Electricity Test

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Electricity Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

Through a metallic conductor, electric current is considered to be due to movement of

A
ions

B
amperes

C
electrons

D
protons

Solution

Through a metallic conductor, electric current is considered to be due to movement of electrons.
Question 2
1 / -0

If 5000 alpha particles per minute are passing through a straight tube of radius r, then the resulting electric current is (approximately)

A
0.27 10^–16 A

B
1 10¹² A

C
0.26 10¹² A

D
1 10^–12 A

Solution

Charge on one alpha particle = q = 3.2 10^–19C
Total charge = 5000 3.2 10^–19
= 16 10³ 10^–19
= 16 10^–16
= 1.6 10^–15
Current = Charge flow per second =
= 0.0267 10^–15A
= 0.27 10^–16 A
Question 3
1 / -0

The work done in moving a unit positive charge across two points in an electric circuit is a measure of

A
current

B
potential difference

C
resistance

D
power

Solution

Electrostatic potential difference between two points is the amount of work done per unit positive charge, by an external agent, in moving it (without acceleration) from one point to another against the electrostatic field along any path.
Question 4
1 / -0

What will be the ratio of resistances of Aluminium and copper wire where resistivity of aluminium wire is twice the resistivity of copper wire and density of aluminium wire is one-third the density of copper wire and mass per unit length is same for aluminium and copper wire?

A
3:2

B
2:3

C
3:4

D
4:3

Solution

Let resistance of aluminium R_a = (where is its resistivity and A_a is the area)
Let R_C be the resistance of copper = (where is its resistivity and Ac is is the area of cross-section)
= 2 (given)
D_a = (where D_a and are the densities of aluminium and copper)
M_a = M_c (masses of aluminium and copper)
According to question,

Since, D_a = and D_c =
By comparing both the equations, we get
(where V_a and V_c are the volumes of aluminium and copper)

V_a = 3V_c
and A_a = 3A_c
Since, Volume = Area Length
V_a = A_a × and V_c = A_c ×
= 2 : 3
Question 5
1 / -0

The heating element of an electric heater should be made with a material, which inherits the property of

A
high specific resistance and high melting point

B
high specific resistance and low melting point

C
low specific resistance and low melting point

D
low specific resistance and high melting point

Solution

A heating element converts electricity into heat through the process of joule heating. Electrical current running through the element encounters resistance, resulting in heating of the element.
( and melting point should be high)
Question 6
1 / -0

Four equal resistances are connected as shown in the figure below. The resistance across AB will be

A
R

B
R/4

C
R/2

D
(3/4)R

Solution

Let the total resistance be R_S.
R_s =
R_s =

=
Question 7
1 / -0

The effective resistance between the points A and B is

A
R

B
2R

C
4R

D
5R

Solution

The circuit given is the balanced Wheatstone bridge.

As

∴ There will be no current through R.

So, the circuit can be redrawn as:

Let the total resistance across AB be R_F.

=

R_F = 2R
Question 8
1 / -0

In the arrangement of resistances, shown in figure, the potential difference between B and D will be zero when the resistance X is

A
4

B
2

C
3

D
zero

Solution

= 1 + 1 = 2
R_p = ohm
Using the condition,

= 8
X = 2 ohms
Question 9
1 / -0

N identical cells are connected to form a battery. When the terminals of the battery are joined directly, current I flows in the circuit. Under what condition can we obtain the maximum value of I?

A
When all the cells should be joined in series.

B
When all the cells should be joined in parallel.

C
When two rows of each of N/2 cells should be joined in parallel.

D
When the current drawn is same for all possible combinations.

Solution

The more cells we connect in series, the higher the circuit resistance will be. Hence, the current will be less.
And, the more cells we connect in parallel, the lower the circuit resistance will be. So, the current will be more.
Question 10
1 / -0

In a series combination of electrical appliances, total electric power

A
increases

B
decreases

C
may increase or decrease according to the situation

D
no definite observation

Solution

P = V²/R where R = Resistance
Now, in series circuit, as the number of appliances increases, the resistance increases and hence, the total power decreases.
Question 11
1 / -0

A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards, the particle will

A
get deflected vertically upwards

B
move in a circular orbit with its speed increased

C
move in a circular orbit with its speed unchanged

D
continue to move due east

Solution

A charge particle `q` moving with a velocity `v`, if enters in a uniform magnetic field `B` acting, then it moves in a circular path of radius `r`.
The speed will be uniform and the direction of F (force) is perpendicular to the plane in which the charged particle is moving,
Question 12
1 / -0

The working of a dynamo is based on the principle of

A
electromagnetic induction

B
magnetic effect of current

C
heating effect of current

D
chemical effect of current

Solution

The generation of electrical energy by a Dynamo is based on a principle of magnetism called electromagnetic induction. When the lines of force that pass from the north to the south pole of a magnet are cut by a wire, there is current induced in the wire. That is, if we take a loop or coil of wire which has no current in it and a magnet which also has no current, and move the loop or coil between the poles, loops or coils are used instead of one loop, current may be generated continuously. This method of generating electric current is called electromagnetic induction.
Question 13
1 / -0

The output of a Transformer is:

A
dc voltage

B
ac voltage

C
both ac and dc voltages

D
none of the above

Solution

A transformer is an electrical device that is used to change the voltage in alternating current (AC) electrical circuits. The ability of this device to change voltage is a major advantage of AC electricity over direct current (DC).
Question 14
1 / -0

To convert a galvanometer into an ammeter, which of the following should be done?

A
A high resistance should be connected in parallel to it.

B
A low resistance should be connected in parallel to it.

C
A high resistance should be connected in series to it.

D
A low resistance should be connected in series to it.

Solution

In order to convert a galvanometer into an ammeter, a very low resistance known as "Shunt" resistance is connected in parallel to galvanometer. The value of shunt is so adjusted that most of the current passes through the shunt. In this way, a galvanometer is converted into an ammeter and can measure heavy currents without fully deflected.
Question 15
1 / -0

A current is flowing north along a power line. The direction of the magnetic field above it, neglecting the earth's field, is directed towards

A
north

B
south

C
east

D
west

Solution

According to the right hand rule, when the current is flowing upwards (north), the magnetic field will be in an anti-clockwise direction. According to this question, magnetic field will be directed towards east.
Question 16
1 / -0

Two particles, each of mass m and carrying charge Q, are separated by some distance. If they are in equilibrium under mutual gravitational and electrostatic forces, what is the order of Q/m?

A
10^–5

B
10^–10

C
10^-15

D
10^–20

Solution

On comparing gravitational force and electrical force, we get so = 10^-10 approximately
Question 17
1 / -0

The meter which measures the consumption of electric energy in houses is called a

A
galvanometer

B
multimeter

C
kilowatt hour meter

D
potentiometer

Solution

Kilowatt hour meter (kWh) is an electric meter that measures the amount of electrical energy in kWh consumed in the houses. The cost of electricity bill is calculated by multiplying the number of kWh that was consumed at the cost of 1 kWh.
Question 18
1 / -0

The peak voltage of the AC supplied in India is

A
220/ Volt

B
/220 Volt

C
× 220 Volt

D
none of the above

Solution

In India, residential power supplied is two-wire 220 volts. Peak value is the maximum value and this is equal to times the RMS value. So, the peak voltage in India is × 220.
Question 19
1 / -0

The value of internal resistance of an ideal cell is

A
zero

B
0.5

C
1

D
infinite

Solution

The reason of the battery failure is typically because it has built up enough internal resistance that it can no longer supply a useful amount of power to an external load. That is why, the value of internal resistance of an ideal battery or a cell is zero.
Question 20
1 / -0

A long magnetic needle of length L, having a magnetic moment M and pole strength m units, is broken into two at the middle. The magnetic moment and pole strength of each piece will be

A
(M/2), m/2

B
M, m/2

C
M/2, m

D
M, m

Solution

A long magnetic needle of length L, having a magnetic moment M and pole strength m units, is broken into two at the middle. The magnetic moment and pole strength of each piece will be M/2 , m respectively