# Physics (2017) | Study Mode

Question 41
A

2.20 $$\times$$ 10$$^3$$m

B

5.0 $$\times$$ 10$$^2$$m

C

5.0 $$\times$$ 10$$^5$$m

D

11.8 $$\times$$ 10$$^{11}$$m

##### Explanation

V = f

= $$\frac{v}{f}$$

= $$\frac{3 \times 10^8}{600 \times 10^3}$$ = 500m

= 5 $$\times$$ 102m

N.B 600KHz Frequency = 600 $$\times$$ 103Hz

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

Question 43
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

Question 44
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

Question 45
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

Question 46
A
wave in closed pipe
B
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

Question 47
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; = $$\frac{OB}{AB}$$ = $$\frac{1}{\sin \theta}$$

Question 48
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

Question 49
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

Question 50
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

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