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The following question based on electric current and magnetism topic of general science mcq
(a) Twice as large
(b) Unchanged
(c) 4 times as large
(d) Half as large
The correct answers to the above question in:
Answer: (d)
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Read more physics gk Based General Science Questions and Answers
Question : 1
A ray of ligh is incident normally on one of the faces of right angled isosceles prism as shhown above. It undergoes total internal reflection from hypotenuse. Which one of the following is the minimum refractive index of the material of the prism?
a) 1.414
b) 1.0
c) 1.33
d) 1.6
Answer »Answer: (a)
Question : 2
Alternating current is converted to direct current by
a) dynamo
b) transformer
c) rectifier
d) motor
Answer »Answer: (c)
Question : 3
Which element is the best conductor of electricity?
a) Aluminium
b) Silver
c) Copper
d) Iron
Answer »Answer: (b)
Question : 4
Out of the following, diamagnetic substance is
a) lead
b) iron
c) copper
d) silver
Answer »Answer: (c)
Copper has negative susceptibility
(i.e. χ = –0.96). Hence, it is diamagnetic.
Question : 5
Consider the following statements :
- Magnetic field produced by current in a straight wire has no poles.
- The magnetic lines of force produced by a straight current carrying conductor are straight in nature.
- To produce a strong magnetic field at its centre, we prefer a current carrying wire loop of larger radius.
a) 1 only
b) 1 and 3
c) 2 only
d) 1, 2 and 3
Answer »Answer: (a)
Magnetic field lines form in concentric circles around a cylindrical current-carrying conductor, such as a length of wire.
The strength of the magnetic field at the centre of a circular coil carrying current is inversely proportional to the radius of the circular coil.
i.e. the field strength reduces as the radius of the coil increases.
Question : 6
An AC generator of 220 V having internal resistance r = 10Ω and external resistance R = 100Ω. What is the power developed in the external circuit?
a) 400 W
b) 441 W
c) 484 W
d) 369 W
Answer »Answer: (a)
V = 200V; r = 10Ω
R' = 10 + 100Ω = 110Ω
I = $V/R' = 220/100 = 2A$
P=$I^2R$ = 4 × 100 = 400 W
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