Science Class 10 Chapter 11 Question Answer

(i)6 V battery in series with 1 Ω 2 Ω resistor 

Total resistance: R=1+2=3Ω

Current in circuit:

Current through 2Ω resistor = 2 A

Power in 2Ω resistor :

P=I²R- (2)² x 2 =8 W

(ii) 4 V battery in parallel with 12 Ω and 2 Ω resistor

Voltage across each resistor in parallel = 4 V

Power in 2 Ω resistor:

Comparison
Power in case (i): 8 W
Power in case (ii): 8 W  

∴ The power used in the 2 Ω resistor is the same in both cases.

Q.15 Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer:-Step 1: Formula
The current drawn by each lamp is given by: I =\(\frac{P}{V}\)

where P = power of the lamp, and V= supply voltage.

Step 2: Current for 100 W lamp 

I₁= \(\frac{100}{220 }\)= 0.4545 A≈0.45 A

Step 3: Current for 60 W lamp

I₂= \(\frac{60}{220 }\)=0.2727 A≈0.27 A

Step 4: Total current

Since the lamps are connected in parallel, the total current is the sum of the currents:

I=I₁+ I₂=0.45+0.27=0.72 A

Q16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer: We know,
Energy consumed (E)=Power (P)×Time (t)

For TV set:
P= 250 W , t  = 1h
E= 250 × 1 = 250 Wh

For toaster: P=1200W,t=10 minutes= \(\frac{10}{60}\)h =\(\frac{1}{6}\)h

E=1200 x \(\frac{1}{6}\) = 200 Wh

Comparison: 250Wh > 200Wh

Therefore, the 250 W TV set in 1 hour uses more energy than the 1200 W toaster in 10 minutes.

Q17. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:

We know,

Power (rate of heat developed)= I²R

Here,

I=15 A, R= 8 Ω
P= I²R (15)² × 18
P=225 × 8= 1800W

Rate of heat developed in the heater = 1800 W (or 1.8 kW).

18. Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity
transmission

Answer: 

(a)Tungsten is used because it has a very high melting point (3380°C) and can withstand high temperatures without melting. It also emits light when heated to high temperature, making it suitable for lamp filaments.

(b) Alloys have higher resistivity than pure metals, so they produce more heat. They also do not oxidise or burn easily at high temperatures, which increases the life of the heating element.

(c) In a series circuit, the current is the same through all devices. If one appliance fails, the circuit is broken and all stop working. Also, different appliances need different current values, which cannot be supplied properly in series. Hence, domestic circuits use parallel arrangement.

(d) Resistance is inversely proportional to the area of cross-section of the wire.
R∝ \(\frac{1}{A}\)

(e) Copper and aluminium have very low resistivity and are good conductors of electricity. This allows easy flow of current with minimum power loss, which makes them ideal for transmission lines.

NCERT Intext Questions  for Class 10 Science Page Number: 172

Q1. What does an electric circuit mean?
Answer: An electric circuit means a continuous and closed path of an electric current.

Q2. Define the unit of current.
Answer: The unit of current is ampere (A). One ampere is the flow of one coulomb of charge per second, i.e.,
1 A = 1 C/1 s.

Q3. Calculate the number of electrons constituting one coulomb of charge.
Answer:

Charge on 1 electron = 1.6 × 10^-19
Therefore, number of electrons in 1 C charge = 1 /(1.6 × 10^-19 ) = 6× 10¹⁸  electrons (approximately)

NCERT Intext Questions  for Class 10 Science Page Number: 174

Q1. Name a device that helps to maintain a potential difference across a conductor.
Answer: A device that helps to maintain a potential difference across a conductor is a cell or a battery. The chemical action within the cell or battery produces the required potential difference.

Q2. What is meant by saying that the potential difference between two points is 1 V?
Answer: If the potential difference between two points is said to be 1 volt, it means that 1 joule of work is done in moving 1 coulomb of charge from one point to the other in an electric circuit.

Q3. 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 calculated as:

Energy= Potential difference × Charge

Here, potential difference = 6 V, charge = 1 C

Energy=6×1=6joules

So, each coulomb of charge receives 6 joules of energy.

NCERT Intext Questions  for Class 10 Science Page Number: 181

Q1. On what factors does the resistance of a conductor depend?
Answer: The resistance of a conductor depends on:
(i) its length,
(ii) its area of cross-section, and
(iii) the nature of its material.

Q2. 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: Current will flow more easily through a thick wire because resistance is inversely proportional to the area of cross-section. A thick wire has less resistance than a thin wire.

Q3. 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:According to Ohm’s law, current (I) = V/R.
If resistance is constant and potential difference is reduced to half, then the current will also become half of its initial value.

Q4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:Alloys have higher resistivity and do not oxidise (burn) readily at high temperatures. Therefore, they are preferred over pure metals in devices like electric toasters and irons.

Q5. Use the data in Table 11.2 to answer the following –

(a) Which among iron and mercury is a better conductor?
Answer: Iron is a better conductor than mercury because it has lower resistivity.

(b) Which material is the best conductor?
Answer: Silver is the best conductor as it has the lowest resistivity.

NCERT Intext Questions  for Class 10 Science Page Number: 185

Q 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:- The circuit shows three 2 V cells in series (total emf = 6 V) connected through a plug key to three resistors of 5 Ω, 8 Ω and 12 Ω arranged end-to-end (in series). The wire returns to the negative terminal of the battery forming a closed loop.

Q 2.Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer: – 

Step 1: Circuit Description

• • A battery of 3 cells of 2 V each → total = 6 V.
  • Three resistors 5 Ω, 8 Ω, 12 Ω connected in series.
  • Ammeter: connected in series with the resistors (so it shows the same current through all).
  • Voltmeter: connected in parallel across the 12 Ω resistor.

Step 2: Total Resistance 

\(R_total\) =5+8+12=25Ω

Step 3: Current in the Circuit (Ammeter Reading)

So, the ammeter reading = 0.24 A.

Step 4: Voltage Across 12 Ω Resistor (Voltmeter Reading)

V₁₂  =I×R=0.24×12= 2.88 V

So, the voltmeter reading = 2.88 V.

Ammeter Reading = 0.24 A
Voltmeter Reading = 2.88 V

NCERT Intext Questions  for Class 10 Science Page Number: 188

Q1. 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: We know, for resistors in parallel:

(a) For 1 Ω and 10⁶ Ω in parallel:

(b) For 1 Ω, 10³ Ω, and 10⁶ Ω in parallel:

(a) Equivalent resistance ≈ 1 Ω
(b) Equivalent resistance ≈ 1 Ω

Q2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer: We are given

Lamp resistance R1 ​=100 Ω
Toaster resistance, R2 ​=50 Ω
Water filter resistance,R3 =500 Ω
Voltage, 𝑉=220

Step 1: Current through each appliance

(using Ohm’s law, I=\(\frac{V}{R}\)

I₁= \(\frac{220}{100}\)=2.2A

I₂= \(\frac{220}{50}\)​=4.4A

I₃= \(\frac{220}{500}\)​=0.44A

Step 2: Total current taken by the three appliances

\( I_total \) = I₁+I₂+I₃
\( I_total \) = 2.2+4.4+0.44=7.04A

Step 3: Resistance of the electric iron
If the iron takes the same current (7.04 A) from the same voltage (220 V):

R= \(\frac{V}{I}\)= \(\frac{220}{7.04}\) ≈31.25Ω

Step 4: Current through the iron
I=7.04A

Resistance of the electric iron = 31.25 Ω
Current through the electric iron = 7.04 A

Q 3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer: When electrical devices are connected in parallel with the battery, each device gets the same potential difference as the battery, so all appliances work properly. If one device fails, the others continue to operate independently. In parallel circuits, each device draws the current according to its resistance without affecting the other devices, and the overall resistance of the circuit is reduced, which allows a proper current supply.

Q 4.How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Answer:-
(a) To get 4 Ω:

• First connect 3 Ω and 6 Ω in parallel.

\(\frac{I}R_p \)=\(\frac{1}{3}\) +\(\frac{1}{6}\) =\(\frac{2+1}{6}\)​= \(\frac{3}{6}\)= \(\frac{1}{2}\)

\( R_p \)=2Ω

• • Now connect this parallel resistance

\( R_p \)=2Ω in series with 2 Ω.

R =\( R_p \)+2-2+2-4 Ω
Thus, required combination is: 2 Ω in series with (3 Ω ∥ 6 Ω).

(b) To get 1 Ω:
Connect 2 Ω, 3 Ω, and 6 Ω all in parallel.

\(\frac{I}R_p \)=\(\frac{1}{2}\) +\(\frac{1}{3}\) +\(\frac{1}{6}\)

\(\frac{I}R_p \)=\(\frac{3+2+1}{6}\) –\(\frac{6}{6}\) – 1

\(\frac{I}R_p \)= 1Ω
Thus, required combination is: 2 Ω ∥ 3 Ω ∥ 6 Ω.

Q 5.What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer:- (a) The highest total resistance is obtained when the four coils are connected in series.

Total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) The lowest total resistance is obtained when the four coils are connected in parallel.

\(\frac{1}{R}\)= \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{12}\) +\(\frac{1}{24}\)=\(\frac{1}{2}\)

So, 𝑅=2Ω
Thus, highest resistance = 48 Ω and lowest resistance = 2 Ω.

NCERT Intext Questions  for Class 10 Science Page Number: 190

Q1: Why does the cord of an electric heater not glow while the heating element does?

Answer: – The cord of an electric heater is made of metals like copper or aluminium, which have very low resistance, so very little heat is produced in them. The heating element is made of an alloy having much higher resistance. Since heat produced is given by 𝐻= I²Rt for the same current, heat is directly proportional to resistance. Therefore, the heating element produces much more heat and glows, while the cord does not.

Q.2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer:
Given:
Charge, 𝑄=96000
Potential difference, 𝑉=50𝑉
Time, 𝑡=1 ℎ𝑜𝑢𝑟= 3600 𝑠

2.We know that:
Work done (or Heat generated),
𝐻=𝑉×𝑄

Substituting the values:
𝐻=50×96000
𝐻=48,00,000𝐽

The heat generated is 4.8 × 10⁶ joules.

Q3: An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30s.

Answer: – The amount of heat (H) produced is given by the joule’s law of heating as H= Vlt

Where,
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current x Resistance = 5 x 20 = 100 V
H= 100 x 5 x 30 = 1.5 x 10⁴ J.

Therefore, the amount of heat developed in the electric iron 1.5 x 10⁴ J.

NCERT Intext Questions  for Class 10 Science Page Number: 192

Q1. What determines the rate at which energy is delivered by a current?

Answer:- The rate at which energy is delivered by a current is determined by the electric power, which is given by the product of current and potential difference
(P = VI).

Q2.An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer:-

Power of the motor:
Formula: P = V × I
Where: V = 220 V (voltage), I = 5 A (current)
P = 220 V × 5 A = 1100 W

Energy consumed in 2 h:
Formula: Energy = P × t
Where: P = 1100 W (power), t = 2 h = 2 × 3600 s = 7200 s
Energy = 1100 W × 7200 s = 7,920,000 J
Converting to kJ: 7,920,000 J / 1000 = 7920 kJ

.: Power = 1100 W, Energy = 7920 kJ