# Free Electricity 01 Practice Test - 10th Grade

### Question 1

Ohm’s Law gives the relation between potential difference and current i.e, voltage is directly and linearly proportional to current.

True

False

#### SOLUTION

Solution :A

According to Ohm's law:

V=I×R; this means if R (resistance) is constant, current increases linearly as the voltage is increased.

At a constant temperature, following is the graph between voltage and current stated as Ohm's law.

From the above graph, we can say that the current varies linearly with potential difference.

### Question 2

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio RR′ is:

125

15

5

25

#### SOLUTION

Solution :D

Given the wire is cut into 5 equal parts.

Let the initial resistance of the wire be R

where R=ρlA→R∝l (length of the wire)

then, resistance of each one of the five parts =(R5)

Equivalent resistance (R') of resistors connected in parallel is given by 1R′=1R1+1R2..1Rn

Equivalent resistance (R') of five parts connected in parallel is given by:

1R′=1(R5)+1(R5)+1(R5)+1(R5)+1(R5)R′=R25

Therefore, RR′=25

### Question 3

1 kg piece of copper is drawn into a wire of cross-sectional area 1 mm^{2}, and then the same wire is remoulded again into a wire of cross-sectional area 2 mm^{2}. Find the ratio of the resistance of the first wire to the second wire.

#### SOLUTION

Solution :B

Volume of a solid is the product of length and cross-sectional area. V=AL

Since volume is same in both the cases, the length of second wire will be half of the length of 1st wire.

Accordingly, let length of 1st wire = L

area of 1st wire = A = 1mm2

So resistance R=ρLA=ρL1=ρL

Length of the second wire =L2 and

area of 2nd wire A=2 mm2

So resistance of 2nd wire R′=ρL2A

R′=ρL2×2

R′=14(ρL)

R′=14(R)

RR′=41

R : R' = 4 : 1

we get the resistance of the 1st wire is 4 times the value of resistance of the 2nd wire.

### Question 4

Resistance of alloys decreases with the increase in temperature.

True

False

#### SOLUTION

Solution :B

In alloys, the increase in temperature results in an increase of its resistance, but unlike pure metals, the increase is relatively small and irregular. As the temperature increases, the molecules in the atom vibrate with more frequency, they collide with each other more frequently which makes it tough for the movement of free electrons. Thus the resistance increase with increasing temperature in the metal.

### Question 5

Which of the following represents electrical power dissipated in an electrical circuit?

I2R

IR2

VI

V2R

#### SOLUTION

Solution :A, C, and D

Electrical power, P can be defined as the rate at which electrical energy (W) is consumed in an electrical circuit.

⇒P=Wt

Potential difference, V across any circuit is defined as the electrical energy W supplied by a source for a unit flow of charge (I×t)

∴V=WI×t

⇒W=V×I×tSo, power, P=V×I×tt=VI

By Ohm's law, V=IR or I=VR

Substituting the value of I in the expression of power, P=VI=V×VR=V2R

Since, V=IR, P=V2R=(IR)2R=I2R

Thus, electrical power P dissipated in a resistor =Wt=VI=V2R=I2R.

### Question 6

For a constant resistance, the correct relation between heat produced (H) and electric current (I) flowing is:

H∝I

H∝1I

H∝I2

H∝1I2

#### SOLUTION

Solution :C

According to Joule's law, the heat produced in a resistor is;

(i) directly proportional to the square of current for a given resistance.

(ii) directly proportional to resistance for a given current.

(iii) directly proportional to the time for which the current flows through the resistor.

That is H=I2×R×t

Hence H∝I2 is the right relation.

### Question 7

The physical quantity that determines the rate at which energy is delivered by an electric current is ___.

potential difference

electric power

resistance

electrical resistivity

#### SOLUTION

Solution :B

The physical quantity that determines the rate at which energy is delivered by an electric current is electric power. The SI unit of electric power is watt (W). 1 W electrical power is equivalent to 1 Joule of electrical energy consumed in 1 second time. It is mathematically expressed as P=Et where E is the electrical energy and t is time.

### Question 8

The resistivity of all pure metals increase with the rise in temperature.

True

False

#### SOLUTION

Solution :A

Resistivity depends on temperature. With the increase in temperature, the random motion of electrons increases. As a result, the number of collisions of electrons with the positive ions increases. Hence, the resistivity of all pure metals increases with the rise in temperature and decreases with the decrease in temperature.

### Question 9

Four wires each of the same length, diameter and material are connected to form a square. If the resistance of each wire is R, then equivalent resistance across the opposite corners is :

#### SOLUTION

Solution :A

The connection of wires is as shown below.

We have to calculate equivalent resistance across the opposite corners i.e. across A and D or B and C. If we take two corners B and C, then we can clearly see that the resistances of AB and AC, and CD and BD are in series.

So the equivalent resistance across CAB is 2R ( CA and AB are in series) and CDB is also 2R ( CD and BD are in series).

The two equivalent resistances across CAB and CDB are in parallel. Hence, the equivalent resistance across the two corners i.e. B and C is 1Req=12R+12R.

On solving we get, Req=R. Similarly, equivalent resistance across A and D will also be R.

Hence the equivalent resistance across the opposite corners is R.

### Question 10

How much electrical energy flows through a wire in 1 second when the power is 1 kW?

1400 J

1000 J

800 J

400 J

#### SOLUTION

Solution :B

We know that, Power=Energytime, the SI unit of power is watt, energy is joule and time is second.

So, 1 W = 1 joule per sec and 1 kW = 1000 joule per sec.

Therefore, 1000 J electrical energy flows through the wire in 1 second when the power is 1 kW.