KCET 2023 Question Paper with Solutions

We know that, magnetic susceptibility,

$$\begin{aligned} & \chi=\frac{C}{T-T_C} \\ \therefore \quad & \chi \propto \frac{1}{T-T_C} \end{aligned}$$

$$\begin{aligned} \Rightarrow \frac{\chi_{\text {cobalt }}}{\chi_{\text {iron }}} & =\frac{T-\left(T_C\right)_{\text {iron }}}{T-\left(T_C\right)_{\text {cobalt }}} \\ & =\frac{1600-1000}{1600-1400}=\frac{600}{200}=3 \end{aligned}$$

Given, $P=60 \mathrm{~W}$

$\lambda=662.5 \mathrm{~nm}=6.625 \times 10^{-7} \mathrm{~m}$

Energy of 1 photon $=h v=h \frac{c}{\lambda}$

$=\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6.625 \times 10^{-7}}=3 \times 10^{-19} \mathrm{~J}$

$\therefore$ Number of photons emitted per second

$$\begin{aligned} & =\frac{\text { Power of source }}{\text { Energy of one photon }} \\ & =\frac{60}{3 \times 10^{-19}} \\ & =20 \times 10^{19} \\ & =2 \times 10^{20} \end{aligned}$$

Convex mirror always produces virtual, erect and diminished image in air.

Let x be the equivalent resistance of the circuit, then

$$\begin{array}{ll} \therefore \quad & R_{A B}=x \\ \Rightarrow \quad & \frac{x \times 2}{2+x}+2+2=x \\ \Rightarrow \quad & \frac{2 x}{x+2}=x-4 \\ \Rightarrow \quad & 2 x=(x-4)(x+2) \\ \Rightarrow \quad & 2 x=x^2+2 x-4 x-8 \\ \Rightarrow \quad & x^2-4 x-8=0 \\ \Rightarrow \quad & x=\frac{-(-4) \pm \sqrt{(-4)^2-4 \times 1 \times(-8)}}{2 \times 1} \\ & =\frac{4 \pm \sqrt{48}}{2}=\frac{4 \pm 4 \sqrt{3}}{2} \\ & =2+2 \sqrt{3}=5.46 ~\Omega \simeq 5.5 ~\Omega \end{array}$$

Potential energy at $B=$ Potential difference $\times$ Charge $+$ Energy spent by moving charge from $A$ to $B$

$=V_A \times q-q \cdot E \cdot x=q\left(V_A-E x\right)$

$$\begin{aligned} \text { Given, } \quad x & =\frac{\beta}{3} \\ I_0 & =2 \times 10^{-2} ~\mathrm{Wm}^{-2} \end{aligned}$$

for YDSE where $D >>>d, \Delta x=d \sin \theta$

Also $\sin \theta \simeq \tan \theta \simeq \frac{x}{D}=\frac{\beta}{3 D}=\frac{\lambda D}{d \times 3 D}=\frac{\lambda}{3 d}$

$$\begin{aligned} & \therefore \Delta =\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} d \sin \theta=\frac{2 \pi}{\lambda} \times \frac{d \times \lambda}{3 d}=\frac{2 \pi}{3}=\Delta \phi \\ & I=4 I_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)=4 I_0 \cos ^2\left(\frac{2 \pi}{3} \times \frac{1}{2}\right) \\ & =4 I_0 \cos ^2\left(\frac{\pi}{3}\right)=4 \times I_0 \times \frac{1}{4} \\ & \therefore I=I_0=2 \times 10^{-2} \mathrm{~W} \mathrm{~m}^{-2} \end{aligned}$$

Given, mass of the body, $m=10 \mathrm{~kg}$

Coefficient of kinetic friction, $\mu_k=0.5$

Force applied in horizontal direction, $F=60 \mathrm{~N}$

Let $f$ be the frictional force on the body.

$$\begin{aligned} & \text { Thus, } \quad m a=F-\mu m g \\ & =60-10 \times(10)(0.5)=60-50 \\ & \Rightarrow 10 \times a=10 \Rightarrow a=1 \mathrm{~m} / \mathrm{s}^2 \\ \end{aligned}$$

Given,

$$\begin{aligned} & S=10 \Omega \\ & G=90 \Omega \\ & i=5 \times 10^{-3} \mathrm{~A} \end{aligned}$$

Number of divisions on one side of zero $=50$

$i_g=\frac{S}{S+G} \times i=\left(\frac{10}{90+10}\right)\left(5 \times 10^{-3}\right)=5 \times 10^{-4} \mathrm{~A}$

Number of divisions per unit current

$=\frac{50}{5 \times 10^{-4}} =1 \times 10^{-5} \mathrm{div} / \mathrm{A}$

In the Rutherford's $\alpha$-scattering experiment, impact parameter $(b)$ and scattering angle $\theta$ are related as,

$$\begin{aligned} b & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{z e^2 \cot \frac{\theta}{2}}{\left(\frac{1}{2} m v^2\right)} \\ \Rightarrow \quad b & \propto \cot \frac{\theta}{2} \end{aligned}$$

Thus, with the increase of impact parameter $(b)$, $\cot \frac{\theta}{2}$ increases. It is possible only when $\theta$ will decrease.

If the metal sphere is connected to insulated mount, then net flow due to induction of charges is zero and hence even after, the net charge on the metal sphere will be zero.

Red light has the longest wavelength and the smallest refractive index within the visible region. Thus, it bends least when passing from glass to air. As a result, its critical angle is maximum.

In full wave rectifier, one diode works for positive half cycle and another diode works for negative half cycle of alternating voltage. Thus, each diode works one time in one complete cycle of input AC signal. Hence, for $50 \mathrm{~Hz}$ AC supply, diode $D_1$ conducts 50 times in 1s.

Electric field due to infinitely charged long wire, $E_1=\frac{2 k \lambda}{r}$.

Electric field at the centre of semicircular ring,

$E_2=\frac{2 k \lambda}{r}$

Therefore, $E_1=E_2$

Colours in sequence,

Gray $\rightarrow 8 \quad$ (1st digit)

Red $\rightarrow 2\quad$ (2nd digit)

Gold $\rightarrow 0.1\quad$ (Multiplier)

Gold $\rightarrow \pm 5 \%\quad$ (Tolerance)

$\Rightarrow(82 \times 0.1) \pm 5 \% \Rightarrow R=8.2 \Omega \pm 5 \%$

Total flux $\phi_E=E \cdot d s$

$$\begin{aligned} & =9 \mathrm{k}-6 \mathrm{k} \times\left(10^{-2} \times 10\right)^2 \\ & =3 \times 10^3 \times 10^{-2}=30 \mathrm{~NC}^{-1} \mathrm{~m}^2 \end{aligned}$$

According to Gauss' law, $\frac{q}{\varepsilon_0}=\phi_E$

$$\begin{aligned} \Rightarrow q_{\text {enclosed }} & =3 \times \varepsilon_0 \\ & =30 \times 9 \times 10^{-12} \mathrm{C}=270 \times 10^{-9} \times 10^{-3} \\ & =0.27 \times 10^{-9} \mathrm{C}=0.27 \mathrm{~nC} \end{aligned}$$

$$\begin{aligned} \text { Given, } \quad { }^a \mu_g & =\frac{3}{2}, f_{\text {air }}=f \\ { }^a \mu_w & =\frac{4}{3} \end{aligned}$$

Using lens Maker's formula, when lens is in air,

$$\begin{aligned} \quad \frac{1}{f_{\text {air }}} & =\left({ }^a \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow \quad \frac{1}{f} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow \quad \frac{1}{f} & =\frac{1}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \quad \text{..... (i)} \end{aligned}$$

When lens is immersed in water, then

$\frac{1}{f_w}=\left({ }^w \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ..... (ii)

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

$$\begin{aligned} \frac{f_w}{f} & =\frac{1}{2\left(^w \mu_g-1\right)} \\ \Rightarrow \quad f_w & =\frac{f}{2\left(\frac{\mu_g}{\mu_w}-1\right)}=\frac{f}{2\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)}=\frac{f}{\frac{1}{4}}=4 f \\ \Rightarrow \quad f_w & =4 f \end{aligned}$$

$\therefore$ Percentage change in focal length

$$\begin{aligned} & =\frac{f_w-f}{f} \times 100 \\ & =\frac{4 f-f}{f} \times 100=300 \% \end{aligned}$$

According to question, moment of inertia of 7 discs will be

$$\begin{aligned} I & =I_{\mathrm{COM}}+M d_2 \\ & =\frac{7}{2} M R^2+6 M(2 R)^2 \end{aligned}$$

( $\because$ radius of all 6 discs will be $2 R$ from centre).

$=\frac{55}{2} M R^2$

Net force due to dipole on equatorial line, $F_{eq}=E.q$

$=\frac{K \mathbf{p}}{r^3} \hat{\mathbf{i}} q=\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}}$

Net torque; $\tau=F \times r_{\perp}$

$=\frac{1}{4} \pi \frac{p q}{r^3} \hat{\mathbf{i}} \times r(\hat{\mathbf{j}})=\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}$

Hence, option (d) is correct.

Given, resistance of carbon resistor

R = 4.7 k$\Omega$ $\pm$ 5% = 47 $\times$ 10$^{2}$ $\Omega$ $\pm$ 5%

The colour of four bands will be

4 $\to$ yellow

7 $\to$ Violet

2 $\to$ Red

$\pm~ 5$ $\to$ Gold

Thus, colour of third band is red.

The given situation is shown in the figure.

According to law of Malus,

$$\begin{aligned} I^{\prime} & =\left(\frac{I}{2}\right) \cos ^2 \theta \\ \Rightarrow \quad \frac{I}{4} & =\frac{I}{2} \cos ^2 \theta \\ \Rightarrow \cos ^2 \theta & =\frac{1}{2} \\ \Rightarrow \cos \theta & =\frac{1}{\sqrt{2}} \\ \Rightarrow \quad \theta & =45^{\circ} \end{aligned}$$

Let the final temperature of mixture be $T$.

Heat absorbed by ice $=$ heat lost by water

$\therefore m L+m c(T-0)=m c(100-T)$

Substituting values,

$$\begin{aligned} & \therefore & (100 \times 80)+(100 \times 1) T & =100 \times 1(100-T) \\ & \therefore & 8000+100 T & =10000-100 T \\ & \therefore & 200 T & =2000 \\ & & T & =10^{\circ} \mathrm{C} \end{aligned}$$

According to given diagram,

$$\begin{aligned} & \omega_1=400 \mathrm{~rad} / \mathrm{s} ; \quad \omega_r=500 \mathrm{~rad} / \mathrm{s} \\ & \omega_2=600 \mathrm{~rad} / \mathrm{s} \end{aligned}$$

$$\begin{aligned} \text { Qualitỵ factor, } \quad Q & =\frac{\omega_r}{\omega_2-\omega_1} \\ & =\frac{500}{600-400}=\frac{500}{200}=2.5 \end{aligned}$$

When $p$-$n$ junction diode is forward biased, then free electrons flow in the connecting wire.

Since, the middle branch conducts no current, therefore the middle capacitor represents open circuit and hence reduced circuit is shown.

$$\begin{aligned} \text { Now } \quad \frac{1}{C_{\text {series }}}=\frac{1}{1}+\frac{1}{1} \\ C_{\text {series }}=0.5 \mu \mathrm{F} \\ C_{\text {parallel }}=0.5+0.5 \\ \text{or} \quad C_{A B}=1 \mu \mathrm{F} \\ \end{aligned}$$

As we known,

$v=\frac{\mathbf{E}}{\mathbf{B}} \text { or } v=\frac{|\mathbf{E}|}{|\mathbf{B}|}$

For an electromagnetic wave, $v=c$

$\therefore \frac{|\mathbf{E}|}{|\mathbf{B}|}=c=3 \times 10^8 \mathrm{~ms}^{-1}$

$\therefore|\mathbf{E}|:|\mathbf{B}|$ is of the order of $10^8 \mathrm{~ms}^{-1}$.

[Hint Area enclosed by $p$-$V$ diagram is equal to work done] Total work done by gas in one carnot cycle is given by.

$W=\mu R T_1 \ln \frac{V_2}{V_1}-\mu R T_2 \ln \frac{V_3}{V_4}$

$$\begin{aligned} & \Delta I=5-2=3 \mathrm{~A} \\ & \Delta t=0.3 \mathrm{~s} \end{aligned}$$

Induced emf, $e=1 \mathrm{~V}$

We know that, $e=\frac{L d I}{d t}$

$$\begin{aligned} & \Rightarrow \quad e=\frac{L \Delta I}{\Delta t} \\ & \Rightarrow \quad L=\frac{e}{\frac{\Delta I}{\Delta t}}=\frac{1}{\frac{3}{0.3}}=0.1 \mathrm{~H}=100 \mathrm{~mH} \end{aligned}$$

Given, $B_H=3 \times 10^{-5} \mathrm{~T}$

$l=1 \mathrm{~m}$

Height travelled by rod in $t=2 \mathrm{~s}$

$h=\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times 2^2=20 \mathrm{~m}$

$\therefore$ Speed of rod, $v=\frac{h}{t}=\frac{20}{2}=10 \mathrm{~m} / \mathrm{s}$

Hence, induced emf

$$\begin{aligned} e & =B_H v l=3 \times 10^{-5} \times 10 \times 1 \\ & =3 \times 10^{-4} \mathrm{~V} \end{aligned}$$

Given, actual length of wire $\left(l_0\right)=3.678 \mathrm{~cm}$

Measurement with instrument, $A=3.5 \mathrm{~cm}$

Measurement with instrument, $B=3.38 \mathrm{~cm}$

Measurement of $A$ is close to $l_0$ so it is more accurate but measurement of $B$ is more precise as it can count upto two decimal places.

As the loop moves, there are three cases.

(i) When the loop moves fully outside the magnetic field, there is no change in flux.

$\therefore \varepsilon=\frac{d \phi}{d t}=0$

(ii) When the loop is entering/leaving the area of magnetic field.

$\begin{gathered} \varepsilon=B L V=(0.5)\left(2 \times 10^{-2}\right)\left(2 \times 10^{-2}\right) \\ =2 \times 10^{-4} \mathrm{~V} \end{gathered}$

As flux is in opposite direction to magnetic field.

$\varepsilon$ when loop enters the field $=-2 \times 10^{-4} \mathrm{~V}$

$$\begin{aligned} \varepsilon \text { when loop exits the field } & =-\left(-2 \times 10^{-4} \mathrm{~V}\right) \\ & =2 \times 10^{-4} \mathrm{~V} \end{aligned}$$

(iii) When moving inside the magnetic field.

$\frac{d \phi}{d t}=0$

$\therefore$ No emf is induced.

Radius of the circle, $r=\frac{m v}{B q}$

$\therefore$ Equation of line, $y=\frac{2 m v}{B q}=2 R$

Time period, $T=\frac{2 \pi m}{B q}$

$t=\frac{\pi m}{B q}=\frac{T}{2}$

$\therefore$ Angular momentum of charged particle about origin

$$\begin{aligned} L & =m v y+m v y=2 m v y \\ & =2 m v \times 2 R \\ & =4 m v \times \frac{m v}{B q}=\frac{4 m^2 v^2}{B q} \end{aligned}$$

From velocity selector,

$$\begin{aligned} & v=\frac{E}{B} \\ \therefore & \quad L=\frac{4 m^2 E^2}{q B^3} \end{aligned}$$

Given, $Q=5.5 \mathrm{~MeV}$

Mass number of nucleus $=220$

By conservation of linear momentum.

$M_\alpha v_\alpha=M_y v_y$

where, $M_\alpha, v_\alpha$ are mass and velocity of $\alpha$-particle, and $M_X v_X$ are mass and velocity of the daughter nuclei $X$.

Since, $Q=(\mathrm{KE})_\alpha+(\mathrm{KE})_X$ .... (i)

Let $Y$ be the nucleus with the mass number 220 and $X$ be the daughter nuclei produced.

Hence, we have ${ }^{220} Y \longrightarrow{ }^{216} X+{ }_4^2 \alpha$

Using conservation of linear momentum.

$M_\alpha v_\alpha=M_X v_X$

$$\begin{aligned} & \Rightarrow \quad v_x=\frac{M_\alpha v_\alpha}{M_X} \\ & \therefore \text { From Eq. (i), } Q=\frac{1}{2} M_\alpha v_\alpha^2+\frac{1}{2} M_X v_X^2 \\ & \Rightarrow \quad 5.5=\frac{1}{2} M_\alpha v_\alpha^2+\frac{1}{2} \frac{M_\alpha^2 v_\alpha^2}{M_x} \\ & \Rightarrow \quad 5.5=\frac{1}{2} M_\alpha v_\alpha^2\left[1+\frac{M_\alpha}{M_X}\right] \\ & \Rightarrow \quad 5.5=(\mathrm{KE})_\alpha\left(1+\frac{4}{216}\right) \\ & \Rightarrow \quad \mathrm{KE}_\alpha=5.4 \mathrm{~MeV} \end{aligned}$$

According to given situation,

Net emf of the combination,

$$\begin{aligned} E_{\text {net }} & =8 \times \text { emf of one cell }-2 \times \text { emf of one cell } \\ & =8 \times 2-2 \times 2=12 \mathrm{~V} \end{aligned}$$

Equivalent resistance of the circuit

$R^{\prime}=10 r+R=10 \times 1+10=20 \mathrm{~V}$

$\therefore$ Current through the resistor,

$I=\frac{E_{\mathrm{nct}}}{R^{\prime}}=\frac{12}{20}=0.6 \mathrm{~A}$