Exams Question Papers General Knowledge Election Directory
Question 1:
Judge the equivalent resistance when the following are connected in parallel −
(a) 1 Ω and 106 Ω, (b) 1 Ω and 103 Ω and 106 Ω.
Answer:
(a) When 1 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.
Therefore, equivalent resistance ≈ 1 Ω
(b) When 1 Ω , 103 Ω, and 106 Ω are connected in parallel:
Let R be the equivalent resistance.
Therefore, equivalent resistance = 0.999 Ω
Question 2:
Redraw the circuit of question 1, putting in an ammeter to measure the current
through the resistors and a voltmeter to measure potential difference across the
12 Ω resistor.What would be the readings in the ammeter and the voltmeter?
Answer:
To measure the current flowing through the resistors, an ammeter should be
connected in the circuit in series with the resistors. To measure the potential
difference across the 12 Ω resistor, a voltmeter should be connected parallel to
this resistor, as shown in the following figure.
The resistances are connected in series.
Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According
to Ohm’s law,
V = IR,
Where,
Potential difference, V = 6 V
Current flowing through the circuit/resistors = I
Resistance of the circuit, R = 5 + 8 + 12 = 25 Ω
Potential difference across 12 Ω resistor = V1
Current flowing through the 12 Ω resistor, I = 0.24 A
V1 = IR = 0.24 × 12 = 2.88 V
Therefore, the reading of the ammeter will be 0.24 A.
The reading of the voltmeter will be 2.88 V.
Question 1:
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V
each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all
connected in series.
Answer:
Three cells of potential 2 V, each connected in series, is equivalent to a battery of
potential 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors
of resistances 5 Ω, 8 Ω and 12 Ω respectively connected in series and a battery of
potential 6V.
Question 5:
Use the data in Table 12.2 to answer the following −
Table 12.2 Electrical resistivity of some substances at 20°C
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer:
(a) Resistivity of iron = 10.0 × 10−8 Ω m
Resistivity of mercury = 94.0 × 10−8 Ω m
Resistivity of mercury is more than that of iron. This implies that iron is a better
conductor than mercury.
(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest
among the listed materials. Hence, it is the best conductor.
Question 4:
Why are coils of electric toasters and electric irons made of an alloy rather than a
pure metal?
Answer:
The resistivity of an alloy is higher than the pure metal. Moreover, at high
temperatures,the alloys do not melt readily. Hence, the coils of heating appliances
such as electric toasters and electric irons are made of an alloy rather than a pure
metal.
Question 3:
Let the resistance of an electrical component remains constant while the potential
difference across the two ends of the component decreases to half of its former
value.What change will occur in the current through it?
Answer:
The change in the current flowing through the component is given by Ohm’s law as,
V = IR
I = V/R
Where,
Resistance of the electrical component = R
Potential difference = V
Current = I
The potential difference is reduced to half, keeping resistance constant.
Let the new resistance be R’ and the new amount of current be I ‘.
Therefore, from Ohm’s law, we obtain the amount of new current.
Therefore, the amount of current flowing through the electrical component is
reduced by half.
Question 2:
Will current flow more easily through a thick wire or a thin wire of the same
material,when connected to the same source? Why?
Answer:
Resistance of a wire, 
Where,
ρ = Resistivity of the material of the wire
l = Length of the wire
A = Area of cross-section of the wire
Resistance is inversely proportional to the area of cross-section of the wire.
Thicker the wire, lower is the resistance of the wire and vice-versa. Therefore,
current can flow more easily through a thick wire than a thin wire.
Question 1:
On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends upon the following factors:
(a) Length of the conductor
(b) Cross-sectional area of the conductor
(c) Material of the conductor
(d) Temperature of the conductor
Question 3:
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
The energy given to each coulomb of charge is equal to the amount of work required
to move it. The amount of work is given by the expression,
Work done = Potential Difference × Charges
Where,
Charge = 1 C
Potential difference = 6 V
Work Done = 6 × 1 = 6 J
Therefore, 6 J of energy is given to each coulomb of charge passing through a
battery of 6v.
Question 2:
What is meant by saying that the potential difference between two points is 1 V?
Answer:
If 1 J of work is required to move a charge of amount 1 C from one point to another,
then it is said that the potential difference between the two points is 1 V.
Question 1:
Name a device that helps to maintain a potential difference across a conductor.
Answer:
A source of electricity such as cell, battery, power supply, etc. helps to maintain a
potential difference across a conductor.
Question 3:
Calculate the number of electrons constituting one coulomb of charge.
Answer:
One electron possesses a charge of 1.6 × 10−19 C, i.e., 1.6 × 10−19 C of charge is
contained in 1 electron.
∴ 1 C of charge is contained in 
Therefore, 6 × 1018 electrons constitute one coulomb of charge.
Question 2:
Define the unit of current.
Answer:
The unit of electric current is ampere (A). 1 A is defined as the flow of 1 C of
charge through a wire in 1 s.
Question 1:
What does an electric circuit mean?
Answer:
An electric circuit consists of electric devices, switching devices, source of
electricity, etc. that are connected by conducting wires.
Question 12:
Why does the sky appear dark instead of blue to an astronaut?
Answer:
The sky appears dark instead of blue to an astronaut because there is no
atmosphere in the outer space that can scatter the sunlight. As the sunlight
is not scattered, no scattered light reach the eyes of the astronauts and the
sky appears black to them.
Question 11:
Why does the Sun appear reddish early in the morning?
Answer:
During sunrise, the light rays coming from the Sun have to travel a greater distance
in the earth’s atmosphere before reaching our eyes. In this journey, the shorter
wavelengths of lights are scattered out and only longer wavelengths are able to
reach our eyes. Since blue colour has a shorter wavelength and red colour has a
longer wavelength, the red colour is able to reach our eyes after the atmospheric
scattering of light.Therefore, the Sun appears reddish early in the morning.
Question 10:
Explain why the planets do not twinkle?
Answer:
Planets do not twinkle because they appear larger in size than the stars as they
are relatively closer to earth. Planets can be considered as a collection of a large
number of point-size sources of light. The different parts of these planets produce
either brighter or dimmer effect in such a way that the average of brighter and
dimmer effect is zero.Hence, the twinkling effects of the planets are nullified and
they do not twinkle.
Question 9:
Why do stars twinkle?
Answer:
Stars emit their own light and they twinkle due to the atmospheric refraction of
light. Stars are very far away from the earth. Hence, they are considered as point
sources of light. When the light coming from stars enters the earth’s atmosphere,
it gets refracted at different levels because of the variation in the air density at
different levels of the atmosphere. When the star light refracted by the atmosphere
comes more towards us, it appears brighter than when it comes less towards us.
Therefore, it appears as if the stars are twinkling at night.
Question 8:
What happens to the image distance in the eye when we increase the distance of
an object from the eye?
Answer:
Since the size of eyes cannot increase or decrease, the image distance remains
constant.
When we increase the distance of an object from the eye, the image distance in
the eye does not change. The increase in the object distance is compensated by
the change in the focal length of the eye lens. The focal length of the eyes changes
in such a way that the image is always formed at the retina of the eye.
Question 7:
Make a diagram to show how hypermetropia is corrected. The near point of a
hypermetropic eye is 1 m. What is the power of the lens required to correct this
defect? Assume that the near point of the normal eye is 25 cm.
Answer:
A person suffering from hypermetropia can see distinct objects clearly but faces
difficulty in seeing nearby objects clearly. It happens because the eye lens focuses
the incoming divergent rays beyond the retina. This defect of vision is corrected by
using a convex lens. A convex lens of suitable power converges the incoming light
in such a way that the image is formed on the retina, as shown in the following
figure.
The convex lens actually creates a virtual image of a nearby object (N’ in the figure)
at the near point of vision (N) of the person suffering from hypermetropia.
The given person will be able to clearly see the object kept at 25 cm (near point of
the normal eye), if the image of the object is formed at his near point, which is
given as 1 m.
Object distance, u = −25 cm
Image distance, v = −1 m = −100 m
Focal length, f
Using the lens formula,
A convex lens of power +3.0 D is required to correct the defect.