KCET 2021 Question Paper with Solutions

In single slit diffraction pattern, the width of central maxima is larger than the width of maxima on either sides and the intensity decreases rapidly on either side, but the light energy remains conserved.

Hence, statements I and II are correct.

As we know, kinetic energy of an object in pure rolling motion is given as

$E=\frac{1}{2} m v_{\mathrm{cm}}^2\left(1+\frac{k^2}{R^2}\right)$

where, $k$ is radius of gyration and $m$ is the mass of the object.

$\Rightarrow \quad \frac{E}{v_{\mathrm{cm}}^2}=\frac{1}{2}\left[1+\frac{k^2}{R^2}\right] \quad \text{.... (i)}$

[$\because$ Given, $m=1 \mathrm{~kg}$]

From the given graph, substituting the value of $\frac{E}{v_{\mathrm{cin}}^2}$ in Eq. (i), we get

$$\begin{aligned} \frac{3}{4} & =\frac{1}{2}\left[1+\frac{k^2}{R^2}\right] \\ \frac{3}{2}-1 & =\frac{k^2}{R^2} \\ \frac{k^2}{R^2} & =\frac{1}{2} \end{aligned}$$

Since, for a

(a) Sphere, $\frac{k^2}{R^2}=\frac{2}{5}$

(b) Ring, $\frac{k^2}{R^2}=1$

(c) Hollow cylinder, $\frac{k^2}{R^2}=1$

(d) Disc, $\frac{k^2}{R^2}=\frac{1}{2}$

So, the given object is disc.

The graph between potential energy of a pair of nucleons (in MeV) and their separation is shown below.

For, $r< r_0$, force is repulsive and force $r> r_0$, the force is attractive between nucleons.

$$\begin{aligned} \text { Given, } m & =2 \mathrm{~g}=2 \times 10^{-3} \mathrm{~kg} \\ \mathbf{E} & =(300 \mathrm{NC}^{-1}) \hat{\mathbf{i}} \end{aligned}$$

At, $x=0,(\mathrm{KE})_1=0$ and $x=0.5 \mathrm{~m},(\mathrm{KE})_2=0.12 \mathrm{~J}$

From work-energy theorem,

work done $=$ change in kinetic energy

$\Rightarrow$ Force $\times$ displacement $=\Delta \mathrm{KE}$

$$\begin{array}{rlrl} Q E \times d & =\mathrm{KE}_2-\mathrm{KE}_1 \\ \Rightarrow \quad Q \times 300 \times 0.5 & =0.12 \\ \text { or } & Q =\frac{0.12}{300 \times 0.5}=800 \mu \mathrm{C} \end{array}$$

The charge $Q$ is moving in the direction of electric field. Hence, it is positive.

When a magnetic field is applied to a charged particle, such as an electron, the force on the particle depends on its velocity relative to the field, and the direction of the force is given by the Lorentz force law. The magnetic component of the Lorentz force is described by the equation:

$\vec{F} = q\vec{v} \times \vec{B}$

where:

• $\vec{F}$ is the force on the particle,
• $q$ is the electric charge of the particle,
• $\vec{v}$ is the velocity of the particle, and
• $\vec{B}$ is the magnetic field.

The symbol $\times$ denotes the cross product, which means that the force is always perpendicular to both the velocity of the charged particle and the direction of the magnetic field.

In the scenario described where the electron is initially stationary, its velocity is zero. Therefore, applying the equation:

$\vec{F} = q\vec{v} \times \vec{B} = q\cdot\vec{0} \times \vec{B} =\vec{0}$

Since both $q$ and $\vec{B}$ are nonzero, but $\vec{v}$ is zero, the cross product yields a zero vector, which means there is no force on the stationary electron due to the magnetic field. Consequently, a stationary electron will not move just because a magnetic field is applied. It will remain stationary until an electric field or some other force acts upon it.

Therefore, the correct answer is:

Option C

remains stationary

In projectile motion, centre of acceleration is directed towards the centre of earth and at peak point, the angle between velocity and acceleration is $90^{\circ}$. because at peak point the vertical component of velocity is zero. Thus, it is minimum and acute at only one point (i.e. at highest point)

Path difference, $\Delta x=\frac{x d}{D}$

So, corresponding phase difference, $\phi=\frac{2 \pi}{\lambda}(\Delta x)$

$=\frac{2 \pi}{\lambda}\left(\frac{x d}{D}\right)$

The resultant intensity at $P$ is

$$\begin{aligned} I_P & =4 I_0 \cos ^2\left(\frac{\phi}{2}\right) \\ & =4 I_0 \cos ^2\left(\frac{2 \pi}{\lambda} \frac{x d}{D} \times \frac{1}{2}\right)=4 I_0 \cos ^2\left(\frac{\pi d x}{\lambda D}\right) \end{aligned}$$

Energy stored in a capacitor is given as

$U=\frac{1}{2} C V^2\quad \text{.... (i)}$

where, $C$ is the capacitance. From the given graph, we get

Slope $=\tan \theta=\frac{Q}{V}=\frac{120 \mu \mathrm{C}}{10 \mathrm{~V}}\quad \text{.... (ii)}$

As we know, capacitance is given as

$C=\frac{Q}{V}$

Substituting the value of $\frac{Q}{V}$ from Eq. (ii) in the above relation, we get

$$\begin{aligned} C & =\frac{120 \mu \mathrm{C}}{10 \mathrm{~V}}=\frac{120 \times 10^{-6}}{10}=12 \times 10^{-6} \mathrm{~F} \\ & =12 \mu \mathrm{F} \end{aligned}$$

Now, substituting the values of $C$ and $V$ in Eq: (i), we get

$$\begin{aligned} U & =\frac{1}{2} \times 12 \times(10)^2 \\ & =600 \mu \mathrm{J} \end{aligned}$$

Given, refractive index from air to glass,

${ }_a \mu_g=\frac{3}{2}$

Refractive index from air to water, ${ }_a \mu_w=\frac{4}{3}$

Using lens Maker's formula,

$$\begin{aligned} \frac{1}{f} & =\left({ }_1 \mu_2-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \therefore \quad \frac{f_{\text {water }}}{f_{\text {air }}} & =\frac{\left({ }_6 \mu_g-1\right)}{\left({ }_w \mu_g-1\right)}=\frac{\left(\mu_g-1\right)}{\left(\frac{{ }_a \mu_g}{{ }_a \mu_w}-1\right)} \\ & =\frac{\left(\frac{3}{2}-1\right)}{\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)}=\frac{1}{2} \times \frac{8}{1}=\frac{4}{1} \text { or } 4: 1 \end{aligned}$$

For proton, $v^2=\frac{c^2}{4}\quad \left(\because v=\frac{c}{2}\right)$

$$\begin{aligned} p_1 & =\sqrt{2 m E_k} \\ & =\sqrt{2 \times m \times \frac{1}{8} m c^2} \quad\left(\because E_k=\frac{1}{8} m c^2\right) \\ p_1 & =\frac{m c}{2} \end{aligned}$$

For photon,

$$\begin{aligned} & E=\mathrm{KE} \\ & \frac{h c}{\lambda}=\frac{1}{8} m c^2 \\ \Rightarrow \quad & p_2=\frac{h}{\lambda}=\frac{m c}{8} \\ \therefore \quad & \frac{p_1-p_2}{p_1}=\frac{\frac{m c}{2}-\frac{m c}{8}}{\frac{m c}{2}}=\frac{4-1}{8} \times 2=\frac{3}{4} \end{aligned}$$

As, we know

Area under $a$-$t$ graph $=v-u$

$$\begin{array}{lll} \Rightarrow & \frac{1}{2} \times 8 \times 10=v & (\because u=0) \\ \Rightarrow & v=40 \mathrm{~ms}^{-1} \end{array}$$

As we know, height of capillary rise or fall of a liquid in a capillary tube is given as

$$\begin{aligned} & h=\frac{2 T \cos \theta}{r \rho g} \\ & \Rightarrow \quad h \propto \frac{1}{r} \quad \text{..... (i)}\\ \end{aligned}$$

where, $r$ is the radius of the capillary tube. Given, height of water level in capillary tube $P=\frac{2}{3}$ height of water in capillary tube $Q$.

$$\begin{array}{ll} \Rightarrow & h_P=\frac{2}{3} h_Q \\ \text { or } & \frac{h_P}{h_Q}=\frac{2}{3} \end{array}$$

From Eq. (i), we can write

$$\begin{array}{rlrl} \frac{h_P}{h_Q} =\frac{r_Q}{r_P} \\ \Rightarrow \quad\frac{2}{3} =\frac{r_Q}{r_P} \\ \text { or } \quad r_P: r_Q =3: 2 \end{array}$$

As diameter of the tube is equals to twice the radius.

So, $\frac{(\text {diameter})_P}{(\text {diameter})_Q}=\frac{3}{2}$

The trajectory of a projectile is a parabola in the absence of air resistance. In the presence of air resistance, range and maximum height of projectile, both will decrease, hence path of projectile will not remain parabolic.

Electromagnetic waves are produced by accelerated charges. In given options, only a charge particle moving in a circular orbit is in accelerated motion. So, it produces electromagnetic wave.

Electric field intensity due to infinity straight uniformly charged wire at some distance $r$ is,

$E=\frac{\lambda}{2 \pi \varepsilon_0 r}$

where, $\lambda=$ the surface charge density.

If $\lambda, \pi$ and $\varepsilon_0$ are constants, then

$E \propto \frac{1}{r}$

Option (a) and (b) represents electric field lines due to positive and negative charges. However, magnetic monopoles does not exist in nature.

So, option (a) and (b) is not correct for magnetic field lines.

Option (c) represents magnetic field lines for current carrying conductor.

Also, electric field lines cannot form closed loop. So, option (c) is incorrect.

Option (d) is possible for magnetic field lines of solenoid and electric field lines of dipole.

Thus, option (d) is valid for both electric and magnetic field.

Total internal energy of a gas is given as

$U=\frac{n}{2} f R T$

where, $n=$ number of moles

and $f=$ degree of freedom.

Given, $n_{\text {diatomic }}=n_{\text {monoatomic }}=2$

As, $f_{\text {monoatomic }}=3$

$$\begin{aligned} f_{\text {diatomic }} & =5 \\ \Rightarrow U_{\text {monoatomic }} & =\frac{2}{2} \times 3 R T=3 R T \\ U_{\text {diatomic }} & =\frac{2}{2} \times 5 R T=5 R T \end{aligned}$$

$\therefore$ Total internal energy of the mixture of gases,

$$\begin{aligned} U_{\text {total }} & =U_{\text {monoatomic }}+U_{\text {diatomic }} \\ & =3 R T+5 R T=8 R T \end{aligned}$$

Magnetic force on the charged particle is given as

$F_B=q(v \times B)=q v B \sin \theta$

Here, $\theta=90^{\circ}$

$F_B=q v B \sin 90^{\circ}=q v B$

Since, the particle is projected perpendicular to he field, so its trajectory will be circular in nature, where necessary centripetal force $=F_B$

$$\begin{aligned} & \Rightarrow \quad \frac{m v^2}{r^{\prime}}=q v B \\ & \frac{m v^2}{\left(\frac{r}{2}\right)}=q v B \quad\left(\because r^{\prime}=\frac{r}{2}\right) \\ & \Rightarrow \quad v=\frac{q B r}{2 m} \quad \text{.... (i)} \end{aligned}$$

Magnetic field inside the solenoid is given as,

$B=\mu_0 n I$

Substituting the value of $B$ in Eq. (i), we get

$v=\frac{\mu_0 n I q r}{2 m}$

Given,

$$\begin{aligned} & m_1=1 \mathrm{~kg} \\ & u_1=12 \mathrm{~ms}^{-1} \\ & m_2=2 \mathrm{~kg} \\ & u_2=-24 \mathrm{~ms}^{-1} \end{aligned}$$

($-$ve sign implies that, it is travelling in the opposite direction)

$e=\frac{2}{3}$

As we know, coefficient of restitution, $e=\frac{v_2-v_1}{u_1-u_2}$.

$$\begin{aligned} & \frac{2}{3}=\frac{v_2-v_1}{12+24} \\ & \frac{2}{3}=\frac{v_2-v_1}{36} \end{aligned}$$

or $\quad v_2-v_1=24\quad \text{.... (i)}$

According to the law of conservation of momentum,

Initial momentum of the system $=$ Final momentum of the system

$\Rightarrow \quad m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2$

Substituting the given values in the above relation, we get

$$\begin{array}{ll} & 1 \times 12+2 \times(-24)=1 \times v_1+2 v_2 \\ \Rightarrow & 12-48=v_1+2 v_2 \\ \Rightarrow & v_1+2 v_2=-36 \quad \text{.... (ii)} \end{array}$$

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

$$\begin{aligned} & v_2=-4 \mathrm{~ms}^{-1} \\ & \Rightarrow \quad-4-v_1=24 \quad \because\text{[from Eq. (i)]}\\ & \Rightarrow \quad v_1=-24-4 \\ & =-28 \mathrm{~ms}^{-1} \\ \end{aligned}$$

Given, $\lambda=6500 \mathop A\limits^o=6500 \times 10^{-10} \mathrm{~m}, \theta=30^{\circ}$

According to the diffraction pattern of single slit,

$a \sin \theta=n \lambda$

where, $n=$ order of secondary minima $=1$

$$\begin{aligned} & \Rightarrow a \sin 30^{\circ}=1 \times\left(6500 \times 10^{-10}\right) \\ & \Rightarrow \quad \frac{a}{2}=6500 \times 10^{-10} \\ \end{aligned}$$

or

$$\begin{aligned} a & =13000 \times 10^{-10} \\ & =1.3 \times 10^{-6} \mathrm{~m}=1.3 \text { micron } \end{aligned}$$

Volume expansion is the expansion of volume of a gas due to increase in its temperature.

$\therefore$ Coefficient of volume expansion is given as

$\alpha_V=\frac{\Delta V}{V \times \Delta T}\quad \text{.... (i)}$

where, $V=$ real volume of gas,

$\Delta V=$ change in volume

and $\Delta T=$ change in temperature.

From ideal gas equation,

$p V=n R T \text { or } T=\frac{p V}{n R}\quad \text{.... (ii)}$

For constant pressure,

$$\begin{aligned} p \Delta V & =n R \Delta T \\ \text{or}\quad\frac{\Delta V}{\Delta T} & =\frac{n R}{p} \quad \text{.... (iii)} \end{aligned}$$

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

$\alpha_V=\frac{n R}{p V}=\frac{1}{T}\quad$ [$\because$ From Eq. (ii)]

Thus, the variation of $\alpha_V$ with temperature is correctly depicted in option (b).

Let $R$ and $r$ be the radii of bigger and each smaller drop, respectively.

Volume of bigger drop $=8 \times$ Volume of smaller drop

$$\begin{aligned} \Rightarrow & \quad \frac{4}{3} \pi R^3 =8 \times \frac{4}{3} \pi r^3 \\ \text { or } \quad & R =2 r \quad \text{.... (i)} \end{aligned}$$

As, capacitance of a smaller spherical drop is

$$\begin{aligned} & C=4 \pi \varepsilon_0 r \\ & \Rightarrow \quad C \propto r \\ \end{aligned}$$

So, $\quad \frac{C_{\text {bigger }}}{C_{\text {smaller }}}=\frac{r_{\text {bigger }}}{r_{\text {smaller }}}=\frac{R}{r}=\frac{2 r}{r}=2\quad \text{[using Eq. (i)]}$

So, capacitance of bigger drop is two times as compared to each smaller drop.

Given, resistance, $R_1=3 \Omega$

If $l$ is the original length of the wire, then after stretching the length of wire is

$l^{\prime}=2 l$

From the relation, resistance of a wire,

$R=\rho \frac{l}{A}$

where, $\rho$ is the resistivity and $A$ is cross-sectional area.

$$\begin{aligned} & R \propto \frac{l}{A} \\ & \Rightarrow \quad R \propto \frac{l^2}{V} \quad\left[\because A=\frac{\text { Volume }}{\text { Length }}\right] \\ \end{aligned}$$

$\Rightarrow \quad \frac{R_1}{R_2}=\frac{l_1^2}{V} \times \frac{V}{l_2^2}$

$$\begin{array}{ll} \text { Here, } l_1=l, l_2 =2 l \\ \Rightarrow \quad \frac{3}{R_2}=\frac{l^2}{4 l^2} \\ \text { or } \quad R_2=3 \times 4=12 \Omega \end{array}$$

The maximum kinetic energy of photoelectron ejected is given by

$\mathrm{KE}=h v-\phi_0=h v-h v_0=h\left(v-v_0\right)$

If the frequency of incident radiation is greater than threshold frequency $v_0$, then the ejection of photoelectrons starts. After that as frequency increases, kinetic energy also increases linearly. It is shown in option (d).

Given, $\mathbf{u}=10 \hat{\mathbf{j}} \mathrm{ms}^{-1}, \mathbf{a}=8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}} \mathrm{ms}^{-2}$

$x=16 \mathrm{~m}$

Using equation of motion,

$\begin{gathered} s=u t+\frac{1}{2} a t^2 \\ x \hat{\mathbf{i}}+y \hat{\mathbf{j}}=10 \hat{\mathbf{j}} \times t+\frac{1}{2}(8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) t^2 \end{gathered}$

$\Rightarrow \quad 16 \hat{\mathbf{i}}+y \hat{\mathbf{j}}=10 \hat{\mathbf{j}}+4 t^2 \hat{\mathbf{i}}+t^2 \hat{\mathbf{j}}$

Comparing both sides, we get

$$\begin{array}{rlrl} & 4 t^2 =16 \\ \Rightarrow & t =2 \mathrm{~s} \\ \text { and } & y =10 t+t^2 \\ & =10 \times 2+(2)^2 \\ & =24 \mathrm{~m} \end{array}$$

In a meter bridge, balancing length of the wire is independent of the area of cross-section of the wire.

So, if the radius of wire $A B$ is doubled, the balancing length will still be $X$ only.

This diagram can be shown as

$$\begin{aligned} Y & =\overline{\overline{A B}} \\ & =A B \end{aligned}$$

$\therefore$ Output represents logic operations of AND gate.

The LED is a heavily doped $p$-$n$ junction with forward biasing. It is a specially designed forward biased $p$-$n$ junction diode that emits light spontaneously when energised.

According to de-Broglie, the circumference of a stationary orbit must be an integral number of wavelengths.

$n \lambda=2 \pi r_n$

Also, angular momentum, $m v_n r_n=\frac{n h}{2 \pi}$

$$\begin{aligned} & \Rightarrow \quad m \frac{e^2}{2 n h \varepsilon_0} r_n=\frac{n h}{2 \pi} \quad\left[\because v_n=\frac{e^2}{2 n h \varepsilon_0}\right] \\ & \Rightarrow \quad r_n=\frac{n^2 h^2 \varepsilon_0}{m e^2 \pi}=\frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \end{aligned}$$

Here, number of standing waves, $n=6$

$\therefore \quad r_n=\frac{(6)^2 h^2 \varepsilon_0}{\pi m e^2}=\frac{36 h^2 \varepsilon_0}{\pi m e^2}$

Given, $E_1=2.5 \mathrm{~eV}, E_2=2 \mathrm{~eV}, E_3=3 \mathrm{~eV}$ and $\lambda=600 \mathrm{~nm}$

As we know,

$$\begin{aligned} E & =\frac{1242}{\lambda}=\frac{1242}{600} \\ & =2.07 \mathrm{~eV} \end{aligned}$$

Since, $E_1$ and $E_3$ is greater than $E$. So, only $D_2$ will able to detect the given light of wavelength 600 $\mathrm{nm}$.

The energy of the electron in the $n$th orbit is given by

$$\begin{aligned} & E_n=-13.6 \mathrm{~eV} \frac{Z^2}{n^2} \\ & E_n \propto \frac{Z^2}{n^2} \end{aligned}$$

For hydrogen atom, $Z=1$

$E_n \propto \frac{1}{n^2}$

In second orbit, $n=2$

$E_2 \propto \frac{1}{(2)^2}$

In case of $\mathrm{He}^{+}, Z=2$ and $n=3$

$$\begin{aligned} & E_3 \propto \frac{(2)^2}{(3)^2} \\ \therefore \quad & \frac{E_2}{E_3}=\frac{1}{4} \times \frac{9}{4}=\frac{9}{16} \\ \Rightarrow \quad & E_3=\frac{16 E_2}{9} \end{aligned}$$

Since, the area under $v$-$t$ graph gives the displacement of object.

i.e., $\quad s=$ Area of $v$-$t$ graph

As area under uniform motion > Area under uniform acceleration

$\therefore$ Displacement during uniform acceleration is less than that during uniform motion.

As we know,

Induced emf, $|e|=L \frac{d I}{d t}$

Here, $L=6 \mathrm{~mH}=6 \times 10^{-3} \mathrm{H}$

From the given graph,

When the coin placed on a rotating table is first about to slips, then till the time the coin rotates along with the table, necessary centripetal force for rotation will be provided by the frictional force,

i.e., $\quad m r \omega^2=\mu m g$

where,

$m=\text { mass of the coin, }$

$r=$ distance of the coin from the centre of the table,

$\omega=$ angular velocity,

$\mu=$ coefficient of friction

and $g=$ acceleration due to gravity.

Since, $m, \mu$ and $g$ remains constant, so $r \omega^2=$ constant

$$\begin{array}{ll} \Rightarrow & r \propto \frac{1}{\omega^2} \\ \text { or } & \frac{r_2}{r_1}=\left(\frac{\omega_1}{\omega_2}\right)^2 \end{array}$$

Given, $\omega_2=2 \omega_1, r_1=4 \mathrm{~cm}$

$$\begin{array}{ll} \Rightarrow & \frac{r_2}{r_1}=\left(\frac{\omega_1}{2 \omega_1}\right)^2 \\ \text { or } & r_2=\frac{r_1}{4}=\frac{4}{4}=1 \mathrm{~cm} \end{array}$$

Polar molecules are the molecules in which the centres of positive and negative charges are separated even there is no external field. Such molecules have a permanent dipole moment. e.g., $\mathrm{HCl}, \mathrm{H}_2 \mathrm{O}$, etc.

Thus, statement given in option (c) is false.

Number of capacitors that can be connected in each row,

$$\begin{aligned} m & =\frac{\text { Desired voltage }}{\text { Voltage across each capacitor }} \\ & =\frac{1.5 \mathrm{kV}}{500 \mathrm{~V}}=\frac{1500}{500}=3 \end{aligned}$$

Effective capacitance when $m$ capacitors are connected in $n$ rows is given as

$C_{\text {eff }}=\frac{n C}{m}$

$$\begin{aligned} & \text { Here, } m=3, C=2 \mu \mathrm{F}, C_{\text {eff }}=6 \mu \mathrm{F} \\ & \Rightarrow \quad n \times \frac{2}{3}=6 \\ & \text { or } \quad n=\frac{18}{2}=9 \end{aligned}$$