Physics (2017) | Study Mode

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Question 41
A

5

B

4

C

3

D

2

Explanation

When two mirror are inclined at an angle \(\theta\), the number of images formed, n = \(\frac{360}{\theta}\) - 1

since \(\theta\) = 90o

\(\frac{360}{\theta}\) - 1

= 4 - 1

= 3


Correct option: C
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Question 42
A
Zener material
B
P-n junction
C
n-type
D
p-type
Explanation
When silicon is doped with a trivalent atom. A p-type semi-conductor will be formed when silicon is doped with a pentavalent atom such as arsenic, a n-type semi-conductor will be formed
Phosphorous is a pentavalent element, so it will produce a o -type semi-conductor

Correct option: C
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Question 43
A
38.00o
B
19.47o
C
16.25o
D
49o
Explanation
For a refractive index () = \(\frac{\sin\frac{1}{2} (A + D)}{\sin\frac{1}{2}A}\)

D = angle of minimum deviation

A = refractive angle of the prism

1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60}\)

1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin 30}\)

Sin \(\frac{1}{2}\) (60 + D) = 1.5 * Sin 30

Sin \(\frac{1}{2}\) (60 + D) = 0.75

\(\frac{1}{2}\) (60 + D) = Sin-1 (0.75)

but Sin-1 (0.75) = 49o

\(\frac{1}{2}\) (60 + D) = 49

60 + D = 2 * 49 = 98o

D = 98o - 60o

D = 38o

Correct option: A
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Question 44
A
N80.00
B
N45.36
C
N65.00
D
N57.12
Explanation
Seven 40W lamps = 7 \(\times\) 40 = 280W

Five 80W lamps = 5 \(\times\) 80 = 400W

Total power = 280W + 400W

= 680W

= \(\frac{680KW}{1000}\) = 0.68KW

The kilowatt hour = 0.68 KW \(\times\) 12 hrs

= 8.16 KWh

Cost is N 7 per KWh

So the cost of running them = 8.16KWh \(\times\) N 7/KWh

= 57.12

Correct option: D
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Question 45
Physics (2017) Q 45 image
A

5.0A

B

3.7A

C

4.5A

D

4.0A

Explanation
V = IR from ohms law

I = \(\frac{V}{R}\)

Since the resistance are in parallel

\(\frac{1}{Reff}\) = \(\frac{1}{8}\) + \(\frac{1}{12}\) + \(\frac{1}{6}\)

= \(\frac{2 + 3 + 6}{36}\) = \(\frac{11}{36}\)

Reff = \(\frac{36}{11}\) = 3.27

I = I1 + I2 + I3

= \(\frac{V}{Reff}\)= \(\frac{12}{3.27}\)

= 3.669A

I = 3.7A

Correct option: B
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Question 46
A

10s

B

5s

C

4s

D

2s

Explanation
u = ?, H = 20, g = 10

From equation of motion under gravity

H = ut \(\frac{1}{2}\)gt2, It is falling so g is +ve

H = 0 + \(\frac{1}{2}\)gt2

H = \(\frac{1}{2}\)gt2

2H = gt2

\(\frac{2H}{g}\) = t2

t = \(\frac{\sqrt{2H}}{g}\)

= \(\frac{\sqrt {2 \times 20}}{10}\)

= \(\sqrt{4}\)

= 2s

Correct option: D
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Question 47
A
wave in closed pipe
B
radio waves
C
water waves
D
sound waves
Explanation
Mechanical waves require material medium for their propagation only radio waves among the options docs require a material medium

Correct option: B
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Question 48
A
Sin?
B
Cos?
C
Tan?
D
\(\frac{1}{\sin \theta}\)
Explanation
Consider an inclined plane shown below, as the effort moves along OB, the load moves is lifted up through a vertical height AB

V.R = \(\frac{\text{distance moved by effort}}{\text{distance moved by load}}\) = \(\frac{OB}{AB}\)

But Sin? = \(\frac{opp}{hyp}\) = \(\frac{AB}{OB}\)

therefore; = \(\frac{OB}{AB}\) = \(\frac{1}{\sin \theta}\)

Correct option: D
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Question 49
A
30cm3
B
40cm3
C
50cm3
D
20cm3
Explanation
Using boyle's law

P1V1 = P2V2

V2 = \(\frac{P_1 V_1}{200}\) = \(\frac{400 \times 10}{200}\)

= 20cm3

Correct option: D
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Question 50
A
\(^{58}_{28}X\)
B
\(^{30}_{28}X\)
C
\(^{28}_{30}X\)
D
\(^{58}_{30}X\)
Explanation
An atom can be represented thus,

\(^{Z}_{A}X\) where Z = mass number

A = atomic number

A = number of proton = number of electrons in a free state = 28

Z = number of protons + number of neutrons = 28 + 30 = 58

\(^{58}_{28}X\)is the representation

Correct option: A

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