10th Science - Book Back Answers - Physics Unit 4 - English Medium Guides

 

SSLC / 10th - Science - Book Back Answers - Physics Unit 4 - English Medium

Tamil Nadu Board 10th Standard  Science - Physics Unit 4: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Unit 4 – Physics from the Tamil Nadu State Board 10th Standard  Science textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Physics Unit 4 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Unit 4 :Electricity

I. Choose the correct 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.
Answer Key:
(b) Rate of change of charge is current.

2. SI unit of resistance is:
(a) mho
(b) joule
(c) ohm
(d) ohm meter
Answer Key:
(c) ohm
 
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.
Answer Key:
(b) Closing the switch completes the circuit

4. Kilowatt hour is the unit of:
(a) resistivity
(b) conductivity
(c) electrical energy
(d) electrical power
Answer Key:
(c) electrical energy

II. Fill in the blanks

1. When a circuit is open, ……….. cannot pass through it.
2. The ratio of the potential difference to the current is known as ……….
3. The wiring in a house consists of ………… circuits.
4. The power of an electric device is a product of ……… and ………..
5. LED stands for ………..
Answer Key:
1. current
2. resistance
3. parallel
4. potential difference, current
5. Light Emitting Diode

III. State whether the following statements are true or false. correct the statement if it is false.

1. Ohm’s law states the relationship between power and voltage.
2. MCB is used to protect house hold electrical appliances.
3. The SI unit for electric current is the coulomb.
4. One unit of electrical energy consumed is equal to 1000 kilowatt hour.
5. The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistances.
Answer Key:
1. False – Ohm’s law states that the relationship between current and voltage.
2. True
3. False – The SI unit for electric current is ampere.
4. False – One unit of electrical energy consumed is equal to 1 kilowatt hour.
5. False – The effective resistance of three resistors connected in series is greater than the highest of the “individual resistance.

IV.  Match the following  

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

Answer Key:

Column - I	Column - II
(i) electric current	(e) ampere
(ii) potential difference	(a) volt
(iii) specific resistance	(b) ohm meter
(iv) electrical power	(c) watt
(v) electrical energy	(d) joule

 

.

V.  Assertion & Reasoning

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 Key:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

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 Key:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

3. Assertion: LED bulbs are far better than incandescent bulbs.
Reason: LED bulbs consume less power than incandescent bulbs.
Answer Key:
(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

Question 1.
Define the unit of current.
Answer Key:
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,
$\mathbf{1}\mathbf{ }\mathbf{a}\mathbf{m}\mathbf{p}\mathbf{e}\mathbf{r}\mathbf{e}\mathbf{=}\frac{\mathbf{1}\mathbf{ }\mathbf{coulomb}}{\mathbf{1}\mathbf{ }\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{d}}$

Question 2.
What happens to the resistance, as the conductor is made thicker?
Answer Key:
If the conductor is made a thicker area of cross-section of conduction increases that will decrease the resistance.

Question 3.
Why is tungsten metal used in bulbs, but not in fuse wires?
Answer Key:
Tungsten metal is used in bulbs because its melting point is the greatest.
But it is not used in fuse wires. When a current more than 5A flows tungsten wire will be melted. Hence tungsten is not used in fuse wire.

Question 4.
Name any two devices, which are working on the heating effect of the electric current.
Answer Key:
The heating effect of electric current is used in many home appliances such as electric iron and electric toaster.

VII. Short answer questions

Question 1.
Define electric potential and potential difference.
Answer Key:
Electric Potential: It is the amount of work done in moving unit positive charge from infinity to that point against the electric force.
Electric potential
 $\mathbf{=}\frac{\mathbf{W}\mathbf{o}\mathbf{r}\mathbf{k}\mathbf{ }\mathbf{d}\mathbf{o}\mathbf{n}\mathbf{e}}{\mathbf{C}\mathbf{h}\mathbf{a}\mathbf{r}\mathbf{g}\mathbf{e}}\mathbf{\Rightarrow }\mathbf{V}\mathbf{=}\frac{\mathbf{W}}{\mathbf{Q}}$
Potential difference: It is the amount of work done in moving a unit positive charge from one point to another against the electric force.
Potential difference V_A – V_B = $\frac{{\mathbf{W}}_{\mathbf{A}}\mathbf{-}{\mathbf{W}}_{\mathbf{B}}}{\mathbf{Q}}$
 

Question 2.
What is the role of the earth wire in domestic circuits?
Answer Key:
1. The earth wire provides a low resistance path to the electric current.
2. The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of the metallic electric appliance.
3. Thus, the earth wire serves as a protective conductor, which saves us from electric shocks.

Question 3.
State Ohm’s law.
Answer Key:
According to Ohm’s law, at a constant temperature, the steady current ‘I’ flowing through a conductor is directly proportional to the potential difference ‘V’ between the two ends of the conductor.
I ∝ V. Hence, $=\frac{1}{\mathrm{v}}$ = constant.
The value of this proportionality constant is found to $=\frac{1}{\mathrm{R}}$
Therefore, I = $=\left(\frac{1}{\mathrm{R}}\right)V$
V = IR

Question 4.
Distinguish between the resistivity and conductivity of a conductor.
Answer Key:
Resistivity    Conductivity
It is the resistance of a coducts of unit length and unit are     

Resistivity	Conductivity
It is the resistance of a coducts of unit length and unit area of cross section.	It is the reciprocal of electrical resistivity
It’s unit is ohm metre	It’s unit is mho metre-1
$\mathbf{\rho }\mathbf{=}\frac{\mathbf{R}\mathbf{A}}{\mathbf{L}}$	$\mathbf{\sigma }\mathbf{=}\frac{\mathbf{1}}{\mathbf{\rho }}$

 
   
Question 5.
What connection is used in domestic appliances and why?
Answer Key:
1. In the domestic appliance, it is used as a parallel connection to avoid short circuit and breakage.
2. It has an alternative current (AC). Not DC current as it is from cables, so high potential flows through this.
3. One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.

VIII. Long answers questions

Question 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
Answer Key:
(a)  Resistors in series : A series circuit connects the components one after the other to form a ‘single loop’. A series circuit has only one loop through which current can pass. If the circuit is interrupted at any point in the loop, no current can pass through the circuit and hence no electric appliances connected in the circuit will work. Series circuits are commonly used in devices such as flashlights. Thus, if
resistors are connected end to end, so that the same current passes through each of them, then they are said to be connected in series.

Let, three resistances R₁, R₂ and R₃ be connected in series. Let the current flowing through theorem be I. According to Ohm’s Law, the potential differences V₁,V₂ and V₃ across R₁, R₂ and R₃ respectively, are given by:
V₁ = I R₁ ………. (1)
V₂ = I R₂ ……… (2)
v₃ = I R₃ ………. (3)
The sum of the potential differences across the ends of each resistor is given by:
V = V₁ + V₂ + V₃
Using equations (1), (2) and (3), we get
V = I R₁ + I R₂ + I R₃ …….. (4)
The effective resistor is a single resistor, which can replace the resistors effectively, so as to allow the same current through the electric circuit. Let, the effective resistance of the series-combination of the resistors, be RS.
Then,
V = I RS ……….(5)
Combining equations (4) and (5), we get,
I R_S = I R₁ + I R₂ + I R₃
R_S = R₁ + R₂ + R₃ ……….. (6)
Thus, you can understand that when a number of resistors are connected in series, their equivalent resistance or effective resistance is equal to the sum of the individual resistances. When ‘n’ resistors of equal resistance R are connected in series, the equivalent resistance is ‘n R’.
i.e., R_S = n R
The equivalent resistance in a series combination is greater than the highest of the individual resistances.
(b) Resistors in Parallel : A parallel circuit has two or more loops through which current can pass. If the circuit is disconnected in one of the loops, the current can still pass through the other loop(s). The wiring in a house consists of parallel circuits.
 

Consider that three resistors R₁, R₂ and R₃ are connected across two common points A and B The potential difference across each resistance is the same and equal to the potential difference between A and B. This is me sured using the voltmeter. The current I arriving at A divides into three branches I1, I2 and I3 passing through R₁, R₂ and R₃ respectively.
According to the Ohm’s law, you have,
${\mathbf{I}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{1}}}$ ....(1)

${\mathbf{I}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{2}}}$ ....(2)

${\mathbf{I}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{3}}}$ ....(3)
The total current through the circuit is given by
I = I₁ + I₂ + I₃
Using equations (1), (2) and (3), you get
I = $=\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{1}}}+\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{2}}}+\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{3}}}$ ....(4)
Let the effective resistance of the parallel combination of
 resistors be RP Then,
I = $=\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{p}}}$ ……… (5)
Combining equations (4) and (5), you have
$\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{p}}}=\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{1}}}+\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{2}}}+\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{3}}}$

$\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{p}}}=\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}+\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}+\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{3}}}$
Thus, when a number of  resistors are connected in parallel, the sum of the reciprocals of the individual resistances is equal to the reciprocal of the effective or equivalent resistance. When ‘n’ resistors of equal resistances R are connected in parallel, the equivalent resistance is $\frac{\mathbf{R}}{\mathbf{n}}$
i.e., $\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{p}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{R}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{R}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{R}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\frac{\mathbf{1}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{R}}$

Hence, ${\mathbf{R}}_{\mathbf{p}}=\frac{\mathbf{R}}{\mathbf{n}}$
The equivalent resistance in a parallel combination is less than the lowest of the individual resistances.

Question 2.
(a) What is meant by electric current? Give its direction?
(b) Name and define its unit.
(c) Which instrument is used to measure the electric current? How should it be r connected in a circuit?
Answer Key:
(a) (i) Electric current is often termed as ‘current’ and it is represented by the symbol ‘I’. It is defined as the rate of flow of charges in a conductor.
(ii) The electric current represents the number of charges flowing in any cross-section of a conductor (say a metal wire) in unit time.

(b) The SI unit of electric current is ampere (A).
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 = $\frac{\mathbf{1}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{u}\mathbf{l}\mathbf{o}\mathbf{m}\mathbf{b}}{\mathbf{1}\mathbf{ }\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{d}}$

$\frac{\mathbf{ }}{}$
(c) (i) The ammeter is used to measure the current.
(ii) An Ammeter is connected in series with the circuit.
(iii) The Ammeter is a low impedance device connecting it in parallel with the circuit would cause a short circuit, damaging the Ammeter or the circuit.

Question 3.
(a) State Joule’s 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?
Answer Key:

 
(a) Joule’s law of heating states that the heat produced in any resistor is:
1. directly proportional to the square of the current passing through the resistor.
2. directly proportional to the resistance of the resistor.
3. directly proportional to the time for which the current is passing through the resistor.

(b) Because,
1. It has high resistivity,
2. It has a high melting point,
3. It is not easily oxidized.

(c) When a large current passes through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage.
 
Question 4.
Explain about domestic electric circuits, (circuit diagram not required)
Answer Key:
1. Electricity is distributed through the domestic electric circuits wired by the electricians.
2. The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer.
3. The important components of the main-box are: (i) a fuse box and (ii) a meter. The meter is used to record the consumption of electrical energy.
4. The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).
5. The function of the fuse wire or an MCB is to protect the household electrical appliances from overloading due to excess current.

Question 5.
(a) What are the advantages of LED TV over the normal TV?
(b) List the merits of LED bulb.
Answer Key:

 
(a) Advantages of LED TV:

 
1. It has brighter picture quality.
2. It is thinner in size.
3. It uses less power and consumes very less energy.
4. Its life span is more.
5. It is more reliable.

(b) Advantages of LED bulb:
1. As there is no filament, there is no loss of energy in the form of heat.
2. It is cooler than the incandescent bulb.
3. In comparison with the fluorescent light, the LED bulbs have significantly low power requirement.
4. It is not harmful to the environment.
5. A wide range of colours is possible here.
6. It is cost-efficient and energy efficient.
7. Mercury and other toxic materials are not required. One way of overcoming the energy crisis is to use more LED bulbs.

IX. Numerical Problems

Question 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?
Answer Key:
(i) When heating is maximum, the power
P₁ = 420 W
Applied voltage V = 220 V
P = VI
Current I = $=\frac{\mathbf{p}}{\mathbf{v}}$
I = $=\frac{\mathbf{420}}{\mathbf{220}}$ = 1.909 A
I = 1.909 A
(ii) When heating is minimum
Power P₂ = 180 W
Applied voltage V = 220 V
P = VI
∴ Current I = $=\frac{\mathbf{p}}{\mathbf{v}}$
I = $=\frac{\mathbf{180}}{\mathbf{220}}$ = 0.8181 A
I = 0.8181 A

 

Question 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.
Answer Key:
100 W = 100 joules per second
1 watt hours = 3600 joules
The electric bulb is lighted for 5 hours daily,
100 W × 5 = 500 watt hours
500 watt hours = 1800000 joules
1 kWh = 3600000 joules
Units consumed per day  $=\frac{\mathbf{1800000}}{\mathbf{3600000}}$ = 0.5 units
Untis consumed in month = 0.5 × 31 = 15.5 units …. (1)
Now, Sum of power of four 60 watt bulbs = 240 W
240 W × 5 hours = 1200 watt hours
1200 watt hours = 4320000 joules
Energy consumed per day = $\frac{\mathbf{4320000}}{\mathbf{3600000}}$ = 1.2 units
Energy consumed in a month = 1.2 × 31 = 37.2 units …. (2)
Total energy consumed in a month = 15.5 + 37.2 = 52.7 units
1 unit = 1 kWh
The energy consumed in the month of January = 52.7 kWh.

 

Question 3.

A torch bulb is rated at 3 V and 600 mA. Calculate it’s
(a) power
(b) resistance
(c) energy consumed if it is used for 4 hour.
Answer Key:
Voltage V = 3 V
Current I = 600 mA
(a) Power = VI
= 3 × 600 × 10^-3
= 1800 × 10^-3
= 1.8 W

(b) Power P=$\frac{{\mathrm{V}}^2}{\mathrm{R}}$

$\therefore \mathrm{Resis}tan\mathrm{ce} \mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{(3{)}^2}{1.8}$

$\mathrm{R}=\frac{9}{1.8}=\frac{1}{0.2}=5\Omega$
(c) Time = 4h
Energy consumed E = P × t
E = 1.8 × 4
= 7.2 W

 

Question 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?
Answer Key:
(a) Original resistance
R = $\mathrm{R}=\frac{\mathrm{l}}{\mathrm{A}}$
∴ R α l
After cutting length of each piece
r = $\mathrm{r}=\frac{\mathrm{l}}{5}$
New resistance
R’ = $\mathrm{R}\text{'}=\frac{\mathrm{l}\text{'}}{\mathrm{A}}$
R’ α l’

$\frac{\mathrm{R}}{\mathrm{R}\text{'}}=\frac{\mathrm{l}}{\mathrm{l}\text{'}}=\frac{\mathrm{l}}{\frac{\mathrm{l}}{5}}=5$
R : R’ = 5 : 1

(b) When five parts of the wire are placed in parallel.
Effective Resistance
 $\frac{1}{R_p}=\frac{1}{\frac{\mathrm{R}}{5}}+\frac{1}{\frac{\mathrm{R}}{5}}+\frac{1}{\frac{\mathrm{R}}{5}}+\frac{1}{\frac{\mathrm{R}}{5}}+\frac{1}{\frac{\mathrm{R}}{5}}$

$\frac{1+1+1+1+1}{\frac{\mathrm{R}}{5}}=\frac{5}{\frac{\mathrm{R}}{5}}$

$\frac{1}{R_p}=\frac{25}{R}$

${\mathrm{R}}_{\mathrm{p}}=\frac{\mathrm{R}}{25}$

Resistance of the combinations
${\mathrm{R}}_{\mathrm{p}}=\frac{R}{25}$

(c) When resistance are connected in series
 $\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}$
R_S = R
When resistance are connected in parallel
${\mathrm{R}}_{\mathrm{p}}=\frac{R}{25}$
R_S : R_P  = R : ${\mathrm{R}}_{\mathrm{p}}=\frac{R}{25}$
= 25 R : R_1
R_S : R_P  = 25 : 1

 

X.  Hot questions

Question 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.
Answer Key:
Let the resistance be R, and R₂ when two resistances are connected in series
R_S = R₁ + R₂
= 9
R₁ + R₂ = 9 ……….(1)
When two resistance are connected in parallel
 $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2}$

$\frac{R_1+R_2}{R_1 R_2}=\frac{1}{2}$
Using (1) equation (2) becomes
$\frac{9}{R_1 R_2}$
R₁R₂ = 18 ……….(3)
(R₁ – R₂)² = (R₁ + R₂)² – 4R₁R₂
= (9)² – 4 × 18
= 81 – 72 = 9
∴ (R₁ – R₂) = √9 = 3 ………(4)
From (1)
R₁ + R₂ = 9
2R₁ = 12
∴ R₁ = 122 = 6 ohm
From (1)
R₂ = 9 – R₁
= 9 – 6 = 3Ω
The values of resistances are
R₁ = 6 ohm
R₂ = 3 ohm

Question 2.
How many electrons are passing per second in a circuit in which there is a current of 5 A?
Answer Key:
Current I = 5A
Time (t) = 1 second
Charge of electron e = 1.6 × 10^-19 C
$=\frac{q}{t}=\frac{ne}{t}$
Number of electron, $\mathrm{n}=\frac{\mathrm{It}}{\mathrm{e}}=\frac{5\times 1}{1.6\times }$10^-19 $\frac{}{}$
n = 3.125 × 10¹⁹..

Question 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.
Answer Key:
Resistance R = 10Ω
Let l be the length of the wire R ∝ 1
When the length is increased to three times,
l’ = 3l
∴ New Resistance
R’ ∝ l’ ∝ 3l
$\therefore \frac{\mathrm{R}}{\mathrm{R}\text{'}}=\frac{\mathrm{l}}{3\mathrm{l}}=\frac{1}{3}$
∴ R’ = 3R
New resistance = 3 times the original resistance.