SEMICONDUCTORS & EXPERIMENTAL

1. In the given circuit, the current (I) through the battery will be

(A) 2.5 A (B) 1 A

(C) 2 A (D) 1.5 A
2. Statement-I: When a Si sample is doped with Boron, it becomes P type and when doped
by Arsenic it becomes N-type semiconductor such that P-type has excess of holes and
N-type has excess of electrons.
Statement-II: When such P-type and N-type semiconductors are fused to make a
junction, a current will automatically flow which can be detected with an externally
connected ammeter.
In light of the above statements, choose the most appropriate answer from the options
given below.
(A) Both Statement-I and statement-Il are incorrect
(B) Statement-l is incorrect but statement-II is correct
(C) Both Statement-I and statement-II are correct
(D) Statement-I is correct but statement-II is incorrect
3. The effect of increase in temperature on the number of electrons in conduction band
(ne) and resistance of a semiconductor will be as:
(A) Both (ne) and resistance decrease
(B) Both (ne) and resistance increase
(C) (ne) increases, resistance decreases
(D) (ne) decreases, resistance increases

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 SEMICONDUCTORS & EXPERIMENTAL
4. Choose the correct statement about Zener diode:
(A) It works as a voltage regulator in reverse bias and behaves like a simple pn junction
diode in forward bias.
(B) It works as a voltage regulator in both forward and reverse bias.
(C) It works as a voltage regulator only in forward bias.
(D) It works as a voltage regulator in forward bias and behaves like a simple pn
junction diode in reverse bias.
5. Given below are two statements: -one is labelled as
Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Diffusion current in a p-n junction is greater than the drift current in
magnitude if the junction is forward biased.
Reason (R): Diffusion current in a p-n junction is from the n-side to the p-side if the
junction is forward biased. In the light of the above statements, choose the most
appropriate answer from the options given below:
(A) Both A and R are correct and R is the correct explanation of A
(B) Both A and R are correct but R is not the correct explanation of A
(C) A is correct but R is not correct
(D) A is not correct but R is correct
6. Given below are two statements: one is labeled as Assertion (A) and the other is labeled
as Reason (R)
Assertion (A): Photodiodes are used in forward bias usually for measuring the light
intensity.
Reason (R): For ap-n junction diode, at applied voltage V the current in the forward
bias is more than the current in the reverse bias for |VZ| > ± V ≥ |V0| where V0 is the
threshold voltage and VZ is the breakdown voltage.
Choose the correct answer from the options given below:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is false but R is true
(D) A is true but R is false

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 SEMICONDUCTORS & EXPERIMENTAL
7. Which of the following Statement-Is not correct in the case of light emitting diodes?
A. It is a heavily doped p-n junction.
B. It emits light only when it is forward biased.
C. It emits light only when it is reverse biased.
D. The energy of the light emitted is equal to or slightly less than the energy gap of the
semiconductor used.
Choose the correct answer from the options given below:
(A) C And D (B) A
(C) C (D) B
8. Statement-I: By doping silicon semiconductor with pentavalent material, the electrons
density increases.
Statement-II: The n-type semiconductor has net negative charge.
In the light of the above statements, choose the most appropriate answer from the
statements from the options given below.
(A) Both statement I and statement II are false.
(B) Statement I is true but statement II is false.
(C) Statement I is false but statement II is true.
(D) Both statement I and statement II are true.
9. For extrinsic semiconductors, when doping level is increased
(A) Fermi-level of p-type semiconductors will go downward and Fermi-level of n-type
semiconductor will go upward.
(B) Fermi-level of p-type semiconductor will go upward and Fermi-level of n-type
semiconductors will go downward.
(C) Fermi-level of both p-type and n-type semiconductors will go upward for T > TF K
and downward for T < TF K, where TF is Fermi temperature.
(D) Fermi-level of p and n-type semiconductors will not be affected.

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 SEMICONDUCTORS & EXPERIMENTAL
10.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 1019
m–3 and their mobility is 1.6 m2/(V-s) then the resistivity of the semiconductor (since it
is an n-type semiconductor contribution of holes is ignored) is close to
(A) 0.29 m (B) 42 m
(C) 22 m (D) 0.42 W m
11.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?
(A) 4/7 (B) 5/8
(C) 4/5 (D) 5/4
12.The energy band gap of semiconducting material to produce violet (wavelength =
4000 Å) LED is ___________ eV. (Round off to the nearest integer).
13.If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in
the depletion region having the width of 6 × 10–6 m, will be ___________ × 105 N/C.
14.In a semiconductor, the number density of intrinsic charge carriers at 27°C is 1.5 ×
1016/m3. If the semiconductor is doped with impurity atom, the hole density increases to
4.5 × 1022/m3. The electron density in the doped semiconductor is ___________ ×
109/m3.
15.For the logic circuit shown, the output waveform at Y is

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 SEMICONDUCTORS & EXPERIMENTAL

(A)

(B)

(C)

(D)

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 SEMICONDUCTORS & EXPERIMENTAL
16.For the following circuit and given inputs A and B, choose the correct option for output
‘Y’

(A)

(B)

(C)

(D)

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 SEMICONDUCTORS & EXPERIMENTAL
17.For using a multimeter to identify diode from electrical component, choose the correct
statement of out of the following about the diode.
(A) It is two terminal device which conducts current in both directions.
(B) It is two terminal device which conducts current in one direction only.
(C) It does not conduct current gives an initial deflection which decays to zero.
(D) It is three terminal device which conducts current in one direction only between
central terminal and either of the remaining two terminals.
18.For the forward biased diode characteristics shown in the figure, the dynamic resistance
at ID = 3 mA will be ___________ .

19.The circuit contains diodes each with a forward resistance of 50 and with infinite
reverse resistance. If the battery voltage is 6 V, the current through the 120 resistance
is ___________ mA.

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 SEMICONDUCTORS & EXPERIMENTAL
20.If each diode has a forward bias resistance of 25 in the below circuit,

Which of the following options is correct?

(A) (B)

(C) (D)

21.Identify the solar cell characteristics from the following options.

(A) (B)

(C) (D)

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 SEMICONDUCTORS & EXPERIMENTAL
22.For the circuit shown below, calculate the value of Iz.

(A) 0.15 A (B) 0.05 A

(C) 25 mA (D) 0.1 A
23.LED is constructed from Ga-As-P semiconducting material. The energy gap of this
LED is 1.9 eV. Calculate the wavelength of light emitted and its colour.
[h = 6.63 × 10–34 J s and c = 3 × 108 m s–1]
(A) 1046 nm and red colour
(B) 654 nm and red colour
(C) 1046 nm and blue colour
(D) 654 nm and orange colour
24.In the given circuit the current through Zener Diode is close to

(A) 4.0 mA (B) 0.0 mA

(C) 6.0 mA (D) 6.7 mA
25.In the circuit shown below, maximum zener diode current will be _____________ mA.

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 SEMICONDUCTORS & EXPERIMENTAL
26.In the given circuit, the value of current IL will be _____________ mA.
(When RL = 1 k )

27.A zener of breakdown voltage VZ = 8 V and maximum zener current, IZM = 10 mA is

subjected to an input voltage Vi = 10 V with series resistance R = 100 . In the given
circuit RL represents the variable load resistance. The ratio of maximum and minimum
value of RL is _____________.

28.The logic performed by the circuit shown in figure is equivalent to

(A) AND (B) NOR

(C) OR (D) NAND
29.The logic operations performed by the given digital circuit is equivalent to:

(A) NOR (B) AND

(C) OR (D) NAND
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 SEMICONDUCTORS & EXPERIMENTAL
30.The truth table of the circuit shown is

(A) A B Y (B) A B Y
0 0 0 0 0 1
0 1 1 0 1 0
1 0 1 1 0 1
1 1 0 1 1 0
(C) A B Y (D) A B Y
0 0 0 0 0 1
0 1 1 0 1 1
1 0 1 1 0 0
1 1 1 1 1 1
31.Truth table for system of four NAND gates as shown in the figure is

(A) A B Y (B) A B Y
0 0 0 0 0 0
0 1 1 0 1 0
1 0 1 1 0 1
1 1 0 1 1 1
(C) A B Y (D) A B Y
0 0 1 0 0 1
0 1 1 0 1 0
1 0 0 1 0 0
1 1 0 1 1 1
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 SEMICONDUCTORS & EXPERIMENTAL
32.In an experiment with vernier callipers of least count 0.1 mm, when two jaws are joined
together the zero of vernier scale lies right to the zero of the main scale and 6th division
of vernier scale coincides with the main scale division, While measuring the diameter of
a spherical bob, the zero of vernier scale lies in between 3.2 cm and 3.3 cm marks and
4th division of vernier scale coincides with the main scale division. The diameter of bob
is measured as
(A) 3.22 cm (B) 3.18 cm
(C) 3.26 cm (D) 3.25 cm
33.The main scale of Vernier callipers reads 1 mm and 10 divisions of Vernier scale is
equal to the 9 divisions on main scale. When the two jaws of the instrument touch each
other, the zero of the Vernier lies to the right of zero of the main scale and its fourth
division coincides with a main scale division. When a spherical bob is tightly placed
between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2 cm
and 6th Vernier division coincides with a main scale division. The diameter of the bob
will be ____ × 10–2 cm.
34.In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth
vernier scale division coincides with ninth main scale division. Then the value of
vernier constant will be ____ × 10–2 mm.
35.The Vernier constant of Vernier callipers is 0.1 mm and it has zero error of –0.05 cm.
While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding
vernier division is 5. The corrected diameter will be ____ × 10–2 cm.
36.A screw gauge of pitch 0.5 mm is used to measure the diameter of uniform wire of
length 6.8 cm, the main scale reading is 1.5 mm and circular scale reading is 7. The
calculated cured surface area of wire to appropriate significant figure is
[Screw gauge has 50 divisions on the circular scale]
(A) 6.8 cm2
(B) 3.4 cm2
(C) 3.9 cm2
(D) 2.4 cm2

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 SEMICONDUCTORS & EXPERIMENTAL
37.In an experiment to find out the diameter of wire using screw gauge, the following

observations were noted:

(a) Screw moves 0.5 mm on main scale in one complete rotation

(b) Total divisions on circular scale = 50

(c) Main scale reading is 2.5 mm

(d) 45th division of circular scale is in the pitch line

(e) Instrument has 0.03 mm negative error

Then the diameter of wire is

(A) 2.92 mm (B) 2.54 mm

(C) 2.98 mm (D) 3.45 mm

38.In a screw gauge, fifth division of the circular scale coincides with the reference line

when the ratchet is closed. There are 50 divisions on the circular scale, and the main

scale moves by 0.5 mm on a complete rotation. For a particular observation the reading

on the main scale is 5 mm and the 20th division of the circular scale coincides with

reference line. Calculate the true reading.

(A) 5.20 mm (B) 5.00 mm

(C) 5.15 mm (D) 5.25 mm

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 SEMICONDUCTORS & EXPERIMENTAL
39.In a Vernier calipers, 10 divisions of vernier scale is equal to 9 divisions of main scale.
When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is
shifted to the left of zero of the main scale and 4th vernier scale division exactly
coincides with the main scale reading. One main scale division is equal to 1 mm. While
measuring diameter of a spherical body, the body is held between two jaws. It is now
observed that zero of the vernier scale lies between 30 and 31 divisions of main scale
reading and 6th vernier scale division exactly coincides with the main scale reading.
The diameter of the spherical body will be
(A) 3.02 cm (B) 3.06 cm
(C) 3.10 cm (D) 3.20 cm
40.A student measuring the diameter of a pencil of circular cross-section with the help of a
vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.54 mm, 5.65
mm. The average of these four readings is 5.5375 mm and the standard deviation of the
data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as
(A) (5.5375 ± 0.0739) mm
(B) (5.5375 ± 0.0740) mm
(C) (5.538 ± 0.074) mm
(D) (5.54 ± 0.07) mm
41.Assertion A: If in five complete rotations of the circular scale, the distance travelled on
main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale,
then least count is 0.001 cm.

Reason R: Least count =

In the light of the above statements, choose the most appropriate answer from the
options given below.
(A) A is not correct but R is correct.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) A is correct but R is not correct.
(D) Both A and R are correct and R is the correct explanation of A.

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 SEMICONDUCTORS & EXPERIMENTAL
42.The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale.
When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions
below the reference line. When a wire is placed between the jaws, the first linear scale
division is clearly visible while 72nd division on circular scale coincides with the
reference line. The radius of the wire is
(A) 0.90 mm
(B) 0.82 mm
(C) 1.64 mm
(D) 1.80 mm
43.A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in
Searle’s apparatus experiment. The increase in length produced in the wire is 4.0 mm.
Now the load is fully immersed in a liquid of relative density 2. The relative density of
the material of load is 8. The new value of increase in length of the steel wire is
(A) 4.0 mm (B) zero
(C) 5.0 mm (D) 3.0 mm
44.In an experiment to verify Stoke’s law, a small spherical ball of radius r and density ρ
falls under gravity through a distance h in air before entering a tank of water. If the
terminal velocity of the ball inside water is same as its velocity just before entering the
water surface, then the value of h is proportional to (ignore viscosity of air)
(A) r4 (B) r
(C) r3 (D) r2
45.A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of
sound (v) in air by resonance tube method. Resonance is observed to occur at two
successive lengths of the air column, 1 = 30 cm and 2 = 70 cm. Then v is equal to
(A) 384 m s–1
(B) 338 m s–1
(C) 379 m s–1
(D) 332 m s–1

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 SEMICONDUCTORS & EXPERIMENTAL
46.In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V

battery and a high resistance of 11 kΩ are used. The figure of merit of the galvanometer

is 60 µA per division. In the absence of shunt resistance, the galvanometer produces a

deflection of θ = 9 divisions when current flows in the circuit. The value of the shunt

resistance that can cause the deflection of θ/2, is closest to

(A) 55 Ω (B) 110 Ω

(C) 220 Ω (D) 550 Ω

47.A travelling microscope has 20 divisions per cm on the main scale while its vernier

scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale

divisions, what is the least count of the travelling microscope?

(A) 0.001 cm (B) 0.002 mm

(C) 0.002 cm (D) 0.005 cm

48.Three students S1, S2 and S3 perform an experiment for determining the acceleration

due to gravity (g) using a simple pendulum. They use different lengths of pendulum and

record time for different number of oscillations. The observations are as shown in the

table.

(Least count of length = 0.1 cm, least count for time = 0.1 s)
If E1, E2 and E3 are the percentage errors in ‘g’ for students 1, 2 and 3 respectively, then
the minimum percentage error is obtained by student no. ______.
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 SEMICONDUCTORS & EXPERIMENTAL
49.A student determined Young’s Modulus of elasticity using the formula . The
value of g is taken to be 9.8 m/s2, without any significant error, his observation are as
following.

Then the fractional error in the measurement of Y is

(A) 0.155 (B) 0.0083
(C) 0.083 (D) 0.0155
50.A thin 1 m long rod has a radius of 5 mm. A force of 50π kN is applied at one end to
determine its Young’s modulus. Assume that the force is exactly known. If the least
count in the measurement of all lengths is 0.01 mm, which of the following statements
is false?
(A) The maximum value of Y that can be determined is 1014 N/m2.

(B) gets minimum contribution from the uncertainty in the length.

(C) gets its maximum contribution from the uncertainty in strain.

(D) The figure of merit is the largest for the length of the rod.

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 SEMICONDUCTORS & EXPERIMENTAL
ANSWER

1. D 2. D 3. C 4. A 5. C
6. C 7. C 8. B 9. A 10. D
11. D 12. 3 13. 1 14. 5 15. A
16. A 17. B 18. 25 19. 20 20. C
21. B 22. C 23. B 24. B 25. 9
26. 5 27. B 28. A 29. B 30. A
31. A 32. B 33. 412 34. 5 35. 180
36. B 37. C 38. C 39. C 40. D
41. A 42. B 43. D 44. A 45. A
46. B 47. C 48. 1 49. D 50. A

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SOLUTIONS
1. Answer (D)

Req = Ω

2. Answer (D)
When a P-N junction diode is formed an electric field is developed from N-side to
P-side due to which barrier potential
is created & majority charge carriers can not flow through the junction due to
barrier potential so current is zero unless
we apply forward bias voltage.
3. Answer (C)
As temperature is increased, more electrons are excited to the conduction band and
hence conductivity increases, therefore resistance decreases.
4. Answer (A)
Works as voltage regulator in reverse bias and as simple p-n junction in forward
bias
5. Answer (C)
Diffusion current takes place from p-side to n-side when p-n junction is forward
bias.
6. Answer (C)
A Photodiode is operated in reverse bias condition.

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 SEMICONDUCTORS & EXPERIMENTAL
For a P-N junction diode current in forward bias (for V ≥ V0) is always greater than
the current in reverse bias (for V Vz )
Hence, Assertion is false but Reason is true.
7. Answer (C)
LED works when it is forward biased.
8. Answer (B)
In the pentavalent material, the electron density increases by doping.
9. Answer (A)
In n-type semiconductor, pentavalent impurity is added which donates a free
electron and in p-type semiconductor, a trivalent impurity is added which creates
hole in the valence band. So, the fermi energy level of p-type semiconductors will
go downward and for the n-type semiconductor will go upward.
10.Answer (D)
j = nevd
Resistivity,

= = 0.39 m 0.4 m
11.Answer (D)
Drift velocity vd =

12.Answer (3)

Given:

13.Answer (1)
Given: V = 0.6; d = 6 × 10–6 m

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14.Answer (5)
Temperature, T = 27°C = 300 K
Number density of intrinsic charge carrier ni = 1.5 × 1016 / m3
Number density of holes, nh = 4.5 × 1022 / m3
Let the number density of electrons is ne.
We know that,

15.Answer (A)
y = (A′ · B′)
=A+B
OR gate
Correct waveform is first.
16.Answer (A)
Y = (A′ B)′
= A + B′
Option 1.
17.Answer (B)
A diode is a two terminal device which conducts when forward biased and do not
conduct when reverse biased.
18.Answer (25)

Dynamic resistance,

19.Answer (20)
D2 is in reverse bias, so no current flows through this branch. The total resistance,
R = 50 + 130 + 120 = 300 ,

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20.Answer (C)

21.Answer (B)
For solar cell, the correct graph is shown in option (B).
22.Answer (C)
Let the total current is I.

0.05 A 50 mA

I = Iz + ; Iz = I – = 50 mA – 25 mA; Iz = 25 mA
23.Answer (B)
Energy gap, E = 1.9eV

As

Wavelength of red colour range from 620 – 750 nm.

24.Answer (B)
Net resistance of the circuit
R =R1 + = 1250

Current drawn from battery,

Voltage across R2 =

= 7.2 V < Zener voltage (10 V)

Hence, zener diode is reverse biased without breakdown. So, current through diode
is 0 mA.

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 SEMICONDUCTORS & EXPERIMENTAL
25.Answer (9)
For maximum current, Here, V = 120 V

I=

IZ = I – IL = 0.015 A – 0.006 A, IZ = 0.009 A = 9 mA

26.Answer (5)
Voltage across RL is 5 V, VZ = 5 V

So, IL = A
= 5 mA
27.Answer (B)
Here, Vi = 10 V and VZ = 8V
Maximum zener current, IZ = 10 mA
For maximum value of RL, current through the load is,

For minimum value of RL, diode is shorted.

28.Answer (A)

( ) ( ) AND gate

29.Answer (B)
(A + B) . (A.B) = A.(AB) + B.(AB)
= AB + AB = (AB)

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 SEMICONDUCTORS & EXPERIMENTAL
30.Answer (A)

Hence, the correct choice is (A).

31.Answer (A)

Thus, the correct option is (A).

32.Answer (B)
L.C. = 0.1 mm
For 6th division = 0.6 mm
Reading = 3.20 + 0.04 – 0.06

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 SEMICONDUCTORS & EXPERIMENTAL
= 3.20 – 0.02 = 3.18 cm
33.Answer (412)
Given, 10 VSD = 9 MSD
or 1 VSD = MSD

or 1 VSD = mm

Least count of the vernier calipers is,

LC 1MSD 1VSD

LC = ( ) mm = 0.1 mm = 0.01 cm

Positive zero error = = 0.4 mm 0.04 cm

Negative zero error = 4.1 cm + 6 × 0.01 cm
= 4.12 cm = 412 10-2 cm
34.Answer (5)
1 cm is divided into 20 divisions
1 MSD = cm; 10 VSD = 9MSD; 1 VSD = MSD

LC = 1 MSD – 1 VSD = 1 MSD – MSD

LC = 1 MSD = cm = cm

LC = mm

35.Answer (180)
Given, LC 0.1 mm = , zero error = −0.05 cm
MSR = = 1.7 cm, n = 5
So, VSR = 5 × LC = 5 × 0.1 = 0.5 mm = 0.05 cm
Diameter = MSR + VSR – zero error
= 1.7 + 0.05 – (-0.05) = 1.8 cm = 180 × 10-2 cm
36.Answer (B)
Given, length of the wire, = 6.8 cm

Least count =

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 SEMICONDUCTORS & EXPERIMENTAL
=
Least count = 0.01 mm
Diametre of the wire, D = + 1.5 mm + 7 × L.C
= 1.5 mm + 7 × 0.01 mm = 1.57 mm
Now, curved surface area of the wire,
A = D = 3.14 × 1.57 × 10-1 × 6.8 = 3.35 cm2 = 3.4 cm2
37.Answer (C)

Least count, LC =

LC = = .01 mm

Diameter of wire = MSR + (CSR × LC) + negative error

= 2.5 mm + (45 × 0.01 mm) + 0.03 mm = 2.98 mm
38.Answer (C)
5th division coincide with reference line, so, it is zero error.
Number of divisions on circular scale, N = 50
Pitch = 0.5 mm
Main scale reading, MSR = 5 mm

Least count, L.C. = = = 0.01 mm

Reading = MSR + CSR × LC – Zero error × L.C.

= 5 + 0.01 × 20 - 0.5 × 0.01 = 5.15 mm
39.Answer (C)
Given: 1 M.S.D = 1 mm

As 10 V.S.D = 9 M.S.D or 1 V.S.D = M.S.D

Vernier constant = 1 M.S.D – 1 V.S.D

= 1 M.S.D - M.S.D = (1 - 0.9) mm = 0.1 mm = 0.01 cm

Zero Error = − (10 – 4) × 0.1 mm = -0.6 mm

Reading = Main scale reading + Vernier divisions
coinciding × Vernier constant – Zero error

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 SEMICONDUCTORS & EXPERIMENTAL
= 3 + 6 × 0.01 – (-0.06)
= 3 + 0.06 + 0.06 = 3.12 cm (Nearest to option (C))
40.Answer (D)
Here dav = 5.5375 mm 5.54 mm
∆d = 0.07395 mm 0.07 mm
Average diameter of the pencil = (5.54 ± 0.07) mm
41.Answer (A)

Pitch =

Least count =

=
So, A is not correct but R is correct.
42.Answer (B)
Least count = 0.01 mm
Positive zero error = 0.08 mm
Reading = 1 mm + 72 × 0.01 = 1.72 mm
True reading = 1.72 - zero error = 1.72 - 0.08 = 1.64 mm
Radius = = 0.82 mm
43.Answer (D)
Let ρ and σ be the density of the liquid and material of the load respectively.
In first case, the extension in the wire is
x = Mg / k = V ρg / k …(i)
When the load is immersed in the liquid,
upthrust + internal force due to extension in wire = weight of the load
Vσg + kx1 = Vρg x1= Vg (ρ−σ) / k …(ii)
Using (i) and (ii),

( ) ( )

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44.Answer (A)
Velocity attained by the spherical ball while falling under gravity through a
distance h, v = √
Velocity attained by the spherical ball after entering into the water tank,

where ρ = density of the spherical ball

r = radius of the spherical ball
σ = density of fluid (water)
and η = coefficient of viscosity
According to the question, v = vT

= i.e., √

h is proportional to r4.
45.Answer (A)
Given, frequency of tuning fork, = 480 Hz
Resonance lengths, l = 30 cm and 2 = 70 cm
Speed of sound ( )air = 2 ( )
=2 480 40 10-2 = 38400 10-2 m s-1 384 m s−1
46.Answer (B)

Initially I =

Finally,

S=
( )

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47.Answer (C)
Travelling microscope has 20 division per cm on main scale.
Each division is cm on main scale.

Also 25VSD = 24MSD; 1VSD = MSD

Least count = − 1MSD 1VSD
=

LC = 0.002 cm
48.Answer (1)

The time period of simple pendulum is given by √

where, L is length of pendulum and g is the acceleration due to gravity.

Also,

Percentage error,

and T is same for all so, is minimum for highest

value of L, n and T.
So, the minimum percentage error is obtained by student 1.
49.Answer (D)

g = 9.8 m/s2 (constant)

So, fractional error is measurement of Y is 0.0155.

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 SEMICONDUCTORS & EXPERIMENTAL
50.Answer (A)

Here, L = 1 m, r = 5 mm = 5 × 10-3 m

F = 50π kN, L.C. of all lengths = 0.01 mm

Y=1

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