Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Radioactivity
Q10: If the binding energy of the electron in a hydrogen atom is $13.6 \mathrm{eV}$, the energy required to remove the electron from the first excited state of $\mathrm{Li}^{\text {t+ }}$ is
Options: (a) $30.6 \mathrm{eV}$
(b) $13.6 \mathrm{eV}$
(c) $3.4 \mathrm{eV}$
(d) $122.4 \mathrm{eV}$
Answer: (a) $30.6 \mathrm{eV}$
Solution: $\mathrm{E}_{2}=\left(-\mathrm{Z}^{2} \mathrm{E}_{0}\right) / \mathrm{n}^{2}$

$\mathrm{E}_{2}=\left(-(3)^{2} \times 13.6\right) /(2)^{2}$

$=-30.6 \mathrm{eV}$

Energy required $=30.6 \mathrm{eV}$
Topic: Ray Optics
Q1: A convex lens is put $10 \mathrm{~cm}$ from a light source and it makes a sharp image on a screen, kept $10 \mathrm{~cm}$ from the lens. Now a glass block (refractive index 1.5 ) of $1.5 \mathrm{~cm}$ thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance $d$. Then $d$ is
Options: (a) $0.55 \mathrm{~cm}$ towards the lens
(b) 0
(c) $1.1 \mathrm{~cm}$ away from the lens
(d) $0.55 \mathrm{~cm}$ away from the lens
Answer: (d) $0.55 \mathrm{~cm}$ away from the lens
Solution: Case I: $u=-10 \mathrm{~cm}, \mathrm{v}=10 \mathrm{~cm}, \mathrm{f}=$ ?

Using lens formula,

$(1 / \mathrm{f})=[(1 / \mathrm{v})-(1 / \mathrm{u})]$

$1 / \mathrm{f}=[(1 / 10)-(1 /-10)]$

$\mathrm{f}=5 \mathrm{~cm}$

Case II: Due to introduction of slab, shift in the source is $=\mathrm{t}[1-(1 / \mu)]=1.5[1-(2 / 3)]=0.5$

Now, $u=-9.5 \mathrm{~cm}, \mathrm{v}=10.55 \mathrm{~cm} \mathrm{~d}=10.55-10=0.55 \mathrm{~cm}$ away from the lens.
Topic: Ray Optics
Q2: The speed of light in the medium is
Options: (a) maximum on the axis of the beam
(b) minimum on the axis of the beam
(c) the same everywhere in the beam
(d) directly proportional to the intensity I
Answer: (b) minimum on the axis of the beam
Solution: Given $\mu=\mu_{0}+\mu_{2} \mathrm{I}$

As $\mu=$ Speed of light in vacuum/Speed of light in the medium

$\mu=\mathrm{c} / \mathrm{v}$
\$mathrm{v}=\mathrm{c} / \mu=\mathrm{c} /\left(\mu_{0}+\mu_{2} \mathrm{I}\right)$

As the intensity is maximum on the axis of the beam, therefore $v$ is minimum on the axis of the beam.
Topic: Ray Optics
Q3: As the beam enters the medium, it will
Options: (a) travel as a cylindrical beam
(b) diverge
(c) converge
(d) diverge near the axis and converge near the periphery
Answer: (c) converge
Solution: As the beam enters the medium, it will converge.
Topic: Ray Optics
Q4: The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea $(7.8 \mathrm{~mm}$ ). This surface separates two media of refractive indices 1 and 1.34 . Calculate the distance from the refracting surface at which a parallel beam of light will come to focus
Options: (a) $4.0 \mathrm{~cm}$
(b) $1 \mathrm{~cm}$
(c) $3.1 \mathrm{~cm}$
(d) $2 \mathrm{~cm}$
Answer: (c) $3.1 \mathrm{~cm}$
Solution: $(1.34 / v)-(1 / \infty)=(1.34-1) / 7.8$

$1.34 / \mathrm{v}=0.34 / 7.8$

$\mathrm{v}=(1.34 \times 7.8) / 0.34$

$\mathrm{v}=30.7 \mathrm{~mm} \approx 3.1 \mathrm{~cm}$
Topic: Ray Optics
Q5: In an interference experiment the ratio of amplitudes of coherent waves is $\left(a_{1} / a_{2}\right)=1 / 3$. The ratio of maximum and minimum intensities of fringes will be
Options: (a) 4
(b) 9
(c) 2
(d) 18
Answer: (a) 4
Solution: $\left(\mathrm{I}_{\max } / \mathrm{I}_{\text {min }}\right)=\left(\mathrm{a}_{1}+\mathrm{a}_{2}\right)^{2} /\left(\mathrm{a}_{1}-\mathrm{a}_{2}\right)^{2}=(1+3)^{2} /(1-3)^{2}=16 / 4=4$
Topic: Ray Optics
Q6: The image formed by an objective of a compound microscope is
Options: (a) virtual and diminished
(b) real and diminished
(c) real and enlarged
(d) virtual and enlarged
Answer: (c) The objective of a compound microscope forms a real and enlarged image.
Solution: The image formed by an objective of a compound microscope is real and enlarged.
Topic: Ray Optics
Q7: Calculate the limit of resolution of a telescope objective having a diameter of $200 \mathrm{~cm}$, if it has to detect light of wavelength $500 \mathrm{~nm}$ coming from a star.
Options: (a) $610 \times 10^{-9}$ radian
(b) $152.5 \times 10^{-9}$ radian
(c) $457.5 \times 10^{-9}$ radian
(d) $305 \times 10^{-9}$ radian
Answer: (d) $305 \times 10^{-9}$ radian
Solution: The limit of resolution,

$\Delta \theta=1.22 \lambda / \mathrm{a}=\left(1.22 \times 500 \times 10^{-9} / 200 \times 10^{-2}\right)=3.05 \times 10^{-7} \mathrm{radian}$

$\Delta \theta=305 \times 10^{-9}$ radian
Topic: Ray Optics
Q8: A concave mirror for face viewing has a focal length of $0.4 \mathrm{~m}$. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is
Options: (a) $0.16 \mathrm{~m}$
(b) $1.60 \mathrm{~m}$
(c) $0.32 \mathrm{~m}$
(d) $0.24 \mathrm{~m}$
Answer: (c) $0.32 \mathrm{~m}$
Solution: Given, $\mathrm{f}=-0.4 \mathrm{~m}, \mathrm{~m}=5=(-\mathrm{v} / \mathrm{u})=(\mathrm{y} /-\mathrm{x}) \Rightarrow \mathrm{y}=5 \mathrm{x}$

$(1 / \mathrm{f})=(1 / \mathrm{v})+(1 / \mathrm{u})$

$\Rightarrow(1 /-0.4)=(1 / 5 x)+(1 /-x)=4 /-5 x$

$\Rightarrow \mathrm{x}=-0.32 \mathrm{~m}$ so $\mathrm{u}=0.32 \mathrm{~m}$
Topic: Ray Optics
Q9: The value of the numerical aperture of the objective lens of a microscope is 1.25. If the light of wavelength $5000 \AA$ is used, the minimum separation between two points, to be seen as distinct, will be
Options: (a) $0.48 \mathrm{~m}$
(b) $0.12 \mathrm{~m}$
(c) $0.38 \mathrm{~m}$
(d) $0.24 \mathrm{~m}$
Answer: (d) $0.24 \mathrm{~m}$
Solution: Numerical aperture of the objective lens of microscope $=0.61 \lambda / \mathrm{d}$

Minimum separation between two points $\mathrm{d}$ to be seen clearly,

$\mathrm{d}=0.61 \lambda /$ Numerical aperture

$\mathrm{d}=\left(0.61 \times 5000 \times 10^{-10}\right) / 1.25$

$\mathrm{d}=0.24 \mathrm{~m}$
Topic: Ray Optics
Q10: A convergent doublet of separated lenses, corrected for spherical aberration, has a resultant focal length of $10 \mathrm{~cm}$. The separation between the two lenses is $2 \mathrm{~cm}$. The focal lengths of the component lenses are
Options: (a) $18 \mathrm{~cm}, 20 \mathrm{~cm}$
(b) $12 \mathrm{~cm}, 14 \mathrm{~cm}$
(c) $16 \mathrm{~cm}, 18 \mathrm{~cm}$
(d) $10 \mathrm{~cm}, 12 \mathrm{~cm}$
Answer: (a) $18 \mathrm{~cm}, 20 \mathrm{~cm}$
Solution: Since the doublet is corrected for the spherical aberration, it satisfies the following condition

$\mathrm{f}_{1}-\mathrm{f}_{2}=\mathrm{d}=2 \mathrm{~cm}$

$\mathrm{f}_{1}=\mathrm{f}_{2}+2$

Equivalent focal length $=\mathrm{F}$

$\mathrm{F}=\left(\mathrm{f}_{1} \mathrm{f}_{2}\right) /\left(\mathrm{f}_{1}+\mathrm{f}_{2}-\mathrm{d}\right)=10 \mathrm{~cm}$

Solving it, we get

$\mathrm{f}_{1}=20 \mathrm{~cm}$

$\mathrm{f}_{2}=18 \mathrm{~cm}$
Topic: Ray Optics
Q11: A single slit of width $b$ is illuminated by coherent monochromatic light of wavelength $\lambda$. If the second and fourth minima in the diffraction pattern at a distance $1 \mathrm{~m}$ from the slit are at $3 \mathrm{~cm}$ and $6 \mathrm{~cm}$, respectively from the central maximum, what is the width of the central maximum? (i.e. the distance between the first minimum on either side of the central maximum)
Options: (a) $6.0 \mathrm{~cm}$
(b) $1.5 \mathrm{~cm}$
(c) $4.5 \mathrm{~cm}$
(d) $3.0 \mathrm{~cm}$
Answer: (d) $3.0 \mathrm{~cm}$
Solution: For single slit diffraction, $\sin \theta=\mathrm{n} \lambda / \mathrm{b}$

Position of $\mathrm{n}^{\text {th }}$ minima from central maxima $=\mathrm{n} \lambda_{\mathrm{D}} / \mathrm{b}$

When $\mathrm{n}=2$, then $\mathrm{x}_{2}=2 \lambda_{\mathrm{D}} / \mathrm{b}=0.03$

When $\mathrm{n}=4$, then $\mathrm{x}_{4}=4 \lambda_{\mathrm{D}} / \mathrm{b}=0.06$

Eqn. (2) - Eqn. (1)

$\mathrm{x}_{4}-\mathrm{x}_{2}=\left(4 \lambda_{\mathrm{D}} / \mathrm{b}\right)-\left(2 \lambda_{\mathrm{D}} / \mathrm{b}\right)=0.03$ or

then width of central maximum $=2 \lambda_{\mathrm{D}} / \mathrm{b}=2 \times(0.03 / 2)=0.03 \mathrm{~m}=3 \mathrm{~cm}$
Topic: Ray Optics
Q12: An observer looks at a distant tree of height $10 \mathrm{~m}$ with a telescope of the magnifying power of 20 . To the observer, the tree appears
Options: (a) 10 times taller
(b) 10 times nearer
(c) 20 times taller
(d) 20 times nearer
Answer: (d) 20 times nearer
Solution: The telescope resolves and brings the objects closer, which are far away from the telescope. Hence, for a telescope with magnifying power 20 , the tree appears 20 times nearer.
Topic: Ray Optics
Q13: To determine the refractive index of a glass slab using a travelling microscope, the minimum number of readings required are
Options: (a) Two
(b) Four
(c) Three
(d) Five
Answer: (c) Three
Solution: To determine the refractive index of a glass slab using a travelling microscope, the minimum number of readings required are three.
Topic: Ray Optics
Q14: You are asked to design a shaving mirror assuming that a person keeps it $10 \mathrm{~cm}$ from his face and views the magnified image of the face at the closest comfortable distance of $25 \mathrm{~cm}$. The radius of curvature of the mirror would then be
Options: (a) $30 \mathrm{~cm}$
(b) $24 \mathrm{~cm}$
(c) $60 \mathrm{~cm}$
(d) $-24 \mathrm{~cm}$
Answer: (c) $60 \mathrm{~cm}$
Solution: $(1 / 15)+(1 /-10)=1 / \mathrm{f}$

$\mathrm{f}=-30 \mathrm{~cm}$

$R=2 f=2(-30)=-60 \mathrm{~cm}$
Topic: Ray Optics
Q15: A green light is incident from the water to the air-water interface at the critical angle ( $\boldsymbol{\theta}$ ). Select the
Options: (a) The entire spectrum of visible light will come out of the water at various angles to the normal
(b) The entire spectrum of visible light will come out of the water at an angle of $90^{\circ}$ to the normal
(c) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium
(d) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium
Answer: (c) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium
Solution: As, $\sin \theta=1 / \mu$

Also, the refractive index $(\mu)$ of the medium depends on the wavelength of the light. $\mu$ is less for the greater wavelength (i.e. lesser frequency).

So, $\theta$ will be more for a lesser frequency of light. Hence, the spectrum of visible light whose frequency is less than that of green light will come out to the air medium.
Topic: Rotational Motion
Q1: Two-point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod which is of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of
Options: 1. 0.42 m from the mass of 0.3 kg
2. 0.70 m from the mass of 0.7 kg
3. 0.98 m from the mass of 0.3 kg
4. 0.98 m from the mass of 0.7 kg
Answer: (3) 0.98 m from the mass of 0.3 kg
Solution: The moment of inertia of the system about the axis of rotation O is I = I1 + I2 = (0.3)x^2 + 0.7(1.4 - x)^2
I = 0.3 x^2 + 0.7(1.96 + x^2 - 2.8 x)
= 0.3 x^2 + 1.372 + 0.7 x^2 - 1.96 x
= x^2 + 1.372 - 1.96 x
The work done in rotating the rod is converted into its rotational kinetic energy.
W = 1 / 2 I ω^2
= 1 / 2 [x^2 + 1.372 - 1.96 x] ω^2
For the work done to be minimum
dW / dx = 0 ⇒ 2 x - 1.96 = 0
x = 1.96 / 2 = 0.98 m
Topic: Rotational Motion
Q2: A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved?
Options: 1. centre of the circle
2. on the circumference of the circle
3. inside the circle
4. outside the circle
Answer: (1) centre of the circle
Solution: The force will pass through the centre of the circle. Therefore, the angular momentum will remain conserved at the centre of the circle.
Topic: Rotational Motion
Q3: A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is
Options: 1. 2 K
2. K / 2
3. K / 4
4. 4 K
Answer: (2) K / 2
Solution: According to the conservation of angular momentum
I ω0 = I1 ω1
So we have I1 = 2 I
ω1 = ω0 / 2
Initial kinetic energy = I ω0^2 / 2 = K
Final Kinetic energy = I1 ω1^2 / 2 = 2 I (ω0 / 2)^2 / 2 = K / 2
Topic: Rotational Motion
Q4: A particle is confined to rotate in a circular path with decreasing linear speed. Then which of the following is correct?
Options: 1. angular momentum is conserved about the centre
2. only the direction of L is conserved
3. it spirals towards the centre
4. its acceleration is towards the centre
Answer: (2) only the direction of L is conserved
Solution: L is not conserved in magnitude since v is changing (decreasing). It is given that a particle is confined to rotate in a circular path, it cannot have a spiral path. Since the particle has two accelerations ac and at, therefore the net acceleration is not towards the centre. The direction of L remains the same even when the speed decreases.
Topic: Rotational Motion
Q5: A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach point B. The angle θ at which the speed of the bob is half of that at A satisfies
Options: 1. θ = π / 2
2. π / 4 < θ < π / 2
3. π / 2 < θ < 3 π / 4
4. 3 π / 2 < θ < π
Answer: (4) 3 π / 2 < θ < π
Solution: This is the case of vertical motion when the body just completes the circle. Here
v = √(5 g L)
Applying energy conservation,
1 / 2 mv0^2 = 1 / 2 mv^2 + mgl(1 - cos θ)
where v0 is the horizontal velocity at the bottom point, v is the velocity of bob where the bob inclined θ with vertical.
Also, we know the relation between the velocity at the topmost and velocity at the bottom point.
mg(2l) = 1 / 2 mv0^2 - 1 / 2 mvtop^2
Since v0 is just sufficient
mvtop^2 / l = T + mg
T = 0
vtop = √(g l)
Then equation 2 becomes
v0 = √(5 g l)
According to the question v = v0 / 2
So from equation (1)
1 / 2 m(5 gl) = 1 / 2 m(5 gl / 4) + mgl(1 - cos θ)
(20 mgl - 5 mgl) / 8 = mgl(1 - cos θ)
(1 - cos θ) = 15 / 8
cos θ = 7 / 8
Hence, 3 π / 2 < θ < π
Topic: Rotational Motion
Q6: A ball of mass (m) 0.5 kg is attached to the end of a string having a length (L) 0.5 m. The ball is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of the angular velocity of ball (in radian/s) is
Options: 1. 9
2. 18
3. 27
4. 36
Answer: (4) 36
Solution: m ω^2 L sin θ cos θ = mg sin θ
cos θ = g / ω^2 L
sin θ = 1 / ω^2 L
sin θ = 1 / (ω^2 L) √((ω^2 L)^2 - g^2)
T = mg cos θ + m ω^2 L sin^2 θ
T = mg (g / ω^2 L) + (m ω^2 / ω^2 L)^2 ((ω^2 L)^2 - g^2)
T = mg / ω^2 L [ g^2 + (ω^2 L)^2 - g^2]
T = m ω^2 L = 324 (given)
ω = √(324 / 0.5 * 0.5)
ω = 36 rad / s
Topic: Rotational Motion
Q8: A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m / s. A small ball of mass 0.1 kg, moving velocity 20 m / s in the opposite direction hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m / s. Immediately after the collision.
Options: 1. The ring has pure rotation about its stationary CM
2. The ring comes to a complete stop
3. Friction between the ring and the ground is to the left
4. There is no friction between the ring and the ground
Answer: (1) The ring has pure rotation about its stationary CM
(3) Friction between the ring and the ground is to the left
Solution: Let's assume that friction between the ground and the ring gives no impulse during the collision with the ball. Using conservation of momentum along the x-axis we get that the CM of the ring will come to rest. Thus option A is correct.
Secondly, the question tells us that the ball gets a velocity in the vertical direction, hence there must be an impulse in the vertical direction. There will be horizontal and a vertical impulse on the ring at the point of contact. These will have components along the tangent of the ring, which will provide angular impulses.
Using angular impulse = change in angular momentum, we get
= 2 cos 30°(1 / 2) - 1 sin 30°(1 / 2) = 2(1 / 4)(ω2 - ω1)
note that we have assumed that the direction of angular velocities is the same before and after and since LHS of the above equation is positive ω2 > ω1 so the ring must be slipping to right and hence the friction will be to the left as it will be opposite to the direction of motion. Thus option C is correct
Topic: Rotational Motion
Q9: A carpet of mass M, made of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on the rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R / 2.
Options: 1. √(14 / 3 g R)
2. √(7 / 3 g R)
3. √(2 g R)
4. √(g R)
Answer: (1) √(14 / 3 g R)
Solution: The radius is reduced to R / 2, the mass of the rolled carpet = (M / π R^2 1) × (R / 2)^2 1 = M / 4
Release of potential energy = MgR - (M / 4) gR / 2 = 7 / 8 MgR
The kinetic energy at this instant is given by = 1 / 2 Mv^2 + 1 / 2 I ω^2 = 1 / 2(M / 4) v^2 + 1 / 2(MR^2 / 32)(2v / R)^2 = (3 / 16) Mv^2
(3 / 16) Mv^2 = 7 / 8 MgR
v = √(14 / 3 g R)
Topic: Rotational Motion
Q10: A bob of mass m attached to an inextensible string of length is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad / s about the vertical. About the point of suspension
Options: 1. Angular momentum changes in direction but not in magnitude
2. Angular momentum changes both in direction and magnitude
3. Angular momentum is conserved
4. Angular momentum changes in magnitude but not in direction.
Answer: (1) Angular momentum changes in direction but not in magnitude
Topic: Rotational Motion
Q14: Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is
Options: 1. (55 / 2) MR^2
2. (73 / 2) MR^2
3. (181 / 2) MR^2
4. (19 / 2) MR^2
Answer: (3) (181 / 2) MR^2
Solution: I0 = Icm + md^2
I0 = (7 MR^2 / 2) + 6 (M * (2 R)^2) = 55 MR^2 / 2
Ip = I0 + md^2
Ip = 55 MR^2 / 2 + 7 M (3 R)^2 = (181 / 2) MR^2
Topic: Rotational Motion
Q15: A pulley of radius 2 m is rotated about its axis by a force F = 20 t - 5 t^2 newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kgm^2, the number of rotations made by the pulley before its direction of motion it reversed, is
Options: 1. more than 6 but less than 9
2. more than 9
3. less than 3
4. more than 3 but less than 6
Answer: (4) more than 3 but less than 6
Solution: Torque is given by τ = FR
Or α = FR / I
Given F = 20 t - 5 t^2
R = 2 m,
I = 10 kgm^2
α = [(20 t - 5 t^2) * 2] / 10
α = 4 t - t^2
ω = ∫0^t α dt = 2 t^2 - t^3 / 3
at ω = 0 ⇒ 2 t^2 - t^3 / 3 = 0
t^3 = 6 t^2
t = 6
θ = ∫0^6 ω dt = ∫0^6 (2 t^2 - t^3 / 3) dt
θ = 36 / 2 π < 6
Topic: Semiconductors
Q1: The energy band gap is maximum in
Options: (a) metals
(b) superconductors
(c) insulators
(d) semiconductors
Answer: (c) The energy band gap is maximum in insulators
Topic: Semiconductors
Q2: A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of
Options: (a) each of them increases
(b) each of them decreases
(c) copper decreases and germanium increases
(d) copper increases and germanium decreases
Answer: (c) copper decreases and germanium increases
Solution: Copper is a conductor. Germanium is a semiconductor. When cooled, the resistance of copper decreases and that of germanium increases.
Topic: Semiconductors
Q3: In the common base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor ( β ) will be
Options: (a) 48
(b) 49
(c) 50
(d) 51
Answer: (b) 49
Solution: β=Ic / Ib=Ic /(Ie-Ic)=5.488 /(5.60-5.488)=5.488 / 0.112=49
Topic: Semiconductors
Q4: Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate?
Options: (a) The number of free electrons for conduction is significant only in Si and Ge but small in C
(b) The number of free conduction electrons is significant in C but small in Si and Ge
(c) The number of free conduction electrons is negligibly small in all the three
(d) The number of free electrons for conduction is significant in all three
Answer: (a) The number of free electrons for conduction is significant only in Si and Ge but small in C
Solution: Carbon (C), silicon (Si) and germanium (Ge) have the same lattice structure and their valence electrons are 4. For C, these electrons are in the second orbit, for Si it is third and for germanium, it is the fourth orbit. In solid-state, the higher the orbit, the greater the possibility of overlapping of energy bands. Ionization energies are also less therefore Ge has more conductivity compared to Si. Both are semiconductors. Carbon is an insulator.
Topic: Semiconductors
Q5: In the ratio of the concentration of electrons that of holes in a semiconductor is 7 / 5 and the ratio of currents is 7 / 4 then what is the ratio of their drift velocities?
Options: (a) 4 / 7
(b) 5 / 8
(c) 4 / 5
(d) 5 / 4
Answer: (d) 5 / 4
Solution: Drift velocity, Vd=I / nAe
(vd)electron /(vd)hole=(Ie / Ih)(nh / ne)=(7 / 4) ×(5 / 7)=5 / 4
Topic: Semiconductors
Q6: At absolute zero, silicon (Si) acts as
Options: (a) non-metal
(b) metal
(c) insulator
(d) none of these
Answer: (c) insulator
Solution: Semiconductors like silicon (Si) and germanium (Ge) act as insulators at low temperature.
Topic: Semiconductors
Q7: Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an n-type semiconductor, the density of electrons is 10^19 m^-3 and their mobility is 1.6 m^2 /(V-s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to
Options: (a) 0.2 m
(b) 4 m
(c) 2 m
(d) 0.4 m
Answer: (d) 0.4 m
Solution: J=neVd
Resistivity, ρ=E / j=E / neVd=1 / ne (vd / E)=1 / ne μe
Resistivity, 1 /(10^19 × 1.6 × 10^-19 × 1.6)=0.39 Ω m=0.4 Ω m
Topic: Semiconductors
Q8: The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than 2480 nm is incident on it. The bandgap in (eV) for the semiconductor is
Options: (a) 0.5 eV
(b) 0.7 eV
(c) 1.1 eV
(d) 2.5 eV
Answer: (a) 0.5 eV
Solution: Band gap = Energy of photon of =2480 nm
Energy =(hc / λ) J=(hc / λe) e V
Band gap =((6.63 × 10^-34) ×(3 × 10^8)) /((2480 × 10^-9) ×(1.6 × 10^-19))=0.5 eV
Topic: Semiconductors
Q9: The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the
Options: (a) crystal structure
(b) variation of the number of charge carriers with temperature
(c) type of bonding
(d) variation of scattering mechanism with temperature
Answer: (b) variation of the number of charge carriers with temperature
Topic: Semiconductors
Q10: A strip of copper and another germanium are cooled from room temperature to 80 K. The resistance of
Options: (a) each of these decreases
(b) copper strip increases and that of germanium decreases
(c) copper strip decreases and that of germanium increases
(d) each of these increases
Answer: (c) copper strip decreases and that of germanium increases
Solution: Copper is conductor and germanium is a semiconductor. When cooled, the resistance of copper strip decreases and that of germanium increases.
Topic: Simple Harmonic Motion
Q1. A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in second is
Options: (a) 8π / 3
(b) 4π / 3
(c) 3π / 8
(d) 7π / 3
Answer: Answer: (a) 8π / 3
Solution: In SHM, speed v = ω √(A^2 - x^2), at x = 4, v = ω √((5)^2 - (4)^2) = 3ω
Acceleration a = -ω^2 x
at x = 4, a = -4ω^2
As |v| = |a|
ω = 3 / 4 ⇒ T = 2π / ω = 8π / 3