TOPIC: CURVED MIRRORS
DIRECTION: Choose the correct answer from the lettered options.
1.Images of real objects formed by a convex mirror are always _.
A.inverted, real and diminished
B.inverted, virtual and diminished
C.erect, virtual and diminished
D.erect, real and magnified
The correct answer is option [C]
2.Light rays parallel to the principal axis are reflected by the curved mirror through
the _.
A.center of curvature
B.pole
C. principal focus
D.principal axis
The correct answer is option [C]
3.An object is placed 32cm from a concave mirror of focal length 16cm. Find the
magnification of the image produced.
A.11.06cm.
B.10.7cm.
C. 12cm.
D.8.09cm.
The correct answer is option [B].
1/32 + 1/u = 1/16
1/u = 1/16 – 1/32
Therefore, u = 10.7cm.
State whether the following statement is true or false.
4.A plane mirror is used as a rearview mirror.
A.True
B.False
The correct answer is option [B]
5.A concave mirror can be used to produce a parallel beam of light if a lighted bulb
is placed _.
A.between its focus and the pole
B.at its focus
C. at its Centre of curvature
D.between its focus and Centre of curvature
The correct answer is option [B].
6.The radius of curvature of a convex mirror is 30cm. Its focal length is __ cm.
A.20
B.15
C. 60
D. 30
The correct answer is option [B]
TOPIC: EFFECT OF ELECTRIC CURRENT
DIRECTION: Choose the correct answer from the lettered options
1.Which of the following is the main effect of electric currents?
A. Chemical effect
B. Heat effect
C. Magnetic effect
D. Electric effect
The correct answer is option [D].
- Find the power operating in a refrigerator when a current of 15A flows through the
refrigerator at a resistance of 5Ω.
A. 1125AΩ.
B. 1125A/Ω.
C. 1125J.
D. 1125A
The correct answer is option [A].
Reason: Power, P = I2R and the unit follows.
- An electric bulb is rated 60W for a 240V supply. Find the resistance of the bulb
and the current it consumes.
A. 960 ohms and 0.25A
B. 340 ohms and 0.35A
C. 0.25 ohms and 1.50A D. 1.50 ohms and 960A
The correct answer is option [A].
Power w = 60W,
Voltage supplied V = 240V
Power = V 2/R,
60 = (240) 2/R R = (240) 2/60
R = 960 and 0.25A
- Water in an electric kettle connected to a 240V supply took 6mins to reach its
boiling point. How long would it have taken if the supply had been one of 210V?
A. 5.25 mins
B. 4 mins
C. 3.75 mins
D. 2.90 mins The correct answer is option [A].
240->360s, 210->360 x 210/240 = 630/2 = 315/60 = 5.25m
- A transformer which supplies 12V when connected to 240V mains takes 0.55A
from the mains when used to light five 12V, 24W lamps in parallel, find its efficiency
and the cost of using it for 12hr, at 60k per kWh.
A. 92.91%, 90.04k.
B. 91.91%, 95.04k.
C. 90.91%, 95.04k.
D. 90.91%, 95.10k.
The correct answer is option [C].
Hint: Use the equation Efficiency = [power output] x 100/ [power input],
Where power output = 5 x 24W = 120W and power input = IV = 0.55 x 240 = 132W.
Therefore, efficiency = [120 x 100]/132 = 90.91%. The cost of using it is power input x
number of hours x cost = 132W = 132/1000 = 0.132kW x 12 x 60k = 95.04k.
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