KCET 2022 Question Paper with Solutions

Given, length of solenoid, $l=50 \mathrm{~cm}=0.5 \mathrm{~m}$

Number of turns, $N=100$

Current, $I=2.5 \mathrm{~A}$

Number of turns per unit length in the solenoid

$n=\frac{N}{l}=\frac{100}{0.5}=200 \text { turns } / \mathrm{m}$

Magnetic field at the one end of solenoid,

$$\begin{aligned} B & =\frac{\mu_0 n I}{2}=\frac{4 \pi \times 10^{-7} \times 200 \times 2.5}{2} \\ & =10 \pi \times 10^{-5}=\pi \times 10^{-4}=3.14 \times 10^{-4} \mathrm{~T} \end{aligned}$$

Given, mass of nucleus of nitrogen,

$m\left[{ }_7 \mathrm{~N}^{14}\right]=14.00307 \mathrm{u}$

Mass of proton, $m_p=1.00783 \mathrm{~amu}$

Mass of neutron, $m_n=1.0087 \mathrm{~amu}$

Mass defect, $\Delta m=\left(7 m_p+7 m_n\right)-m\left({ }_7 \mathrm{~N}^{14}\right)$

$$\begin{aligned} & =7 \times 1.00783+7 \times 1.00807-14.00307 \\ & =0.11243 \mathrm{~amu} \end{aligned}$$

$\therefore$ Binding energy of nitrogen nucleus

$$\begin{aligned} & =\Delta m \times 931=0.11243 \times 931 \\ & =104.67 \simeq104.7 \mathrm{~MeV} \end{aligned}$$

The given situation is shown below

Equation of motion for body $m_2$,

$$\begin{aligned} T_2-m_2 g & =m_2 a \\ T_2-3 g & =3 \times 2 \\ T_2 & =3 g+6=3 \times 9.8+6=35.4 \mathrm{~N} \end{aligned}$$

Equation of motion of mass of $5 \mathrm{~kg}$,

$$\begin{aligned} & & T_1-T_2-m_1 g & =m_1 a \\ \Rightarrow & & T_1-35.4-5 \times 9.8 & =5 \times 2 \\ \Rightarrow & & T_1-84.4 & =10 \\ \Rightarrow & & T_1 & =10+84.4=94.4 \mathrm{~N} \end{aligned}$$

de-Broglie wavelength $(\lambda)$ of the particle in terms of kinetic energy $K$ is given as

$\lambda=\frac{h}{\sqrt{2 m K}}$ .... (i)

where, $m$ is the mass of the particle and $h$ is Planck's constant.

Let at wavelength $\lambda^{\prime}$, kinetic energy of particle becomes $K^{\prime}=\frac{K}{4}$

$$\begin{aligned} \text { i.e } \quad \lambda^{\prime} & =\frac{h}{\sqrt{2 m K^{\prime}}}=\frac{h}{\sqrt{2 m \cdot \frac{K}{4}}} \\ & =\frac{h}{\frac{1}{2} \sqrt{2 m K}}=2 \frac{h}{\sqrt{2 m K}} \\ \Rightarrow \quad \lambda^{\prime} & =2 \lambda \quad \text { [from Eq. (i)] } \end{aligned}$$

Given, initial angular frequency of wheel,

$f_0=1200 \mathrm{rpm}=\frac{1200}{60} \mathrm{rps}=20 \mathrm{~rps}$

$\therefore$ Initial angular velocity,

$\omega_0=2 \pi f_0=2 \pi \times 20=40 \pi \mathrm{rad} / \mathrm{s}$

Similarly, final angular speed,

$\omega=2 \pi \times\left(\frac{3120}{60}\right) \mathrm{rad} / \mathrm{s}=104 \pi \mathrm{rad} / \mathrm{s}$

$\text { Time, } \quad t=16 \mathrm{~s}$

If $\alpha$ be the angular acceleration of the wheel, then

$$\begin{aligned} \omega & =\omega_0+\alpha t \\ \Rightarrow \quad \alpha & =\frac{\omega-\omega_0}{t}=\frac{104 \pi-40 \pi}{16} \\ & =\frac{64 \pi}{16}=4 \pi \mathrm{~rad} / \mathrm{s}^2 \end{aligned}$$

When a charged capacitor is disconnected from the battery and a glass slab (dielectric material) is introduced between the plates, then charge $(Q)$, capacitance $(C)$, electric potential $(V)$ and stored energy $(U)$ is given as

$Q=Q_0, C=K C_0, V=\frac{V_0}{K} \text { and } U=\frac{U_0}{K}$

where, $K=$ dielectric constant

$V=$ electric potential

$U=$ potential energy

$V_0=$ initial potential of charged capacitor

$U_0=$ Initial energy of charged capacitor

From above, we see that $V$ and $U$ decreases after introducing glass plate of dielectric constant $k$ whereas capacitance increases and charge remains same because battery is disconnected after introducing dielectric material.

Ge is a semiconductor material which behaves as an insulator at $0 \mathrm{~K}$. We know that, forbidden energy gap of an insulator is more than $3 \mathrm{~eV}$. Hence, among the given options, option (c) is correct.

Given, $E=\frac{81 \pi}{7} \times 10^5 \mathrm{~V} / \mathrm{m}$

Terminal velocity , $v=2 \times 10^{-3} \mathrm{~m} / \mathrm{s}$

$g=9.8 \mathrm{~m} / \mathrm{s}^2$

Viscosity, $\eta=1.8 \times 10^{-5} \mathrm{~N}-\mathrm{s} / \mathrm{m}^2$

Density, $\rho=900 \mathrm{~kg} / \mathrm{m}^3$

Since, $q E=m g$ ..... (i)

In the absence of electric field,

$$\begin{aligned} & m g=6 \pi \eta r v \\ & \Rightarrow \quad q E=6 \pi \eta r v \text { [from Eq. (i)] } \\ & \Rightarrow \quad r=\frac{q E}{6 \pi \eta v} \quad \text{.... (ii)}\\ \end{aligned}$$

From Eq. (i), we get

$m=\frac{q E}{g}$

$$\begin{aligned} & \Rightarrow \frac{4}{3} \pi r^3 d=\frac{q E}{g} \Rightarrow \frac{4}{3} \pi\left(\frac{q E}{6 \pi \eta v}\right)^3 d=\frac{q E}{g} \\ & \begin{aligned} \Rightarrow \quad q & =\sqrt{\frac{3 \times 6^3 \pi^2 \eta^3 v^3}{4 E^2 g}} \\ & =\sqrt{\frac{3 \times 6^3 \times(314)^2 \times\left(1.8 \times 10^{-5}\right)^3 \times\left(2 \times 10^{-3}\right)^3}{4 \times\left(\frac{81 \pi}{7} \times 10^5\right)^2 \times 9.8}} \\ & =8 \times 10^{-19} \mathrm{C} \end{aligned} \end{aligned}$$

The intensity of cosmic rays stream of charged particles is move at the poles than at the equator because the Earth's magnetism is strongest at poles. Therefore, cosmic rays are deflected away from the equator. In order to reach at equator, they need to pass greater kinetic energy which shows that the Earth has a magnetic field.

Equation of alternating current,

$i=i_1 \sin \omega t+i_2 \cos \omega t$

$$\begin{aligned} \text { Let } i_1 & =I \cos \phi \quad \text{.... (i)}\\ \text { and } i & =I \sin \phi \quad \text{.... (ii)}\\ & i=I \cos \phi \sin \omega t+I \sin \phi \cos \omega t \\ \Rightarrow \quad i & =I \sin (\omega t+\phi) \end{aligned}$$

$\therefore$ rms value of $i_1$,

$i_{\mathrm{rms}}=\frac{I}{\sqrt{2}}$ .... (iii)

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

$\begin{gathered} i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \sin ^2 \phi=I^2\left(\cos ^2 \phi+\sin ^2 \phi\right) =I^2\end{gathered}$

$\Rightarrow \quad \sqrt{i_1^2+i_2^2}=I$ .... (iv)

From Eqs. (iii) and (iv), we get

$i_{\mathrm{rms}}=\sqrt{\frac{i_1^2+i_2^2}{2}}$

For an equilateral glass prism, Angle of prism, $A=60^{\circ}$

If $i$ and $e$ be the angle of incidence and angle of emergence respectively, then

$i=\frac{3}{4} A=\frac{3}{4} \times 60^{\circ}=45^{\circ} \text { and } e=i=45^{\circ}$

For a prism, we know that

$$\begin{aligned} A+\delta & =i+e \Rightarrow \delta=i+e-A \\ \delta & =45^{\circ}+45^{\circ}-60^{\circ}=90^{\circ}-60^{\circ}=30^{\circ} \end{aligned}$$

According to given situation, force on charge particle in uniform electric field, $F=q E$ Acceleration of charge particle,

$a=\frac{F}{m}=\frac{q E}{m}$

Velocity of charge particle after time $t$ is given as

$v=u+a t=0+\frac{q E}{m} t \Rightarrow v=\frac{q E t}{m}$

$\therefore$ Kinetic energy of the charged particle

$K=\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{q E t}{m}\right)^2=\frac{E^2 q^2 t^2}{2 m}$

Given, power delivered by nuclear reactor,

$$\begin{aligned} & P=10^9 \mathrm{~W} \\ & c=3 \times 10^8 \mathrm{~m} / \mathrm{s} \\ & t=1 \mathrm{~h}=60 \times 60 \mathrm{~s} \end{aligned}$$

$$\begin{aligned} \text { Since, } \quad P & =\frac{E}{t}=\frac{m c^2}{t} \quad\left[\because E=m c^2\right] \\ \Rightarrow \quad m & =\frac{P t}{c^2}=\frac{10^9 \times 60 \times 60}{\left(3 \times 10^8\right)^2} \\ & =\frac{10^9 \times 3600}{9 \times 10^{16}}=4 \times 10^{-5} \mathrm{~kg} \\ & =0.04 \times 10^{-3} \mathrm{~kg}=0.04 \mathrm{~g} \end{aligned}$$

Radius of circular track, $r=10 \mathrm{~m}$

speed of car, $v=10 \mathrm{~m} / \mathrm{s}$

The given situation is shown below

$$\begin{aligned} T \sin \theta & =\frac{m v^2}{r} \quad \text{.... (i)}\\ T \cos \theta & =m g \quad \text{.... (ii)} \end{aligned}$$

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

$$\begin{aligned} & \frac{T \sin \theta}{T \cos \theta}=\frac{m v^2 / r}{m g} \Rightarrow \tan \theta=\frac{v^2}{r g}=\frac{10^2}{10 \times 10}=1 \\ & \Rightarrow \quad \tan \theta=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}=\frac{\pi}{4} \mathrm{rad} \end{aligned}$$

By lens Maker's formula, focal length of lens is given as

$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

where, $\mu$ is refractive index.

$$\begin{array}{ll} \Rightarrow \frac{1}{f} \propto(\mu-1) \Rightarrow f \propto \frac{1}{(\mu-1)} \quad \text{.... (i)}\\ \text { Since, } \quad \mu \propto \frac{1}{\lambda} \quad \text{.... (ii)} \end{array}$$

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

$f \propto \lambda$

But, wavelength of red colour of light is maximum as all the other constituent colours of white light. Hence, focal length of a convex lens is maximum for red light.

Given, distance between object and screen for convex lens $=x$

i.e $\quad u+v=x$ .... (i)

but magnification, $m=\frac{v}{u}$ ..... (ii)

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

$$\begin{aligned} & u+m u & =x \Rightarrow u(m+1)=x \\ \Rightarrow & u & =\frac{x}{m+1} \quad \text{..... (iii)} \end{aligned}$$

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

$\frac{v}{m}+v=x \Rightarrow v\left(\frac{1}{m}+1\right)=x$

$\Rightarrow \quad v=\frac{m x}{m+1}$ .... (iv)

Using lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{\left(\frac{m x}{m+1}\right)}-\frac{1}{\left(\frac{-x}{m+1}\right)}$

[from Eqs. (iii) and (iv)]

$$\begin{aligned} & =\frac{m+1}{m x}+\frac{m+1}{x}=\frac{m+1}{x}\left[\frac{1}{m}+1\right] \\ & =\frac{m+1}{x}\left(\frac{1+m}{m}\right)=\frac{(m+1)^2}{m x} \Rightarrow f=\frac{m x}{(m+1)^2} \end{aligned}$$

When a compact disc (CD) is illuminated by small source of white light, coloured bands are observed. This happens due to the phenomenon of diffraction, in which small ripples on the surface of it, break up white light into the colour of rainbow.

Let initial length of the wire be $l$. After stretching new length of the wire $r^{\prime}$ is given as

$l^{\prime}=l+10 \% \text { of } l=l+\frac{10}{100} l=\frac{11 l}{10}=1.1l$

i.e. length is increased by 1.1 times

$\therefore \quad n=1.1$

We know that, when length of a wire is increased by $n$ times, then its new resistance is increased by $n^2$ times i.e. $R^{\prime}=n^2 R=(1.1)^2 R\quad$ [$\because$ Here, $n=1.1$]

$=1.21 \mathrm{R}$

Specific resistance (resistivity) of a wire does not depends on its dimensions (length, width etc.) because it is a characteristic property of a material and depends only nature of material of wire. Hence, specific resistance remains same.

The resistivity of a material is a measure of how strongly the material resists the flow of electric current. Semiconductors have resistivity values that are between those of good conductors (like metals) and good insulators (like ceramics).

Option A states that the resistivity range is from $10^{-3}$ to $10^6 \Omega-\mathrm{cm}$.

Option B suggests a range from $10^6$ to $10^8 \Omega-\mathrm{cm}$.

Option C offers a range from $10^{10}$ to $10^{12} \Omega-\mathrm{cm}$.

Option D proposes a range from $10^{-2}$ to $10^{-5} \Omega-\mathrm{cm}$.

The correct resistivity range for semiconductors at room temperature is typically closer to Option A: $10^{-3}$ to $10^6 \Omega-\mathrm{cm}$. This range is wide because it can be affected by several factors, including the type of semiconductor material, the presence of impurities (doping), and the temperature of the material.

Increasing order of wavelength of given electromagnetic wave is given as

$\lambda_{\text {X-rays }}<\lambda_{\text {UV-rays }}<\lambda_{\text {IR-rays }}<\lambda_{\text {microwaves }}$

Hence, microwaves has highest wavelengths.

Given,

$$\begin{aligned} & P=100 \mathrm{~W} \\ & V=220 \mathrm{~V} \end{aligned}$$

We know that, $P=\frac{V^2}{R}$

where, $R$ is resistance of bulb.

$\Rightarrow \quad R=\frac{V^2}{P}=\frac{220 \times 220}{100}=484 \Omega$

Now, according to question, voltage drops by $2.5 \%$ of its rated value.

New voltage, $V^{\prime}=220-2.5 \%$ of 220

$=220-\frac{2.5}{100} \times 220=214.5 \mathrm{~V}$

$\therefore$ New power, $P^{\prime}=\frac{\left(V^{\prime}\right)^2}{R}=\frac{214.5 \times 214.5}{484}=95.06 \mathrm{~W}$

$\%$ decrease in power $=\frac{P-P^{\prime}}{P} \times 100$

$=\frac{100-95.06}{100} \times 100 \simeq5 \%$

Given, for an electric dipole, dipole moment,

$p=4 \times 10^{-9} \mathrm{C}$

$\theta=30^{\circ}$

Magnitude of electric field, $E=5 \times 10^4 \mathrm{~N} / \mathrm{C}$

$\therefore$ Torque, $\tau=p E \sin \theta$

$=4 \times 10^{-9} \times 5 \times 10^4 \sin 30^{\circ}=10^{-4} \mathrm{~N}-\mathrm{m}$

Given, change in kinetic energy of photoelectrons,

$$\begin{aligned} \Delta K & =0.52 \mathrm{~eV} \\ \lambda_1 & =500 \mathrm{~nm}, \lambda_2=? \end{aligned}$$

We know that, kinetic energy of emitted photoelectron (using Einstein's equation)

$K=\frac{h c}{\lambda}-\phi$

$\therefore$ At wavelength $\lambda_1$,

$K_1=\frac{h c}{\lambda_1}-\phi$ .... (i)

At wavelength $\lambda_2$,

$K_2=\frac{h c}{\lambda_2}-\phi$ .... (ii)

From Eqs. (i) and (ii), we have

$$\begin{array}{llrl} & & K_2-K_1 & =\frac{h c}{\lambda_2}-\frac{h c}{\lambda_1} \\ \Rightarrow & & \Delta K & =h c\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right) \\ \Rightarrow & & 0.52 & =1242\left(\frac{1}{\lambda_2}-\frac{1}{500}\right) \\ \Rightarrow & & \frac{0.52}{1242} & =\frac{1}{\lambda_2}-\frac{1}{500} \\ \Rightarrow & & \frac{1}{\lambda_2} & =\frac{0.52}{1242}+\frac{1}{500} \\ \Rightarrow & & \lambda_2 & \approx 400 \mathrm{~nm} \end{array}$$

In case of Fraunhoffer diffraction at a single slit, the diffraction pattern on the screen, central bright band having alternate dark, and bright bands of decreasing intensity on either side.

We know electric field and potential due to electric dipole on axial position is given as

$E=\frac{1}{4 \pi \varepsilon_0} \frac{2 p}{r^3} \text { and } V=\frac{1}{4 \pi \varepsilon_0} \frac{p}{r^2}$

i.e. $\quad E \propto \frac{1}{r^3}$ and $V \propto \frac{1}{r^2}$

Charge on first sphere, $q_1=1.8 \mu \mathrm{C}=1.8 \times 10^{-6} \mathrm{C}$

Charge on second sphere, $q_2=2.8 \mu \mathrm{C}=2.8 \times 10^{-6} \mathrm{C}$

Distance between the two spheres, $r=40 \mathrm{~cm}$

Distance between mid-point and each sphere

$r_1=r_2=20 \mathrm{~cm}=0.2 \mathrm{~m}$

$\therefore$ Electric potential at mid-point due to both charged sphere

$$\begin{aligned} V & =V_1+V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2} \\ & =9 \times 10^9 \times \frac{1.8 \times 10^{-6}}{0.2}+9 \times 10^9 \times \frac{2.8 \times 10^{-6}}{0.2} \\ & =9 \times 10^3(9+14)=2.07 \times 10^5 \mathrm{~V} \simeq 2.1 \times 10^5 \mathrm{~V} \end{aligned}$$

The given logic gate is represented as

Output of logic gate is written as

$y=\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}$

[According to Demorgan's law $\overline{x+y}=\bar{x} \cdot \bar{y}$]

$=A \cdot B=$ Output of AND gate

We know that, maximum vertical height in projectile motion,

$H=\frac{u^2 \sin ^2 \phi}{2 g}$ [where, $\phi$ is angle of projection.]

$$\begin{array}{ll} \Rightarrow & H \propto \sin ^2 \phi \\ \Rightarrow & \frac{H_1}{H_2}=\frac{\sin ^2 \phi}{\sin ^2 \phi_2}=\frac{\sin ^2 \theta}{\sin ^2(90-\theta)} \end{array}$$

[Given, $\phi_1=\theta$ and $\phi_2=90^{\circ}-\theta$]

$=\frac{\sin ^2 \theta}{\cos ^2 \theta}=\tan ^2 \theta$

$\therefore \quad H_1: H_2=\tan ^2 \theta: 1$

According to given situation, total voltage of 10 cells $=10 E$

Total resistance of 10 cells $=10 r$

$\therefore$ Current in the circuit, $I=\frac{10 E}{10 r}=\frac{E}{r}$

Potential difference across 3 cells,

$V=I \times(3 r)=\frac{E}{r} \times 3 r=3 E$

Hence, ideal voltmeter will read 3E.

Given, Bohr radius, $r_1=0.529\mathop A\limits^o$

We know that, radius of revolving electron in $n$th orbit is given as

$r \propto n^2$

$$\begin{aligned} & \Rightarrow \quad \frac{r_1}{r_2}=\left(\frac{n_1}{n_2}\right)^2 \\ & \Rightarrow \quad \frac{r_1}{r_2}=\left(\frac{1}{3}\right)^2=\frac{1}{9} \quad\left[\because n_1=1 \text { and } n_2=3\right] \\ & \Rightarrow \quad r_2=9 r_1=9 \times 0.529=4.761\mathop A\limits^o \\ \end{aligned}$$

Concept of field is a model used to explain the influence that a massive body or charged particle extends into the space around itself, producing a force on another massive body or charged body placed in space. Therefore, force concept of field is used to explain gravitational and electrostatic phenomena.

In photoelectric effect, saturation photocurrent is directly proportional to the intensity of incident radiation but independent of frequency of incident radiation. Hence, saturation current in photoelectric effect becomes double when both intensity and frequency of the incident light are doubled.

The least count of a measuring instrument is the smallest value that can be measured with the instrument and is calculated by the formula:

$\text{Least Count} = \frac{\text{Value of one main scale division}}{\text{Number of divisions on the Vernier scale}}$

In the given travelling microscope, there are 50 divisions on the Vernier scale which coincides with 49 main scale divisions. Each main scale division is of $0.5\ \mathrm{mm}$.

First, we need to find out the value of one main scale division:

Given that one main scale division is $0.5\ \mathrm{mm}$, we keep this value for the calculation.

Next, to calculate the least count, we divide this value by the number of divisions on the Vernier scale.

$\text{Least Count} = \frac{0.5\ \mathrm{mm}}{50}$

To calculate the value that the Vernier scale can measure beyond the main scale, we consider the discrepancy between the 49 main scale divisions and the 50 Vernier divisions. Since 49 main scale divisions coincide with 50 Vernier divisions, the length covered by 50 Vernier divisions is equal to the length covered by 49 main scale divisions.

Therefore, one Vernier division is equivalent to $\frac{49}{50}$ of a main scale division. The difference between the value of one main scale division and one Vernier division gives us the least count:

$\text{Main scale division value} - \text{Vernier scale division value} = \text{Least Count}$

$\text{Least Count} = \frac{0.5\ \mathrm{mm}}{50} \times 49$

Since there is a difference of 1 Vernier division in 50 divisions, the Vernier scale is more precise by:

$\text{Least Count} = \frac{0.5\ \mathrm{mm}}{50}$

Simplifying the fraction:

$\text{Least Count} = \frac{0.5}{50}\ \mathrm{mm}$

$\text{Least Count} = 0.01\ \mathrm{mm}$

So the correct answer is:

$0.01\ \mathrm{mm}$

The given situation is shown below

Since, charges kept at point $A$ and $C$ are in same magnitude and same nature (i.e. both positive), hence net force due these two charges on unit charge at centre $O$ will be zero.

Force due to charge $(+2 q)$ at unit positive charge at centre will be towards $B O$ and also force due to charge $(-2 q)$ at unit positive charge at centre will be towards $O D$.

Therefore, net force on the unit positive charge kept at centre $O$ will be towards $B D$.