KCET 2021 Physics Questions with Answers Key Solutions

Q1. The physical quantity which is measure in the unit of $w b A^{-1}$ is

A

Self inductance

B

Mutual inductance

C

Magnetic flux

D

Both $A$ and $B$

Solution:

$\phi= LI$ or $\phi= MI$
$L =\frac{\phi}{ I } $ or $ M =\frac{\phi}{ I }$

Q2. What will be the reading in the voltmeter and ammeter of the circuit shown ?

A

90 V, 2 A

B

0 V, 2 A

C

90 V, 1 A

D

0 V, 1 A

Solution:

$X_{L}=X_{C} $
Hence, the resultant voltage across $L$ and $C$ is $0$.
$Z = R$
$I =\frac{ V }{ Z }=\frac{ V }{ R }$
$=\frac{90}{45}=2 A$

Q3. LC osillations are similar and analogous to the mechanical oscillations of a block attached to a spring. The electrical equivalent of the force constant of the spring is

A

reciprocal of capacitive reactance

B

capacitive reactance

C

reciprocal of capacitance

D

capacitance

Solution:

$k \to \frac{1}{C}$

Q4. In an oscillating LC circuit, $L =3.00\, mH$ and $C =2.70 \,\mu F .$ At $t =0$ the charge on the capacitor is zero and the current is $2.00\, A$. The maximum charge that will appear on the capacitor will be

A

$1.8 \times 10^{-5} C$

B

$18 \times 10^{-5} C$

C

$9 \times 10^{-5} C$

D

$90 \times 10^{-5} C$

Solution:

$\frac{ Q ^{2}}{2 C }=\frac{ LI ^{2}}{2}$
$\frac{ Q ^{2}}{2 \times 2.7 \times 10^{-6}} =\frac{3 \times 10^{-3} \times 4^{2}}{2}$
$Q ^{2}=12 \times 10^{-3} \times 2.7 \times 10^{-6}$
$=32.4 \times 10^{-9}$
$Q =18 \times 10^{-5} C$

Q5. Suppose that the electric field amplitude of electromagnetic wave is $E _{0}=120\, NC ^{-1}$ and its frequency if $f =50 \,MHz$. Then which of the following value incorrectly computed ?

A

Magnetic field amplitude is $400 \,nT$.

B

Angular frequency of EM wave is $\pi \times 10^{8} rad / s$

C

Propagation constant (angular wave number) is $2.1\, rad / m$

D

Wavelength of EM wave is $6\, m$.

Solution:

$c =\frac{ E _{0}}{ B _{0}}$
$B _{0}=\frac{ E _{0}}{ c }$
$=\frac{120}{3 \times 10^{8}}=400\, nT$
$k =\frac{2 \pi}{\lambda}=\frac{2 \times 3.14}{6}$
$k \neq 2.1 \,rad \,m ^{-1}$

Q6. The source of electromagnetic wave can be a charge.

A

Moving with a constant velocity

B

Moving in a circular orbit

C

At rest

D

Moving parallel to the magnetic field

Solution:

Charges moving in a circular orbit undergo centripetal acceleration hence emit em waves. Or when electron jumps from one higher stationary orbit to another lower stationary orbit emits em radiation.

Q7. In refraction, light waves are bent on passing from one medium to second medium because, in the second medium.

A

frequency is different

B

speed is different

C

coefficient of elasticity is different

D

amplitude is smaller

Solution:

Refraction takes because of change in speed of light when it moves from one medium to another

Q8. If the refractive index from air to glass is $\frac{3}{2}$ and that from air to water is $\frac{4}{3}$, then the ratio of focal lengths of a glass lens in water and in air is

A

$1: 2$

B

$2: 1$

C

$1: 4$

D

$4: 1$

Solution:

$n _{ w }=\frac{4}{3} \,\,\, n _{ g }=\frac{3}{2}$
$f _{ a }\left( n _{ g }-1\right)= f _{ w }\left(\frac{ n _{ g }}{ n _{ w }}-1\right)$
$ \frac{ f _{ w }}{ f _{ a }}=\frac{ n _{ g }-1}{\frac{ n _{ g }}{ n _{ w }}-1}=\frac{\frac{3}{2}-1}{\frac{\frac{3}{2}}{\frac{4}{3}}-1} $
$=\frac{\frac{3-2}{2}}{\frac{9}{8}-1}=\frac{\frac{1}{2}}{\frac{9-8}{8}}=\frac{\frac{1}{2}}{\frac{1}{8}}=\frac{8}{2}$
$ \therefore \frac{ f _{ w }}{ f _{ a }}=\frac{4}{1} $

Q9. Two thin biconvex lenses have focal lengths $f _{1}$ and $f _{2} .$ A third thin biconcave lens has focal length of $f_{3} .$ If the two biconvex lenses are in contact, the total power of the lenses is $P _{1}$. If the first convex lens is in contact with the third lens, the total power is $P _{2}$. If the second lens is in contact with the third lens, the total power is $P _{3}$ then

A

$P _{1}=\frac{ f _{1} f _{2}}{ f _{1}- f _{2}}, P _{2}=\frac{ f _{1} f _{3}}{ f _{3}- f _{1}}$ and $P _{3}=\frac{ f _{2} f _{3}}{ f _{3}- f _{2}}$

B

$P _{1}=\frac{ f _{1}- f _{2}}{ f _{1} f _{2}}, P _{2}=\frac{ f _{3}- f _{1}}{ f _{3}+ f _{1}}$ and $P _{3}=\frac{ f _{3}- f _{2}}{ f _{2} f _{3}}$

C

$P _{1}=\frac{ f _{1}- f _{2}}{ f _{1} f _{2}}, P _{2}=\frac{ f _{3}- f _{1}}{ f _{1} f _{3}}$ and $P _{3}=\frac{ f _{3}- f _{2}}{ f _{2} f _{3}}$

D

$P_{1}=\frac{f_{1}+f_{2}}{f_{1} f_{2}}, P_{2}=\frac{f_{3}-f_{1}}{f_{1} f_{3}}$ and $P_{3}=\frac{f_{3}-f_{2}}{f_{2} f_{3}}$

Solution:

$f _{1}=+ f _{1}$
$f _{2}=+ f _{2}$
$f _{3}=- f _{3}$
$\frac{1}{F_{R}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$P_{R}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$P_{1}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{f_{2}+f_{1}}{f_{1} f_{2}}$
$P_{2}=\frac{1}{f_{1}}+\frac{1}{f_{3}}=\frac{f_{3}-f_{1}}{f_{1} f_{3}}$
$P_{2}=\frac{1}{f_{2}}+\frac{1}{f_{3}}=\frac{f_{3}-f_{2}}{f_{2} f_{3}}$

Q10. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length $30\, cm$ is $2 \, cm .$ If a concave lens of focal length $20 \, cm$ is placed between the convex lens and the image at a distance of $26\, cm$ from the lens, the new size of the image is :

A

1.25 cm

B

2.5 cm

C

1.05 cm

D

2 cm

Solution:

$A 'B'$ is the real image due to convex lens and it is at Focus of convex lens.
$A'B'$ acts as virtual object for concave lens and object distance is $+4 \,cm$
$\frac{1}{f} =\frac{1}{v}-\frac{1}{u} v=\frac{20}{4} $
$\frac{1}{v} =\frac{1}{f}+\frac{1}{u} v=5 \,cm$
$\frac{1}{v} =\frac{1}{-20}+\frac{1}{4} $
$\frac{1}{v} =\frac{-1+5}{20}=\frac{4}{20} $
$m =\frac{h_{i}}{h_{o}}=\frac{v}{u} $
$h_{i} =\frac{v h_{o}}{u} $
$=\frac{5 \times 2}{4}=\frac{10}{4}=2.5\, cm$

Q11. A slit of width 'a' is illuminated by red light of wavelength $6500 \mathring{A}$. If the first diffraction minimum falls at $30 \mathring{A}$, then the value of ' $a$ ' is

A

$6.5 \times 10^{-4} mm$

B

$1.3$ micron

C

$3250 \mathring{A}$

D

$2.6 \times 10^{-4} cm$

Solution:

For diffraction minima
$a \sin \theta= n \lambda$
$n =1 $ for $ I$ minimum
$a =\frac{\lambda}{\sin \theta}=\frac{6.5 \times 10^{-7}}{\sin 30} $
$=\frac{6.5 \times 10^{-7}}{0.5} $
$=13 \times 10^{-7} $
$=1.3 \times 10^{-6} m$
$=1.3$ micron

Q12. Which of the statements are correct with reference to single slit diffraction pattern?
(i) Fringes are of unequal width
(ii) Fringes are of equal width
(iii) Light energy is conserved
(iv) Intensities of all bright fringes are equal

A

(i) and (iii)

B

(i) and (iv)

C

(ii) and (iv)

D

(ii) and (iii)

Solution:

Option (i) and (iii) are correct.

Q13. In the Young's double slit experiment a monochromatic source of wavelength $\lambda$ is used. The intensity of light passing through each slit is $I _{0}$. The intensity of light reaching the screen $S _{ C }$ at a point $P$, a distance $X$ from $O$ is given by (Take $d < < D$ )

A

$I _{ o } \cos ^{2}\left(\frac{\pi D }{\lambda d } x \right)$

B

$4 I _{ o } \cos ^{2}\left(\frac{\pi d }{\lambda D } x \right)$

C

$I_{o} \sin ^{2}\left(\frac{\pi d}{2 \lambda D} x\right)$

D

$4 I _{ o } \cos \left(\frac{\pi d }{2 \lambda D } x \right)$

Solution:

$\phi=\left(\frac{2 \pi}{\lambda}\right)$ path difference
Path difference $= d \sin \theta$
$= d \left[\frac{ x }{ D }\right]$
$\phi=\frac{2 \pi}{\lambda}\left[\frac{ dx }{ D }\right]$
$I=4 I_{o} \cos ^{2} \frac{\phi}{2}$
$=4 I _{ o } \cos ^{2} \frac{\pi d }{\lambda D } \cdot x$

Q14. The work function of a metal is $1\, eV$. Light of wavelength $3000 \,\mathring{A}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

A

$10 \,ms ^{-1}$

B

$1 \times 10^{3} ms ^{-1}$

C

$1 \times 10^{4} ms ^{-1}$

D

$1 \times 10^{6} ms ^{-1}$

Solution:

$E _{ k \max } =\left( E _{ photon }- W \right) eV$
$=\left(\frac{12.42 \times 10^{-7}}{3 \times 10^{-7}}-1\right) eV$
$=(4-1) eV $
$E _{ k \max } =3 \,eV =3 \times 1.6 \times 10^{-19} J $
$ \frac{1}{2} mv _{\max }^{2} =3 \times 1.6 \times 10^{-19} $
$ v _{\max }^{2} =\frac{3 \times 2 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$
$v _{\max }^{2} =10^{12} $
$ v _{\max } =10^{6} m s ^{-1} $

Q15. A proton moving with a momentum $P _{1}$ has a kinetic energy $\frac{1}{8}$ th of its rest mass energy. Another light photon having energy equal to the kinetic energy of the possesses a momentum $P _{2} .$ Then the ratio $\frac{ P _{1}- P _{2}}{ P _{1}}$ is equal to

A

1

B

$\frac{1}{4}$

C

$\frac{1}{2}$

D

$\frac{3}{4}$

Solution:

For proton $v ^{2}=\frac{ c ^{2}}{4} v =\frac{ c }{2}$
$P =\sqrt{2 mE _{ k }} $
$=\sqrt{\not 2 m \frac{1}{\not 8_{4}} mc ^{2}} $
$P_{1} =\frac{ mc }{2}$
For photon
$E _{\text {photon }}= KE _{\text {proton }}$
$\frac{ h \not c}{\lambda}=\frac{1}{8} mc ^{\not 2}$
$P _{\text {photon }}=\frac{ h }{\lambda}=\frac{ mc }{8}= P _{2}$
$\frac{P_{1}-P_{2}}{P_{1}}=\frac{\frac{\not m c}{2}-\frac{\not m c}{8}}{\frac{\not m c}{2}}$
$=\frac{\frac{1}{2}-\frac{1}{8}}{\frac{1}{2}}=\frac{\frac{4-1}{8}}{\frac{1}{2}}$
$=\frac{3 \times 2}{8}=\frac{3}{4}$

Q16. According to Einstein’s photoelectric equation to the graph between kinetic energy of photoelectrons ejected and the frequency of incident radiation is

A

B

C

D

Solution:

$ K _{\max }= h v-\phi_{o} $
$ y = m x - C$
Graph is a straight line with slope ' $h$ '
Till $v=v_{ o }$, no photoelectric emission occurs

Q17. Energy of an electron in the second orbit of hydrogen atom is $E _{2} .$ The energy of electron in the third orbit of $He ^{+}$will be

A

$\frac{9}{16} E _{2}$

B

$\frac{16}{9} E _{2}$

C

$\frac{3}{16} E _{2}$

D

$\frac{16}{3} E _{2}$

Solution:

For $2^{\text {nd }}$ orbit of H-atom
$E _{2}=\frac{-13 \cdot 6}{2^{2}} eV$
For $3^{\text {rd }}$ orbit of $He ^{+}$
$E _{3}^{ He ^{+}} =\frac{-13 \cdot 6}{3^{2}} 2^{2}$
$=\frac{-13 \cdot 6}{4} \times \frac{4}{9} \times 2^{2} $
$= E _{2} \times \frac{16}{9}=\frac{16}{9} E _{2}$

Q18. The figure shows standing de Broglie waves due to the revolution of electron in a certain orbit of hydrogen atom. Then the expression for the orbit radius is (all notations have their usual meanings)

A

$\frac{ h ^{2} \in_{0}}{\pi me ^{2}}$

B

$\frac{4 h^{2} \epsilon_{0}}{\pi m e^{2}}$

C

$\frac{9 h^{2} \in_{0}}{\pi m e^{2}}$

D

none

Solution:

Number of stationary waves is $6$ i.e., $n =6$
$r=\frac{n^{2} \varepsilon_{0}^{2} h^{2}}{\pi m e^{2}}$
$=\frac{36 \varepsilon_{0}^{2} h^{2}}{\pi m e^{2}}$

Q19. An electron in an excited state of $Li ^{2+}$ ion has angular momentum $\frac{3 h }{2 \pi}$. The de Broglie wavelength of electron in this state is $P \pi a _{0}$ (where $a_{0}=$ Bohr radius ). The value of $P$ is

A

3

B

2

C

1

D

4

Solution:

$L =\frac{ nh }{2 \pi}=\frac{3 h }{2 \pi} $
$n =3 $
$\lambda =\frac{ h }{ mv }=\frac{ h \cdot r }{ mvr }=\frac{ hr }{\frac{3 h }{2 \pi}}$
$=\frac{2}{3} \pi r$
For $Li ^{2+}$ atom radius of orbit.
$r=r_{0} \frac{n^{2}}{z} $
$r=a_{0} \times \frac{3^{2}}{3} $
$(a_0 = $ Bohr radius)
$\therefore \lambda =\frac{2}{3} . \pi \times a_{0} \times \frac{3^{2}}{3} $
$=2 \pi a_{0}$
Comparing
$\therefore P =2$

Q20. Which graph in the following diagram correctly represents the potential energy of a pair of nucleons as a function of their separation?

A

B

C

D

Q21. In a nuclear reactor heavy nuclei is not used as moderators because

A

They will break up

B

Elastic collision of neutrons with heavy nuclei will not slow them down

C

The net weight of the reactor would be unbearably high

D

Substance with heavy nuclei do not occur in liquid or gaseous state at room temperature

Q22. The circuit given represents which of the logic operations?

A

OR

B

AND

C

NOT

D

NOR

Solution:

$ Y =\overline{\overline{ AB } \cdot \overline{ AB }}$
$=\overline{ AB }+\overline{ AB } $
$= AB $

Q23. Identify the incorrect statement:

A

when a P-N junction diode is forward biased, the width of the depletion region decreases.

B

when a P-N junction diode is reverse biased, the barrier potential increases

C

a photon diode is operated in the reverse bias

D

an LED is a lightly doped P-N junction diode which emits spontaneous radiation of forward biasing.

Solution:

LED is heavily doped PN-junction

Q24. Three phtodiodes $D _{1}, D _{2}$ and $D _{3}$ are made of semiconductors having band gaps of $2.5 \,eV$, $2\, eV$ and $3\, eV$ respectively. Which one will be able to detect light of wavelength $600 \,nm?$

A

$D _{1}$ only

B

both $D _{1}$ and $D _{3}$

C

$D _{2}$ only

D

All the three diodes

Solution:

$ E _{1}=2.5\, eV$
$ E _{2}=2 \,eV $
$ E _{3}=3\, eV$
$\lambda=600\, nm $
$ \therefore E =\frac{1242}{\lambda_{ nm }}=\frac{1242}{600} eV $
$=2.07\, eV $
$( E _{1} $ and $E _{3}$ is $ > E $
Only $D _{2}$ will detect

Q25. For a body moving along a straight line, the following $v-t$ graph is obtained, According to the graph, the displacement during

A

Uniform acceleration is greater than that during uniform motion

B

Uniform acceleration is less than that during uniform motion

C

Uniform acceleration is equal to that during uniform motion

D

Uniform motion zero

Solution:

$S=$ Area under $v-t$ graph
Area under uniform motion > Area under triangle

Q26. A particle starts from rest. Its acceleration ‘$a$’ versus time ‘$t$’ is shown in the figure. The maximum speed of the particle will be:

A

$80\, ms ^{-1}$

B

$40\, ms ^{-1}$

C

$18 \,ms ^{-1}$

D

$2\, ms ^{-1}$

Solution:

Area under a $-t$ graph $=v-u$
$\frac{1}{2} \times 8 \times 10= v $
$v =40 \,m s ^{-1}$

Q27. The maximum range of a gun on horizontal plane is $16 \,km$. If $g =10 \,ms ^{-2}$, then muzzle velocity of a shell is

A

$160 \,ms ^{-1}$

B

$200 \sqrt{2} \,ms ^{-1}$

C

$400\, ms ^{-1}$

D

$800\, ms ^{-1}$

Solution:

$\frac{ u ^{2}}{ g }=16 \times 10^{3} m$
$u ^{2}=16 \times 10^{3} \times 10$
$u ^{2}=16 \times 10^{4}$
$u =400 \,m s ^{-1}$

Q28. The trajectory of projectile is

A

Semicircle

B

An ellipse

C

A parabola always

D

A parabola in the absence of air resistance

Q29. For a projectile motion, the angle between the velocity and acceleration is minimum and acute at

A

Only one point

B

Two points

C

Three points

D

Four points

Q30. A particle starts from the origin at $t =0 \,s$ with a velocity of $10\, \hat{ j }\, ms ^{-1}$ and move in the $x - y$ plane with a constant acceleration of $(8 \hat{ j }+2 \hat{ j }) ms ^{-2}$. At an instant when the $x$-coordinate of the particle is $16\, m , y$-coordinate of the particle is:

A

16 m

B

28 m

C

36 m

D

24 m

Solution:

$\vec{ r }=\vec{ u } t +\frac{1}{2} \vec{ a } t ^{2}$
$x \hat{ i }+ y \hat{ j }=10 t \hat{ j }+\frac{1}{2}(8 \hat{ i }+2 \hat{ j }) t ^{2}$
$x \hat{ i }+ y \hat{ j }=10 t \hat{ j }+(4 \hat{ i }+\hat{ j }) t ^{2}$
$16 \hat{ i }+ yj =10 t \hat{ j }+4 t ^{2} \hat{ i }+ t ^{2} \hat{ j }$
$16 \hat{ i }+\hat{ j }=4 t ^{2} \hat{ i }+\left(10 t + t ^{2}\right) \hat{ j }$
$4 t ^{2}=16$
$t =4 s$
$y =10 t + t ^{2}$
$=20+4=24 \,m .$

Q31. A coin placed on a rotating turn table just slips if it is placed at a distance of $4\, cm$ from the centre. If the angular velocity of the turn table is doubled it will just slip at a distance of

A

1 cm

B

2 cm

C

4 cm

D

8 cm

Solution:

$r =4 \,cm$
$f = mr \omega^{2}$
$f =\mu mg$
$\mu mg = m \omega^{2} r$
$\mu g =\omega^{2} r$
$\mu g =\omega^{2} \times 4$ ... (1)
$\omega^{1}=2 \omega$
$\mu g =(2 \omega)^{2} r$
$\mu g =4 \omega^{2} r$ ... (2)
$\omega^{2} \times 4=4 \omega^{2} r$
$r =1 \,cm$

Q32. A $1\, kg$ ball moving at $12\, ms ^{-1}$ collides with a $2\, kg$ ball moving in opposite direction at $24 \,ms ^{-1} .$ If the coefficient of restitution is $\frac{2}{3}$, the their velocities after the collision are

A

$-4 \,ms ^{-1},-28 \,ms ^{-1}$

B

$-28 \,ms ^{-1},-4 \,ms ^{-1}$

C

$4 \,ms ^{-1}, 28 \,ms ^{-1}$

D

$28\, ms ^{-1}, 4\, ms ^{-1}$

Solution:

$m _{1}=1 \,kg , \,\,u _{1}=12\, m s ^{-1}$
$m _{2}=3\, kg , \,\,u _{2}=-24\, m s ^{-1}$
$e =\frac{ v _{2}- v _{1}}{ u _{1}- u _{2}}$
$\frac{2}{3}=\frac{ v _{2}- v _{1}}{36}$
$v _{2}- v _{1}=\frac{36 \times 2}{3}$
$v _{2}- v _{1}=24 $ ... (1)
$m _{1} u _{1}+ m _{2} u _{2}= m _{1} v _{1}+ m _{2} v _{2}$
$1 \times 12+2 \times(-24)= v _{1}+2 v _{2} $
$12-48= v _{1}+2 v _{2}$
$v _{1}+2 v _{2}=-36 $
$\frac{- v _{1}+ v _{2}=24}{3 v _{2}=-12} $
$v _{2}=-4 \,m s ^{-1}$ [from equation (1)]
$-4- v _{1}=24$
$-4-24= v _{1} $
$v _{1}=-28 \,m s ^{-1}$

Q33. A ball hits the floor and rebounds after an inelastic collision. In this case

A

The momentum of the ball is conserved

B

The mechanical energy of the ball is conserved

C

The total momentum of the ball and the earth is conserved

D

The total mechanical energy of the ball and the earth is conserved

Q34. In figure $E$ and $V _{ cm }$ represent the total energy and speed of centre of mass of an object of mass $1 \,kg$ in pure rolling. The object is:

A

Sphere

B

Ring

C

Disc

D

Hollow Cylinder

Solution:

$ E =\frac{1}{2} mv _{ cm }^{2}\left[1+\frac{ K ^{2}}{ R ^{2}}\right] $
$\frac{ E }{ v _{ cm }^{2}}=\frac{1}{2}\left[1+\frac{ K ^{2}}{ R ^{2}}\right]$
From graph, $\frac{ E }{ v _{ cm }^{2}}=\frac{3}{4}$
$\frac{3}{4}=\frac{1}{2}\left[1+\frac{ K ^{2}}{ R ^{2}}\right] $
$\frac{3}{2}-1=\frac{ K ^{2}}{ R ^{2}} $
$\frac{ K ^{2}}{ R ^{2}}=\frac{1}{2}$
For disc, $I =\frac{ mR ^{2}}{2}$
$\frac{m R^{2}}{2}=m K^{2} $
$\frac{K^{2}}{R^{2}}=\frac{1}{2}$

Q35. Two bodies of masses $8\, kg$ are placed at the vertices $A$ and $B$ of an equilateral triangle $A B C$. A third body of mas $2\, kg$ is placed at the centroid $G$ of the triangle. If $AG = BG = CG =1 \,m$, where should a fourth body of mass $4 \,kg$ be placed so that the resultant force on the $2\, kg$ body is zero?

A

At C

B

At a point $P$ on the line CG such that $PG =\frac{1}{\sqrt{2}} m$

C

At a point P on the line CG such that PG = 0.5 m

D

At a point P on the line CG such that P G = 2 m

Solution:

$F _{ A } = F _{ B }=\frac{ G m _{1} m _{2}}{ r ^{2}}=\frac{ G 8 \times 2}{1^{2}}= G (16) $
$F _{ AB } =\sqrt{ F _{ A }^{2}+ F _{ B }^{2}+2 F _{ A } F _{ B } \cos \theta} $
$=\sqrt{ F _{ A }^{2}+ F _{ B }^{2}+2 F _{ A } F _{ B } \cos 120} $
$=\sqrt{ F _{ A }^{2}+ F _{ B }^{2}+2 F _{ A } F _{ B }\left(-\frac{1}{2}\right)} $
$F _{ AB } = F _{ A }= G (16)$
For resultant force on $2\, kg$ to be zero
$\overrightarrow{ F }_{ CG }=-\overrightarrow{ F }_{ AB }$
$\Rightarrow \frac{ G 2 \times 4}{ X ^{2}}= G (16)$
$X ^{2}=\frac{2 \times 4}{16}=\frac{1}{2}$
$X =\frac{1}{\sqrt{2}}$

Q36. Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2^{ rd }}{3}$ of the height in capillary tube $Q$. The ratio of their diameter is

A

$2: 3$

B

$3: 2$

C

$3: 4$

D

$4: 3$

Solution:

$r=\frac{2 T \cos \theta}{\rho g h}$
As we know $r \propto \frac{1}{h}$
$\frac{ r _{ P }}{ r _{ Q }}=\frac{ h _{ Q }}{ h _{ P }}=\frac{ h }{\frac{2}{3} h } $
$\Rightarrow \frac{ r _{ p }}{ r _{ Q }}=\frac{3}{2}$

Q37. Which of the following curves represent the variation of coefficient of volume expansion of an ideal gas at constant pressure?

A

B

C

D

Solution:

We know PV $= nRT$ or $\frac{1}{ T }=\frac{ nR }{ PV }$
Since, $P$ is constant, $P \Delta V = nR \Delta T$
(or) $\frac{\Delta V }{\Delta T }=\frac{ nR }{ P }$
Coefficient of volume expansion $\alpha_{v}=\frac{\Delta V}{V \Delta T}$
$\alpha_{v}=\frac{ nR }{ PV }=\frac{1}{ T }$

Q38. A number of cannot engines are operated at identical cold reservoir temperatures $\left( T _{ L }\right)$. However, their hot reservoir temperatures are kept different. A graph of the efficiency of the engines versus hot reservoir temperature $\left( T _{ H }\right)$ is plotted. The correct graphical representation is

A

B

C

D

Solution:

$n =1-\frac{ T _{ L }}{ T _{ H }}$ At constant $T _{ L }$, efficiency increases with the increase in $T _{ H }$.

Q39. A gas mixture contains monoatomic and diatomic molecules of 2 moles each. The mixture has a total internal energy of (symbols have usual meanings)

A

$3 RT$

B

$5 RT$

C

$8 RT$

D

$9 RT$

Solution:

$C _{ v }$ for monoatomic gas is $\frac{3 R }{2}$
$C _{ v }$ for diatomic gas is $\frac{5 R }{2}$
Total internal energy for mixture
$U =2 \times\left(\frac{3}{2}\right) RT +2\left(\frac{5}{2}\right) RT$
$U =8 \,RT$

Q40. A pendulum oscillates simple harmonically and only if
i. The sizer of the bob of pendulum is negligible in comparison with the length of the pendulum
ii. The angular amplitude is less than $10^\circ$

A

Both i and ii are correct

B

Both i and ii are incorrect

C

Only i is correct

D

Only ii is correct

Q41. To propagate both longitudinal and transverse waves, a material must have

A

Bulk and shear moduli

B

Only bulk modulus

C

Only shear modulus

D

Young’s and Bulk modulus

Q42. A copper rod $AB$ of length $l$ is rotated about end A with a constant angular velocity $\omega$. The electric field at a distance $x$ from the axis of rotation is

A

$\frac{\operatorname{m} \omega^{2} x }{ e }$

B

$\frac{m \omega x}{e l}$

C

$\frac{m x}{\omega^{2} l}$

D

$\frac{m e}{\omega^{2} x}$

Solution:

We know in circular motion net force acting on it must be $\frac{ mv ^{2}}{ r }$
When the rod rotates, electrons in it also rotates which produce electric field $E$ at distance $x$.
Force on the electron, $Fe = eE = m \omega^{2} x$
$\therefore Ee =\frac{ mv ^{2}}{ x } \quad( v = x \omega) $
$ Ee =\frac{ m ( x \omega)^{2}}{ x }$
$ \Rightarrow E =\frac{ m \omega^{2} x }{ e }$

Q43. Electric field due to infinite, straight uniformly charged wire varies with distance '$r$' as

A

$r$

B

$\frac{1}{r}$

C

$\frac{1}{r^{2}}$

D

$r ^{2}$

Solution:

$\vec{ E }=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \lambda}{ r }$
(or) $\vec{ E }=\frac{\lambda}{2 \pi \varepsilon_{0} r } \hat{ r } $
$\Rightarrow |\vec{ E }| \propto \frac{1}{ r }$

Q44. A $2$ - gram object, located in a region of uniform electric field $\overrightarrow{ E }=\left(300\, N C ^{-1}\right) \hat{ i }$ carries a charge $Q .$ The object released from rest at $x=0$, has a kinetic energy of $0.12\, J$ at $x =0.5$ $m$. Then $Q$ is

A

$400\, \mu c$

B

$-400\, \mu c$

C

$800\, \mu c$

D

$-800\, \mu c$

Solution:

Work done $( W )=$ Change in $K E$
Force $\times$ displacement $=$ change in $KE$
$QE \times x =0.12\, J$
$ Q (300) \times(0.5)=0.12 $
$ Q =\frac{0.12}{300 \times 0.5}=800\, \mu C$
The charge $Q$ is moving in the direction of electric field, hence it is positive.

Q45. If a slab of insulating material (conceptual). $4 \times 10^{-3} m$ thick is introduced between the plates of a parallel plate capacitor, the separation between the plates has to be increased by $3.5 \times 10^{-3} m$ to restore the capacity to original value. The dielectric constant of the material will be

A

6

B

8

C

10

D

12

Solution:

If $k$ be the dielectric constant, the distance increased due to introduced dielectric is $x = t -\frac{ t }{ k } \Rightarrow x = t \left(1-\frac{1}{ k }\right)$
Where, $t \rightarrow$ thickness of the dielectric.
According to the question,
$ x =3.5 \times 10^{-3} m $
$ t \left(1-\frac{1}{ k }\right)=3.5 \times 10^{-3} $
$1-\frac{1}{ k }=\frac{3.5 \times 10}{4 \times 10}=\frac{3.5}{4} $
$\frac{1}{ k }=1-\frac{35}{4}=\frac{0.5}{4}=0.125$
$k =8$

Q46. Eight drops of mercury of equal radii combine to form a big drop. The capacitance of a bigger drop as compared to each smaller drop is

A

2 times

B

8 times

C

4 times

D

16 times

Solution:

$C_{B}=n^{\frac{1}{3}} C_{S}$
$C_{B}=8^{\frac{1}{3}} C_{S}$
$C_{B}=2 C_{S}$

Q47. Which of the statements is false in the case of polar molecules?

A

Centers of positive and negative charges are separated in the absence of external electric field.

B

Centers of positive and negative charges are separated in the presence of external electric field.

C

Do not possess permanent dipole moments

D

Ionic molecule HCl is the example of polar molecule.

Solution:

Polar molecules have permanent dipole moment.

Q48. An electrician requires a capacitance of $6 \,\mu F$ in a circuit across a potential difference of $1.5\, kV$. A large number of $2\, \mu F$ capacitors which can withstand a potential difference of not more than $500 \,V$ are available. The minimum number of capacitors required for the purpose is

A

3

B

9

C

6

D

27

Solution:

Number of capacitors to be connected in each row,
$m =\frac{1500}{500}=3$
Effective capacitance when connected in $n$ rows with $m$ capacitors in each row is,
$C_{\text {eff }}=n \frac{C}{m} $
$\therefore n \times \frac{2}{3}=6$
$n=\frac{18}{2}=9$
$ \Rightarrow n=9$
$\therefore $ Total number of capacitors required
$N = mn =3 \times 9=27$

Q49. In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively.

A

$12\, \mu F , 1200\, \mu J$

B

$12\, \mu F , 600 \,\mu J$

C

$24\, \mu F , 600 \,\mu J$

D

$24\, \mu F , 1200 \,\mu J$

Solution:

$q = CV $
$ C =\frac{ q }{ V }=\frac{120 \times 10^{-6}}{10}=12 \times 10^{-6} F $
$ C =12 \,\mu F $
$ U =\frac{1}{2} CV ^{2}$
$=\frac{1}{2} \times 12 \times 10^{-6} \times 10 \times 10 $
$ U =600 \times 10^{-6} J $
$ U =600 \,\mu J $

Q50. A wire of resistance $3 \,\Omega$ is stretched to twice its original length. The resistance of the new wire will be

A

$1.5\, \Omega$

B

$3 \,\Omega$

C

$6\, \Omega$

D

$12\, \Omega$

Solution:

$R _{2}= n ^{2} R _{1}$
$R _{2}=2^{2} \times 3$
$R _{2}=12 \,\Omega$

Q51. In the given arrangement of experiment on metre bridge, if AD corresponding to null deflection of the galvanometer is $X$, what would be its value if the radius of the wire $AB$ is doubled?

A

$X$

B

$\frac{ X }{4}$

C

$4 X$

D

$2 X$

Solution:

The balancing length is independent of radius of the bridge wire.

Q52. A copper wire of length $1\, m$ and uniform cross sectional area $5 \times 10^{-7} m ^{2}$ carries a current of $1\, A$. Assuming that there are $8 \times 10^{28}$ free electrons per $m ^{3}$ in copper, how long will an electron take to drift from one end of the wire to the other?

A

$0.8 \times 10^{3} S$

B

$1.6 \times 10^{3} S$

C

$3.2 \times 10^{3} S$

D

$6.4 \times 10^{3} S$

Solution:

$t =\frac{l}{ v _{ d }}$
$I =neAv_d$
${ v _{ d }=\frac{ I }{ neA }}$
$t =\frac{l neA }{ I }$
$t =\frac{1 \times 8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}}{1}$
$t =6.4 \times 10^{3} s$

Q53. Consider an electrical conductor connected across a potential difference $V$. Let $\Delta q$ be a small charge moving through it in time $\Delta t .$ If I is the electric current through it,
(i) the kinetic energy of the charge increases by $IV \Delta t$
(ii) the electric potential energy of the charge decreases by IV $\Delta t$.
(iii) the thermal energy of the conductor increases by IV $\Delta t$.
Then the correct statements is / are

A

(i)

B

(i), (ii)

C

(i) and (iii)

D

(ii), (iii)

Solution:

As the charges flow from higher to lower potential, its potential energy decreases and increase in kinetic energy is transferred to conductor which increases the thermal energy.

Q54. A strong magnetic field is applied on a stationary electron. Then the electron

A

Moves in the direction of the field

B

Moves in an opposite direction of the field

C

Remains stationary

D

Starts spinning

Solution:

A stationary charge in a magnetic field donot experience any force.

Q55. Two parallel wires in free space are $10 \,cm$ apart and each carries a current of $10 \,A$ in the same direction. The force exerted by one wire on the other [per unit length] is

A

$2 \times 10^{-4} Nm ^{-1}$ [attractive]

B

$2 \times 10^{-7} Nm ^{-1}$ [attractive]

C

$2 \times 10^{-4} Nm ^{-1}$ [repulsive]

D

$2 \times 10^{-7} Nm ^{-1}$ [repulsive]

Solution:

$\frac{ F }{l} =\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I _{1} I _{2}}{ d } $
$=\frac{10^{-7} \times 2 \times 10 \times 10}{10 \times 10^{-2}}$
$\frac{ F }{l}=2 \times 10^{-4} N m ^{-1}$
[attractive since the currents are in the same direction]

Q56. A toroid with thick windings of $N$ turns has inner and outer radii $R_{1}$ and $R_{2}$ respectively. If it carries certain steady current I, the variation of the magnetic field due to the toroid with radial distance is correctly graphed in

A

B

C

D

Solution:

$B =\mu_{0} nI$
where $n =\frac{ N }{2 \pi r }$
$B=\frac{\mu_{0} N I}{2 \pi\left(\frac{R_{1}+R_{2}}{2}\right)} $
$B=\frac{\mu_{0} N I}{\pi\left(R_{1}+R_{2}\right)}$

Q57. A tightly wound long solenoid has '$n$' turns per unit length, a radius ' $r$ ' and carries a current $I$. A particle having charge ' $q$ ' and mass '$m$' is projected from a point on the axis in a direction perpendicular to the axis. The maximum speed of the particle for which the particle does not strike the solenoid is

A

$\frac{\mu_{0} nIqr }{ m }$

B

$\frac{\mu_{0} nIqr }{2 m }$

C

$\frac{\mu_{0} nIqr }{4 m }$

D

$\frac{\mu_{0} nIqr }{8 m }$

Solution:

$F _{ B } = qvB \sin \theta \,\, \theta = 90^{\circ}$
$ F _{ B } = qvB $
$ F _{ C } =\frac{ mv ^{2}}{ r } $
$ F _{ B } = F _{ C } $
$qvB =\frac{ mv ^{2}}{\left(\frac{ r }{2}\right)}$
$ v =\frac{ qrB }{2 m } $
$ B =\mu_{0} nI $
$ v =\frac{ qr \mu_{0} nI }{2 m }$

Q58. Earth's magnetic field always has a horizontal component except at

A

equator

B

magnetic poles

C

a latitude of $60^{\circ}$

D

an altitude of $60^{\circ}$

Solution:

Magnetic dip at poles, $\theta=90^{\circ}$
$B_{H}=B \cos \theta=0$

Q59. Which of the field pattern given below is valid for electric field as well as for magnetic field?

A

B

C

D

Solution:

No monopoles exist and electric field lines donot form closed loops

Q60. The current following through an inductance coil of self inductance $6 \,mH$ at different time instants is as shown. The emf induced between $t =20\, s$ and $t =40\, s$ is nearly

A

$2 \times 10^{-2} V$

B

$3 \times 10^{-4} V$

C

$4 \times 10^{-3} V$

D

$30 \times 10^{2} V$

Solution:

$|e|=L \frac{d I}{d t}$
$|e|=6 \times 10^{-3}\left[\frac{I_{2}-I_{1}}{t_{2}-t_{1}}\right]$
$|e|=6 \times 10^{-3}\left[\frac{1}{40-20}\right]$
$e=3 \times 10^{-4} V .$

KCET 2021 Physics Questions with Answers Key Solutions

Q1. The physical quantity which is measure in the unit of $w b A^{-1}$ is

Self inductance

Mutual inductance

Magnetic flux

Both $A$ and $B$

$\phi= LI$ or $\phi= MI$ $L =\frac{\phi}{ I } $ or $ M =\frac{\phi}{ I }$

Q2. What will be the reading in the voltmeter and ammeter of the circuit shown ?

90 V, 2 A

0 V, 2 A

90 V, 1 A

0 V, 1 A

$X_{L}=X_{C} $ Hence, the resultant voltage across $L$ and $C$ is $0$. $Z = R$ $I =\frac{ V }{ Z }=\frac{ V }{ R }$ $=\frac{90}{45}=2 A$

Q3. LC osillations are similar and analogous to the mechanical oscillations of a block attached to a spring. The electrical equivalent of the force constant of the spring is

reciprocal of capacitive reactance

capacitive reactance

reciprocal of capacitance

capacitance

$k \to \frac{1}{C}$

Q4. In an oscillating LC circuit, $L =3.00\, mH$ and $C =2.70 \,\mu F .$ At $t =0$ the charge on the capacitor is zero and the current is $2.00\, A$. The maximum charge that will appear on the capacitor will be

$1.8 \times 10^{-5} C$

$18 \times 10^{-5} C$

$9 \times 10^{-5} C$

$90 \times 10^{-5} C$

$\frac{ Q ^{2}}{2 C }=\frac{ LI ^{2}}{2}$ $\frac{ Q ^{2}}{2 \times 2.7 \times 10^{-6}} =\frac{3 \times 10^{-3} \times 4^{2}}{2}$ $Q ^{2}=12 \times 10^{-3} \times 2.7 \times 10^{-6}$ $=32.4 \times 10^{-9}$ $Q =18 \times 10^{-5} C$

Q5. Suppose that the electric field amplitude of electromagnetic wave is $E _{0}=120\, NC ^{-1}$ and its frequency if $f =50 \,MHz$. Then which of the following value incorrectly computed ?

Magnetic field amplitude is $400 \,nT$.

Angular frequency of EM wave is $\pi \times 10^{8} rad / s$

Propagation constant (angular wave number) is $2.1\, rad / m$

Wavelength of EM wave is $6\, m$.

$c =\frac{ E _{0}}{ B _{0}}$ $B _{0}=\frac{ E _{0}}{ c }$ $=\frac{120}{3 \times 10^{8}}=400\, nT$ $k =\frac{2 \pi}{\lambda}=\frac{2 \times 3.14}{6}$ $k \neq 2.1 \,rad \,m ^{-1}$

Q6. The source of electromagnetic wave can be a charge.

Moving with a constant velocity

Moving in a circular orbit

At rest

Moving parallel to the magnetic field

Charges moving in a circular orbit undergo centripetal acceleration hence emit em waves. Or when electron jumps from one higher stationary orbit to another lower stationary orbit emits em radiation.

Q7. In refraction, light waves are bent on passing from one medium to second medium because, in the second medium.

frequency is different

speed is different

coefficient of elasticity is different

amplitude is smaller

Refraction takes because of change in speed of light when it moves from one medium to another

Q8. If the refractive index from air to glass is $\frac{3}{2}$ and that from air to water is $\frac{4}{3}$, then the ratio of focal lengths of a glass lens in water and in air is

$1: 2$

$2: 1$

$1: 4$

$4: 1$

$P _{1}=\frac{ f _{1} f _{2}}{ f _{1}- f _{2}}, P _{2}=\frac{ f _{1} f _{3}}{ f _{3}- f _{1}}$ and $P _{3}=\frac{ f _{2} f _{3}}{ f _{3}- f _{2}}$

$P _{1}=\frac{ f _{1}- f _{2}}{ f _{1} f _{2}}, P _{2}=\frac{ f _{3}- f _{1}}{ f _{3}+ f _{1}}$ and $P _{3}=\frac{ f _{3}- f _{2}}{ f _{2} f _{3}}$

$P _{1}=\frac{ f _{1}- f _{2}}{ f _{1} f _{2}}, P _{2}=\frac{ f _{3}- f _{1}}{ f _{1} f _{3}}$ and $P _{3}=\frac{ f _{3}- f _{2}}{ f _{2} f _{3}}$

$P_{1}=\frac{f_{1}+f_{2}}{f_{1} f_{2}}, P_{2}=\frac{f_{3}-f_{1}}{f_{1} f_{3}}$ and $P_{3}=\frac{f_{3}-f_{2}}{f_{2} f_{3}}$

Q10. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length $30\, cm$ is $2 \, cm .$ If a concave lens of focal length $20 \, cm$ is placed between the convex lens and the image at a distance of $26\, cm$ from the lens, the new size of the image is :

1.25 cm

2.5 cm

1.05 cm

2 cm

Q11. A slit of width 'a' is illuminated by red light of wavelength $6500 \mathring{A}$. If the first diffraction minimum falls at $30 \mathring{A}$, then the value of ' $a$ ' is

$6.5 \times 10^{-4} mm$

$1.3$ micron

$3250 \mathring{A}$

$2.6 \times 10^{-4} cm$

For diffraction minima $a \sin \theta= n \lambda$ $n =1 $ for $ I$ minimum $a =\frac{\lambda}{\sin \theta}=\frac{6.5 \times 10^{-7}}{\sin 30} $ $=\frac{6.5 \times 10^{-7}}{0.5} $ $=13 \times 10^{-7} $ $=1.3 \times 10^{-6} m$ $=1.3$ micron

Q12. Which of the statements are correct with reference to single slit diffraction pattern? (i) Fringes are of unequal width (ii) Fringes are of equal width (iii) Light energy is conserved (iv) Intensities of all bright fringes are equal

(i) and (iii)

(i) and (iv)

(ii) and (iv)

(ii) and (iii)

Option (i) and (iii) are correct.

Q13. In the Young's double slit experiment a monochromatic source of wavelength $\lambda$ is used. The intensity of light passing through each slit is $I _{0}$. The intensity of light reaching the screen $S _{ C }$ at a point $P$, a distance $X$ from $O$ is given by (Take $d < < D$ )

$I _{ o } \cos ^{2}\left(\frac{\pi D }{\lambda d } x \right)$

$4 I _{ o } \cos ^{2}\left(\frac{\pi d }{\lambda D } x \right)$

$I_{o} \sin ^{2}\left(\frac{\pi d}{2 \lambda D} x\right)$

$4 I _{ o } \cos \left(\frac{\pi d }{2 \lambda D } x \right)$

$\phi=\left(\frac{2 \pi}{\lambda}\right)$ path difference Path difference $= d \sin \theta$ $= d \left[\frac{ x }{ D }\right]$ $\phi=\frac{2 \pi}{\lambda}\left[\frac{ dx }{ D }\right]$ $I=4 I_{o} \cos ^{2} \frac{\phi}{2}$ $=4 I _{ o } \cos ^{2} \frac{\pi d }{\lambda D } \cdot x$

Q14. The work function of a metal is $1\, eV$. Light of wavelength $3000 \,\mathring{A}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

$10 \,ms ^{-1}$

$1 \times 10^{3} ms ^{-1}$

$1 \times 10^{4} ms ^{-1}$

$1 \times 10^{6} ms ^{-1}$

$\phi= LI$ or $\phi= MI$
$L =\frac{\phi}{ I } $ or $ M =\frac{\phi}{ I }$

$X_{L}=X_{C} $
Hence, the resultant voltage across $L$ and $C$ is $0$.
$Z = R$
$I =\frac{ V }{ Z }=\frac{ V }{ R }$
$=\frac{90}{45}=2 A$

$\frac{ Q ^{2}}{2 C }=\frac{ LI ^{2}}{2}$
$\frac{ Q ^{2}}{2 \times 2.7 \times 10^{-6}} =\frac{3 \times 10^{-3} \times 4^{2}}{2}$
$Q ^{2}=12 \times 10^{-3} \times 2.7 \times 10^{-6}$
$=32.4 \times 10^{-9}$
$Q =18 \times 10^{-5} C$

$c =\frac{ E _{0}}{ B _{0}}$
$B _{0}=\frac{ E _{0}}{ c }$
$=\frac{120}{3 \times 10^{8}}=400\, nT$
$k =\frac{2 \pi}{\lambda}=\frac{2 \times 3.14}{6}$
$k \neq 2.1 \,rad \,m ^{-1}$

For diffraction minima
$a \sin \theta= n \lambda$
$n =1 $ for $ I$ minimum
$a =\frac{\lambda}{\sin \theta}=\frac{6.5 \times 10^{-7}}{\sin 30} $
$=\frac{6.5 \times 10^{-7}}{0.5} $
$=13 \times 10^{-7} $
$=1.3 \times 10^{-6} m$
$=1.3$ micron

Q12. Which of the statements are correct with reference to single slit diffraction pattern?
(i) Fringes are of unequal width
(ii) Fringes are of equal width
(iii) Light energy is conserved
(iv) Intensities of all bright fringes are equal

$\phi=\left(\frac{2 \pi}{\lambda}\right)$ path difference
Path difference $= d \sin \theta$
$= d \left[\frac{ x }{ D }\right]$
$\phi=\frac{2 \pi}{\lambda}\left[\frac{ dx }{ D }\right]$
$I=4 I_{o} \cos ^{2} \frac{\phi}{2}$
$=4 I _{ o } \cos ^{2} \frac{\pi d }{\lambda D } \cdot x$

$ K _{\max }= h v-\phi_{o} $
$ y = m x - C$
Graph is a straight line with slope ' $h$ '
Till $v=v_{ o }$, no photoelectric emission occurs

For $2^{\text {nd }}$ orbit of H-atom
$E _{2}=\frac{-13 \cdot 6}{2^{2}} eV$
For $3^{\text {rd }}$ orbit of $He ^{+}$
$E _{3}^{ He ^{+}} =\frac{-13 \cdot 6}{3^{2}} 2^{2}$
$=\frac{-13 \cdot 6}{4} \times \frac{4}{9} \times 2^{2} $
$= E _{2} \times \frac{16}{9}=\frac{16}{9} E _{2}$

Number of stationary waves is $6$ i.e., $n =6$
$r=\frac{n^{2} \varepsilon_{0}^{2} h^{2}}{\pi m e^{2}}$
$=\frac{36 \varepsilon_{0}^{2} h^{2}}{\pi m e^{2}}$

$ Y =\overline{\overline{ AB } \cdot \overline{ AB }}$
$=\overline{ AB }+\overline{ AB } $
$= AB $

$ E _{1}=2.5\, eV$
$ E _{2}=2 \,eV $
$ E _{3}=3\, eV$
$\lambda=600\, nm $
$ \therefore E =\frac{1242}{\lambda_{ nm }}=\frac{1242}{600} eV $
$=2.07\, eV $
$( E _{1} $ and $E _{3}$ is $ > E $
Only $D _{2}$ will detect

$S=$ Area under $v-t$ graph
Area under uniform motion > Area under triangle

Area under a $-t$ graph $=v-u$
$\frac{1}{2} \times 8 \times 10= v $
$v =40 \,m s ^{-1}$