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NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current. LearnInsta.com  provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 13 – Magnetic Effects of Electric Current solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 13 – Magnetic Effects of Electric Current Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
Why does a compass needle get deflected when brought near a bar magnet ?
Answer:
Compass needle is a small magnet which experiences a force in the magnetic field of a bar magnet. Due to this force, it gets deflected.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Draw magnetic field lines around a bar magnet. (CBSE 2011, 2012, 2013, 2014)
Answer:

Question 3.
List the properties of magnetic field lines. (CBSE 2015)
                                            Or
Write four properties of magnetic field lines. (CBSE (Delhi) 2010, 2011, 2012, 2013)
Answer:

1. Magnetic field lines are closed continuous curves.
2. The tangent at any point on the magnetic field line gives the direction of the magnetic field at that point.
3. No two magnetic field lines can cross each other.
4. Magnetic field lines are crowded in a region of strong magnetic field and magnetic field lines diverge in a region of weak magnetic field.

Question 4.
Why do not two magnetic field lines intersect each other ? (CBSE 2011, 2012, 2015)
                                         Or
No two magnetic field lines can intersect each other. Explain. (CBSE 2010, 2014)
Answer:
The tangent at any point on a magnetic field line gives the direction of magnetic field at that point. If two magnetic field lines cross each other, then at the point of intersection, there will be two tangents. Hence, there will be two directions of the magnetic field at the point of intersection. This is not possible. Hence, no two magnetic field lines can cross each other.

Question 5.
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 magnetic field inside and outside the loop.
(CBSE (Delhi) 2009)
Answer:
Magnetic field inside the loop is perpendicular to the plane of table and in the downward direction. However, outside the loop, magnetic field is perpendicular to the plane of the table and in the upward direction.

Question 6.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
(CBSE 2011, 2012, 2015)
Answer:

Question 7.
Choose the correct option :
The magnetic field inside a long straight solenoid carrying current is :
(a) zero
(b) decreases as we move towards its ends .
(c) increases as we move towards its ends
(d) is the same at all points.
Answer:
(b).

Question 8.
Which of the property a proton can change when it moves freely in a magnetic field ? (There may be more than one correct answer).
(a) mass
(b) speed
(c) velocity
(d) momentum.
Answer:
A force acts on a proton when it moves freely in a magnetic field. Hence its velocity and momentum can change.

Question 9.
In activity 13-7 (NCERT book), how do we think the displacement of rod AB will be affected if

1. current in rod AB is increased
2. a stronger horse shoe magnet is inserted
3. length of the rod AB is increased.

Answer:
Force acting on a current carrying conductor of length l placed perpendicular to magnetic field B is given by F = B I l

1. When I increase, F also increases. Hence the displacement of the rod increases.
2. When a stronger horse shoe magnet is inserted, magnetic field at B increases. So force also increases. Hence displacement increases.
3. When l increases, force increases and hence displacement increases.

Question 10.
A positively charged partical (alpha partical) 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).

Question 11.
State Fleming’s left hand rule with a labelled diagram. (CBSE 2005, 2010, 2011, 2012)
Answer:
Fleming’s left hand rule
Statement : Stretch the left hand such that the thumb, first finger and the central finger are mutually perpendicular to each other. If the First finger points in the direction of the magnetic Field and the Central finger points in the direction of Current, then the thumb will point in the direction of Motion (or Force) as shown in figure 24.

Thus, in Fleming’s left hand rule, first finger shows the direction of the magnetic field. The central finger shows the direction of electric current flowing in the conductor. Thumb shows the direction of force on the conductor or the direction of motion of the conductor

Question 12.
What is the principle of electric motor ? (CBSE 2012)
Answer:
Electric motor works on the principle that a current carrying conductor placed perpendicular to a magnetic field experiences a force.

Question 13.
What is the role of the split ring in an electric motor ?
Answer:
The split-ring in an electric motor reverses the direction of current in the armature coil of the motor. Therefore, the direction of the force acting on the two arms of the coil is also reversed. As a result of this, the coil of d.c. motor continues to rotate in the same direction.

Question 14.
Explain different ways to induce current in a coil. (CBSE 2010, 2011, 2012)
Answer:

1. By moving a bar magnet towards or away from the coil.
2. By placing a closed coil near another coil connected across a battery.

Question 15.
State the principle of electric generator.
Answer:
It is based on the principle of electromagnetic induction. That is, the changing magnetic field induces current in the coil.

Question 16.
Name some sources of direct current. (Bihar Board 2012, 2015)
Answer:
A dry cell, a battery, a solar cell, d.c. generator etc. are some sources of direct current.

Question 17.
Which source produces alternating current ?
Answer:
AC generator (which converts mechanical energy into alternating current or electricity) and an oscillator (a device which converts D.C. into A.C.) are the sources which produce alternating current.

Question 18.
Choose the correct option :
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution,
Answer:
(c).

Question 19.
Name two safety measures commonly used in electric circuit and appliances.
(CBSE 2010, 2012, 2014, 2015)
Answer:

1. Electric fuse and
2. earthing.

Question 20.
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. (CBSE 2010, 2011, 2012, 2013, 2014)
Answer:
P = 2 kW = 2000 W and V = 220 V

This shows that current flowing through the oven is more than the current rating (5 A). Hence, the fuse in the circuit melts and oven is saved from damage.

Question 21.
What precautions should be taken to avoid the overloading of domestic electric circuit ? (CBSE 2010, 2011, 2012, 2015)
                                                                         Or
Write any two precautions to be taken to avoid overloading of a domestic electric circuit.
Answer:

1.  We should not connect many appliances in the same socket.
2. Electrical appliances of high power rating should not be switched on simultaneously.

NCERT Chapter End Exercises

Question 1.
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 centered on the wire.
Answer:
(d).

Question 2.
The phenomena of electromagnetic induction is
(a) the process of charging a body
(b) the process of generating magnetic field due to current passing through a coil
(c) producing induced current in a coil due to relative motion between a magnet and the coil
(d) the process of rotating a coil of an electric motor.
Answer:
(c).

Question 3.
The device used for producing electric current is called a
(a) generator
(b) galvanometer
(c) ammeter
(d) motor. (Bihar Board 2012)
Answer:
(a).

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electromagnet while a DC generator has permanent magnet
(b) DC generator will generate a higher voltage
(c) AC generator will generate & higher voltage
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(d).

Question 5.
At the time of short circuit, the current in the ciruit
(a) reduces substantially
(b) does not change
(c) increases heavily
(d) vary continuously. (Bihar Board 2012)
Answer:
(c).

Question 6.
State whether the following statements are true or false :
(a) an electric motor converts mechanical energy into electrical energy ‘
(b) an electric generator works on the principle of electromagnetic induction
(c) the field at the centre of a long circular coil carrying current will be parallel straight lines
(d) a wire with a green insulation is usually the live wire.
Answer:
(a) False. It converts electrical energy into mechanical energy.
(b) True.
(c) True.
(d) False. Five wire has red insulation cover.

Question 7.
List three sources of magnetic fields. (CBSE 2013)
Answer:

1. a permanent magnet
2. a current carrying conductor
3. a current carrying solenoid.

Question 8.
How does a solenoid behave like a magnet ? Can you determine the north and south poles of a current carrying solenoid with the help of a bar magnet. Explain. (CBSE 2012, 2013)
Answer:
When electric current flows through a solenoid, magnetic field is set up around the solenoid. The pattern of the magnetic field is same as that of the magnetic field of a bar magnet. One end of the solenoid behaves as north pole and the other end of the solenoid behaves as south pole.
To determine the north and south poles of a current carrying solenoid with the help of a bar magnet, suspend it with a strong thread. Now bring the north pole of a bar magnet towards one end of the solenoid. If the solenoid attracts towards the magnet, then that face of the solenoid is south pole. If the bar magnet moves away from the solenoid, then that face of the solenoid is the north pole.

Question 9.
When is the force experienced by a current carrying conductor placed in magnetic field is the largest ?
[CBSE, (All India) 2009, 2010]
                                                                              Or
Under what condition does a current carrying conductor kept in a magnetic field experience maximum force ? (CBSE 2011, 2012, 2014, 2015)
Answer:
When current carrying conductor is placed perpendicular to the magnetic field.

Question 10.
Think 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:
Movement from electron beam from back wall to the front wall is equivalent to the flow of electric current from front wall to the back wall. The deflection of the beam means, the force is acting towards our right side. According to Fleming’s Left Hand Rule, the direction of magnetic field is vertically downward. That is, the magnetic field is perpendicular to the plane of the paper and directed inward.
Such magnetic field is shown by ⊗.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor ?
Answer:
Electric motor converts electrical energy into mechanical energy.
Principle: Electric Motor is based on the fact that a current carrying conductor placed perpendicular to the magnetic field experiences a force.

1. Armature coil: It consists of a single loop of an insulated copper wife in the form of a rectangle. Rectangle ABCD shown in figure 25 is an armature coil.
   
2. Strong field magnet: Armature coil is placed between two pole pieces (N and S poles) of a strong magnet. This magnet provides a strong magnetic field.
3. Split-ring type Commutator. It consists of two halves (R₁ and R₂) of a metallic ring. The two ends of the armature coil are connected to these two halves of the ring. Commutator reverses the direction of current in the armature coil.
4. Two carbon brushes B₁ and B₂ press against the commutator. These brushes act as the contacts between the commutator and the terminals of the battery.
5. A battery is connected across the carbon brushes. This battery supplies the current to the armature coil.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric cars, rolling mills, electric fans, hair dryers, mixers, blenders etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is

1. pushed into the coil,
2. withdrawn from inside the coil,
3. held stationary inside the coil ?

Name the phenomenon involved in the above cases.
(CBSE 2010, Term I, 2011, 2012, 2013, 2014)

Answer:

1. When a bar magnet is pushed into the coil, induced current flows through the coil due to the phenomenon of electromagnetic induction. This induced current is indicated by the deflection of the needle of the galvanometer as shown in figure (a).
2. When a bar magnet is withdrawn from inside the coil, again induced current flows through the coil due to the phenomenon of electromagnetic induction. In this case, the direction of induced current is opposite to the direction of the current in case (i) as shown in figure (b).
3. When the bar magnet is held stationary inside the coil, there is no change in magnetic field around the coil. Hence, no induced current flows through the coil. Therefore, galvanometer shows no deflection as shown in figure (c).
   Phenomenon involved in electromagnetic induction.

Question 14.
Two circular coils A and B are placed close to each other. If the current in the coil A is changed, will some current be induced in the coil B 1 Give reason. (CBSE 2010, 2012)
Answer:
When current in coil A is changed, a changing magnetic field is set up around it. This changing magnetic field also links with coil B and hence some current will be induced in coil B due to electromagnetic induction.

Question 15.
State the rule to determine the direction of a

1. magnetic field produced around a straight conductor carrying current,
2. force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
3. current induced in a coil due to its rotation in a magnetic field.
   (CBSE 2010, 2011, 2014)

Answer:

1. Right hand thumb rule,
2. Fleming’s left hand rule,
3. Fleming’s right hand rule.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes ?
Answer:
An electric device used to convert mechanical energy (kinetic energy) into electrical energy (electricity) is called an electric generator.
Principle: Electric generator works on the principle of electromagnetic induction. When the coil of electric generator rotates in a magnetic field, induced current flows in the circuit connected with the coil.
types of electric generator

1. AC generator
2. DC generator

AC generator: AC generator converts mechanical energy into electrical energy in the form of alternating current or AC.
DC generator: DC generator converts mechanical energy into electrical energy in the form of direct current or DC.                                                AC Generator Construction : The main components of AC generator are (Figure 33) :

1. Armature : Armature coil (ABCD) consists of large number of turns of insulated copper wire wound over a soft iron core.
2. Strong field magnet : A strong permanent magnet or an electromagnet whose poles (N and S) are cylindrical in shape is a field magnet. The armature coil rotates between the pole pieces of the field magnet. The uniform magnetic field provided by the field magnet is perpendicular to the axis of rotation of the coil.
3. Slip Rings : The two ends of the armature coil are connected to two brass slip rings R₁ and R₂. These rings rotate along with the armature coil. Rings R₁ and R₂ are at different heights.
4. Brushes : Two carbon brushes (B₁ and B₂), are pressed against the slip rings. The brushes are fixed while slip rings rotate along with the” armature. These brushes are connected to the external circuit across which the output is obtained.

Working : When the armature coil ABCD rotates in the magnetic field provided by the strong field magnet, it cuts the magnetic field lines. Thus, the changing magnetic field produces induced current in the coil. The direction  of the induced current in the coil is determined by the Fleming’s right hand rule.
The current flows out through the brush B₁ in one direction in the first half of the revolution and through the brush B₂ in the next half revolution in the reverse direction. This process is repeated. Therefore, induced current produced is of alternating nature. Such a current is called alternating current.
DC generator or Dynamo Construction:

1. Armature coil. It consists of large number of turns of insulated copper wire wound on iron core in the form of a rectangle coil. Rectangle coil ABCD shown in figure 34 is an armature coil.
   
2. Strong field magnet. Armature coil is placed between two pole pieces (N and S poles) of a strong magnet. This magnet provides a strong magnetic field.
3. Split-ring Type Commutator. It consists of two halves (R₁ and R₂) of a metallic ring. The two ends of the armature coil are connected to these two halves of the ring.
4. Two carbon brushes B₁ and B₂ press against the commutator.
5. The output is shown by the glowing bulb connected across the carbon brushes.

Working of d.c. generator: When the coil of d.c. generator rotates in the magnetic field, induced potential difference is produced in the coil. This induced potential difference gives rise to the flow of current through the bulb and hence the bulb glows.
In d.c. generator, the flow of current in the circuit is in the same direction as long as the coil rotates in the magnetic field. This is because one brush is always in contact with the arm of the armature moving up and the other brush is in contact with the arm of the armature moving downward in the magnetic field.
Note: AC generator can be converted into DC generator by replacing slip rings used in AC generator by a split ring type commutator.

Question 17.
When does an electric short circuit occur ? (CBSE 2013)
Answer:
When live wire and neutral wire touch each other (i.e. come in direct contact.), the resistance of the circuit becomes small and hence large amount of current flows through it. As a result, large amount of heat is produced and the circuit catches fire.

Question 18.
What is the function of an earth wire ? Why is it necessary to earth metallic casings of electric appliances ? (CBSE 2010, 2011, 2012, 2013, 2015)
Answer:
Earth wire acts as a safety measure. When the live wire touches the metallic casing of an electric appliance, the electric current flows from the casing of the appliance to the earth through the copper wire. As the earth offers very’ low or almost no resistance to the flow of current, so large current passes through the copper wire instead of human body. This large current heats the circuit and hence the fuse in the circuit melts. As a result of this, the circuit is switched off automatically and hence the electric appliance is saved from burning and the human body suffers no electric shock.

NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

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NCERT Solutions for Class 10 Science Chapter 12 Electricity

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 12 Electricity. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 12 – Electricity solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 12 – Electricity Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register for our free webinar class with best Science tutor in India.

NCERT Questions

In Text Questions

Question 1.
What does an electric circuit mean ? (CBSE 2011, 2013, 2014)
Answer:
An electric circuit is a closed conducting path containing a source of potential difference or electric energy (i.e. a cell or battery) and a device or element utilizing the electric energy.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Define the unit of current. (CBSE 2008)
Answer:
Unit of electric current is ampere. Electric current in a conductor is said to be 1 A if 1 coulomb charge flows through the cross-section of the conductor in 1 second.

Question 3.
Calculate the number of electrons consisting one coulomb of charge. (CBSE 2015)
Answer:

Question 4.
Name a device that helps to maintain a potential difference across a conductor. (CBSE 2008)
Answer:
A cell or battery.

Question 5.
What is meant by saying that a potential difference between two points is 1 V ?
[CBSE (Delhi) 2008 ; CBSE (All India) 2008, 2010 Term I]
 Or
Define the term “Volt”. (CBSE 2009, 2013)
Answer:
Potential difference between two points is 1 V if 1 joule work is done in moving 1 coulomb charge from one point to another point.

Question 6.
How much energy is given to each coulomb of charge passing through a 6 V battery ?
Answer:
Energy = Charge x Potential difference =1 C X 6 V = 6 J.

Question 7.
On what factors does the resistance of a conductor depend ? (CBSE 2009, 2011, 2012, 2013, 2014)
Answer:
Resistance of a conductor depends on

1. length (l) of the conductors,
2. area of corss-section of the conductor,
3. nature of the material of the conductor and
4. temperature of the conductor.

Question 8.
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:
Area of cross-section

therefore, resistance of thin wire is more than the resistance of thick wire. Hence, current in thick wire flows easily than in thin wire.
Let the resistance of an electrical component remains constant while the potential difference across the ends of the component decreases to half of its former value.

Question 9.
What change will occur with current through it ?
Answer:

Thus, current in the component becomes half of its former value.

Question 10.
Why are coil of electric toasters and electric irons made of an alloy rather than a pure metal ?
(CBSE 2011, 2012)
Answer:
This is because the resistivity of an alloy is more than the resistivity of a pure metal and hence more heat is produced in an alloy than in pure metal due to the flow of current. Moreover, alloy does not burn (or oxidise) easily even at higher temperature.

Question 11.
(a) Which among, iron and mercury is a better conductor ? (resistivity of iron = 10.0 x 10^-8 Ω m and resistivity of mercury = 94 x 10^-8 Ω m)
(b) Which material is the best conductor ?
Answer:
(a) A material whose resistivity is low is a good conductor of electricity. Therefore, iron is better conductor than mercury.
(b) Silver is the best conductor of electricity.

Question 12.
Draw a schematic diagram of a circuit consisting of a batteries of three of 2 V each, a 5 Ω resistor, 8 Ω resistor and a 12 Ω resistor and a plug key, all connected in series. (CBSE 2011)
Answer:

Question 13.
Redraw the circuit of question 12, putting an ammeter to measure the current through the resistor and a voltmeter to measure the potential difference across 12 Ω resistor. What would be the reading in the ammeter ?
Answer:

Question 14.
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.
(CBSE 2013)
Answer:
(a) When resistors are connected in parallel, then equivalent resistance of the combination is less than the least resistance in the combination. Therefore, equivalent resistance of 1 Ω and 10⁶ Ω connected in parallel is approximately 1 Ω but less than 1 Ω.
(b) The equivalent resistance is approximately 1 Ω but less than 1 Ω.

Question 15.
An electric lamp of 100 W, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to 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:

Question 16.
What is
(a) highest
(b) lowest resistance that can be secured by combining four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω ? (CBSE 2010 Term I, 2012, 2013)
Answer:

Question 17.
Why does the connecting cord of an electric heater not glow while the heating element does ? [CBSE (Delhi) 2008, 2010, 2011, 2013]
Answer:
This is because resistance of cord of electric heater is less than the resistance of heating element. So more heat is produced in the heating element and less heat is produced in the cord. Due to more heat, heating element glows.

Question 18.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V. (CBSE 2012, 2013)
Answer:

Question 19.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 seconds. (CBSE 2010)
Answer:

Question 20.
What are the advantages of connecting electrical devices in parallel with a battery instead of connecting them in series ?
                                                                           Or
Give two advantages of connecting electrical devices in parallel with battery. (CBSE 2010 Term I)
                                                                           Or
In a house-hold electric circuit different appliances are connected in parallel to one another. Give two reasons. (CBSE Sample Papers, CBSE 2012)
                                                                          Or
Why is parallel arrangement used in domestic ? (NCERT Question Bank, CBSE 2015)
Answer:

1. If any one of the electric devices in parallel fuses, then the working of other devices will not be affected.
2. When different devices are connected in parallel, they draw the current as per their requirement and hence they work properly.

Question 21.
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 ? (CBSE 2012)
Answer:
(a) We get 4 Ω resistance if 3 Ω and 6 Ω resistors are connected in parallel and this parallel combination is . connected in series with 2 Ω as shown in figure.

Question 22.
What determines the rate at which energy is delivered by an electric current ? [CBSE (All India) 2008]
Answer:
Electric power determines the rate at which energy is delivered by an electric current.

Question 23.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and energy consumed in 2 h.
[CBSE (All India) 2008]
Answer:
Power, P = VI = 220 x 5 = 1100 W
Energy consumed = Power x Time
= 1100 W x 2 h = 2200 Wh = 2.2 kWh.

NCERT Chapter End Exercises

Question 1.
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
Answer:

∴ (d) is correct answer.

Question 2.
Which of the following terms does not represent electrical power in a circuit ?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer:
(b) is correct answer.

Question 3.
An electric bulb is rated as 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
Answer:

∴ (d) is the correct answer.

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in an electric circuit. The ratio of heat produced in series and parallel combinations would be
(a) 1 :2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1.
Answer:

(c) is the correct answer.

Question 5.
How is voltmeter connected in circuit to measure the potential difference between two points ? (CBSE 2011, 2012)
Answer:

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 x 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 diameter is doubled ?
Answer:

When D is doubled and length remains the same, resistance becomes 1/4 th of the original resistance.

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :

Plot a graph between V and I and calculate the resistance of that resistor. (CBSE Sample Paper 2017-18)
Answer:

Question 8.
When a 12 V battery is connected across an unknown resistance, there is a current of 2.5 mA in the circuit. Find the value of resistance of the resistor.
Answer:

Question 9.
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 12 Ω resistor ? (CBSE 2010 Term I)
Answer:

Question 10.
How many 176  Ω resistors (in parallel) are required to carry 5 A on 220 V line ?
Answer:

Practical Skills Based Questions (Two Marks Questions)

Question 1.
In,a voltmeter, there are 20 divisions between the 0 mark and 0.5 mark. Calculate the least count of the voltmeter.
(CBSE 2015)
Answer:
Least count of voltmeter= value of 1 division
Here, 20 divisions = 0.5 V

Thus, least count of voltmeter = 0.025 V.

Question 2.
To verify Ohm’s law in the laboratory. Name the following circuit components used. (CBSE 2015)

Answer:
(a) Variable resistance
(b) Battery
(c) Wire crossing without joining
(d) Rheostat or adjustable resistance.

Question 3.
A student prepared a circuit diagram to study the dependence of potential difference (V) on electric current (I) across a resistor.

In this circuit diagram X, Y, Z, P are represented for the components.
(a) X = ……………..
(b) Y = ……………..
(c) Z = ………………
(d) P = ………………
(CBSE 2015)
Answer:
X = Ammeter
Y = Resistor
Z = Voltmeter
P = Battery.

Question 4.
Least count of a voltmeter and an ammeter in given figures would respectively be : (CBSE 2015)

Answer:
Least count of a voltmeter or an ammeter is equal to the value of 1 division on the scale.
In a voltmeter, 5 divisions = IV
or 1 division = 1/5 V = 0.2 V
Thus, least count of voltmeter = 0.2 mA
In an ammeter, 5 division = 1 mA
or 1 division = 1/5 mA = 0.2 mA
Thus, least count of ammeter = 0.2 mA.

Question 5.
To verify Ohm’s law, a student drew a circuit diagram which is given below :

1. Name two components in this circuit diagram which are connected in series.
2. Name two components in this circuit diagram which are connected in parallel. (C. B. S. E. 2015)

Answer:

1. Key, rhoestat, battery, ammeter and resistor are connected in series.
2. Resistor of resistance R and voltmeter (V) are connected in parallel.

Question 6.
In the Ohm’s law experiment, it is advised to take out the key from the plug when the observations are not being taken. Why it is essential l (CBSE 2015)
Answer:
Ohm’s law is valid only if the resistance of a resistor remains constant ‘during the experiment. If the key from the plug is not taken out when the observations are not taken, then the resistance of the resistor increases due to heat produced as a result of continuous flow of current through it. Hence, Ohm’s law cannot be verified.

Question 7.
The rest positions of the pointers of a milliammeter and volt-meter not in use are as shown in fig. A. When a student uses these in his experiment, the reading of pointers are in positions shown in fig. B.

Calculate the corrected value of current and voltage in this experiment.  (CBSE 2015)
Answer:
Least count of milliammeter = 10mA/5 = 2 mA
Least count of voltmeter = 1V/5 = 0.2 V
Zero error of milliammeter = 2 x 2 mA = 4 mA
Zero error of voltmeter = 3 x 0.2 V = 0.6 V
Correct value of current = 38 mA + 4 mA = 42 mA
Correct value of voltmeter = 3.6V – 0.6 V = 3.0 V.

Question 8.
In an experiment to study the dependence of current on potential difference across a resistor, a student obtained a graph as shown

Calculate the value of resistance of the resistor. (CBSE 2015)
Answer:

Question 9.
In a given ammeter, a student sees that needle indicates 17 divisions in ammeter while performing an experiment to verify Ohm’s law. If ammeter has 10 divisions between 0 and 0.5 A, then what is the value corresponding to 17 divisions ?  (CBSE Sample Paper 2017-18)
Answer:
10 Div = 0.5 A
1 Div = 0.5 A/10 = 0.05 A
∴ 17 Div = 0.05 A x 17 = 0.85 A

Question 10.
While performing an experiment to verify Ohm’s law, what precautions are to be taken ?
Answer:

1. Connections should be tight.
2. The conductor used should be such that its resistance does not change much with increase in temperature.
3. The plug of the key must only be inserted while reading ammeter and voltmeter. There after, the plug of the key must be taken out to avoid heating of conductor with the continuous flow of current through it.

Question 11.
In a given voltmeter, a student sees that the needle indicates 12 divisions in voltmeter while performing an experiment to verify Ohm s law. If voltmeter has 10 divisions between 0 and 1.0 V, then what is the value corresponding to 12 divisions ?
Answer:

Question 12.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :

Plot a graph between V and I and calculate the resistance of the resistor. (CBSE Sample Paper 2017-18)
Answer:
Scale : Along x-axis, 1 div = 0.1 V
Along y-axis, 1 div = 0.1 A
Graph between V and I is shown in figure:

Question 13.
An ammeter has a range of 0—3 ampere and there are 30 divisions on the scale. Calculate the least count of the ammeter. (CBSE 2015)
Answer:
30 divisions = 3 A
1 division = 3 A/30 = 0.1 A
Thus, least count of ammeter = 0.1 A

NCERT Solutions for Class 10 Science Chapter 12 Electricity

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NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 11 – Human Eye and Colourful World solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 11 – Human Eye and Colourful World Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1.
What is meant by power of accommodation of the eye ?
(CBSE Sample Paper 2010, 2012, 2014, 2015, 2016, 2017)
Answer:
The ability of the eye lens to adjust its focal length to see nearby and distant objects clearly.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
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 ?
Answer:

Question 3.
What is the far point and near point of the human eye with normal vision ? (CBSE 2011, 2012, 2014)
Answer:
The farthest position of an object from the human eye so that its sharp image is formed on the retina is at infinite distance from the eye.
The nearest position of an object from a human eye so that its sharp image is formed on the retina is at 25 cm from the eye.

Question 4.
A student has difficulty in reading the black board while sitting in the last row. What could be the defect the child is suffering from ? How can it be corrected ? (CBSE 2012)
Answer:
Near sightedness or myopia. This defect can be corrected by using a concave lens of suitable focal length.

NCERT Chapter End Exercises

Question 1.
Choose the correct option :
Human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia
(b) near sightedness
(c) accommodation
(d) far sightedness.
Answer:
(c).

Question 2.
Human eye forms the image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina.
Answer:
(d).

Question 3.
The least distance of distinct vision for a young adult with normal vision is about (CBSE 2012, Bihar Board 2012)
(a) 25 m
(b) 2.5 cm
(c) 25 cm
(d) 50 cm. (CBSE 2011)
Answer:
(c).

Question 4.
The change in focal length of an eye lens is caused by the action of
(a) pupil
(b) retina
(c) ciliary muscles
(d) iris. (CBSE 2011)
Answer:
(c).

Question 5.
A person needs a lens of power – 5.5 diopters for correcting distant vision. For correcting his near vision, he needs a lens of power +1.5 diopter. What is the focal length of the lens required for correcting
(i) distant vision and
(ii) near vision ?
Answer:

Question 6.
The far point of a myopic person is 150 cm in front the eye. What is the nature and power of the lens required to correct the problem ?
Answer:

The lens is concave lens.

Question 7.
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.
(CBSE 2011)
Answer:
For diagram,

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm ? (CBSE 2011)
Answer:
The nearest position of an object from a normal human eye so that its sharp image is formed on retina is 25 cm. If the object is placed at a distance less than 25 cm, then the blurred image of the object is formed on retina as the focal length of eye lens cannot be decreased below a certain limit. Hence, eye cannot see it clearly.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye ?
[CBSE (Delhi) 2008, 2011]
Answer:
The image distance remains the same in the eye because the eye has the ability to change the focal length of its lens to make the image always on the retina when the object distance increases from the eye.

Question 10.
Why do stars twinkle ? (CBSE 2011, 2012, 2015, 2016)
                                     Or
Explain with the help of a labelled diagram, the cause of twinkling of stars. (CBSE 2014)
Answer:
Twinkling of Stars:
Light emitted by distant stars (act as point sources of light) passes through the atmosphere of the earth before reaching our eyes. The atmosphere of the earth is not uniform but consists of many layers of different densities. The layers close to the surface of the earth are optically denser. As we go higher and higher, the density of layers and refractive index decreases progressively. As the light from a star enters the upper­most layer of the atmosphere, it bends towards the normal as it enters the next layer. This process continues till the light enters our eyes. So due to refraction of light, the apparent position of the star is different from the actual position of the star (Figure 13).

Question 11.
Explain why the planets do not twinkle. (CBSE 2011, 2012, 2015)
Answer:
Planets do not twinkle:
Planets are very close to the earth as compared to the stars. The planets act as extended sources of light. So the intensity of light we receive from the planets is very large. Therefore, the variation in the brightness of the planets is not detected. Hence, planets do not twinkle.

Question 12.
Why does the Sun appear reddish early in the morning ? (CBSE 2011, 2012)
Answer:
When sunlight enters the atmosphere of the earth, the atoms and molecules of different gases present in the atmosphere absorb this light. Then these atoms and molecules of the gases re-emit light in all directions. This process is known as scattering of light. The atoms or particles scattering light are known as scatters.
The intensity of scattered light is inversely proportional to the fourth power of the wavelength of incident light, if the size of the particles (say atoms or molecules) scattering the light is less than the wavelength of the incident light.
That is, intensity of scattered light,                I ∝ 1/λ⁴.
We know, wavelength of red light is greater than the wavelength of blue or violet light. Therefore, the intensity of scattered red light is less than the intensity of the scattered blue or violet light.
The blue colour of sky, greenish blue colour of sea water, red colour of sunset and sunrise and white colour of clouds are due to the scattering of sun light by the particles present in the atmosphere of the earth.

Question 13.
Why does the sky appear dark instead of blue to an astronaut ?
[CBSE (Delhi) 2008, 2011, 2012, CBSE (Foreign) 2016]
                                                     Or
What will the colour of the sky be for an astronaut staying in international space station orbiting the earth 1 Justify your answer giving reason. (CBSE 2014)
Answer:
The blue colour of sky is due to the scattering of sunlight. The scattering of sunlight in the atmosphere is due to the presence of atoms and molecules of gases, droplets and dust particles. When the astronaut is in space, then there is no atmosphere (or atoms and molecules of gases, droplets and dust particles) around him. Therefore, sunlight does not scatter and hence sky appears dark.

Practical Skills Based Questions ( 2 Marks)

Question 1.
Draw a path of Light ray passing through a prism.
Label angle of incidence and angle of deviation in the ray diagram. (CBSE Sample Paper 2017-18)
Answer:

Question 2.
Draw the path of light ray passing through a glass prism. Label angle of incidence, angle of deviation and angle of emergence.
Answer:

Question 3.
The path of a ray of light through a glass prism is shown below :

Write the names of the angles represented by 1, 3, 4 and 5 respectively.
Answer:
1. represents angle of incidence.
3. represents angle of prism.
4. represents angle of deviation
5. represents angle of emergence

Question 4.
The path of a ray of light through a glass prism is shown below :

Name the incident ray of light, refracted ray of light and emergent ray of light.
Answer:

1. represents incident ray of light
2. represents refracted ray of light
3. represents emergent ray of light

Question 5.
The path of a ray of light passing through a glass prism is shown below :

Name the angles X, Y, Z and O
Answer:
∠X = Incident angle
∠Y = Angle of prism
∠Z = Angle of emergence
∠O = Angle of deviation

Question 6.
A beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 on a screen as shown in figure.

Name the colours similar to the colour of

1. danger or stop signal light
2. core of hard boiled egg
3. colour of clear sky
4. solution of potassium permanganate

Answer:

1. 1 is red colour which corresponds to the colour of danger or stop signal light.
2. 3 is yellow colour which corresponds to the colour of hard boiled egg.
3. 5 is blue colour which corresponds to the colour of clear sky.
4. 7 is violet colour which corresponds to the colour of the solution of potassium permanganate.

NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World

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If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 9 – Heredity and Evolution solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 9 – Heredity and Evolution Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same species, which trait is likely to have arisen earlier ?
Answer:
In asexually reproducing population, there is no reshuffling of traits. New traits do develop due to small inaccuracies produced during DNA copying. They will be in smaller proportion than the traits already present. Therefore, trait B which exists in 60% of population must have arisen earlier than the trait A which occurs in 10% of the population.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
How does creation of variations in a species promote survival ?
Answer:
A number of different types of variations develop in a population. All of them do not have survival value. However, some of them are pre-adaptations which can be beneficial under certain environmental conditions. For example, in a heat wave most of the bacteria will die but a few having pre-adaptation or variation to tolerate heat wave will survive and multiply. Actually selection of variants by different environmental factors constitutes the basis for evolution.

Question 3.
How do Mendel’s experiments show that traits may he dominant or recessive ?
(CCE 2012, CBSE Delhi 2016, 2017)
Answer:
Mendel crossed Garden Pea plants having contrasting visible traits, e.g., tall and dwarf, violet and white flowered.
In F₁ generation there were no halfway characteristics. A cross between pure tall and pure dwarf plants yielded only tall plants in F₁ generation. There were no medium height plants. When F₁ plants were self bred, the F₂ plants were not all tall plants. Instead, both tall and dwarf plants appeared in ratio of 3 : 1. It means that the trait for dwarfness was present in F₁ generation but was not expressed while the trait for tallness expressed itself.

The trait of tallness which expresses itself in the presence of its contrasting form is called dominant. The other trait of dwarfness which is unable to express its effect in the presence of its contrasting trait is known as recessive.

Question 4.
How do Mendel’s experiments show that traits are inherited independently ? (CCE 2012, CBSE A.I. 2016, Delhi 17)
Answer:
Independent inheritance of traits is proved by employing dihybrid crosses and obtaining dihybrid ratios. Mendel crossed pure breeding tall plants having round seeds (TTRR) with pure breeding short plants having wrinkled seeds (ttrr). The plants of F₁ generation were all tall and with rounded seeds (TtRr) indicating that the characteristics of tallness and round seededness were dominant. Self breeding of F₁ yielded plants in the ratio of 9 tall round seeded, 3 tall wrinkled seeded, 3 short round seeded and one short wrinkled seeded. Tall wrinkled seeded and short round seeded plants are new combinations which can develop only if the traits are inherited independently. If the two traits are considered individually, F₂ ratio would be same as for monohybrid crosses, i.e., 12 tall : 4 short, 12 round seeded : 4 wripkled seeded.

TR, Tr, rR, tr x TR, Tr, rR, tr Gametes 9 tall rounded : 3 tall wrinkled : 3 short rounded : 1 short wrinkled

Question 5.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits—blood group A or O, is dominant ? Why or why not ?
Answer:
No. The information is not enough to tell whether the trait of the blood group A (I^A) or blood group
0(I⁰) is dominant. Either can be possible. Each individual carries two alleles. A recessive trait appears only when the two alleles are similar.
Possibility I: Blood Group A is Dominant and O Recessive. The trait of blood group O can appear only when both the recessive alleles occur together as in mother and daughter (I⁰I⁰). A group father should carry both the alleles of A and O (I^AI⁰).
Possibility 2: Blood Group O is Dominant and A Recessive. In this case the father should carry the alleles of A(I^AI^A) while the mother can be homozygous or heterozygous (I⁰I⁰, I⁰I^A). The daughter will have one dominant alleles of 0(I⁰I^A).
As both the possibilities can occur, the given information is unable to tell whether allele for blood group A or O is dominant.

Question 6.
How is the sex of the child determined in human beings ? (CCE 2013)
Answer:
Sex of the child is determined by the gametes that fuse to form zygote which later grows into offspring. Human females (44 + XX) produce only one type of ova (22 + X). Human males (44 + XY) form two types of sperms, androsperms (22 + Y) and gynosperms (22 + X). Both are formed in equal number. It is a chance factor whether an androsperm or a gynosperm fuses with egg to form 44 + XY or 44 + XX child. A child that obtains an X-chromosome from father will be girl and the one who inherits a Y-chromosome will be boy.

Question 7.
Draw Fig. 4.12. What are the different ways in which individuals with a particular trait may increase in a population ?
Answer:
There are three different ways in which individuals with a particular trait can increase in a population.

1. Survival Value (Natural Selection): The trait has survival value. It is picked up by natural selection. Through differential reproduction, it increases in population, e.g., green colour in beetles instead of red providing camouflage in bushes against being picked up by crows.
2. Genetic Drift: There is seasonal or accidental decline in population. The survivors have certain combination of traits which increase in number with the increase in population. The traits may not give any extra benefit to population.
3. Food: Individuals with particular trait may have extra abundance of food in their environment. They will naturally increase in number.

Question 8.
Why are traits acquired during the life time of an individual not inherited ?
(CCE 2011, 2012, 2013, CBSE A.I. 2017)
Answer:
Acquired traits are structural, functional and behavioural changes that an individual develops during its life time due to a particular environment, disease, trauma, use and disuse, conditioning or learning. The traits are not passed on to DNA of germ cells. They remain restricted to somatic cells. They are destroyed with the death of the individual. Therefore, intelligence, experiences and structural changes acquired during life time of an individual cannot pass to the progeny. Weismann (1892) cut the tails of mice for 21 generations but a tail still developed in 22nd generation.

Question 9.
Why are the small number of surviving tigers a cause of worry from the point of view of genetics ?
Answer:
A small population is always at a risk of degeneration and extinction due to

1. Excessive inbreeding that brings about inbreeding depression or degeneration,
2. Fewer recombinations and variations which are otherwise essential for maintaining vitality and vigour of the species.
3. Lesser adaptability to changes in the environment,
4. Increased threat to survival due to poaching, habitat destruction and environmental change.

Question 10.
What factors could lead to the rise of a new species ? (CCE 2012, 2014, CBSE Foreign 2017)
Answer:

1. Absence of gene flow amongst sub-populations due to the presence of physical barriers, long distance, differences in habitats, environmental and climatic conditions.
2. Accumulation of different variations in the different sub-populations of the species.
3. Natural selection of particular traits in a particular environment.
4. Genetic Drift. Separation of a small population, changes in its allele frequency, new mutations and adaptations to new habitat.
5. Reproductive Isolation. Accumulation of different variations and genetic drift result in absence of interbreeding in the previous subpopulations of a species. This results in the formation of new species. e.g., Finches of Galapogos islands.

Question 11.
Will geographical isolation be a major factor in the spéciation of a self pollinating plant species ? Why or why not ?
(CCE 2012)
Answer:
No. Geographical isolation has little role in spéciation of self pollinating plant species because there is already no gene flow among members of the species. Pea or Wheat which is self pollinated (due to pollination in bud condition) is not affected by any type of isolation. However, self pollinated plants can accumulate variations due to mutations and other factors and form new species.

Question 12.
Will geographical isolation be a major factor in the spéciation of an organism that reproduces asexually ? Why or why not ?
(CCE 2012)
Answer:
Recombination of genes is absent in asexually reproducing organisms. Therefore, variations originating in them do not get diluted but spread to all the» subsequent generations. Geographical isolation, which helps in spéciation due to formation of a separate gene pool, has no role in spéciation of asexually reproducing organisms.

Question 13.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
(CBSE Foreign 2017)
Answer:

1. Closeness of species is determined by presence or absence of fundamental characteristics and correlated characters. Two species of bacteria are closely related as they possess fundamental similarities of occurrence of nucleoid (instead of nucleus), absence of membrane covered cell organelles and presence of 70 S ribosomes. Human beings are close to monkeys because they possess similar eukaryotic multicellular body with vertebrate characters, mammalian traits and primate characters.
2. These days DNA matching is undertaken to find out the degree of closeness of the species.

Question 14.
Can the wing of a butterfly and the wing of a bat be considered homologous ? Why or why not ?
Answer:
No. Wings of butterfly and bat are fundamentally different in their origin and structure. In butterfly they are integumentary outgrowths having hollow tubes. In bat they are modified fore limbs which are covered by skin. Such organs which have a different origin and basic structure but are functionally similar are called analogous organs.

Question 15.
What are fossils ? What do they tell us about the process of evolution ? (CCE 2013, CBSE Foreign 2017)
Answer:
Fossils are remains or impressions of the past organisms that are found in the rocks of the old ages. They are often called written documents of evolution because they directly indicate the presence of different types of organisms in different ages. The path of evolution is known by arranging the fossils in a proper sequence age-wise. The early fossils are of simple organisms. Later on different complex forms arose, flourished and died down. They were replaced by newer forms. Study of fossils can also indicate the evolutionary stages of organisms. For example, modern horse (Equus) arose from a fossil animal Eohippus that existed on earth 60 million years back as a 30 cm high small animal. It evolved into 60 cm high goat sized Mesohippus about 40 million years back. Mesohippus gave rise to Merychippus (16-18 million years back) that formed Pliohippus (100-120 cm high, 10 million years ago). The modern horse evolved only 0-5 million years ago from Pliohippus.

Question 16.
Why are human being who look so different from each other in terms of size, colour and looks said to belong to same species ? (CBSE A.I. 2009 C, CBSE Foreign 2017)
Answer:
Delimitation of a species is based on the presence of a common gene pool, free inbreeding and reproductive isolation. Differences in size, colour and looks are based on preponderance of specific alleles and their interactions with the environment. All human beings, despite presence of different races, belong to same species {Homo sapiens) because they share the same gene pool, can marry amongst themselves and produce fertile offspring.

Question 17.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzee have a better body design ? Why or why not ?
Answer:
A better body design is the one which has more complexity, more elaboration and more controls which gives the organism a better competitive edge over others. There is no doubt that out of the four (bacteria, spiders, fish and chimpanzee), chimpanzee has a more elaborate body design or organisation. However, since body design is meant for competitive survival in their environment, all the four organisms or for that all living organisms, have a good body design that is suited to their environment.

NCERT Chapter End Exercises

Question 1.
A Mendelian experiment consisted of breeding tall Pea plants bearing violet flowers with short Pea plants bearing white flowers. The progeny all bore violet flowers but almost half of them were short. This suggests that genetic make up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) T_tWW
(d) T_tWw.
Answer:
(c) T_tWW.

Question 2.
An example of homologous organ is
(a) Our arm and a dog’s fore leg
(b) Our teeth and elephant tusks
(c) Potato and runners of grass
(d) All the above.
Answer:
(d) All the above.

Question 3.
In evolutionary terms we have more in common with
(a) A Chinese school boy
(b) A chimpanzee
(c) A spider
(d) A bacterium.
Answer:
(a) A Chinese school boy.

Question 4.
A study found that children with light coloured eyes are likely to have parents with light coloured eyes. On this basis can we say anything about whether the light eye colour is dominant or recessive ? Why or why not ?
Answer:
No. We cannot say with certainity whether the light eye colour is dominant or recessive. But since both the parents as well as the children have light eye colour, the probability is that it is a recessive trait. A recessive trait appears only when an individual possesses both the recessive alleles. As the parents are pure for the trait, the children also possess the trait and are pure for the same. Had the light eye colour been a domination trait, the recessive dark colour trait will have the chance to segregate and appear in some of the children.

Question 5.
How are the two areas of study, evolution and classification, interlinked ? (CBSE A.I. 2016)
Answer:
Classification is based on similarities and differences amongst organisms. The more characteristics two species have in common, the more closely related they are. They must have evolved from a common ancestor. Similarly more differences mean different adaptations and divergence from common ancestor in the remote past.

Question 6.
Explain the terms analogous and homologous organs with examples. (CBSE A.I. 2008 C)
Answer:
Analogous Organs: They are organs which have similar appearance and function but are quite different in their origin, development and anatomy.
Examples: Wings of Butterfly (integumentary outgrowths) and bird (modified fore-limbs).
Homologous Organs: They are organs which have similar origin, similar development and similar internal structure but have different forms and functions.
Examples: Fore-limbs of Horse, human hand, flipper of whale, wing of bird or bat.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:

1. Survey the dog population in and around your locality. Find out the percentage of different colours.
2. Observe the lineages where same colour is present in both parents and offspring. There is possibility that in these lineages both the alleles of coat colour are similar.
3. Allow crossing between two lineages having different coat colours.
4. Find the colour of F₁ individuals. It is probably the dominant coat colour.
5. Cross the F₁ dogs with the one having the other probably recessive colour. Is the ratio similar to test cross, Le., 1 : 1 ?

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
Fossils are remains or impressions of past organisms that are found in the rocks. Fossils of lower strata belong to early periods while those of upper strata are of later periods. Arranging the fossils stratumwise will indicate the occurrence of different forms of life at different times. It is found that the early fossils generally belong to simple organisms. Complexity and elaboration increased gradually with evolution. Evolution has never been linear or straight. A number of variants or branches appeared, some of which were more complex while others were less complex.

1. Fossils indicate the path of evolution of different groups.
2. They can indicate the phylogeny of some organisms, e.g, Horse, Elephant.
3. Some fossils have characteristics intermediate between two groups,
   e.g., toothed bird Archaeopteryx. They indicate how one group has evolved from another.

Question 9.
What evidence do we have for the origin of life from inanimate matter ?
Answer:
Miller and Urey (1953) assembled an apparatus which had a spark chamber (for producing lightning), a flask for boiling and a condenser. They introduced a mixture of methane, ammonia, hydrogen and water into the apparatus. The gaseous mixture was exposed to electric discharges, boiling (800°C) and condensation with the temperature kept just below 100°C. The experiment was continued for a few days. At the end of one week, 15% of carbon (from methane) had been converted into simple organic compounds of amino acids, organic acids, sugars and nitrogen bases. It clearly proved that organic compounds or building blocks of life developed from inanimate matter in the remote past when the hot earth was cooling.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually reproducing organism?
Answer:
Variations arising during sexual reproduction occur due to

1. Chance separation of homologous chromosomes during gametogenesis.
2. Crossing over between homologous chromosomes.
3. Chance coming together of chromosomes during fertilisation,
4. Errors or mutations occurring during DNA replication.

Only the last method of variations is found in asexually reproducing organisms. Therefore, rate of appearance of variations is quite high in sexually reproducing organisms as compared to asexually reproducing organisms. Further, variations developed in sexually reproducing organisms are quite viable as most of them are due to reshuffling of genes. It is not so in asexually reproducing organisms. Here, most of the changes are harmful. They have a negative impact on evolution except when changing environment finds them useful. Because of the abundance and viability of variations, the rate of evolution is also high in sexually reproducing organisms.

Question 11.
How is equal genetic contribution of male and female parents ensured in the progeny ?
Answer:
Most organisms are diploid. Their genetic material consists of two sets of chromosomes. Gametes carry single set of chromosomes, i.e., they are haploid. Sexual reproduction involves the formation and fusion of two types of gametes, male and female. Male gamete brings one set of chromosomes from the male parent. Female gamete also brings one set of chromosomes from the female parent. When two gametes fuse during sexual reproduction, the normal diploid chromosome complement is restored. It consists of 50% chromosomes from male parent and 50% chromosomes from female parent. Therefore, both the parents contribute equal genetic material to the offspring through formation and fusion of gametes.

Question 12.
Only variations that confer an advantage to an individual will survive in a population. Do you agree with this statement ? Why or why not ?
Answer:
No. Alongwith advantageous variations, a number of indifferent variations remain in the populations. Only the disadvantageous variations which are either lethal or extremely harmful are eliminated. All other variations persist in the population. Many of them function as preadaptations.

Selection Type Questions

Alternate Response Type Questions
(True/False (T/F), Right (√)/Wrong (x), Yes/No)

Question 1.
Mendel studied science and mathematics at the university of Amsterdam.
Question 2.
Both the parents contribute DNA equally to the offspring.
Question 3.
A factor which shows its effect in the hybrid is called recessive.
Question 4.
Sex of the child is determined by the type of ovum provided by the mother.
Question 5.
A recessive trait can also be common as blood group O.
Question 6.
Dromaesaurs were the first to fly.
Question 7.
Attached ear lobe is recessive trait.
Question 8.
Charles Darwin discovered the law of independent assortment,

Matching Type Questions

Question 9.
Match the articles given in columns A and B (single matching) :

Question 10.
Match the contents of columns I, II and III (double matching) :

Question 11.
What type of similarity, homologous (H) and analogous (A) occurs in the pairs of organs ?

Question 12.
Match stimulus with appropriate Response

Fill In the Blanks

Question 13.The term genetics was coined by ……………
Question 14.Mendel chose ………….. characters in Pea for his experiments.
Question 15. Broccoli has been developed from ……………… cabbage through artificial selection.
Question 16. ……………… Speciation occurs in geographically separated populations.
Question 17. Fossils are written documents of ………………..

Answers:

NCERT Solutions for Class 10 Science Chapter 9 – Heredity and Evolution

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NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 10 – Light Reflection and Refraction solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 10 – Light Reflection and Refraction Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1.
Define the principal focus of a concave mirror.
Answer:
A point on the principal axis where the parallel rays of light after reflecting from a concave mirror meet.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is focal length ? (Bihar Board 2012)
Answer:
Radius of curvature, R= 20 cm.

Question 3.
Name a mirror that can give an erect and magnified image of an object.
Answer:
A concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles ?
[CBSE (All India) 2007, 2011, 2012]
Answer:
This is because a convex mirror forms an erect and diminished (small in size) images of the objects behind the vehicle and hence the field of view behind the vehicle is increased.

Question 5.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
R = +32 cm. Therefore, f = R/2 = +32/2 = +16 cm.
Thus, focal length of the convex mirror = +16 cm.

Question 6.
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 ?
Answer:
m – -3, But m = -v/u, so v = 3u
u = -10 cm
v = 3 (-10 cm) =-30 cm
Thus, the image is located at a distance of 30 cm to the left side of the concave mirror.

Question 7.
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 ?
Answer:
The ray of light bends towards the normal because the speed of light decreases when it goes from air (rarer medium) into water (denser medium).

Question 8.
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 x 10⁸ m s^-1 (CBSE 2011)
Answer:

Question 9.
You are given kerosene, turpentine and water. In which of these does the light travel faster ?
Answer:
We know, v = c/n. Refractive index (n) of water is 1.333, whereas refractive index of kerosene is 1.44 and that of turpentine is 1.47. As refractive index of water is least, so speed of light in water is more than in kerosene and turpentine. Hence, light travels faster in water.

Question 10.
The refractive index of diamond is 2.42. What is the meaning of this statement ? [CBSE (Delhi) 2008, 2012, 2013; Bihar Board 2012]
Answer:

Question 11.
Define 1 dioptre of power of a lens.
Answer:
Power = I/f (in m).
Power of a lens is 1 dioptre if focal length of the lens is 1 metre or 100 cm.

Question 12.
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.
Answer:

Question 13.
Find the power of concave lens of focal length 2m?
Answer:

NCERT Chapter End Exercises

Question 1.
Which one of the following materials cannot be tised to make a lens 1
(a) water
(b) glass
(c) plastic
(d) clay.
Answer:
(d). This is because clay is opaque (i.e. light cannot pass through it).

Question 2.
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.
Answer:
(d).

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object ? (Bihar Board 2012)
(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.
Answer:
(b).

Question 4.
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 are concave
(b) both are convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex but the lens is concave.
Answer:
(a).

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane only
(b) concave only
(c) convex only
(d) either plane or convex.
Answer:
(d).

Question 6.
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.
Answer:
(c). Magnifying power of a reading glass (Convex lens) = 1/f.

Question 7.
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.
Answer:
A concave mirror produces an erect image if the object is placed between the pole and the focus of the concave mirror. Thus, object may be placed at any position whose distance is less than 15 cm from the concave mirror. The image is virtual and erect. The image is larger than the object. For a ray diagram, see figure 24.

Question 8.
Name the type of mirror used in the following situations :
(a) head lights of a car
(b) side rear view mirror of a vehicle
(c) solar furnace.
Support your answer with reason. (CBSE 2012, 2013)
Answer:
(a) Concave mirror. When a bulb is placed at the focus of a concave mirror, then the beam of light from the bulb after reflection from the concave mirror goes as a parallel beam which lights up the front road.
(b) Convex mirror. Image formed by a convex mirror is erect and small in size. The field of view behind the vehicle is large.
(c) Concave mirror. Concave mirror focuses rays of light coming from the sun at its focus. So, the temperature at the focus is raised.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object ?
(CBSE 2015)
Answer:
A complete image of the object is formed as shown in figure.

Question 10.
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.
Image is real and inverted.
Answer:

Here, u = -25 m, f = 10 m

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. Hou> far is the object placed from the lens ? Draw the ray diagram.
Answer:

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image. (CBSE 2014)
Answer:

Since, m is positive, so the orientation of both object and image is same. Thus,image is erect and virtual.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean ? (CBSE 2011, 2014)
Answer:

It means, size of the image formed by plane mirror is equal to the size of the object. Positive sign with m tells that both object and image are erect.

Question 14.
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. (Similar CBSE 2011)
Answer:
h = 5 cm, u = -20 cm


Since h’ is positive, so image is erect and virtual.

Question 15.
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.
Answer:

Since h’ is negative, so image is inverted.

Question 16.
Find the focal length of a lens of power -2.0 D. What type of lens is this ?
Answer:

The lens is concave.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging ? (CBSE 2012)
Answer:

Since focal length is positive, so the lens is converging.

Practical Skills Based Questions (2 Marks)

Question 1.
A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again ? How will the magnification of the image be affected ? (CBSE 2015)
Answer:

When distance of object u increases, V must decrease so that focal length of the lens remains the same. Thus, screen on which sharp image of object is formed is moved towards the lens.
As the distance of object u increases, the size of the image decreases. Hence, magnification of the imag decreases.

As u increases, v decreases and hence m also decreases.

Question 2.
In the given incomplete ray diagram for image A’B’ of a convex lens, what is the position of object AB ? Also complete the ray of diagram? (CBSE 2015)

Answer:
The position of object AB is between F₁ and 2F₁

Question 3.
To find the image-distance for varying object-distances in case of a convex lens, a student obtains on a screen a sharp image of a bright object placed very far from the lens. After that he gradually moves the object towards the lens and each time focuses its image on the screen.
(a) In which direction towards or away from the lens, does he move the screen to focus the object ?
(b) What happens to the size of image does it increase or decrease ?
(c) What happens when he moves the object very close to the lens ?
Answer:

As distance of object decreases, distance of image V must increase so that focal length of lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases, therefore, magnification increases. In other words, size of the image increases.
(c) When object is very close to convex lens, virtual and magnified image is formed.

Question 4.
A student performed an experiment for the image formation by a convex lens at different positions of an object. If focal length of lens is 15 cm
Answer:

Question 5.
A student performed an experiment with convex lens and found the virtual image of an object. Find :
(a) position of the object.
(b) draw ray diagram for the above situation.
(CBSE 2015)
Answer:
(a) A convex lens forms a virtual image if the object lies between the optical centre and focus- of the convex lens. That is, the distance of the object from the less must be less than the focal length of the lens.
(b)

Question 6.
A student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that he moves gradually the flame towards the lens and each time focuses its image on the screen.
(A) In which direction does he move the lens to focus the flame on the screen 1
(BJ What happens to the size of the image of the a formed on the screen 1
(C) What difference is seen in the intensity (brightness) of the image of the a screen ?
(D) What is seen on the screen when the flame is very close (at about 5 cm) to the lens ? (CBSE 2017)
Answer:

As the distance of the candle flame (object) u decreases, distance of image u must increase so that the focal length of the lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases therefore, magnification increases, other words, size of the image increases.
(C) As the size of the image increases, so the intensity (brightness) of the image of the flame on the screen decreases.
(D) When the flame is at about 5 cm, which is less than the focal length (10 cm) of the lens.
In this case, magnified, erect and virtual image of the flame is formed. Since, virtual image cannot be obtained on the screen, so nothing is seen on the screen.

Question 7.
A student places a candle flame at a distance of about 60 cm from a convex lens of focal length 10 cm and focuses the image of the flame on a screen. After that he gradually moves the flame towards the lens and each time focuses the image on the screen.
(A) In which direction-toward or away from the lens, does he move the screen to focus the image ?
(B) How does the size of the image change ?
(C) How does the intensity of the image change as the flame moves towards the lens ?
(D) Approximately for what distance between the flame and the lens, the, image formed on the screen is inverted and of the same size ? (CBSE 2017)
Answer:

As the distance of the candle flame (object) u decreases, distance of image u must increase so that the focal length of the lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases therefore, magnification increases, other words, size of the image increases.
(C) As the size of the image increases, so the intensity (brightness) of the image of the flame on the screen decreases.
(D) When object is placed at 2F, a real and inverted image of same size as that of the object is formed at 2F on the other side of the lens. Therefore, in this case, the distance between flame and the lens is approximately 20 cm so that the image formed on the screen is inverted and of the same size.

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

Hope given NCERT Solutions for Class 10 Science Chapter 10 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.