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# Book Back Questions with Answers

I. Choose the best answer II. Fill in the blanks III. State whether the following statements are true or false: If false correct the statement IV. Match the items in column-I to the items in column-II V. Assertion and reason type questions VI. Very short answer questions VII. Short answer questions VIII. Long answer questions. IX. Numerical problems XI. HOTS

### I. Choose the best answer

1. Which of the following is correct?

a) Rate of change of charge is electrical power.

b) Rate of change of charge is current.

c) Rate of change of energy is current.

d) Rate of change of current is charge.

2. SI unit of resistance is

a) mho

b) joule

c) ohm

d) ohm meter

3. In a simple circuit, why does the bulb glow when you close the switch?

a) The switch produces electricity.

b) Closing the switch completes the circuit.

c) Closing the switch breaks the circuit.

d) The bulb is getting charged.

4. Kilowatt hour is the unit of

a) resistivity

b) conductivity

c) electrical energy

d) electrical power

### II. Fill in the blanks

1. When a circuit is open, current cannot pass through it.

2. The ratio of the potential difference to the current is known as resistance.

3. The wiring in a house consists of parallel circuits.

4. The power of an electric device is a product of voltage and current.

5. LED stands for Light Emitting Diode.

### III. State whether the following statements are true or false: If false correct the statement.

1. OhmŌĆÖs law states the relationship between power and voltage. - False

Ohm's law states the relationship between current and voltage.

2. MCB is used to protect house hold electrical appliances. - True

3. The SI unit for electric current is the coulomb. - False

The SI unit for electric current is the ampere

4. One unit of electrical energy consumed is equal to 1000 kilowatt hour. - False

One unit of electrical energy consumed is equal to kilowatt hour

5. The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistances. - False

The effective resistance of three resistors connected in series is greater than the highest of the individual resistances.

### IV. Match the items in column-I to the items in column-II: Column - I Column - II

(i) electric current (a) volt

(ii) potential difference (b) ohm meter (iii) specific resistance (c) watt

(iv) electrical power (d) joule

(v) electrical energy (e) ampere

(i) electric current - ampere

(h) potential difference - volt

(iii) specific resistance - ohm meter

(iv) electrical power - watt

(v) electrical energy - joule

### V. Assertion and reason type questions:

Mark the correct choice as

a) if both the assertion and the reason are true and the reason is the correct explanation of the assertion.

b) if both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

c) if the assertion is true, but the reason is false.

d) if the assertion is false, but the reason is true.

1. Assertion: Electric appliances with a metallic body have three wire connections.

Reason: Three pin connections reduce heating of the connecting wires

Answer: c) if the assertion is true, but the reason is false.

2. Assertion: In a simple battery circuit the point of highest potential is the positive terminal of the battery.

Reason: The current flows towards the point of the highest potential

Answer: c) if the assertion is true, but the reason is false.

3. Assertion: LED bulbs are far better than incandescent bulbs.

Reason: LED bulbs consume less power than incandescent bulbs.

Answer: a) if both the assertion and the reason are true and the reason is the correct explanation of the assertion.

### VI. Very short answer questions.

1. Define the unit of current.

The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross section of a conductor, in one second. Hence,

1 ampere = 1 coulomb / 1 second

2. What happens to the resistance, as the conductor is made thicker?

(i) Decreases : The resistance decreases as the conductor is made thicker.

(ii) Reason: Resistance is inversely proportional to area of cross section A.

i. e., R ╬▒ 1/A

here, A = ŽĆr2

Where, r is the radius which determines the thickness.

3. Why is tungsten metal used in bulbs, but not in fuse wires?

Tungsten has high melting point, it can bear high heat for glowing. But in fuse wire, the wire used in it should melt. So a metal (wire) which has low melting point should be used in a fuse wire, but not tungsten wire.

4. Name any two devices, which are working on the heating effect of the electric current.

Electric iron, and electric toaster, or electric oven, and electric heater.

1. Define electric potential and potential difference.

Electric potential : The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.

Potential difference : The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.

2. What is the role of the earth wire in domestic circuits?

The earth wire provides a low resistance path to the electric current. The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of metallic electric appliance. Thus, the earth wire serves as a protective conductor, which saves us from electric shocks.

3. State OhmŌĆÖs law.

OhmŌĆÖs law states that, at a constant temperature, the steady current 'I' flowing through a conductor is directly proportional to the potential difference 'V', between two ends of the conductor. I ŌłØ V

I = (1/R) V

V =IR

4. Distinguish between the resistivity and conductivity of a conductor.

Resistivity

(i) Electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section.

(ii) Its unit is ohm metre.

(iii) Electrical resistivity of a conductor is a measure of the resisting power of a specified material to the passage of an electric current. It is a constant for a given material.

Conductivity

(i) The reciprocal of electrical resistivity of a material is called its electrical conductivity.

(ii) It's unit is mho metre-1.

(iii) Electrical conductivity of a conductor is a measure of its ability to pass the current through it.

5. What connection is used in domestic appliances and why?

(i) All the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit.

(ii) One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.

1. With the help of a circuit diagram derive the formula for the resultant resistance of three resistances connected: a) in series and b) in parallel Resistors in series

1. 2. R1, R2 and R3 are connected in series. Here current through them is same, but voltage is different

3. i.e., V = V1+ V2 + V3

Here, V = IR

V1 = IR1

V2 = IR2

V3=IR3

Ōł┤IR = IR1 + IR2 + IR3

IR = I [R1 + R2 + R3]

i.e., RS = R1 + R2 + R3

4. When a number of resistors are connected in series, their effective resistance is equal to the sum of the individual resistances.

Ie., RS = nR

5. The effective resistance in a series combination is greater than the highest of the individual resistances.

Resistors in parallel

1. 2. R1, R2 and R3 are connected in parallel. Here current through them is different, but voltage is same.

3. i.e., I = I1 + I2 + I3

Here, I = V / R

I1 = V / R1

I2 = V / R2

I3 = V / R3

Ōł┤V/R = V/R1 + V/R2 + V/R3

V(1/R) = V(1/R1 + 1/R2 + 1/R3)

i.e., 1/RP = 1/R1 + 1/R2 + 1/R3

4. When a number of resistors are connected in parallel, the reciprocal of the effective resistance is equal to the sum of the reciprocals of the individual resistances

i.e., 1/RP = n/R

or

RP = R/n

5. The equivalent (or) effective resistance in a parallel combination is less than the lowest of the individual resistances.

2. a) What is meant by electric current?

b) Name and define its unit.

c) Which instrument is used to measure the electric current? How should it be connected in a circuit?

a) Electric current is defined as the rate of flow of charges in a conductor. If a net charge 'Q' passes through any cross section of a conductor in time 't', then the current flowing through the conductor

is I = Q / t.

b) The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross section of a conductor, in one second. Hence,

1 ampere = 1 coulomb / 1 second

c) Ammeter is used to measure electric current. It should be connected in series in a circuit.

3. a) State JouleŌĆÖs law of heating.

b) An alloy of nickel and chromium is used as the heating element. Why?

c) How does a fuse wire protect electrical appliances?

a) Joule's law of heating states that the heat produced in any resistor is:

(i) Directly proportional to the square of the current passing through the resistor.

(ii) Directly proportional to the resistance of the resistor.

(iii) Directly proportional to the time for which the current is passing through the resistor.

b) Alloy of nickel and chromium have the following properties:

(i) It has high resistivity,

(ii) It has a high melting point,

(iii) It is not easily oxidized.

So it is used as the heating element.

c) The fuse wire is connected in series, in an electric circuit. When a large current passes through the circuit, the fuse wire melts due to Joule's law of heating and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage.

4. Explain about domestic electric circuits. (circuit diagram not required)

(i) The first stage of domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer. The important components of the main - box are (i) a fuse box and (ii) a meter.

(ii) The meter is used to record the consumption of electrical energy. The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).

(iii) The function of the fuse wire or a MCB is to protect the house hold electrical appliances from overloading due to excess current.

(iv) The electricity is brought to houses by two insulated wires, out of these two wires, one wire has a red insulation and is called the 'live wire'. The other wire has a black insulation and is called the 'neutral wire'.

(v) The electricity supplied to house is actually an alternating current having an electric potential of 220 V. Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire.

(vi) After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required.

(vii) After the main switch, these wires are connected to live wires of two separate circuits. Out of these two circuits, one circuit is of a 5A rating, which is used to run the electric appliances with a lower power rating, such as tube lights, bulbs and fans. The other circuit is of a 15A rating, which is used to run electric appliances with a high power rating. Such as A/C. refrigerators, electric iron and heaters.

5. a) What are the advantages of LED TV over the normal TV?

b. List the merits of LED bulb.

a) Advantages of LED TV :

(i) It has brighter picture quality.

(ii) It is thinner in size

(iii) It uses less power and consumes very less energy.

(iv) Its life span is more.

(v) It is more reliable.

b) Merits of LED bulb :

(i) As there is no filament, there is no loss of energy in the form of heat. It is cooler than the incandescent bulbs.

(ii) In comparison with the fluorescent light, the LED bulbs have significantly low power requirement.

(iii) It is not harmful to the environment.

### IX. Numerical problems:

1. An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case?

Given

In the first case

The power consumed by an electric iron P = 420 W

Applied voltage

i.e potential difference, V = 220 V

To find : Current, I = ? (in each case)

Solution

P = V I =>  I = P/ V = 420/220 = 21/11

When heating at maximum rate, I = 21/11 ampere.

(or) I = 1.909A.

Heating at minimum rate, the rate at which energy is consumed, P = 180 W

Applied voltage, V = 220 V

I = P/V

I = 180/220

9/11 = 0.818 A

When heating at minimum rate,

1= 9/11 A (or) I = 0.818 A.

2. A 100 watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January.

Energy consumed by Electric bulb of 100 W = ?

Time, t = 5 hr

For month January = 31 days

Number of 60W bulbs = 4

60W used for a time = 5 hr

To find : Total Energy consumed (in kWh) = ?

Solution

Energy consumed by 100W bulb

= Power of the bulb x time

= 100 x 31 x 5 = 15500

Energy consumed by 100 W = 15.5 kwh

Energy consumed by four 60 W bulb = 4x 60 x5x31 = 37,200 wh

Four 60 W bulb = 37.2 kwh

Total energy consumed = Energy consumed by 100 W + Energy consumed by four 60 W

= 15.5 + 37.2

Total energy consumed = 52.7 kwh

3. A torch bulb is rated at 3 V and 600 mA. Calculate itŌĆÖs

a) power

b) resistance

c) energy consumed if it is used for 4 hour.

Given

The Potential of torch bulb V = 3V.

Current I = 600ma

= 600 x 10-3 = 0.6A

To find :

a) Power, P      = ?

b) Resistance, R = ?

c) Energy consumed = ?

Solution

a) P = VI =3x0.6= 1.8W

V = I R => R = V/I

b) Resistance, R = 3/06 = 50

c) Energy consumed for 4hr

= power of the bulb x time = 1.8 x 4

Energy, E = 7.2 wh

4. A piece of wire having a resistance R is cut into five equal parts.

a) How will the resistance of each part of the wire change compared with the original resistance?

b) If the five parts of the wire are placed in parallel, how will the resistance of the combination change?

c) What will be ratio of the effective resistance in series connection to that of the parallel connection?

Given

a) Resistance of a piece of wire is proportional to its length i.e R ŌłØ l

Each piece has a resistance = R

Solution

Wire is cut into five equal parts

Resistance of each part = R/5

Resistance = R/5

b) All the five parts are connected in parallel. For parallel connection, the effective resistance is

1/Rp = 1/R1 + 1/R2

1/Rp = 5/R + 5/R + 5/R + 5/R + 5/R

1/Rp = 25/R

Rp = R/25

c) The ratio of the effective resistance in series to parallel connection.

1/Rp = 25/R

Rs/Rp = 25/1

Ratio of Rs and Rp = 25:1

Rs:Rp = 25:1

### XI. HOTS:

1. Two resistors when connected in parallel give the resultant resistance of 2 ohm; but when connected in series the effective resistance becomes 9 ohm. Calculate the value of each resistance.

Effective resistance in parallel, Rp = 2╬®

Effective resistance in series, R. =9 ╬® R12 - 9R1 + 18 = 0

(R1 - 3) (R1 - 6) = 0.

R1 = 3 (or) 6 ╬®

When R1 = 3 ╬®, R2 = 9-3=6 ╬®

When R1 = 6 ╬®, R2= 9-6=3 ╬®

2. How many electrons are passing per second in a circuit in which there is a current of 5 A?

Given

Current I = 5 A

To find :

Number of electrons passing per second, for 5 A, n = ?

Solution

I = Q/t

Charge of an electron, e = 1.6 x 10-19 coulomb

Then 1 coulomb = 1 / 1.6x10-19 electrons

i.e., 1 coulomb = 6.25 x 1018 electrons

For 1 ampere current, no. of electrons will be 6.25 x 1018 electrons. For 5 ampere current, it is = 5 x 6.25 x 1018 electrons

= 3.125 x 1019 electrons

Method 2:

No. of electrons passing

per second, n = ?

I = Q/t

Where e -> Charge of electron which is equal to 1.6 x 10-19 C

Q is ŌĆśnŌĆś no of charges i.e Q = ne

Here t = 1S

I = ne/t

n = [ I x t ]/ e = [5/1.6]  x 1019

Number of Electrons, n = 3.125 x 1019 electrons

3. A piece of wire of resistance 10 ohm is drawn out so that its length is increased to three times its original length. Calculate the new resistance.

Resistance of wire, R = 10 ╬®

Length is increased to thrice = 3 L

To find : New resistance = ?

Solution

Resistance (R) = [ resistivity (Žü) * length (L)  ] / area (A)

R = ŽüL/A

10 = ŽüL/A

When length is increased by three times (3L) the area of cross section is reduced by three times (A/3).

New length = 3L

New Area = A/3

New resistance R = [ Žü.3L ] / [A/3] = 9 (ŽüL/A)

R = 9.R

9 x 10 = 90 ╬®

R= 90 ╬® Tags : Electricity | Science , 10th Science : Chapter 4 : Electricity
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