Figure 13.1: Food chain in nature (a) in forest, (b) in grassland and (c) in a pond.Figure 13.2: Trophic levels pyramid.
Q2 (Page 212): What is the role of decomposers in the ecosystem?
The decomposers, which comprise microorganisms like bacteria and fungi, play the following crucial roles in an ecosystem:
They break down complex organic substances from dead remains and waste products of organisms into simple inorganic substances.
These simple inorganic substances go back into the soil and are reused by plants, thereby facilitating the natural replenishment of soil nutrients.
They act as natural cleansing agents by decomposing garbage and dead plants and animals.
Section 13.2.1: Ozone Layer and How It Is Getting Depleted
Q1 (Page 214): Why are some substances biodegradable and some non-biodegradable?
Substances are categorized as biodegradable or non-biodegradable based on the specificity of biological catalysts called enzymes:
Biodegradable substances: These are substances that can be broken down into simpler substances by biological processes carried out by bacteria and other saprophytes (decomposers) using specific enzymes. Examples include plant waste, paper, and food leftovers.
Non-biodegradable substances: These are substances that cannot be broken down by biological processes because decomposers do not possess specific enzymes to break them down. They persist in the environment for a long time and are only acted upon by physical processes like heat and pressure. Examples include plastics and synthetic chemicals.
Q2 (Page 214): Give any two ways in which biodegradable substances would affect the environment.
Two ways in which biodegradable substances can affect the environment are:
Foul smell and health hazards: During decomposition, large amounts of biodegradable waste produce foul odors and serve as breeding grounds for flies and insects that spread diseases.
Nutrient replenishment: Upon decomposition, they break down into simple inorganic nutrients that enrich the soil and maintain natural nutrient cycling in the ecosystem.
Q3 (Page 214): Give any two ways in which non-biodegradable substances would affect the environment.
Two ways in which non-biodegradable substances affect the environment are:
Environmental pollution and persistence: Non-biodegradable waste (like plastics) persists in the environment for a long time, causing soil degradation and clogging drains and water bodies.
Biological Magnification: Harmful non-biodegradable chemicals, such as pesticides, enter the food chain, accumulate progressively at each trophic level, and cause severe toxic effects in organisms, particularly top consumers like humans.
Section 13.2.2: Managing the Garbage We Produce
Q1 (Page 216): What is ozone and how does it affect any ecosystem?
What is ozone: Ozone ($\text{O}_3$) is a molecule formed by three atoms of oxygen. While oxygen gas ($\text{O}_2$) is essential for aerobic life, ozone is a deadly poison at ground level.
Effects on an ecosystem:
Protective Shield (Stratosphere): At higher levels of the atmosphere, ozone performs an essential function by shielding the Earth's surface from harmful ultraviolet (UV) radiation from the Sun. UV radiation is damaging to organisms and causes skin cancer in humans.
Harmful Effects (Ground level): At ground level, ozone acts as a toxic pollutant and is hazardous to living organisms.
Q2 (Page 216): How can you help in reducing the problem of waste disposal? Give any two methods.
Two methods to help reduce the problem of waste disposal are:
Segregation of Waste: Separating waste at the source into biodegradable and non-biodegradable categories. Biodegradable waste can be converted into compost, while non-biodegradable materials can be sent for recycling.
Using Eco-Friendly Materials: Replacing single-use non-biodegradable items with biodegradable alternatives, such as using cloth/paper bags instead of plastic bags and using paper cups instead of plastic cups.
End of Chapter Exercises
Q1: Which of the following groups contain only biodegradable items?
(a) Grass, flowers and leather
(b) Grass, wood and plastic
(c) Fruit-peels, cake and lime-juice
(d) Cake, wood and grass
Answer: Both (a) Grass, flowers and leather, (c) Fruit-peels, cake and lime-juice, and (d) Cake, wood and grass contain only biodegradable items.
Note: (b) is incorrect because plastic is non-biodegradable.
Q2: Which of the following constitute a food-chain?
(a) Grass, wheat and mango
(b) Grass, goat and human
(c) Goat, cow and elephant
(d) Grass, fish and goat
Answer:(b) Grass, goat and human
Explanation: Grass is the producer eaten by the primary consumer (goat), which in turn is eaten by the secondary consumer (human).
Q3: Which of the following are environment-friendly practices?
(a) Carrying cloth-bags to put purchases in while shopping
(b) Switching off unnecessary lights and fans
(c) Walking to school instead of getting your mother to drop you on her scooter
(d) All of the above
Answer:(d) All of the above
Explanation: All three practices help conserve resources, reduce non-biodegradable waste, and minimize pollution.
Q4: What will happen if we kill all the organisms in one trophic level?
If all organisms in one trophic level are killed, it will cause an ecological imbalance in the ecosystem:
Organisms at higher trophic levels: They will suffer from starvation and die or migrate due to the unavailability of food.
Organisms at lower trophic levels: Their population will increase abnormally because their natural predators have been removed.
This disruption breaks the food chain and disturbs the entire ecosystem balance.
Q5: Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
1. Will the impact be different for different trophic levels?
Yes, the impact will be different depending on which trophic level is removed:
Removing producers collapses the entire ecosystem immediately, as no energy will be captured from sunlight for any subsequent level.
Removing primary consumers causes producers to overgrow and higher carnivores to starve.
Removing top carnivores leads to overpopulation of herbivores, causing overexploitation of plant producers.
2. Can any trophic level be removed without causing damage?
No, organisms of no trophic level can be removed without causing damage. Each trophic level plays an indispensable role in maintaining energy flow and ecosystem stability.
Q6: What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Biological Magnification: It is the phenomenon of progressive accumulation of non-biodegradable chemicals (such as pesticides and insecticides) at each higher trophic level in a food chain.
Levels at different ecosystem stages:
Yes, the concentration level of these harmful chemicals varies across trophic levels. Since these chemicals are non-degradable, their concentration increases progressively as we move up through the trophic levels. Consequently, the lowest concentration is at the producer level, and the maximum concentration is accumulated at the top trophic level (e.g., in human beings).
Figure 13.3: Food web consisting of many food chains.Figure 13.4: Flow of energy in an ecosystem.
Q7: What are the problems caused by the non-biodegradable wastes that we generate?
The main problems caused by non-biodegradable waste include:
Soil and Water Pollution: They persist in the environment for a very long time, polluting soil and aquatic bodies.
Biological Magnification: Toxic chemicals like pesticides enter the food chain and harm top-level organisms including humans.
Drainage Blockage: Plastic waste clogs drains, leading to waterlogging and breeding grounds for disease vectors.
Harm to Wildlife: Animals like cows or aquatic organisms swallow plastic waste along with food, leading to severe health complications or death.
Q8: If all the waste we generate is biodegradable, will this have no impact on the environment?
No, it will still have an impact on the environment if generated in huge quantities:
Overburdening Decomposers: Decomposers might not be able to break down enormous amounts of organic waste quickly enough.
Pollution & Odour: Accumulation of undecomposed organic matter produces foul gases and creates unhygienic conditions.
Disease Transmission: Stagnant decaying organic waste provides ideal breeding conditions for mosquitoes, flies, and harmful microbes.
Oxygen Depletion in Water: Excessive organic waste reaching aquatic bodies leads to high biological oxygen demand during decomposition, causing depletion of oxygen and suffocation of aquatic life.
Q9: Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?
Why it is a cause for concern:
The ozone layer shields the Earth from harmful ultraviolet (UV) radiation coming from the Sun. Depletion of this layer allows UV radiation to reach the Earth's surface, which causes serious damage to organisms, such as skin cancer in human beings, cataracts, damage to immune systems, and harm to plants and aquatic organisms.
Steps taken to limit damage:
In 1987, the United Nations Environment Programme (UNEP) forged an agreement among nations to freeze chlorofluorocarbon (CFC) production at 1986 levels.
Manufacturing companies globally are now mandated to produce CFC-free refrigerators and fire extinguishers.
Q1: Why does a compass needle get deflected when brought near a bar magnet?
A compass needle is a small bar magnet. When brought near a bar magnet, it experiences a magnetic force due to the magnetic field exerted by the bar magnet in the surrounding region. This magnetic force causes the needle to deflect.
Q1 (Page 200): Draw magnetic field lines around a bar magnet.
The magnetic field lines emerge from the north pole and merge at the south pole outside the magnet, while inside the magnet, they move from the south pole to the north pole to form closed curves.
Figure 12.4: Field lines around a bar magnet.
Q2 (Page 200): List the properties of magnetic field lines.
The properties of magnetic field lines are:
They emerge from the north pole and merge at the south pole outside the magnet. Inside the magnet, their direction is from its south pole to its north pole.
Magnetic field lines are continuous closed curves.
The relative strength of the magnetic field is shown by the degree of closeness of the field lines; the field is stronger where the lines are crowded.
No two field lines intersect each other.
Q3 (Page 200): Why don't two magnetic field lines intersect each other?
Two magnetic field lines do not intersect each other because if they did, it would mean that at the point of intersection, a compass needle would point towards two different directions at the same time, which is impossible.
Section 12.2: Magnetic Field due to a Current-Carrying Conductor
Q1 (Page 201): Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.
Applying the right-hand thumb rule:
Inside the loop: The magnetic field lines point vertically downwards into the plane of the table.
Outside the loop: The magnetic field lines point vertically upwards out of the plane of the table.
Figure 12.8: Magnetic field lines of the field produced by a current-carrying circular loop.
Q2 (Page 201): The magnetic field in a given region is uniform. Draw a diagram to represent it.
A uniform magnetic field is represented by parallel, equidistant straight lines with arrowheads indicating the direction of the field.
Figure 12.10: Parallel straight field lines inside a solenoid representing a uniform magnetic field.
Q3 (Page 202): Choose the correct option.
The magnetic field inside a long straight solenoid-carrying current
(a) is zero.
(b) decreases as we move towards its end.
(c) increases as we move towards its end.
(d) is the same at all points.
Answer:(d) is the same at all points.
Explanation: The field lines inside a current-carrying solenoid are in the form of parallel straight lines, which indicates that the magnetic field is uniform and the same at all points inside it.
Section 12.3: Force on a Current-Carrying Conductor in a Magnetic Field
Q1 (Page 203): Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:(c) velocity and (d) momentum
Explanation: When a proton moves in a magnetic field, it experiences a magnetic force that acts perpendicular to its direction of motion. This force changes the direction of motion of the proton without changing its speed or mass. Since velocity depends on direction, velocity changes. Consequently, momentum (mass × velocity) also changes.
Q2 (Page 204): In Activity 12.7, how do we think the displacement of rod AB will be affected if (i) current in rod AB is increased; (ii) a stronger horse-shoe magnet is used; and (iii) length of the rod AB is increased?
The displacement of rod AB reflects the magnitude of force acting on it:
(i) If current in rod AB is increased: The displacement of the rod increases because the force acting on a current-carrying conductor increases with an increase in current.
(ii) If a stronger horse-shoe magnet is used: The displacement of the rod increases because the force increases with an increase in the strength of the magnetic field.
(iii) If length of the rod AB is increased: The displacement of the rod increases because a longer conductor in a magnetic field experiences a greater magnetic force.
Figure 12.12: A current-carrying rod AB experiencing force in a magnetic field.
Q3 (Page 204): A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:(d) upward
Explanation: The direction of current is the same as the direction of motion of positively charged particles (towards West). The force acts towards North. According to Fleming's left-hand rule, stretching the forefinger, middle finger (pointing West), and thumb (pointing North) places the forefinger pointing vertically upwards. Therefore, the magnetic field direction is upward.
Section 12.4: Domestic Electric Circuits
Q1 (Page 205): Name two safety measures commonly used in electric circuits and appliances.
Two safety measures commonly used in electric circuits and appliances are:
Electric Fuse: Prevents damage to appliances and circuits caused by overloading or short-circuiting by melting and breaking the circuit when excessive current flows.
Earthing (Earth Wire): Protects users from severe electric shocks by providing a low-resistance path for any leaking current from appliances with metallic bodies to the ground.
Q2 (Page 205): An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Calculations:
Power of the electric oven, $P = 2\text{ kW} = 2000\text{ W}$
Supply voltage, $V = 220\text{ V}$
Current drawn by the oven, $I = \frac{P}{V} = \frac{2000}{220} \approx 9.09\text{ A}$
Result & Explanation:
The current required by the electric oven ($9.09\text{ A}$) is significantly higher than the circuit's current rating of $5\text{ A}$. This causes overloading of the circuit. Due to excessive current, the Joule heating effect will cause the electric fuse to melt and break the circuit, or it may heat up the wiring and cause damage.
Q3 (Page 205): What precaution should be taken to avoid the overloading of domestic electric circuits?
The precautions to avoid overloading include:
Do not connect too many high-power electrical appliances to a single socket.
Avoid using high-power appliances simultaneously on the same circuit.
Ensure that electrical wires with appropriate current ratings and high-quality insulation are used, and damaged insulation is replaced promptly.
Use an electric fuse or circuit breaker of appropriate capacity in the circuit.
End of Chapter Exercises
Q1: Which of the following correctly describes the magnetic field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.
Answer:(d) The field consists of concentric circles centred on the wire.
Q2: At the time of short circuit, the current in the circuit
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) vary continuously.
Answer:(c) increases heavily.
Explanation: A short circuit occurs when a live wire comes into direct contact with a neutral wire, leading to an abrupt and heavy increase in current.
Q3: State whether the following statements are true or false.
The field at the centre of a long circular coil carrying current will be parallel straight lines.
A wire with a green insulation is usually the live wire of an electric supply.
(a) True — At the centre of a circular coil, the arcs of field lines appear as straight parallel lines representing a uniform magnetic field.
(b) False — A wire with green insulation is the earth wire; the live wire usually has red insulation cover.
Q4: List two methods of producing magnetic fields.
Two methods of producing magnetic fields are:
Using permanent magnets (e.g., a bar magnet).
Passing an electric current through a conductor, such as a straight wire, a circular loop, or a solenoid (electromagnet).
Q5: When is the force experienced by a current-carrying conductor placed in a magnetic field largest?
The force experienced by a current-carrying conductor placed in a magnetic field is largest when the direction of the current is at right angles ($90^\circ$) to the direction of the magnetic field.
Q6: Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer: The direction of the magnetic field is vertically downward.
Explanation:
The electron beam moves from the back wall to the front wall. Since current flows opposite to the direction of motion of electrons, the conventional current is directed from the front wall to the back wall.
The force (deflection) acts towards the right side.
Applying Fleming's left-hand rule: align the middle finger pointing towards the back wall (current) and the thumb pointing to the right (force). The forefinger then points vertically downward. Thus, the magnetic field is directed downwards.
Q7: State the rule to determine the direction of a:
magnetic field produced around a straight conductor-carrying current
force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it
current induced in a coil due to its rotation in a magnetic field
(i) Right-Hand Thumb Rule: Imagine holding a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your fingers will wrap around the conductor in the direction of the magnetic field lines.
Figure 12.7: Right-hand thumb rule.
(ii) Fleming's Left-Hand Rule: Stretch the thumb, forefinger, and middle finger of your left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field and the middle finger in the direction of current, the thumb will point in the direction of motion or force acting on the conductor.
Figure 12.13: Fleming's left-hand rule.
(iii) Fleming's Right-Hand Rule: Stretch the thumb, forefinger, and middle finger of your right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field and the thumb points in the direction of motion of the conductor, then the middle finger points in the direction of the induced current.
Q8: When does an electric short circuit occur?
An electric short circuit occurs when the live wire and the neutral wire come into direct contact with each other. This can happen when the insulation cover of wires is damaged or when there is a technical fault in an electrical appliance, causing the current in the circuit to increase abruptly.
Q9: What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Function of an earth wire: The earth wire acts as a safety measure by providing a low-resistance conducting path for electric current into the earth.
Necessity of earthing metallic appliances: Metallic body appliances (e.g., electric press, toaster, refrigerator) are earthed so that if any current leaks to the metal body due to faulty insulation, the potential of the body stays equal to that of the earth. The current flows harmlessly to the ground, protecting the user from receiving a severe electric shock.
Figure 12.15: Schematic diagram of a common domestic circuit.
An electric circuit is a continuous and closed path along which an electric current flows.
Q2. Define the unit of current.
The SI unit of electric current is the ampere (A). One ampere is defined as the current constituted by the flow of one coulomb of charge per second through a conductor ($1\text{ A} = 1\text{ C/s}$).
Q3. Calculate the number of electrons constituting one coulomb of charge.
The charge on one electron, $e = 1.6 \times 10^{-19}\text{ C}$.
Total charge, $Q = 1\text{ C}$.
Using the formula $Q = n \cdot e$, where $n$ is the number of electrons:
n = Q / e
n = 1 C / (1.6 × 10⁻¹⁹ C)
n = 6.25 × 10¹⁸ electrons (approximately 6 × 10¹⁸ electrons)
Thus, nearly $6 \times 10^{18}$ electrons constitute one coulomb of charge.
In-Text Questions (Page 173)
Q1. Name a device that helps to maintain a potential difference across a conductor.
A cell or a battery (consisting of one or more electric cells) helps to maintain a potential difference across a conductor.
Q2. What is meant by saying that the potential difference between two points is 1 V?
The potential difference between two points is said to be 1 volt (1 V) when 1 joule of work is done to move a charge of 1 coulomb from one point to the other ($1\text{ V} = 1\text{ J/C}$).
Q3. How much energy is given to each coulomb of charge passing through a 6 V battery?
Given:
Potential difference, $V = 6\text{ V}$
Charge, $Q = 1\text{ C}$
Energy given is equal to the work done ($W$):
W = V × Q
W = 6 V × 1 C = 6 J
Therefore, 6 joules of energy is given to each coulomb of charge.
In-Text Questions (Page 181)
Q1. On what factors does the resistance of a conductor depend?
The resistance of a uniform conductor depends on the following factors:
Its length ($l$) – resistance is directly proportional to length ($R \propto l$).
Its area of cross-section ($A$) – resistance is inversely proportional to cross-sectional area ($R \propto \frac{1}{A}$).
The nature of its material.
Temperature (resistance varies with temperature).
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?
Current will flow more easily through a thick wire.
Reason: Resistance of a conductor is inversely proportional to its area of cross-section ($R \propto \frac{1}{A}$). A thick wire has a larger area of cross-section and therefore lower resistance compared to a thin wire of the same material and length. Lower resistance allows current to flow more easily.
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?
According to Ohm's law, $I = \frac{V}{R}$.
If the resistance $R$ remains constant and the potential difference $V$ decreases to half ($V' = \frac{V}{2}$), the new current $I'$ will be:
I' = V' / R = (V / 2) / R = (1 / 2) × (V / R) = I / 2
Therefore, the current flowing through the component will also decrease to half of its former value.
Q4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Coils of electric toasters and electric irons are made of an alloy rather than a pure metal for the following reasons:
The resistivity of an alloy is generally higher than that of its constituent metals.
Alloys do not oxidize (burn) readily at high temperatures.
Q5. Use the data in Table 11.2 to answer the following:
Which among iron and mercury is a better conductor?
Which material is the best conductor?
From Table 11.2:
Resistivity of Iron = $10.0 \times 10^{-8}\ \Omega\text{m}$
Resistivity of Mercury = $94.0 \times 10^{-8}\ \Omega\text{m}$
(a)Iron is a better conductor than mercury because it has a lower resistivity.
(b)Silver is the best conductor because it has the lowest resistivity ($1.60 \times 10^{-8}\ \Omega\text{m}$) among all listed materials.
In-Text Questions (Page 185)
Q1. 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.
A battery of three cells of $2\text{ V}$ each gives a total potential difference of $2\text{ V} + 2\text{ V} + 2\text{ V} = 6\text{ V}$. All three resistors ($5\ \Omega$, $8\ \Omega$, $12\ \Omega$), the battery, and the plug key are connected in a single series loop.
Figure 11.14: Circuit diagram showing a battery of three 2V cells, key, and 5 Ω, 8 Ω, 12 Ω resistors in series.
Q2. 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?
Figure 11.15: Circuit diagram with ammeter connected in series and voltmeter in parallel across the 12 Ω resistor.
Calculations for Readings:
1. Total resistance in series combination ($R_s$):
R_s = R₁ + R₂ + R₃ = 5 Ω + 8 Ω + 12 Ω = 25 Ω
2. Total potential difference ($V$) = $6\text{ V}$.
3. Ammeter reading ($I$): Since components are in series, current remains the same throughout the circuit:
I = V / R_s = 6 V / 25 Ω = 0.24 A
The ammeter reading is 0.24 A.
4. Voltmeter reading ($V_{12}$): Potential difference across the $12\ \Omega$ resistor:
V₁₂ = I × R = 0.24 A × 12 Ω = 2.88 V
The voltmeter reading is 2.88 V.
In-Text Questions (Page 188)
Q1. Judge the equivalent resistance when the following are connected in parallel:
1 Ω and 10⁶ Ω
1 Ω and 10³ Ω, and 10⁶ Ω
When resistors are connected in parallel, the equivalent resistance ($R_p$) is always less than the smallest individual resistance.
(a) Among $1\ \Omega$ and $10^6\ \Omega$, the smallest resistance is $1\ \Omega$. Therefore, the equivalent resistance is less than 1 Ω (approximately equal to $1\ \Omega$).
(b) Among $1\ \Omega$, $10^3\ \Omega$, and $10^6\ \Omega$, the smallest resistance is $1\ \Omega$. Therefore, the equivalent resistance is less than 1 Ω (approximately equal to $0.999\ \Omega$).
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?
Given resistances in parallel:
$R_1 = 100\ \Omega$
$R_2 = 50\ \Omega$
$R_3 = 500\ \Omega$
Supply voltage, $V = 220\text{ V}$
1. Equivalent resistance ($R_p$) of the three appliances:
2. Total current ($I$) drawn from the source by all three appliances:
I = V / R_p = 220 V / 31.25 Ω = 7.04 A
3. An electric iron connected to the same source that takes as much current as all three appliances must have a resistance equal to the equivalent resistance of the three appliances:
Resistance of the electric iron = 31.25 Ω
Current through the electric iron = 7.04 A
Q3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
The advantages of connecting electrical devices in parallel are:
In a parallel circuit, each electrical device gets the full line voltage (potential difference).
If one component fails or stops working, the circuit is not broken and other devices continue to work independently.
Parallel circuit divides the current among gadgets according to their requirement, which is useful when devices have different resistances and require different currents to operate properly.
The total resistance of the circuit decreases, which keeps the current supply high.
Q4. 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?
(a) To get total resistance of 4 Ω:
Connect $3\ \Omega$ and $6\ \Omega$ resistors in parallel, and then connect this combination in series with the $2\ \Omega$ resistor.
Q1. Why does the cord of an electric heater not glow while the heating element does?
The heating element of an electric heater is made of an alloy (like nichrome) having high resistance. When current flows through it, according to Joule's law of heating ($H = I^2Rt$), a large amount of heat is produced, making it red-hot and glowing.
On the other hand, the cord connecting the heater is made of copper or aluminum, which has very low resistance. Therefore, negligible heat is produced in the cord, and it does not glow.
Q2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Given:
Charge, $Q = 96000\text{ C}$
Time, $t = 1\text{ hour} = 3600\text{ s}$
Potential difference, $V = 50\text{ V}$
Heat generated ($H$) is given by $H = V \times Q$:
H = 50 V × 96000 C
H = 4800000 J = 4.8 × 10⁶ J
The heat generated is $4.8 \times 10^6\text{ J}$.
Q3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Given:
Resistance, $R = 20\ \Omega$
Current, $I = 5\text{ A}$
Time, $t = 30\text{ s}$
Using Joule's law of heating, $H = I^2Rt$:
H = (5 A)² × 20 Ω × 30 s
H = 25 × 20 × 30 J
H = 15000 J = 1.5 × 10⁴ J
The heat developed in $30\text{ s}$ is $15000\text{ J}$ (or $1.5 \times 10^4\text{ J}$).
In-Text Questions (Page 192)
Q1. What determines the rate at which energy is delivered by a current?
The electric power determines the rate at which energy is delivered by a current. It depends on the potential difference across the device and the current flowing through it ($P = VI$).
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.
E = P × t = 1100 W × 7200 s = 7,920,000 J = 7.92 × 10⁶ J
(or in commercial units: E = 1.1 kW × 2 h = 2.2 kWh)
The energy consumed is $7.92 \times 10^6\text{ J}$ (or $2.2\text{ kWh}$).
Exercises (Pages 193-194)
Q1. 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 R/R' is —
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Correct Answer: (d) 25
Explanation:
Resistance of a wire is directly proportional to its length. When cut into five equal parts, the resistance of each part becomes $R_1 = \frac{R}{5}$.
When these five parts are connected in parallel, the equivalent resistance $R'$ is given by:
Q2. Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Correct Answer: (b) IR²
Explanation: Electric power is given by $P = VI$. By Ohm's law ($V = IR$), we can substitute $V$ to get $P = I^2R$ and substituting $I = \frac{V}{R}$ gives $P = \frac{V^2}{R}$. Thus, $IR^2$ does not represent electrical power.
Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be —
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Correct Answer: (d) 25 W
Explanation:
1. Calculate the resistance of the bulb ($R$), which remains constant:
P = V² / R
R = V² / P = (220)² / 100 = 48400 / 100 = 484 Ω
2. Calculate power consumed at $110\text{ V}$:
P' = (V')² / R = (110)² / 484 = 12100 / 484 = 25 W
Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be —
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Heat produced across constant potential difference $V$ in time $t$ is $H = \frac{V^2}{R} t$:
H_series = (V² / R_s) × t = (V² / 2R) × t
H_parallel = (V² / R_p) × t = (V² / (R/2)) × t = (2V² / R) × t
Ratio = H_series / H_parallel = (V² / 2R) / (2V² / R) = 1 / 4
Thus, the ratio is 1:4.
Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
A voltmeter is always connected in parallel across the points between which the potential difference is to be measured.
Q6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
R = ρ × (l / A) = ρ × (l / (π × r²))
l = (R × π × r²) / ρ
l = [10 × (22 / 7) × (0.25 × 10⁻³)²] / (1.6 × 10⁻⁸)
l = [10 × 3.1416 × 0.0625 × 10⁻⁶] / (1.6 × 10⁻⁸)
l = 122.72 m
The length of the wire required is 122.72 m.
2. If the diameter is doubled:
Since $R = \frac{\rho l}{A} = \frac{\rho l}{\frac{\pi d^2}{4}}$, resistance is inversely proportional to the square of diameter ($R \propto \frac{1}{d^2}$).
If diameter is doubled ($d' = 2d$), the new resistance $R'$ becomes:
R' = R / (2)² = R / 4 = 10 Ω / 4 = 2.5 Ω
The resistance decreases to one-fourth of its original value (becomes $2.5\ \Omega$).
Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
I (amperes)
0.5
1.0
2.0
3.0
4.0
V (volts)
1.6
3.4
6.7
10.2
13.2
Plot a graph between V and I and calculate the resistance of that resistor.
Figure 11.16: V-I graph for the given resistor.
Calculation of Resistance ($R$):
The slope of the $V-I$ graph gives the resistance ($R = \frac{\Delta V}{\Delta I}$).
Taking two points on the graph, for example $(I_1 = 0.5\text{ A}, V_1 = 1.6\text{ V})$ and $(I_2 = 4.0\text{ A}, V_2 = 13.2\text{ V})$:
R = (V₂ - V₁) / (I₂ - I₁)
R = (13.2 V - 1.6 V) / (4.0 A - 0.5 A)
R = 11.6 / 3.5 = 3.31 Ω
The resistance of the resistor is approximately 3.31 Ω.
Q8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
R = 12 V / (2.5 × 10⁻³ A)
R = (12 / 2.5) × 10³ Ω
R = 4.8 × 10³ Ω = 4800 Ω = 4.8 kΩ
The value of resistance is 4800 Ω (or 4.8 kΩ).
Q9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Connect two $6\ \Omega$ resistors in series, and connect this series group in parallel with the third $6\ \Omega$ resistor.
Series combination of two 6 Ω resistors:
R_s = 6 Ω + 6 Ω = 12 Ω
Connecting 12 Ω in parallel with third 6 Ω resistor:
1/R_eq = 1/12 + 1/6 = (1 + 2) / 12 = 3 / 12 = 1 / 4
R_eq = 4 Ω
Q12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Given:
Power of each bulb, $P = 10\text{ W}$
Voltage, $V = 220\text{ V}$
Maximum allowable current, $I = 5\text{ A}$
1. Current drawn by one bulb ($I_1$):
I₁ = P / V = 10 W / 220 V = 1 / 22 A
2. Let $n$ be the total number of bulbs connected in parallel:
Total Current = n × I₁
5 A = n × (1 / 22) A
n = 5 × 22 = 110
Therefore, 110 lamps can be connected in parallel.
Q13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Given:
Voltage, $V = 220\text{ V}$
Resistance of coil A = Resistance of coil B = $24\ \Omega$
Case 1: When coils are used separately:
I = V / R = 220 V / 24 Ω ≈ 9.17 A
Current in each coil is 9.17 A.
Case 2: When coils are used in series:
R_s = 24 Ω + 24 Ω = 48 Ω
I = V / R_s = 220 V / 48 Ω ≈ 4.58 A
Current in series combination is 4.58 A.
Case 3: When coils are used in parallel:
1/R_p = 1/24 + 1/24 = 2/24 = 1/12 => R_p = 12 Ω
I = V / R_p = 220 V / 12 Ω ≈ 18.33 A
Current in parallel combination is 18.33 A.
Q14. Compare the power used in the 2 Ω resistor in each of the following circuits:
a 6 V battery in series with 1 Ω and 2 Ω resistors, and
a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
(i) Circuit 1: 6 V battery in series with 1 Ω and 2 Ω resistors:
Total resistance, R_s = 1 Ω + 2 Ω = 3 Ω
Circuit current, I = V / R_s = 6 V / 3 Ω = 2 A
In series, current through $2\ \Omega$ resistor is $I = 2\text{ A}$. Power used in $2\ \Omega$ resistor ($P_1$):
P₁ = I² × R = (2 A)² × 2 Ω = 4 × 2 = 8 W
(ii) Circuit 2: 4 V battery in parallel with 12 Ω and 2 Ω resistors:
In parallel, the potential difference across $2\ \Omega$ resistor is equal to source voltage = $4\text{ V}$. Power used in $2\ \Omega$ resistor ($P_2$):
P₂ = V² / R = (4 V)² / 2 Ω = 16 / 2 = 8 W
Comparison: The power used in the $2\ \Omega$ resistor is 8 W in both cases (the ratio is 1:1).
Q15. 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?
Given:
Lamp 1: $P_1 = 100\text{ W}, V = 220\text{ V}$
Lamp 2: $P_2 = 60\text{ W}, V = 220\text{ V}$
1. Current drawn by Lamp 1 ($I_1$):
I₁ = P₁ / V = 100 W / 220 V = 5 / 11 A
2. Current drawn by Lamp 2 ($I_2$):
I₂ = P₂ / V = 60 W / 220 V = 3 / 11 A
3. Total current drawn from the line ($I$) in parallel:
I = I₁ + I₂ = 5/11 + 3/11 = 8/11 A ≈ 0.727 A
The total current drawn from the line is approximately 0.73 A.
Q16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
1. Energy consumed by TV set:
Power, $P_{TV} = 250\text{ W} = 0.25\text{ kW}$
Time, $t_{TV} = 1\text{ h}$
E_TV = P × t = 250 W × 3600 s = 900,000 J = 9 × 10⁵ J (or 0.25 kWh)
E_toaster = P × t = 1200 W × 600 s = 720,000 J = 7.2 × 10⁵ J (or 0.20 kWh)
Comparing the two energy values, $9 \times 10^5\text{ J} > 7.2 \times 10^5\text{ J}$. Therefore, the 250 W TV set in 1 hr uses more energy than the 1200 W toaster in 10 minutes.
Q17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Given:
Resistance, $R = 44\ \Omega$
Current, $I = 5\text{ A}$
Time, $t = 2\text{ h}$
The "rate at which heat is developed" refers to power ($P = \frac{H}{t} = I^2R$):
P = I² × R
P = (5 A)² × 44 Ω
P = 25 × 44 = 1100 J/s (or 1100 W)
The rate at which heat is developed is 1100 J/s (or 1100 W).
Q18. Explain the following.
Why is the tungsten used almost exclusively for filament of electric lamps?
Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Why is the series arrangement not used for domestic circuits?
How does the resistance of a wire vary with its area of cross-section?
Why are copper and aluminium wires usually employed for electricity transmission?
(a) Tungsten is used almost exclusively for filaments of electric lamps because it has a very high melting point ($3380^\circ\text{C}$) and high resistivity. It can retain heat to get very hot and emit light without melting.
(b) Conductors of electric heating devices are made of an alloy rather than a pure metal because alloys have higher resistivity than constituent metals and do not oxidize (burn) easily at high temperatures.
(c) Series arrangement is not used in domestic circuits because:
In series, the voltage gets divided among appliances.
If one appliance fused or fails, the whole circuit breaks and all other appliances stop working.
All appliances would operate on a single switch, making independent control impossible.
(d) The resistance of a wire is inversely proportional to its area of cross-section ($R \propto \frac{1}{A}$). Thus, as cross-sectional area increases, resistance decreases.
(e) Copper and aluminium wires are usually employed for electricity transmission because they have very low electrical resistivity, making them efficient conductors that minimize energy loss as heat.
Q1: What is meant by power of accommodation of the eye?
The ability of the eye lens to adjust its focal length to focus on both near and distant objects clearly on the retina is called the power of accommodation of the eye. This adjustment is carried out by the ciliary muscles, which contract or relax to alter the curvature and thickness of the eye lens.
Q2: A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
A person suffering from myopia (near-sightedness) requires a concave lens of suitable power to restore proper vision.
Explanation: A concave lens diverges incoming light rays from distant objects so that they appear to originate from the person's far point (1.2 m), allowing the eye lens to focus the final image correctly onto the retina.
Q3: What is the far point and near point of the human eye with normal vision?
Near Point: The minimum distance at which an object can be seen clearly without strain is called the near point (or least distance of distinct vision). For a normal human eye, it is about 25 cm.
Far Point: The farthest distance up to which the eye can see objects clearly is called the far point. For a normal eye, it is at infinity.
Q4: A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
The student is suffering from myopia (or near-sightedness).
Correction: This defect can be corrected by using spectacles fitted with a concave lens of suitable power.
Figure 10.2: (a), (b) The myopic eye (image formed in front of retina), and (c) correction for myopia using a concave lens.
Chapter End Exercises (Page 170)
Q1: The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia.
(b) accommodation.
(c) near-sightedness.
(d) far-sightedness.
Correct Answer:(b) accommodation.
Q2: The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.
Correct Answer:(d) retina.
Q3: The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm.
(c) 25 cm.
(d) 2.5 m.
Correct Answer:(c) 25 cm.
Q4: The change in focal length of an eye lens is caused by the action of the
(a) pupil.
(b) retina.
(c) ciliary muscles.
(d) iris.
Correct Answer:(c) ciliary muscles.
Q5: A person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?
The relation between focal length ($f$) and power ($P$) is given by $f = \frac{1}{P}$.
Q6: The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
To correct myopia, a distant object at infinity ($u = -\infty$) must form a virtual image at the person's far point ($v = -80\text{ cm} = -0.8\text{ m}$).
Conclusion: The corrective lens required is a concave lens of power -1.25 D.
Q7: Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Figure 10.3: (a) Near point of a hypermetropic eye, (b) Hypermetropic eye, and (c) Correction for hypermetropic eye using a convex lens.
Q8: Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
The ciliary muscles of a normal human eye can contract only up to a maximum limit to increase the curvature and thickness of the eye lens. Beyond this limit, the focal length of the eye lens cannot be decreased further. Therefore, if an object is placed closer than 25 cm, the light rays cannot be focused sharply on the retina, causing a blurred image and eye strain.
Q9: What happens to the image distance in the eye when we increase the distance of an object from the eye?
The image distance inside the eye remains constant. It is fixed at the distance between the eye lens and the retina. When the object distance increases, the ciliary muscles relax to make the eye lens thinner, thereby increasing its focal length so that the image is always formed right on the retina.
Q10: Why do stars twinkle?
The twinkling of stars is caused by atmospheric refraction of starlight.
As starlight passes through the Earth's atmosphere, it continuously undergoes refraction through layers of air with gradually changing refractive indices. Since the physical conditions of the atmosphere keep changing constantly, the path of light and the apparent position of the point-sized star fluctuate continuously. This causes the amount of starlight entering the eye to flicker, making the star appear brighter and fainter alternately.
Figure 10.9: Apparent star position due to atmospheric refraction.
Q11: Explain why the planets do not twinkle.
Planets are much closer to the Earth than stars and appear as extended sources of light (a collection of a large number of point-sized light sources). The variations in the intensity of light coming from all individual point sources average out to zero, nullifying the overall twinkling effect.
Q12: Why does the sky appear dark instead of blue to an astronaut?
The sky appears dark to an astronaut because there is no atmosphere at very high altitudes to scatter sunlight. Since scattering of light (specifically shorter blue wavelengths by atmospheric molecules) does not take place in outer space, no scattered light reaches the astronaut's eyes, making the sky look dark.
Q1: Define the principal focus of a concave mirror.
Light rays that are parallel to the principal axis and fall on a concave mirror meet or intersect at a single point on its principal axis after reflection. This specific point on the principal axis is called the principal focus of the concave mirror.
Figure 9.2 (a): Rays parallel to the principal axis meeting at the principal focus (F) of a concave mirror.
Q2: The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
The radius of curvature ($R$) of a spherical mirror is related to its focal length ($f$) by the formula:
Therefore, the focal length of the spherical mirror is 10 cm.
Q3: Name a mirror that can give an erect and enlarged image of an object.
A concave mirror can give an erect and enlarged (magnified) image of an object when the object is placed between its pole ($P$) and principal focus ($F$).
Q4: Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Convex mirrors are preferred as rear-view mirrors in vehicles due to the following reasons:
They always form an erect, though diminished, image of objects behind the vehicle.
They provide a wider field of view because they are curved outwards, enabling the driver to see a much larger area of traffic than a plane mirror would allow.
Section 9.2.4: In-Text Questions (Page 145)
Q1: Find the focal length of a convex mirror whose radius of curvature is 32 cm.
The relationship between focal length ($f$) and radius of curvature ($R$) is:
$$f = \frac{R}{2}$$
Given: $R = +32\text{ cm}$
$$f = \frac{+32\text{ cm}}{2} = +16\text{ cm}$$
The focal length of the convex mirror is +16 cm.
Q2: A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Magnification ($m$) = $-3$ (Negative for real and inverted image)
Formula:
$$m = -\frac{v}{u}$$
Substituting the given values:
$$-3 = -\frac{v}{-10}$$
$$-3 = \frac{v}{10}$$
$$v = -30\text{ cm}$$
The image is located at a distance of 30 cm in front of the mirror (on the same side as the object).
Section 9.3.2: In-Text Questions (Page 150)
Q1: A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
The light ray bends towards the normal.
Reason: Water is optically denser than air. When a ray of light travels obliquely from an optically rarer medium (air) to an optically denser medium (water), its speed decreases, causing it to slow down and bend towards the normal.
Q2: Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is $3 \times 10^8\text{ m s}^{-1}$.
Given:
Refractive index of glass ($n_g$) = $1.50$
Speed of light in vacuum ($c$) = $3 \times 10^8\text{ m s}^{-1}$
Formula:
$$n_g = \frac{c}{v}$$
where $v$ is the speed of light in glass.
$$v = \frac{c}{n_g} = \frac{3 \times 10^8\text{ m s}^{-1}}{1.50} = 2 \times 10^8\text{ m s}^{-1}$$
The speed of light in glass is $2 \times 10^8\text{ m s}^{-1}$.
Q3: Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.
According to Table 9.3:
Medium with the highest optical density:Diamond, as it has the highest refractive index of $2.42$.
Medium with the lowest optical density:Air, as it has the lowest refractive index of $1.0003$.
Q4: You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.
From Table 9.3, the refractive indices ($n$) of the given media are:
Water: $n = 1.33$
Kerosene: $n = 1.44$
Turpentine oil: $n = 1.47$
Light travels fastest in a medium with the lowest refractive index because speed of light is inversely proportional to the refractive index ($v = c/n$). Therefore, light travels fastest in water.
Q5: The refractive index of diamond is 2.42. What is the meaning of this statement?
This statement means that the ratio of the speed of light in air (or vacuum) to the speed of light in diamond is equal to $2.42$. It also indicates that light travels $2.42$ times slower in diamond than in air/vacuum.
Section 9.3.8: In-Text Questions (Page 158)
Q1: Define 1 dioptre of power of a lens.
1 dioptre is the power of a lens whose focal length is $1\text{ metre}$ ($1\text{ D} = 1\text{ m}^{-1}$).
Q2: A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
A convex lens forms a real, inverted image equal to the size of the object when the object is placed at twice the focal length ($2F_1$), and the image is formed at $2F_2$.
1. Position of the needle:
Image distance ($v$) = $+50\text{ cm}$
Since size of image = size of object, $u = -v = -50\text{ cm}$.
Thus, the needle is placed at a distance of 50 cm in front of the lens.
Q1: Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Correct Answer:(d) Clay
Explanation: A lens must be made of a transparent material to transmit light rays. Clay is opaque and does not allow light to pass through it.
Q2: The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Correct Answer:(d) Between the pole of the mirror and its principal focus.
Q3: Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Correct Answer:(b) At twice the focal length
Q4: A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Correct Answer:(a) both concave.
Explanation: By the New Cartesian Sign Convention, focal length is negative for both a concave mirror and a concave lens.
Q5: No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
Correct Answer:(d) either plane or convex.
Explanation: Both plane mirrors and convex mirrors always produce virtual and erect images regardless of the distance of the object.
Q6: Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Correct Answer:(c) A convex lens of focal length 5 cm.
Explanation: A convex lens is used as a magnifying glass. A shorter focal length provides higher power and greater magnification.
Q7: We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Range of object distance: Between $0\text{ cm}$ and $15\text{ cm}$ (i.e., between pole $P$ and focus $F$, or less than $15\text{ cm}$).
Nature of image: Virtual and erect.
Relative size: Larger than the object (magnified).
Figure 9.7 (f): Image formation by a concave mirror when the object is placed between P and F.
Q8: Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
(a) Headlights of a car: Concave mirror. Reason: When a light source is placed at the focus of a concave mirror, it produces a powerful parallel beam of light to illuminate the road ahead clearly.
(b) Side/rear-view mirror of a vehicle: Convex mirror. Reason: It always produces an erect, diminished image and offers a wider field of view to see background traffic safely.
(c) Solar furnace: Concave mirror. Reason: Large concave mirrors concentrate parallel rays of sunlight to a sharp focal point, producing high heat energy.
Q9: One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer: Yes, the lens will still produce a complete image of the object.
Experimental Verification: Take a convex lens and cover its lower half with a black paper. Mount it on a stand and place a lighted candle in front of it. Adjust a paper screen on the other side to capture the image. A full image of the candle flame appears on the screen.
Explanation: Light rays from every point of the object pass through the uncovered upper portion of the lens and refract to intersect and form the complete image. However, since the total intensity of light passing through the lens is halved, the brightness/intensity of the image will be reduced.
Q10: An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Position: Image is formed at a distance of $6\text{ cm}$ behind the mirror.
Nature: Virtual and erect.
Q13: The magnification produced by a plane mirror is +1. What does this mean?
The magnification value of +1 means:
Positive sign (+): Indicates that the image is virtual and erect.
Numerical value (1): Indicates that the size of the image is exactly equal to the size of the object.
Q14: An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Position: Image is formed $8.57\text{ cm}$ behind the mirror.
Nature: Virtual and erect.
Size: Diminished, height is $2.14\text{ cm}$.
Q15: An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
