\frac{F_2 V}{A_2^2 \Delta l} \Rightarrow \frac{F_1}{F_2}=\frac{A_1^2}{A_2^2}=\frac{A^2}{9 A^2}=\frac{1}{9}\quad (\because \quad A_1=A \text { and } A_2=3 A)$

or $\quad F_2=9 F_1=9 F\quad$ (given, $F_1=F$)

The magnetic field induction at $P$ due to currents through both the wires is

$B=\frac{\mu_0}{4 \pi} \frac{2 i}{(r / 2)}+\frac{\mu_0}{4 \pi} \frac{2(2 i)}{(r / 2)}=\frac{\mu_0}{4 \pi} \cdot \frac{12 i}{r}$

acting perpendicular to plane of wire inwards. Now, $B$ and $v$ are acting in the same direction, i.e. $\theta=0^{\circ}$

Force on charged particle is $F=q v B \sin \theta=q v B \times 0=0$.

Current gain factor, $\alpha=0.96$

$R=1000 \Omega$, Voltage drop across collector resistor $=0.5 \mathrm{~V}$

$\therefore \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.96}{1-0.96}=24$

$\therefore$ Collector current,

$i_c=\frac{\text { Voltage drop across collector resistor }}{R}$

$=\frac{0.5}{1000}=0.5 \times 10^{-3} \mathrm{~A}$

The base current, $i_b=\frac{i_c}{\beta}=\frac{0.5 \times 10^{-3}}{24}=20.8 \mu \mathrm{A}$

The dielectric is introduced such that half of its area is occupied by it.

In the given case, the two capacitors are in parallel.

$\therefore \quad C^{\prime}=C_1+C_2$

But $C_1=\frac{A \varepsilon_0}{2 d} \text { and } C_2=\frac{K A \varepsilon_0}{2 d}$

Thus, $C^{\prime}=\frac{A \varepsilon_0}{2 d}+\frac{K A \varepsilon_0}{2 d} \Rightarrow C^{\prime}=\frac{C}{2}(1+K)$

The energy of light of wavelength $\lambda$ is given by

$$\begin{aligned} & E=h v=\frac{h c}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{h c}{E} \quad \text{... (i)} \end{aligned}$$