HOTS Questions for Class 10 Science Chapter 14 Sources of Energy

HOTS Questions for Class 10 Science Chapter 14 Sources of Energy

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 14 Sources of Energy

Question 1.
Following model demonstrates the process of thermo-electric production.
HOTS Questions for Class 10 Science Chapter 14 Sources of Energy image - 1

  1. Identify and label the device marked as 1.
  2. Why is tennis ball fitted with metal sheets rotated ?
  3. Name the device, which depends on the principle demonstrated by this model.
  4. Which form of energy is converted into electrical energy ?

Answer:

  1. Device 1 is a dynamo.
  2. Tennis ball fitted with metal sheets rotates due to the force exerted on it by the steam.
  3. Thermal power plant works on the principle demonstrated by the given model.
  4. Mechanical energy of tennis ball fitted with metal sheets is converted into electrical energy.

More Resources

Question 2.
A diagram shown below is a biogas plant.
HOTS Questions for Class 10 Science Chapter 14 Sources of Energy image - 2

  1. Identify the parts indicated by question marks and labelled 1, 2, 3 and 4 in the diagram.
  2. Name the micro organisms responsible for the fermentation of the slurry in the digester.
  3. Name two chemical elements present in the manure or spent slurry.

Answer:

    1. Slurry of cattle dung and water
    2. Gas tank containing bio-gas.
    3. Digester.
    4. Spent slurry or manure
  1. Anaerobic micro organisms,
  2. Nitrogen and phosphorous are present in the manure.

Question 3.
A student constructed a box type solar cooker. He found that it did not work efficiently. What could this be due to ? Give any four possible mistakes in the solar construction and operation of the cooker. What maximum temperature can ordinarily be reached inside a solar cooker ? (CBSE 2010)
Answer:
He might be committing the following mistakes :

  1. He might have not blackened the interior of the solar cooker,
  2. He might be using a plastic cover instead of a glass cover,
  3. He might have not made it insulated,
  4. He might have not used black containers. Maximum temperature attained in a solar cooker is about 140° C.

Question 4.
A student has set up a solar cooker in a box by using a black painted aluminium sheet, a black cooking vessel, some glass wool, a glass sheet and a mirror plate. What is the role of each item used in the solar cooker ?
Answer:
Black painted aluminium sheet absorbs heat radiation.
Black cooking vessel also absorbs heat radiation to cook food.
Glass wool prevents the loss of heat.
Glass sheet prevents the escape of heat radiation from the box.
Mirror plate reflects the sunlight to fall on the glass sheet.

Hope given HOTS Questions for Class 10 Science Chapter 14 Sources of Energy 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation (Histograms, Frequency Polygon and Ogives) Ex 23

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 23 Graphical Representation (Histograms, Frequency Polygon and Ogives) Ex 23.

Question 1.
Draw histograms for the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.4

Question 2.
Draw a cumulative frequency curve (ogive) for each of the following distributions
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q2.2

Question 3.
Draw an ogive for each of the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.2
Now plotting the points (10, 8), (20, 25), (30, 38), (40, 50) and (50, 67) on the graph and join them with free hand to obtain an ogive as shown in the graph.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.3
Now plot the points (10, 0), (20, 17), (30, 32), (40, 37), (50, 53), (60, 58) and (70, 65) on the graph and join them with free hand to get an ogive as shown in the graph.

Question 4.
Construct a frequency distribution table for the numbers given below, using the class intervals 21-30,31-40…. etc.
75, 67, 57, 50, 26, 33, 44, 58, 67,75, 78, 43, 41, 31, 21, 32, 40, 62, 54, 69, 48, 47,51,38, 39,43,61, 63, 68, 53, 56, 49, 59, 37, 40, 68, 23, 28, 36 and 47.
Use the table obtained to draw:
(i) a histogram
(ii) an ogive
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.3
(ii) Ogive :
We plot the points (30, 4), (40, 13), (50, 22), (60, 29); (70, 37) and (80, 40) on the graph and join them in free hand to obtain an ogive.

Question 5.
(a) Use the information given in the adjoining histogram to c construct a f frequency table.
(b)Use this table to construct an ogive.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.1
Solution:
From the histogram given, the required frequency table will be as given below.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.2
Plot the points (12, 9), (16, 25), (20, 47), (24, 65), (28, 77), (32, 81) on the graph and join them with free hand to get an ogive as shown.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.3

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.1
(a) From the distribution given above, construct a frequency table, (b) Use the table obtained in part (a) to draw :
(i) a histogram
(ii) an ogive.
Solution:
Difference in consecutive class marks. = 17.5 – 12.5 = 5
∴ first class interval will be : 10-15 and so on
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.2
(ii) We plot the point (15. 12), (20. 29), (25. 51), (30, 78). (35, 108), (40. 129) and (45, 145) on the graph and join them in free hand to obtain the ogive.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.3

Question 7.
Use graph paper for this question.
The table given beiow shows the monthly wages of some factory workers.
(i) Using the table calculate the cumulative frequencies of workers.
(ii) Draw the cumulative frequency curve.
Use 2 cm = Rs. 500, starting the origin at Rs. 6500 on x-axis, and 2 cm = 10 workers on the y-axis
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.3
(ii) We plot the points (7000, 10), (7500, 28), (8000, 50), (8500, 75), (9000, 92) (9500, 102) and (10000, 110) on the graph and join them in free hand to obtain an ogive.

Question 8.
The following table shows the distribution of the heights of a group of factory workers :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.1
(i) Determine the cumulative frequencies.
(ii) Draw the ‘less than’ cumulative frequency curve on a graph paper. Use 2 cm = 5cm height on one axis and 2 cm = 10 workers on the other.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.3
We plot points (155, 6), (160, 18), (165, 36),
(170, 56), (175, 69), (180, 77) and (185, 83) on the graph and join them in free hand to obtain an ogive.

Question 9.
Construct a frequency distribution table for each of the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q9.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23  are helpful to you.

If you have any doubts, please comment below. We try to provide online math tutoring for you.

NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity

NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity

Question 1.
A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of the figure given below. The current recorded in the ammeter will be
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 1
(a) maximum in (i)
(b) maximum in (ii)
(c) maximum in (iii)
(d) the same in all the cases.
Answer:
(d).
Explanation : Same current flows through every part of the circuit having resistance connected in series to a cell or battery.

More Resources

Question 2.
In the following circuits shown in figure, power produced in the resistor or combination of resistors connected to a 12V battery will be
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 2
(a) same in all cases
(b) minimum in case (i)
(c) maximum in case (ii)
(d) maximum in case (iii)
Answer:
(d).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 3

Question 3.
Electrical resistivity of a given metallic wire depends upon
(a) its length
(b) its thickness
(c) its shape
(d) nature of material.
Answer:
(d).
Explanation : Resistivity of a given metallic wire does not depend on its dimensions and shape but depends on the nature of the material of the wire.

Question 4.
A current of 1A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly
(a) 1020
(b) 1016
(c) 1018
(d) 1023
Answer:
(a).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 4

Question 5.
Identify the circuit (Figure) in which electrical components have been properly connected.
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 5
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
(CBSE 2010)
Answer:
(b).
Explanation : Voltmeter is always connected in parallel and ammeter is connected in series in an electric circuit. Moreover, positive terminals of voltmeter and ammeter are connected with the +ve terminal of a cell or battery and negative terminals of voltmeter and ammeter are connected with -ve terminal of a cell or battery.

Question 6.
What is the maximum resistance which can be made using five resistors each 1/5 Ω ?
(a) 10 Ω
(b) 10 Ω
(c) 5 Ω
(d) I Ω (CBSE 2010)
Answer:
(d).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 6

Question 7.
What is the minimum resistance which can be made using 1 five resistors each of 1/5 Ω ?
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 7
Answer:
(d).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 8

Question 8.
The proper representation of series combination of cells obtaining maximum potential is
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 9
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
(CBSE 2011,2012)
Answer:
(d).
Explanation : Cells are connected in series if -ve terminal of one cell is connected to the +ve terminal of other cell.

Question 9.
Which of the following represents voltage ?
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 10
Answer:
(a).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 11

Question 10.
A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross-section.
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
Answer:
(c).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 12

Question 11.
A student carries out an experiment and plots the V—I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively as shown in figure. Which of the following is true ?
(a) R1 = R2 = R3
(b) R1 > R2 > R3
(c) R3 > R2 > R1
(d) R2 > R3 > R1
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 14
Answer:
(c).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 15

Question 12.
If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be
(a) 100 %
(b) 200%
(c) 300%
(d) 400%
Answer:
(c).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 16

Question 13.
The resistivity does not change if
(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed.
Answer:
(c).
Explanation : Resistivity of material does not depend on its di¬mensions and shape. However, resistivity depends upon the tem¬perature and the nature of material.

Question 14.
In an electrical circuit, three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness ?
(a) Brightness of bulb A will be maximum
(b) Brightness of bulb B will be more than that of A
(c) Brightness of bulb C will be less than that of B
(d) Brightness of all the bulbs will be same
Answer:
(b).
Explanation : Brightness of bulb depends upon the heat produced per second. Since resistance of bulb is inversely proportional to the power, so resistance of bulb A is more than that of B and resistance of bulb B is more than that of C. Heat produced per second = V2/R
where V is same for all bubls.
Therefore, brightness of bulb B is more than that of A.

Question 15.
In an electrical circuit two resistors of 2Ω and 4Ω respectively are connected in series to a 6V battery. The heat dissipated by 4Ω resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J
Answer:
(c).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 17

Question 16.
An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it ?
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A
Answer:
(d).
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 18

Question 17.
Two resistors of resistance 2Ω and 4Ω when connected to a battery will have
(a) same current flowing through them when connected in parallel
(b) same current flowing through them when connected in series.
(c) same potential difference across them when connected in series
(d) different potential difference across them when connected in parallel
Answer:
(b)
Explanation : Resistors are said to be connected in series if same current flows through them as series combination has a single path for the flow of current. Resistors are said to be connected in parallel if potential difference across them is

Question 18.
Unit of electric power may also be expressed as
(a) Volt ampere
(b) kilowatt hour
(c) watt second
(d) joule second
Answer:
(a).
Explanation : P = VI.

Question 19.
A child has drawn the electric circuit to study Ohm’s law as shown in figure. His teacher told that the circuit diagram needs correction. Study the circuit and redraw it after making all corrections.
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 19
Answer:
Ammeter is connected in series in an electric circuit and volt¬meter is connected in parallel to the resistor R. Positive terminals of ammeter and voltmeter are connected with the positive terminal of battery and negative terminals of ammeter and voltmeter are connected with the negative terminal of the battery. The correct circuit is shown below.
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 20

Question 20.
Three 20 resistors, A, B and C are connected as shown in figure. Each of them dissipates energy that can withstand a maximum power of 18W without melting. Find the maximum current that can flow through three resistors.
(CBSE, 2010)
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 21
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 22
Since resistors B and C are connected in parallel, so potential difference across B and C is same. Let I1, be the current flowing through resistor.B and I2 be the current flowing through resistor C
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 23

Question 21.
Should the resistance of an ammeter be low or high ? Give reason.
Answer:
Ammeter is connected in series in an electric circuit to measure electric current. If its resistance is high, then the net resistance of the electric circuit will increase and hence current in the electric circuit will decrease. Hence, ammeter will not read the actual value of the current in the circuit. If resistance of the ammeter is low, then the net resistance of the circuit will not be effected much. Hence, the current in the circuit is not affected. Ideal ammeter has zero resistance.

Question 22.
Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a reistor of 2Ω in series with a combination of two resistors (4Ω each) in parallel and a voltmeter across the energy. 1 kWh = 1000 Watt x 3600 s parallel combination. Will the potentional difference across the 2Ω resistor be the same as that across the parallel combination of 4Ω resistors ? Give reason.
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 24
Therefore, potential difference across 2Ω resistor will be same ; as that of across the parallel combination of 4Ω resistors. V = IR.
As R and I in both the cases is same so V = same.

Question 23.
How does use of a fuse wire protect electrical appliances ?
(CBSE 2012)
Answer:
When large current flows in an electric circuit, fuse wire melts due to the large heat produced in it. Therefore, current stops flowing in the circuit and electrical appliances connected in the circuit are protected from burning due to large current in j the circuit.

Question 24.
In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5A. The reading of ammeter decreases to half when the length of the wire is doubled. Why ?
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 25

Question 25.
What is the commercial unit of electrical energy ? Represent it in terms of joule.
Answer:
kilowatt hour (kWh) is the commercial unit of electrical energy.
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 26

Question 26.
A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5Ω when connected to a 10V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5Ω condcutor and potentional difference across the lamp will take place ? Give reason. Draw circuit diagram.
(CBSE 2010)
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 27
Net resistance of lamp and conductor connected in series = 5 + 5 = 10Ω
Net resistance of the combination oflamp and conductor and resistance 10Ω connected in parallel,
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 28
It means 1A current will flow through 10Ω resistance and 1A current will flow through the lamp and conductor of 5Ω resistance. Hence, there is no change in the current through 5Ω conductor.
Now potential difference across the lamp in case (i)
V = IR = 1 x 5 = 5V
Potential difference across the lamp in case (ii)
V = IR = 1 x 5 = 5V
Hence, there is no change in the potential difference across the lamp.

Question 27.
Why is parallel arrangement used in domestic wiring ?
(CBSE 2012)
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 28.
B1, B2 and B3 are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

  1. What happenes to the glow of other bulbs when the bulb B1 gets fused ?
  2. What happens to the reading of A1, A2, A3 and A when the bulb B2 gets fused ?
  3. How much power is dissipated in the circuit when all the three bulbs glow together ?
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 29

Answer:

  1. The glow of bulb depends upon the energy disspated per second i.e. P =V2/R.
    Since V and R of both the bulbs B2 and B3 remain the same even if bulb B, gets fused so glow of B2 and B3 remain the same.
  2. Since bulbs are identical, so their resistance is equal (i.e. resistance of each bulb = R ohm).
    When all bulbs glow, net resistnace of the circuit is given by
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 30
    When B2 gets fused, only two bulbs B1 and B2 in parallel are in the circuit.
    .’. Net resistance of the circuit is given by
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 31
    Thus, reading of ammeter A = 2A
    Since B1 and B3 are in parallel and have same resistance, so 2A current will be equally distributed between B1 and B3. Therefore, reading of ammeter A1 = 1A Reading of ammeter A3 = 1A Circuit containing B2 is broken, so no current flows through this circuit. Hence reading of ammeter A2 = zero.
  3. Power dissipated in the circuit,
    P =V x I
    = 4.5 x 3 = 13.5 W

Question 29.
Three incandescent bulbs of 100W each are connected in series in an electric circuit. In another circuit, another set of three bulbs of the same voltage are connected in parallel to the same source.
(a) Will the bulb in the two circuits glow with the same brightness ? Justify your answer.
(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit ? Give reason.
(CBSE 2012)
Answer:
(a) Power dissipated in a circuit = V2/R. Since resistance of the circuit containing bulbs connected in series is more than the resistnace of the circuit containing bulbs in parallel, therefore, power dissipated in parallel combination is more than that in series combination. Hence, bulbs connected in parallel will glow more brightly.
(b) In series combination, there is only one path for the flow of current. So when one bulb gets fused, circuit is broken and hence the bulbs stop glowing.
In parallel combination, each bulb has its own path for the flow of current. So when one bulb gets fused, other bulbs will continue to glow as the current is flowing in the circuits of these bulbs.

Question 30.
State Ohms law. How can it be verified experimentally ? Does it hold good under all conditions ? Comment.
(CBSE 2010)
Answer:
For Ohm’s law: Ohm’s law states that the electric current flowing through a conductor is directly proportional to the potential difference across the ends of the conductor, provided the temperature and . other physical conditions of the conductor remain the same.
For experimental verification: Verify Ohm’s law
Apparatus : A conductor of resistance R, an ammeter, a voltmeter, a battery, a variable resistance (or rheostat used to change the current in the circuit), connecting wires, a key and sand paper.
Procedure:

    1. Connect the various components as shown in figure 12.
      NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 32
    2. Close the key, so that current begins to flow in the circuit.
    3. Note down the potential difference (V) across the conductor PQ of resistance R shown by the voltmeter and the corresponding current (I) shown by the ammeter.
    4. Now move the knob of rheostat so that the current in the circuit increases.
    5. Again note down the potential difference (V) across the conductor PQ of resistance R in the voltmeter and current in the circuit shown by ammeter.
    6. Repeat the experiment at least five times by increasing the current in the circuit by moving the knob of the rheostat in steps.

Observations:

S.No

Potential Difference
 (V)
Current
(I)

V/I

1.

2.

3.

4.

5.

6.

7.

Plot a graph between V and I by taking V along X-axis and I along Y-axis. We get a straight line passing through origin as shown in figure 11.
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 33
Conclusion: From the graph between V and I, we conclude that I x V, which is Ohm’s law. Hence Ohm’s law is verified experimentally.
Precautions: While verifying Ohm’s law experimentally, the following precautions should be taken :

  1. Current should not be allowed to pass through the circuit continuously for a long time, which may cause the increase in temperature of the conductor. Therefore, the plug of the key must be taken out every time after noting the readings of ammeter and voltmeter.
  2. Connections should be tight.
  3. The conductor used in the experiment should be such that its resistance is not changed with increase in temperature of the conductor.

Ohm’s law holds good if the temperature of the conductor remains the same.

Question 31.
What is electrical resistivity of a material ? What is its unit ?
Describe an experiment to study the factors on which the resistance of conducting wire depends. (CBSE 2012)
Answer:
For Electrical resistivity: Electrical resistivity of a material is defined as the resistance of an object (made of the material) of unit length and unit area of cross-section.
Unit of electrical resistivity is ohm-metre.
For experiment:
Connect the various electrical components as shown in figure 15.
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 34

  1. Dependence of length of a conductor: 
    Take a copper wire of length l and connect it between the terminals A and B. Note the reading of ammeter. Now take another copper wire of same area of cross-section but of length 2l. Connect it between the terminals A and B by disconnecting the previous wire. Again, note the reading of ammeter. It will be found that the reading of ammeter (i.e., electric current) in the second case is half of the reading of ammeter in the first case. Since R = V/I, so resistance of second wire is double than the resistance of the first wire. Thus, resistance of a conductor is directly proportional to the length of the conductor,
    i.e      resistance ∝ length of the conductor
    Thus, more is the length of a conductor, more is its resistance.
    Thus, the resistance of’a conductor is ‘inversely proportional to the area of cross -section of the conductor.
  2. Dependence on area of cross-section of a conductor:
    Now take two copper wires of same length but of different area of cross-sections. Let area of cross-section of first wire is more than the area of cross-section of the second wire. Connect first wire between the terminals A and B in the circuit shown in figure 15. Note the reading of ammeter. Now disconnect the first wire and connect the second wire between the terminals A and B. Again note the reading of the ammeter. It will be found that the reading of ammeter (i.e. electric current) is more when first wire (i.e. thick wire) is connected between A and B than the reading of the ammeter when second wire (i.e. thin wire) is connected between the terminals A and B.
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 35
    Thus, the resistance of a thin wire is more than the resistance of a thick wire.
  3. Effect of the Nature of material:
    Take two identical wires, one of copper and other of aluminium. Connect the copper wire between the terminals A and B. Note the reading of ammeter. Now, connect the aluminium wire between the terminals A and B. Again note the reading of ammeter. It is found that the reading of ammeter when copper wire is connected in the circuit is more than the reading of the ammeter when aluminium wire is connected in the circuit.
    Therefore, resistance of copper wire is less than the resistance of aluminium wire. Hence, resistance of a wire or a conductor depends upon the nature of the material of the conductor.
  4. Effect of temperature of conductor:
    If the temperature of a metallic conductor connected in the circuit increases, its resistance increases.
    Thus, factors on which resistance of a conductor depends are :

    1. its length,
    2. its area of cross-section,
    3. the nature of its material and
    4. its temperature.

The various factors affecting resistance of a conductor are given in table 1.
Table 1. Factors affecting resistance of a conductor:
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 36

Question 32.
How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery ?
Answer:
Perform an activity to show that in series combination of resistors, same current flows through each resistor.

  1. Connect three resistors of resistances R1 = 1Ω, R2 = 2Ω and R3 = 3Ω in series. (C.B.S.E. 2014)
  2. Connect the series combination of resistors with a battery of 6 V, a plug key K and an ammeter A as shown in figure 17.
  3. Note the reading of ammeter after plugging the key. Let it be I.
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 37
  4. Now disconnect ammeter and connect it in between the resistors R1 and R2 as shown in figure 18.
  5. Again plug the key and note the reading of ammeter. It is found that again it is I.
  6. Now disconnect the ammeter and connect it in between the resistors R2 and R3.
  7. Plug the key and note the reading of ammeter. Again it is found to be I.

Conclusion : Same amount of current flows through each resistor or element connected in series combination.

Question 33.
How will you conclude that the same potential difference (Voltage) exists across three resistors connected in parallel arrangement to a battery ?
Answer:
Perform an activity to investigate the relation between potential difference across parallel combination of resistors and the potential difference across each individual resistors,

  1. Connect three resistors of resistances R1, R2 and R3 in parallel. One end of each resistor is joined at a common point ‘a’ and the other end of each resistor is connected at another common point ‘b’.
  2. Connect the parallel combination of resistors with a battery, a plug key K and an ammeter A as shown in figure 22(A).
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 38
  3. Now connect a voltmeter across the parallel combination of resistors between a and b points.
  4. Note the reading of voltmeter. Let it be V. This is the potential difference across the parallel combination of resistors.
  5. Now, disconnect the voltmeter and connect it across R1 as shown in figure 22(B).
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 39
  6. Note the reading of voltmeter. It is found to be V.
  7. Disconnect the voltmeter and connect it across R2. Note the reading of voltmeter. It is found to be V.
  8. Again disconnect the voltmeter and connect it across R3. Note the reading of voltmeter. It is found to be V.

Conclusion : When resistors are connected in parallel to each other, potential difference across each resistor is equal to the potential difference across the parallel combination of resistors.

Question 34.
What is Joule’s heating effect ? How can it be demonstrated experimentally ? List its four applications in daily life.
(CBSE 2012)
Answer:
Joule’s law can be stated as : The amount of heat produced in a conductor is
(i) directly proportional to the square of the electric current flowing through it.
This is                  H ∝ I2                                                           –
(ii) directly proportional to the resistance of the conductor or resistor.
That is,                H ∝ R                                          ‘
(iii) directly proportional to the time for which the electric current flows through the conductor or resistor.
That is,               H ∝ t
Combining (i), (ii) and (iii), we get H ∝ I2 Rt.
or         H = KI2Rt, where K is constant of proportionality
If          K = 1, then H = I2Rt joule
This is known as Joule’s law of heating.
Four Applications:

  1. When electric appliances like electric heater, electric iron and water heater etc. are connected to the main supply of electricity, these appliances become hot but the connecting wires remain cold.
    We know, heat produced is directly proportional to the resistance of the material through which current flows. The element of electric heater is made of nichrome. Since, resistance of nichrome is high, so a large amount of heat is produced in the element of the electric heater. Thus, filament of electric heater becomes red hot. However, heat produced in connecting wires made of copper or aluminium is very small and hence they are not heated up.
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 40
  2. Filament of an electric bulb is made of a thin wire of tungsten. The melting point of filament is high i.e., about 3380 °C. The filament of the bulb is enclosed in a glass envelope fixed over an insulated support as shown in figure 28. The glass envelope of electric bulb is filled with inactive gases like nitrogen and argon to increase the life of the tungsten filament.
    Since resistance of thin filament is very high, so a large heat is produced as the electric current flows through the filament. Due to this large amount of heat produced, filament of the bulb becomes white hot. Hence, the filament of the bulb emits light and heat.
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 41
  3. Electric fuse is a safety device connected in series with the electric circuit. Electric fuse is a wire made of a material whose melting point is very low. Examples of the materials of making fuse wire are copper or tin-lead alloy. When large current flows through a circuit and hence through a fuse wire, a large amount of heat is produced. Due to this large amount of heat, the fuse wire melts and the circuit is broken so that current stops flowing in the circuit. This saves the electric circuit from burning due to the passage of large current through it.
    NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 42

Question 35.
Find out the following in the circuit given in the figure.
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 43
(a) Effective resistance of two 8Ω resistors in the combination.
(b) Current flowing through 40 resistor
(c) Potentional difference across 40 resistor
(d) Power dissipated in 4Ω resistor
(e) Difference in ammeter readings, if any.
(CBSE 2010,2012)
Answer:
(a) Two 80 resistors are in parallel, so their effective resistance is given by
NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity image - 44
(e) Since same current flows through every part in a series circuit and both the ammeters are connected in series, so there will be no difference in ammeter readings.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity 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.

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

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

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

Question 1.
Choose the incorrect statement from the following regarding magnetic lines of field.
(a) The direction of magnetic field at a point is taken to be the direction in which the north pole of a magnetic compass needle points.
(b) Magnetic field lines are closed curves.
(c) If magnetic field lines are parallel and equidistant, they rep-resent zero field strength.
(d) Relative strength of magnetic field is shown by the degree of closeness of the field lines.
Answer:
(c).
Explanation : Magnetic field lines which are parallel and equidistant represent uniform magnetic field.

More Resources

Question 2.
If the key in the arrangement shown in figure is taken out (the circuit is made open) and magnetic field lines are drawn over the horizontal plane ABCD, the lines are
(a) concentric circles
(b) elliptical in shape
(c) straight lines parallel to each other
(d) concentric circles near the point O but of elliptical shapes as we go away from it
Answer:
(c).
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 1
Explanation : When no current flows through the wire, mag¬netic field lines over the plane ABCD will represent magnetic field of the earth. These field lines are straight lines parallel to each other.

Question 3.
A circular loop placed in a plane perpendicular to the plane of paper carries a current when the key is ON. The current is seen from points A and B (in the plane of the paper and on the axis of the coil) is anticlockwise and clockwise respectively. The magnetic field lines point from B to A. The N-pole of the resultant magnet is on the face close to
(a) A
(b) B
(c) A if current is small, and B if current is large.
(d) B if the current is small, and A if current is large.
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 2
Answer:
(a).
Explanation : If current seems to flow in anticlockwise direction in a coil, then that face of the coil acts as N-pole.

Question 4.
In a long straight solenoid, N-and S-poles are created at the two ends of the solenoid. Which of the following statements is incor¬rect ?
(a) The field lines inside the solenoid are in the form of straight lines indicating that magnetic field is same at ail points inside the solenoid.
(b) The pattern of the magnetic field associated with the solenoid is different from the pattern of the magnetic field around a bar magnet.
(c) N-and S-poles exchange positions when the direction of the current through the solenoid is reversed.
(d) Magnetic field produced inside the solenoid can be used to magnetise a bar of magnetic material like soft iron, when placed inside the solenoid.
Answer:
(b).
Explanation : The pattern of magnetic field produced by a cur¬rent carrying solenoid resembles with the pattern of magnetic field of a bar magnet.

Question 5.
A uniform magnetic field exists in the plane of paper pointing from left to right as shown in figure. In the field, an electron and a proton move as shown. The electron and proton experience
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 3
(a) forces, both pointing into the plane of paper
(b) forces, both pointing out of the plane of paper
(c) forces, pointing into the plane of paper and out of the plane of paper respectively.
(d) forces, pointing opposite and along the direction of the uniform magnetic field respectively.
Answer:
(a).
Explanation : Apply Fleming’s left hand rule and considering the flow of current in the direction of the movement of proton and in the direction opposite to the movement of electron.

Question 6.
Commercial electric motors do not use
(a) an electromagnet to rotate the armature
(b) effectively large number of turns of conducting wire in the current carrying coil
(c) a permanet magnet to rotate the armature id a soft iron on which the coil is wound.
Answer:
(c).
Explanation : In commercial motor, armature rotates in the mag-netic field produced by an electromagnet. The strength of such magnetic field can be increased to the desired value. However, a permanent magnet provides a fixed and weak magnetic field.

Question 7.
Figure shows the arrangment of two coils wound on a non-conducting cylindrical rod. The plugh of the key is then in¬serted and later removed.
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 4
(a) Deflection in the galvanometer remains zero throughout
(b) There is momentary deflection in the galvanometer that dies out shortly and deflections in both the cases are in the same direction
(c) There is momentary deflection in the galvanometer that dies out shortly and there is no effect when the key is re¬moved
(d) There are momentary deflections in both the cases that die out shortly but the deflections are in apposite directions.
Answer:
(a).
Explanation : Due to electromagnetic induction, induced cur¬rent flows in the coil with which galvanometer is connected, when the key is inserted (i.e. current grows in the primary coil) and when the key is removed (i.e. current decays in the primary coil).

Question 8.
Choose the incorrect statement
(a) Fleming’s right hand rule is a simple rule to know the direction of induced current.
(b) The right-hand thumb rule is used to find the direction of magnetic fields due to current carrying conductors.
(c) The difference between the direct and alternating currents is that the direct current always flows in one direction, whereas the alternating current reverses its direction periodically.
(d) In India, the AC changes direction after every 1/60 seconds
Answer:
(d).
Explanation : Frequency of AC in India = 50 Hz
∴ AC completes 1 cycle in 1/50 seconds.
In one cycle, AC changes direction two times. Hence, time after which AC changes direction
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 5

Question 9.
A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in figure. The direction of magnetic field at a point will be north to south.
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 6
(a) directly below the wire
(b) directly above the wire
(c) at a point located in the plane of the paper, on the north side of the wire
(d) at a point located in the plane of the paper, on the south side of the wire.
Answer:
(d).
Explanation : Applying right hand thumb rule to know the direction of the magnetic field.

Question 10.
The strength of magnetic field inside a long current carrying straight solenoid is
(a) more at the ends than at the centre
(b) minimum in the middle
(c) same at all points
(d) found to increase from one end to the other.
Answer:
(c).
Explanation : Magnetic field inside a long current carrying straight solenoid is uniform.

Question 11.
To convert AC generator to DC generator
(a) slip rings and brushes must be used ‘
(b) split ring type commutator must be used
(c) rectangular wire loop is to be used
(d) stronger magnetic field is to be used.
Answer:
(b).

Question 12.
The most important safety device method used for protecting electrical appliances from short circuiting or overloading is
(a) earthing
(b) use of stabilizers
(c) use of electric meter
(d) use of fuse.
Answer:
(d).
Explanation : Fuse melts and the circuit breaks as soon as the current in the circuit exceeds safe limit.

Question 13.
A magnetic compass needle is placed in a plane of paper near point A as shown in figure.
In which plane should a straight current carrying conductor be placed so that it passes through A and there is no change in the deflection of the compass ? Under what condition is the deflection maximum and why?
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 7
Answer:
The straight current carrying conductor should be placed in the plane of the paper. The deflection of the compass is maxi¬mum when the conductor passes through A and perpendicular to the plane of the paper. The magnetic field due to current carrying conductor is maximum in the plane of the paper.

Question 14.
Under what conditions permanent electromagnet is obtained if a current carrying solenoid is used ? Support your answer with the help of a labelled circuit diagram.
Answer:
Permanent electromagnet is obtained if

  1. direct current flows through the solenoid and
  2. steel rod (a magnetic substance) is placed inside the solenoid.

In other words, a steel rod can be made a permanent electro-magnet if it is placed inside a solenoid carrying direct current as shown in figure.
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 8

Question 15.
AB is a current carrying conductor in the plane of the paper as shown in figure. What are the directions of magnetic fields produced by it at points P and Q ? Given, r1 > r2, where will the strength of the magnetic field be larger ?
(CBSE 2010, 2012)
NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 9
Answer:
According to Right hand thumb rule, the direction of the mag¬netic field is into the plane of the paper at P and the direction of the magnetic field is out of the plane of the paper at Q.
The strength of the magnetic field due to a current carrying conductor decreases with increase in distance (i.e. B ∝ 1/r ).
Therefore, strength of the magnetic field will be larger at point Q than at point P.

Question 16.
A magnetic compass shows a deflection when placed near a current carrying wire. How will the deflection of the compass get affected if the current in the wire is increased ? Support your answer with a reason. (CBSE 2012)
Answer:
A magnetic compass is a tiny magnet. It gets deflected when placed in a magnetic field. When the current in the wire increases, then the magnetic field produced by the wire also increases ( B ∝ I). Hence, deflection of the compass increases.

Question 17.
It is established that an electric current through a metallic conductor produces a magnetic field around it. Is there a similar magnetic field produced around a thin beam of moving

  1. alpha particles,
  2. neutrons ? Justify your answer.
    (CBSE 2012)

Answer:

  1. A thin beam of moving alpha particles constitute an electric current. Thus, beam of moving alpha particles behaves as a current carrying conductor. Hence, magnetic field is produced around the thin beam of moving alpha particles.
  2. Neutrons are neutral particles i.e. they do not have any charge. Therefore, moving neutrons do not constitute an electric cur-rent. Hence, no magnetic field is produced around the thin beam of moving neutrons.

Question 18.
What does the direction of thumb indicate in the right-hand thumb rule. In what way this rule is different from Fleming’s left-hand rule ? (CBSE 2010,2012)
Answer:
In the right hand rule, the direction of thumb indicates the direction of flow of current in the straight conductor.
Right-hand thumb rule is used to determine the direction of the magnetic field produced by a current carrying straight conductor. However, Flemings left hand rule is used to determine the direction of the force experienced by a current carrying conductor placed in a uniform magnetic field.

Question 19.
Meena draws magnetic field lines of field close to the axis of a current carrying circular loop. As she moves away from the centre of the circular loop, she observes that the lines keep on diverging. How will you explain her observation ? (CBSE 2010)
Answer:
Magnetic field due to current carrying circular loop decreases as we move away from the centre of the loop. Diverging field lines represent weak magnetic field.

Question 20.
What does the divergence of magnetic field lines near the ends of a current carrying straight solenoid indicate ?
Answer:
Divergence of magnetic field lines means the magnetic field lines are far apart. Such field lines indicate weak magnetic field. There¬fore, divergence of magnetic field lines near the ends of the current carrying solenoid indicates that the magnetic field near the ends of the current carrying solenoid is weak and non-uniform.

Question 21.
Name four appliances wherein an electric motor, a rotating device that converts electrical energy to mechanical energy is used as an important component. In what respect motors are different from generators ?
Answer:
Appliances wherein an electric motor is used are

  1. electric fan,
  2. electric mixer,
  3. washing machine and
  4. a toy car.

Motor converts electrical energy into mechanical energy. How-ever, generator converts mechanical energy into electrical energy.

Question 22.
What is the role of two conducting stationary brushes in a simple electric motor ?
Answer:
These brushes act as the contacts between the commutator and the terminals of the battery.

Question 23.
What is the difference between direct current and an alternating current ? How many times does AC used in India change direction in one second ?
Answer:
Difference between Direct current and an Alternating current:

Direct current

Alternating current

1. The magnitude of direct current is constant and flows in one direction only.

2. Direct current cannot be used for large scale supply of electricity for household purpose.

3. The frequency of direct current is zero.

1. The magnitude of AC changes with time and direction of alternating current reverses periodically.

2. Alternating current is used to run electrical appliances like electric bulb, electric heater, electric iron, refrigerator etc.

3. Frequency of alternating current in India is 50 Hz.

Since frequency is 50 Hz, so a.c. completes 50 cycles in 1 second. Therefore, it reverses its direction 100 times in one second.

Question 24.
What is the role of fuse, used in series with any electrical appli-ances ? Why should a fuse with defined rating not be replaced by one with a larger rating ?
Answer:
Electric fuse is a safety device. It protects the electrical appliance due to short circuiting and overloading in the electric circuit.
Fuse with defined rating means the maximum current that can flow’ through the fuse wire without melting it. As and when current exceeds this maximum value, the fuse melts and the circuit is switched off. If fuse with defined rating is replaced by one with a larger rating, then large current will flow through the circuit without melting the fuse. This large current may burn the electrical appliance connected in the circuit and the fuse will not act as a safety device.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current 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.

Value Based Questions in Science for Class 10 Chapter 13 Magnetic Effects of Electric Current

Value Based Questions in Science for Class 10 Chapter 13 Magnetic Effects of Electric Current

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 13 Magnetic Effects of Electric Current

Question 1.
A fuse wire is a safety device to protect the electric circuit. Ashok used a fuse rated as 5 A in a circuit in which an electric heater of 2 kW was connected with the main supply of 220 V. The heater was not operating as the fuse wire melted on switching. Ashok was puzzled and he discussed the problem with his friend Ram. Ram suggeted him to use a fuse rated 10 A in the circuit. The problem of Ashok was solved as the heater was now working.
(i) What values are shown by Ram ?
(ii) How much current was flowing in the circuit with the fuse rated 10 A ?
Answer:
(i) Helpful, high degree of awareness.
Value Based Questions in Science for Class 10 Chapter 13 Magnetic Effects of Electric Current image - 1

More Resources

Question 2.
Overloading in an electric circuit causes fire. Common people are unware of this fact. Akshil, an intelligent student told people of his locality about overloading. He told them to take various precautions to avoid overloading. What values are shown by Akshil ?
He is helpful and has high degree of awareness. Interested in educating people.

Question 3.
Students of a school ‘X’ went to a forest for a picnic. They went deep into the forest and started playing on a flat field in the forest. A stranger came there and asked for help. He had forgotten the direction of his village situated in south direction outside the forest. All the students except Sobit said that they had no idea about south direction. They asked stranger to find the south direction himself. However, Sobit asked him to wait for a moment. Sobit had a magnetic compass. He placed the magnetic compass on the ground and told the stranger about south direction.

  1. How the magnetic compass determines the direction from the given place ?
  2. What values are shown by Sobit ?
  3. Is the behaviour of other students desirable ?

Answer:

  1. When a magnetic compass is placed at a place, it always points towards north-south direction.
  2. Sobit is helpful, caring and has a high degree of awareness.
  3. The behaviour of other students is undesirable. They did not bother about the problem of the stranger.

Question 4.
Aayan was going to his home from his school. On his way, he heard that people in a house were crying and weeping. He went to the house to know the reason. He found that a boy is lying on a floor and people around him were crying and weeping. He came to know that boy suffered a severe shock, when he was working with electric iron. He asked people to stop crying and help him to take the boy to the nearby dispensary for Medical aid. The boy was rushed to the dispensary. Medical aid was given to him and his life was saved.
Answer the following questions based on above information.

  1. Why the boy suffered severe shock ?
  2. Comment on the values shown by Aayan.

Answer:

  1. The boy might have touched the metallic casing of electric iron through which current was passing. Hence boy suffered a severe shock.
  2. Aayan is helpful. He feels concerned about the welfare of others. He is quick in his actions and has high degree of general awareness.

Question 5.
What would you suggest to the people of your area regarding the safety measures in using electricity ?
Answer:

  1. Use of electric fuses of proper rating,
  2. Electricity wires should not be touched bare footed.
  3. Electricity should be switched off during lightning and thunder,
  4. Electric circuits should be properly earthed,
  5. Avoid overloading i.e. electric appliances of high power rating should not be connected in a single socket or circuit.

Hope given Value Based Questions in Science for Class 10 Chapter 13 Magnetic Effects of Electric Current 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.

HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

Question 1.
One end of a bar magnet is brought near the south pole of a magnetic compass needle. It was noticed that the needle of the magnetic compass deflected away from the end of the bar magnet. Name the pole of the bar magnet at its end pointing towards the south pole of the magnetic compass. Justify your answer.
Answer:
The pole of the bar magnet at the end pointing towards the south pole of the magnetic compass is south pole. This is because, like magnetic poles repel each other.

More Resources

Question 2.
Magnetic field lines around a bar magnet are shown in figure. A student makes a statement that magnetic field at point A is stronger than at point B. State, whether the statement is correct or incorrect. Explain.
(CBSE 2012)
HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 1
Answer:
Statement is correct. This is because field lines are crowded in a region of strong magnetic field and field lines diverge in a region of weak magnetic field.

Question 3.
Magnetic lines of force of two pairs of magnets are shown in figure A and B. Out of these two figures, which one represents the correct pattern of field lines. Name the poles of magnets facing each other. (CBSE 2012)
HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 2
Answer:
Figure B represents the correct pattern of magnetic field lines of a pair of magnets. Figure A does not represent the correct pattern of field lines because magnetic field lines never cross each other.
HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current image - 3
Poles of magnets facing each other are north poles as the magnetic field lines emerge from a magnet at N-pole.

Question 4.
A student performs an experiment to study the magnetic effect of current around a current carrying straight conductor. He reports that

  1. the direction of deflection of the north pole of a compass needle kept at a given point near the conductor remains unaffected even when the terminals of the battery sending current in the wire are interchanged.
  2. for a given battery, the degree of deflection of a N-pole decreases when the compass is kept at a point farther away from the conductor. Which of the above observations of the student is incorrect and why ?
    (CBSE Sample Paper Question)

Answer:

  1. The first statement is incorrect because the direction of the magnetic field around the conductor is reversed when the direction of the current flowing in the conductor changes. Hence, the direction of deflection of N-pole of the compass needle cannot remain unaffected when the terminals of the battery sending current in the wire or conductor are interchanged.
  2. The second statement is correct because the magnetic field decreases with the increase in the distance from the conductor.

Question 5.
A current carrying conductor is placed perpendicular to the magnetic field of horse-shoe magnet. The
conductor is displaced upward. What will happen to the displacement of the conductor if

  1. current in the conductor is increased,
  2. a horse-shoe magnet is replaced by another stronger horse-shoe magnet and
  3. the length of the conductor is increased. (Similar Text Book Question)

Answer:

  1. The force acting on a current carrying conductor placed perpendicular to a magnetic field increases with the increase in the current flowing through a conductor. Thus, the displacement of the conductor will increase if the current in the conductor is increased.
  2. When a horse-shoe magnet is replaced by a stronger magnet, then magnetic field increases. Since, force acting a conductor increases with the increase in the magnetic field, therefore, the displacement of the conductor will increase.
  3. Since, the force acting on the conductor increases with the increases in the length of the conductor, therefore, the displacement of the conductor will increase.

Hope given HOTS Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B

Other Exercises

Question 1.
In the figure given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m. ∠ADB = 30° and ∠ACB = 45°. Without using tables, find X.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q1.2

Question 2.
Find the height of a tree when it is found that on walking away from it 20m, in a horizontal line through its base, the elevation of its top Changes from 60° to 30°.
Solution:
Let AB be the tree and its height be x DC = 20 m.
Now in right ∆ADB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q2.2

Question 3.
Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30° to 45°.
Solution:
Let AB be the building and its
height be x and DC =40m
Now in right ∆ADB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q3.2

Question 4.
From the top of a light house 100 m high, the angles of depression of two ships are observed as 48° and 36° respectively. Find the distance between the two ships (in the nearest metre) if :
(i) the ships are on the same side of the light house,
(ii) the ships are on the opposite sides of the light house.
Solution:
AD = 100 m
In right angled ∆ABD
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q4.2

Question 5.
Two pillars of equal heights stand 011 either side of a roadway, which is 150m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60° and 30”; find the height of the pillars and the position of the point.
Solution:
Let AB and CD be the two pillars which stand on either side of a road BD, then BD = 150m
Let AB = CD = h
Let P be the point on the road such that the angles of elevation from P to the top of the pillars are 60° and 30° respectively.
Let BP = x then PD = 150 – x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q5.2

Question 6.
From the figure given below, calculate the length of CD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q6.2

Question 7.
The angle of elevation of the top of a tower is observed to be 60°. At a point 30 m vertically above the first point of observation, the elevation is found to be 45°. Find :
(i) the height of the tower,
(ii) its horizontal distance from the points of observation.
Solution:
Let height of tower = x i.e. AB = x
In the right ∆ADB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q7.3

Question 8.
From the top of a cliff. 60 metre high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.
Solution:
Height of the cliff AB = 60m
Let the height of tower = x
Draw TR \(\parallel\) SB. meeting AB in R
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q8.2

Question 9.
A man on a cliff observes a boat, at an angle of depression 30°, which is sailing towards the shore to the point immediately beneath him. Three minutes later the angle of depression of the boat is found to be 60°. Assuming that the boat sails at a uniform speed, determine :
(i) how much more time it will take to reach the shore.
(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.
Solution:
Height of cliff = 500 m.
In right ∆ACD,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q9.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q9.3

Question 10.
A man in a boat rowing away from a light house 150m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.
Solution:
Height of the lighthouse AB = 150m
Now is right ∆ACB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q10.2

Question 11.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find :
(i) the height of the tree, correct to 2 decimal places,
(ii) the width of the river.
Solution:
TR is the height of tree and RQ is the width of the river.
Let TR = x
Now in right ∆TQR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q11.2

Question 12.
The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160m high, is 45°. Find the height of the first tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q12.1
Let the height of first tower CD = x
and height of second tower AB = 160 m
Distance between them DB = 75 m.
In right ∆ACE,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q12.2

Question 13.
The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun’s altitude is 30″ than when it was
45°. Prove that the height of the tower is y (\(\sqrt { 3 }\) +1) metres.
Solution:
Let AB be the tower and AB = x
Distance CD = 2y
In right ∆ADB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q13.2

Question 14.
An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°, find the uniform speed of the aeroplane in km per hour.
Solution:
Height of aeroplane = 1 km = 1000 m.
In right ∆ACB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q14.2

Question 15.
From the top of a hill, the angles of depression of two consecutive kilometre stones, due east, are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill. (2007)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q15.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B are helpful to complete your math homework.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A

Other Exercises

Question 1.
The height of a tree is \(\sqrt { 3 }\) times the length of its shadow. Find the angle of elevation of the sun.
Solution:
Let AB be the tree and BC be its shadow.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q1.1
∴ θ = 60°
∵ Angle of elevation of the sun = 60°

Question 2.
The angle of elevation of the top of a tower, from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.
Solution:
Let AB be the tower and C is the point which is 160 m away from the foot of the tower,
i.e. CB = 160 m
Let height of the tower be x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q2.1

Question 3.
A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68“ with the ground. Find the height, upto which the ladder reaches.
Solution:
Let AB be the wall and
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q3.1
AC be the ladder, which is placed against the wall. If foot is 2.4 m away from the wall i.e. CB = 2.4m1.
Let AB =x m.
In right ∆ ABC,
tan θ = \(\frac { AB }{ BC }\) ⇒ tan 68° = \(\frac { X }{ 2.4 }\)
∴ x = 2.4 x tan 68° = 2.4 x 2.4751
= 5.94 m

Question 4.
Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50 m.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q4.1
Two persons A and B are standing on the opposite side of the tower TR and height of tower TR = 50 m and angles of elevation with A and B are 30° and 38° respectively. Let AR = x and RB = y
Now in right ∆TAR,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q4.2

Question 5.
A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m. and the string makes an angle 30° with the ground.
Solution:
Let KT be the height of kite and PK is the string which makes an angle of 30° with the ground.
∴ KT = 60 m
Let KP = xm.
Now in right ∆PKT,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q5.1

Question 6.
A boy 1.6m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45° (ii) 60°. Find the height of the tower in each case.
Solution:
(i) Let AB be the tower and MN be the boy who is 20m away from the foot of the tower.
Let AB = x and angle of elevation = 45°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q6.3

Question 7.
The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15m. What was the height of the tree before it was broken ?
Solution:
Let AB be the tree which was broken at the point C which makes an angle of elevation of 45°, with the ground at a distance of 15m.
BD = 15m
AC = CD
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q7.2

Question 8.
The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60° ?
Solution:
Distance of a point from the tower = 80 m
Angle of elevation = 30°
In second case the elevation of lower = 60°
In first case,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q8.3

Question 9.
At a particular time, when the sun’s altitude is 30°, the length of the shadow of’C vertical tower is 45 m. Calculate :
(i) height of the tower.
(ii) the length of the shadow of the same tower, when the sun’s altitude is (a) 45° (b) 60°.
Solution:
Shadow of the tower = 45 m and angle of elevation = 30°
Let AB be the lower and BC is its shadow.
∴ CB = 45 m.
Now in right ∆ABC,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q9.1
(ii) In second case,
(a) Angle of elevation = 45°
and height of tower = 25.98 m or 15\(\sqrt { 3 }\) m
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q9.2
(b) Angle of elevation = 60°
and height of tower = 25.98 m or 15\(\sqrt { 3 }\) m.
Let shadow of the tower DB = xm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q9.3

Question 10.
Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32°24′ with the pole and when it is turned to rest against another pole, it makes angle 32°24′ with the road. Calculate the width of the road.
Solution:
Two poles AB and CD which are at the either end of a road BD. A ladder 30 m long subtends an angle of 32° 24′ with the first pole AB and 32°24′ with the road when it is turned to rest against the second pole CD.
Now in right ∆ABE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q10.1

Question 11.
Two climbers are at points A and Bona vertical cliff face. To an observer C, 40 m from the foot of the cliff on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers ?
Solution:
A and B are two climbers on the cliff and ob-server is at C, 40 m from the foot of the cliff while the angles of elevations of each climber is 48° and 57° respectively.
In right ∆ACD,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q11.2

Question 12.
A man stands 9m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28° and the angle of depression of the bottom of the pole is 13°. Calculate the height of the pole.
Solution:
Let PL is the pole and MN is the man The angle of elevation of the top of the pole = 28°
arid the angle of depression of the bottom of the pole =13°
Man is 9 m away from the pole,
i.e. MQ = 9 m
Now in right ∆PMQ,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q12.1

Question 13.
From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate to the nearest metre the distance of the buoy from the foot of the cliff.
Solution:
Let CD be the cliff and CD = 92m, B is the buoy,
then from C ,
the angle of depression is 20°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q13.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A are helpful to complete your math homework.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D

Other Exercises

Question 1.
Find the sum of G.P. :
(i) 1 + 3 + 9 + 27 +….to 12 terms.
(ii) 0.3 + 0.03 + 0.003 + 0.0003 +….to 8 terms.
(iii) \(1-\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } ….to\quad 9\quad terms\)
(iv) \(1-\frac { 1 }{ 3 } +\frac { 1 }{ { 3 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 3 } } ….to\quad n\quad terms \)
(v) \(\frac { x+y }{ x-y } +1+\frac { x-y }{ x+y } +….upto\quad n\quad terms\)
(vi) \(\sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ 3\sqrt { 3 } } +….to\quad n\quad terms\)
Solution:
(i) 1 + 3 + 9 + 27 +….to 12 terms.
Here a = 1, r = 3 and n = 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.8

Question 2.
How many terms of the geometric progression 1 + 4 + 16 + 64 +…. must be added to get sum equal to 5461 ?
Solution:
Sn = 5461 and G.P. is
1 + 4 + 16 + 64 +…..
Here, a = 1, r = 4 (r > 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q2.1

Question 3.
The first term of a G.P. is 27 and its 8th term is \(\\ \frac { 1 }{ 81 } \). Find the sum of its first 10 terms.
Solution:
First term of a G.P (a) = 27
T8 = \(\\ \frac { 1 }{ 81 } \), n = 10
a = 27
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q3.2

Question 4.
A boy spends Rs 10 on first day, Rs 20 on second day, Rs 40 on third day and so on. Find how much, in all, will he spend in 12 days?
Solution:
A boy spends Rs 10 on first day,
Rs 20 on second day
Rs 40 on third day and so on
G.P. is 10 + 20 + 40 +…. 12 terms
Here a = 10, r = 2 and n = 12 (r > 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q4.2

Question 5.
The 4th and the 7th terms of a G.P. are \(\\ \frac { 1 }{ 27 } \) and \(\\ \frac { 1 }{ 729 } \) respectively. Find the sum of n terms of this G.P.
Solution:
In a G.P.
T4 = \(\\ \frac { 1 }{ 27 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q5.3

Question 6.
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728 ; find its first term.
Solution:
In a G.P.
Common ratio (r) = 3
Last term (l) = 486
Sum of its terms (Sn) = 728
Let a be the first term, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q6.1

Question 7.
Find the sum of G.P. : 3, 6, 12, …… 1536.
Solution:
G.P. is 3, 6, 12,….1536
Here a = 3, r = \(\\ \frac { 6 }{ 3 } \) = 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q7.2

Question 8.
How many terms of the series 2 + 6 + 18 +…. must be taken to make the sum equal to 728 ?
Solution:
G.P. is 2 + 6 + 18 +….
Here a = 2, r = \(\\ \frac { 6 }{ 2 } \) = 3, Sn = 728
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q8.1

Question 9.
In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152. Find its common ratio.
Solution:
In a G.P.
Sum of first 3 terms : Sum of 6 terms = 125 : 152
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q9.2

Question 10.
Find how many terms of G.P.\(\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 } \)… must be added to get the sum equal to \(\\ \frac { 55 }{ 72 } \) ?
Solution:
\(\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 } \)…
Let n terms be added
Now,Sn = \(\\ \frac { 55 }{ 72 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q10.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q10.4

Question 11.
If the sum of 1 + 2 + 22 +…..+ 2n – 1 is 255, find the value of n.
Solution:
1 + 2 + 22 +…..+ 2n – 1 = 255
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q11.2

Question 12.
Find the geometric mean between :
(i) \(\\ \frac { 4 }{ 9 } \) and \(\\ \frac { 9 }{ 4 } \)
(ii) 14 and \(\\ \frac { 7 }{ 32 } \)
(iii) 2a and 8a3
Solution:
(i) G.M between \(\\ \frac { 4 }{ 9 } \) and \(\\ \frac { 9 }{ 4 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q12.2

Question 13.
The sum of three numbers in G.P. is \(\\ \frac { 39 }{ 10 } \) and their product is 1. Find the numbers.
Solution:
Sum of three numbers in G.P. = \(\\ \frac { 39 }{ 10 } \)
and their product = 1
Let number be \(\\ \frac { a }{ r } \), a, ar, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q13.2

Question 14.
The first term of a G.P. is – 3 and the square of the second term is equal to its 4th term. Find its 7th term.
Solution:
In G.P.
T1 = – 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q14.1

Question 15.
Find the 5th term of the G.P. \(\\ \frac { 5 }{ 2 } \), 1,…..
Solution:
Given G.P is \(\\ \frac { 5 }{ 2 } \), 1,…..
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q15.1

Question 16.
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
Solution:
Given, First term = a = 125….(i)
and Second term = ar = 25…..(ii)
Now, Divide eq. (ii) by eq (i), we get
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q16.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q16.2

Question 17.
Find the sum of the sequence \(– \frac { 1 }{ 3 } \), 1, – 3, 9,….upto 8 terms.
Solution:
Here, First Term, a = \(– \frac { 1 }{ 3 } \)…(i)
and Second Term, ar = 1 …(ii)
Dividing eq. (i) by eq. (ii), we get
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q17.1

Question 18.
The first term of a G.P. in 27. If the 8th term be \(\\ \frac { 1 }{ 81 } \), what will be the sum of 10 terms ?
Solution:
Given, First term (a) = 27, n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q18.2

Question 19.
Find a G.P. for which the sum of first two terms is – 4 and the fifth term is 4 times the third term.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q19.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D are helpful to complete your math homework.

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

NCERT Exemplar Solutions for Class 10 Science Chapter 16 Management of Natural Resources

NCERT Exemplar Solutions for Class 10 Science Chapter 16 Management of Natural Resources

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 16 Management of Natural Resources

NCERT Exemplar Solutions for Class 10 Science Chapter 16 Short Answer Questions

Question 1.
Prepare a list of items that you use daily in the school. Identify from the list five such items that can be recycled.
Answer:
Items. Rexin bag, steel lunch box, steel spoon, steel compass, steel dividers, paper, plastic box, pen, pencil, blade, eraser, handkerchief.
Recycleable Items. Steel lunch box, steel spoon, steel compass, steel dividers, blade, paper, plastic box.

More Resources

Question 2.
List the advantages associated with water harvesting at the community level. (CCE 2012)
Answer:
Water harvesting at the community level is capturing, collection and storage of rain water and surface run off for filling either small water bodies or recharging ground water. This is carried out through water shed management, check dams, earthen dams, roof top harvesting and filter wells in flood drains.
Benefits:

  1. It ensures water availability in non-rainy season,
  2. It reduces the chances of flooding during rainy season,
  3. Ground water level does not fall as it is regularly recharged,
  4. Ground water recharge is the best form of water harvesting as the water is filtered and free from contaminations. It also does not evaporate,
  5. Water becomes available for drinking as well as irrigation.

Question 3.
In a village in Karnataka people started cultivating crops all around a lake which was always filled with water. They added fertilizers to their field in order to enhance the yield. Soon they discovered that the water body was completely covered with green floating plants and fishes started dying in large numbers.
Analyse the situation and give reasons for excessive growth of plants and death of the fish in the lake.
Answer:
Fertilizer rich run off from fields must have passed into the lake. It caused nutrient enrichment of lake water. The result is excessive growth of algae and other aquatic plants which float on the water surface and produce water bloom. Old dead plants produce a lot of organic matter. The submerged plants are also killed due to shading. BOD of water increases. As more and more oxygen is consumed by decomposers little is left for respiration of aquatic animals. Therefore, fish begin to die. The phenomenon of nutrient enrichment of water body that causes formation of water bloom and subsequent killing of aquatic life is called eutrophication.

Question 4.
What measures would you take to conserve electricity in your house ? (CCE 2012)
Answer:

  1. Judicious use of electricity by switching off lights and electrical appliances not required,
  2. Replacement of incandescent bulbs with fluorescent, compact fluorescent ones and LED bulbs.
  3. Replacement of electricity or gas operated geysers with solar water heaters,
  4. Replacement of electricity generating sets with solar light,
  5. Having more natural light and ventilation with design supporting warming during winters and cooling during summer.

Question 5.
Although coal and petroleum are produced by degradation of biomass, yet we need to conserve them. Why ? (CCE 2012)
Answer:
Coal and petroleum have been produced from large amounts of biomass entrapped inside the earth under high temperature, pressure and anaerobic conditions. Such a situation develops only rarely like big upheavals on earth. At present no more coal or petroleum is being formed. All that is available has been formed millions of years ago. Being rich source of energy, coal and petroleum are being consumed in ever increasing amount in industry, transport, kitchens, etc. If the trend continues, soon they will be exhausted. Therefore, they must be conserved by developing more efficient machines, hybrid engines and using hydrogen as a fuel.

Question 6.
Suggest a few measures for controlling carbon dioxide levels in the atmosphere.
Answer:

  1. Increasing Vegetation Cover. It will increase utilisation of atmospheric CO2 in photosynthesis.
  2. Seeding of Oceans With Phytoplankton. Increased photosynthetic activity of oceans will result in decreasing CO2 concentration.
  3. Carbonation. CO2 released during combustion should not be allowed to pass into atmosphere. Instead, it can be changed into carbonates.
  4. Alternate Sources of Energy. Instead of fossil fuels, hydrogen fuel and solar energy should be used.
  5. Burning of Litter. Litter and crop residue should not be burnt but instead converted into manure.

Question 7.
(a) Locate and name the water reservoirs in figures (i) and (ii).
(b) Which has advantage over the other and why ?
NCERT Exemplar Solutions for Class 10 Science Chapter 16 Management of Natural Resources image - 1
Answer:
(a) Water reservoir in figure (i) is pond while it is underground water body (ground water) in figure (ii).
(b) Ground water is more advantageous than pond water.
For Benefits: 

  1. Prevents flooding,
  2. Checks soil erosion.
  3. Retains water underground and prevents drought,
  4. Increases life of downstream reservoirs and dams,
  5. Higher biomass production and income of water shed community,
  6. Maintenance of ecological balance.

NCERT Exemplar Solutions for Class 10 Science Chapter 16 Long Answer Questions

Question 8.
In the context of conservation of natural resources, explain the terms reduce, recycle and reuse. From among the materials that we use in daily life, identify two materials for each category.
Answer:
Three Rs — reduce, recycle and reuse.

  1. Reduce: It is to reduce consumption by preventing wastage.
    1. Switching off unnecessary lights, fans and other electrical appliances,
    2. Repair of leaky taps.
    3. Reducing food wastage,
    4. Walking down to nearby market instead of using vehicle.
  2. Recycle: Separation of recyclable wastes from non-recyclable wastes. The former are taken by rag pickers for sending them to industries involved in recycling, e.g., paper, plastic, metal, glass.
  3. Reuse: Carry bags, packing material, plastic containers and other reusable articles should not be thrown away if the same are uncontaminated. For example, plastic bottles and jars containing various food items brought from market can be washed and used for storing things in the kitchen.

Question 9.
Prepare a list of five activities that you perform daily in which natural resources can be conserved or energy utilisation can be minimised.
Answer:

  1. Judicious use of electricity by switching off lights and electrical appliances not required,
  2. Replacement of incandescent bulbs with fluorescent, compact fluorescent ones and LED bulbs.
  3. Replacement of electricity or gas operated geysers with solar water heaters,
  4. Replacement of electricity generating sets with solar light,
  5. Having more natural light and ventilation with design supporting warming during winters and cooling during summer,
  6. Reducing wastage of water, food and other articles.
  7. Separation of recyclable waste from non-cyclable waste prior to disposal.
  8. Increasing reuse of containers,
  9. Using cloth bags instead of polythene, plastic or paper bags.

Question 10.
Is water conservation necessary ? Give reasons.
Answer:

  1. Distribution of fresh water is highly uneven. Large tracts are deficient in rain as well as ground water,
  2. At most places more water is withdrawn from reservoir and underground source than their recharging
  3. Requirement in urban and industrial areas is nearly always higher than the availability,
  4. Further demand for water is rising by 4 – 8% annually in all fields, whether agriculture, industry or domestic use.

Therefore, water conservation is necessary. Wastage of the resource should be prevented. Waste water should be recycled. Water harvesting involving recharging of ground water should be practised.

Question 11.
Suggest a few useful ways of utilising waste water.
Answer:
Waste or used water can also become a resource.

  1. Treated municipal water can be poured in irrigation channels for supply to crop fields,
  2. Treated waste water can be used in urban areas for watering gardens, lawns and washing vehicles,
  3. Industries can treat their waste water and recycle the same,
  4. Waste water passed into ponds recharges the ground water,
  5. Sewage sludge, separated from waste water is a source of manure, compost and biogas.

Question 12.
What is the importance of forests as a resource ?
Answer:
Economic Reasons:

  1. Food: Tribals obtain most of their food requirements from the forests, e.g., fruits, tubers, fleshy roots, leaves.
  2. Nuts: Pine Nut (Chilgoza), Almond, Walnut and Cashewnut are obtained from forests trees.
  3. Spices: Cardamom, Cinnamon, Nutmeg and Cloves are spices obtained from forest plants.
  4. Commercial Products: A number of forest products are of commercial importance, g., rubber, resin, tannins, tendu, lac, cork, camphor, essential oils, soap pod and drugs.
  5. Fuel Wood: Nearly two billion persons depend upon forests for fuel wood.
  6. Timber: Wood for the manufacture of furniture, household fitments and several other articles mostly comes from forests. Bamboo is called poorman’s timber as it is used in thatching huts, preparing baskets and a number of other articles including furniture.
  7. Paper: It is prepared from cellulose rich plants like bamboos, Boswellia, Eucalyptus, grasses and several

Protective Functions:                                                                                  i

  1. Forests provide shelter to wild animals. Over 40 million tribals and villagers live in forests.
  2. Plant roots hold the soil firmly. Vegetation protects the soil from action of wind and water. Forests, therefore, protect the soil from erosion and landslides.
  3. Pollution. Forests reduce atmospheric pollution by absorbing gases, collecting suspended particles and reducing noise.

Regulative Functions:

  1. Absorption and Retention of Water. Forests reduce run off, hold water like a sponge and allow slow percolation to form perennial springs and rivulets.
  2. Forests increase atmospheric humidity, increase frequency of rainfall and moderate temperature.
  3. Atmospheric Gases. Forests absorb large quantity^ of C02 from the atmosphere, reducing the threat of global warming. They also release a lot of oxygen.

Question 13.
Why are Arabari forests of Bengal known to be good example of conserved forests.
Answer:
Regeneration of Sal Forests — An Example of People’s Participation in the Management of Forests Despite best efforts, the West Bengal Forest Department could not revive the degraded Sal forests of Southwestern districts of the state. Excessive surveillance and policing of the degraded forests not only alienated the people but also resulted in frequent clashes between villagers and forest officials. This also fueled the militant peasant movement led by Naxalites. Realising the failure, the forest department revised its strategy in 1972. It allowed forest officer A.K. Banerjee of Arabari forest range of Midnapore to involve villagers in regeneration of 1272 hectares of badly degraded Sal forest. Banerjee provided employment to villagers in silviculture (cultivation of trees) and harvesting, 25% of final harvest and allowed collection of fuel wood as well as fodder at nominal fee. By 1983, the Arabari forest had been revived and was then valued at 12-5 crores.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 16 Management of Natural Resources 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 25 Probability Ex 25C.

Other Exercises

Question 1.
A bag contains 3 red balls, 4 blue balls and one yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it : find the probability that die ball is :
(i) yellow
(ii) red
(iii) not yellow
(iv) neither,yellow nor red
Solution:
In a bag. there are 8 balls in which there are Red balls = 3
blue balls = 4
and yellow ball = 1
Total possible out come = 8
(i) Yellow balls = 1
∴ Number of favourable outcome = 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q1.1
(ii) Red balls = 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q1.2
(iii) Not yellow balls = 3+4 = 7
∴ Number of favourable outcome = 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q1.3
(iv) Neither yellow nor red can be blue ball.
∴ Number of favourable outcome = 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q1.4

Question 2.
A dice is thrown once. What is the probability of getting a number :
(i) greater than 2 ?
(iii) less than or equal to 2 ?
Solution:
A die is thrown once
Numbers marked on the faces of a die are 1, 2,3,4, 5, 6
∴ Number of possible outcome = 6
(i) Greater than 2 number = 3, 4, 5, 6 = 4 numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q2.1
(ii) Less than or equal to 2 are 2, 1 which are 2 in numbers.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q2.2

Question 3.
From a well-shuffled deck of 52 cards* one card is drawn. Find the probability that the card drawn is :
(i) a face card.
(ii) not a face card. ,
(iii) a queen of black colour.
(iv) a card with number 5 or 6.
(v) a card with number less than 8.
(vi) a card with number between 2 and 9.
Solution:
A deck of playing cards has 52 cards
∴ Number of possible outcome = 52
(i) A face card ; face cards in the deck are = 3 x 4 = 12
∴ Number of favourable outcome = 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q3.1.
(ii) Not a face card which are 52 – 12 = 40
∴ Number of favourable outcome = 40
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q3.2
(iii) A queen of black color which are 2 in numbers in the deck
∴ Number of favourable outcome = 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q3.3
(iv) A card with number 5 or 6 are 2 x 4 numbers
∴ Number of favourable outcome = 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q3.4
(v) A card with number less than 8 which can be 2, 3, 4, 5, 6, 7 = 6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q3.5
(vi) A card with number between 2 and 9 can be 3, 4, 5, 6, 7, 8 = 6
∴ Number of favourable outcome = 6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q3.6

Question 4.
In a match between A and B.
(i) the probability of winning of A is 0.83. What is the probability of winning of B ?
(ii) the probability of losing the match is 0.49 for
B. What is the probability of winning of A?
Solution:
A match is played between two persons A and B
∴ Number of possible outcome = 1
(i) The probability of winning of A is 0.83
∴ Probability of winning of B = 1 – 0.83 = 0.17 [∵ P(E) + P(\(\bar { E }\) )=1]
(ii) The probability of losing the match is 0.49 by B.
∴ Probability of losing of B or winning of A = 0.49

Question 5.
A and B are friends. Ignoring the leap year, find the probability that both friends will have :
(i) different birthdays ?
(ii) the same birthdays ? (Ignore a leap year)
Solution:
Number of days in a year = 365
and birthday of a person can be on one day only.
(i) Different birthdays can be 365 – 1 = 364
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q5.1

Question 6.
A man tosses two different coins (one of Rs. 2 and another of Rs. 5) simultaneously. What is the probability that he gets :
(i) at least one head ?
(ii) at most one head?
Solution:
There are two coins : one of two rupees and
other is of 5-rupees
∴ Number of Heads =1 + 1=2
and number of tails are 2 i.e., 2 and 5
∴ Number of possible outcome = 2 x 2 = 4
(i) Atleast one head.
Number of favourable outcome = 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q6.1
(ii) At the most one head = 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q6.2

Question 7.
A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is :
(i) white
(ii) neither red nor white
Solution:
Number of balls in a box are 20 in which 7 are red, 8 are green and 5 are white
∴ Number of possible outcome = 20
(i) White = 5
∴ Number of favourable outcome = 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q7.1
(ii) Neither red not white
i.e., all the green which are 8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q7.2

Question 8.
All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting :
(i) a black face card
(ii) a queen
(iii) a black card.
Solution:
A deck of playing cards = 52
Cards which are removed = 3 (3 face cards of spades)
Balance cards in the deck = 52 – 3 = 49
(i) A black face card which are 6 – 3 = 3 in number
∴ Number of favourable outcome = 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q8.1
(ii) A queen : In the deck there are 4 – 1 = 3 queen
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q8.2
(iii) A black cards : which are 26 – 3 = 23 cards in number
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q8.3

Question 9.
In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start.
What is the probability that the music will stop within the first 15 seconds ?
Solution:
Total time for the musical race = 0 to 40 seconds = 40 seconds.
Time taken by a player =15 seconds. (0 to 15 seconds)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q9.1

Question 10.
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that :
(i) it is acceptable to a trader who accepts only a good shirt ?
(ii) it is acceptable to a trader who rejects only a shirt with major defects ?
Solution:
Total number of shirts in a bundle = 50
No. of good shirts = 44
Minor defected = 4
Major defected = 2
∴ Number of possible outcome = 50
(i) Acceptance of only for a good shirt = 44
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q10.1
(ii) Rejecting of totally defected shirts, number of remaining shirts = 44 + 4 = 48
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q10.2

Question 11.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is :
(i) 8
(ii) 13
(iii) less than or equal to 12
Solution:
Two dice are thrown at the same time and each dice has 6 numbers 1, 2, 3, 4, 5, 6 on its faces
∴Number of possible outcome = 6 x 6 = 36
(i) Sum of two numbers on the top is 8 i.e..(2, 6), (3, 5), (4,4), (5, 3), (6, 2)
∴ Number of favourable outcome = 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q11.1
(ii) Sum of two number on the top is 13.
At the most, then sum can be (6, 6) = 12
∴ number of favourable outcome = 0
∴ P(E) = 0
(iii) Sum is less than or equal to 12 i.e.(1, 1), (1,2), (1,3), (1,4), (1,5), (1,6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3). (3, 4), (3. 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,
6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) = 36
∴ Number of favourable outcome = 36
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q11.2

Question 12.
Which of the following cannot be the probability of an event ?
(i) \(\frac { 3 }{ 7 }\)
(ii) 0.82
(iii) 37%
(iv) -2.4
Solution:
(i) ∵ \(\frac { 3 }{ 7 }\) is between 0 and 1
∴ It is a probability event
(ii) ∵ 0.82 is between 0 and 1
∴ It is a probability event.
(iii) 37% = \(\frac { 37 }{ 100 }\)
∵It is between 0 and 1
∴ It is a probability event.
(iv) -2.4
∵It is less than 0.
∴ It is not a probability event.

Question 13.
If P(E) = 0.59; find P(not E).
Solution:
P(E) = 0.59
But P(E) + P(\(\bar { E }\) )= 1
or P(E) + P(not E) = 1
⇒ 0.59 + P(not E) = 1
⇒ ∴ P(not E) = 1 – 0.59 = 0.41

Question 14.
A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is :
(i) black
(ii) red.
Solution:
In a bag, there are certain number red balls. Let it be x balls.
One ball is drawn out
(i) A black
∵ There is no black ball in the bag
∴ Probability of black ball = 0
(ii) A red ball
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q14.1

Question 15.
The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday ?
Solution:
Probability of two boys do not have the same birthday [P (E)] = 0.897
Let Probability of those boys having the same birthday = P(not E)
= P(\(\bar { E }\) )
But P(E) + P(\(\bar { E }\) )= 1
⇒ 0.897 +P(\(\bar { E }\) ) = 1
⇒P(\(\bar { E }\) )= 1 -0.897 = 0.103
Hence probability having the same birthday = 0.103

Question 16.
A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability tharthe ball drawn will be :
(i) not red ?
(ii) neither red nor green ?
(iii) white or green ?
Solution:
In a bag, there are 10 red balls, 16 white and 8 green balls
∴ Total balls =10 + 16 + 8= 34 ball
∴ Number of possible outcome = 34
(i) Not red ball
Number of favourable outcome = 16 + 8 = 24
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q16.1
(ii) Neither red nor green
∴ Number of outcome = 34 – (10 + 8)
= 34-18=16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q16.2
(iii) White or green
∴ Number of outcome = 16 + 8 = 24
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q16.3

Question 17.
A bag contains twenty Rs. 5 coins, fifty Rs. 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin :
(i) will be a Re 1 coin ?
(ii) will not be a Rs. 2 coin ?
(iii) will neither be a Rs. 5 coin nor be a Re 1 coin ?
Solution:
In a bag, there are
5-rupee coins = 20
2-rupee coins = 50
1-rupee coin =30
Total number of coins in the bag = 20 + 50 + 30 = 100
∴ Number of possible outcome =100
(i) one-rupee coin = 30
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q17.1
(ii) When there is no 2-rupee coins
∴ Number of coins = 20 + 30 = 50
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q17.2
(iii) Neither be Rs. 5 coins nor be Re-one coin = 100 – (20 + 30) = 100 – 50 = 50
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q17.3

Question 18.
A game consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q18.1
If the outcomes are equally likely, find the probability that the pointer will point at :
(i) 6
(ii) an eve., number.
(iii) a prime number.
(iv) a number greater than 8.
(v) a number less than or equal to 9.
(vi) a number between 3 and 11.
Solution:
There are 12 numbers on the spinning game.
∴ Number of possible outcome = 12
(i) 6 which is one
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q18.2
(ii) an even number which are 6 i.e. 2, 4, 6, 8, 10, 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q18.3
(iii) A prime number which are 2, 3, 5, 7, 11 and 5 in number
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q18.4
(iv) A number greater than 8 are 9, 10, 11, 12
which are 4 in number.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q18.5
(v) A number less than or equal to 9 are
1, 2, 3, 4, 5, 6, 7, 8, 9 which are 9 in number
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q18.6
(vi) A number between 3 and 11 are 4, 5, 6, 7, 8, 9, 10 which are 7 in number
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q18.7

Question 19.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting :
(i) a queen of red colour.
(ii) a black face card.
(iii) the jack or the queen of hearts.
(iv) a diamond.
(v) a diamond or a spade.
Solution:
Number of cards in a deck of playing card = 52
∴ Number of possible outcome = 52
(i) A queen of red colour
Number of favourable outcome = 2
(As there are 2 red queens in the deck)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q19.1
(ii) A black face card.
There are 3 + 3 = 6 black face cards in the deck.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q19.2
(iii) The jack or the queen of hearts which are two in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q19.3
(iv) A diamond.
There are 13 cards of diamond in the deck.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q19.4
(v) A diamond or a spade.
There are 13 cards of diamond and 13 cards of spade in the deck
∴ Number of favourable outcome = 13 + 13 = 26
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q19.5

Question 20.
From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is :
(i) a black card.
(ii) 8 of red color.
(iii) a king of black color.
Solution:
Number of cards is a deck of playing card = 52
All face cards are removed. Which are 3 x 4 = 12
(i) A black card which are 10 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q20.1
(ii) 8 of red colors
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q20.2
(iii) A king of black color.
∵ In the deck of playing cards, face cards are removed
∴ There is no face cards.
∴ P(E) = 0.

Question 21.
Seven cards : – the eight, the nine, the ten, jack, queen, king and ace of diamonds are well shuffled. One card is then picked up at random.
(i) What is the probability that the card drawn is the eight or the king ?
(ii) If the king is drawn and put aside, what is the probability that the second card picked up is
(a) an ace ?
(b) a king ?
Solution:
There are 7 cards which are the eight, the nine, the ten, the jack, the’ queen, the king and the ace of diamond.
∴ Number of possible outcome = 7
(i) card having the eight or a king which are 2.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q21.1
(ii) If king is drawn, then number of remaining playing cards = 7-1=6
(a) An ace.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q21.2
(b) A king
There is no card of king
∴ P(E) = 0

Question 22.
A box contains 150 bulbs out of which 15 are defective. It is not possible to just look at a bulb and tell whether or not it is defective. One bulb is taken out at random from this box. Calculate the probability that the bulb taken out is :
(i) a good one
(ii) a defective one.
Solution:
Number of bulbs in a box = 150
No. of defective bulbs = 15
∴ No of good bulbs = 150 – 15 = 135
∴ Number of possible outcome = 150
(i) A good bulb and number of good bulbs =135
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q22.1
(ii) A defective bulb. Number of defective bulb = 15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q22.2

Question 23.
(i) 4 defective pens are accidentally mixed with 16 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is drawn at random from the lot. What is the probability that the pen is defective ? (ii) Suppose the pen drawn in (i) is defective and is not replaced. Now one more pen is drawn at random from the rest. What is the probability that this pen is :
(a) defective
(b) not defective ?
Solution:
Number of defective pens = 4
(i) and number of good pens = 16
∴ Total pens = 4 + 16 = 20
∴ Number of possible outcome = 20
One defective pen, no. of defective pens = 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q23.1
(ii) One defective pen is drawn
∴ Remaining pens = 20 – 1 = 19
(a) Defective one
Number of remaining defective pens =4-1=3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q23.2
(b) Not defective, number of good pens =16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q23.3

Question 24.
A bag contains 100 identical marble stones which are numbered from 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears :
(i) a perfect square number.
(ii) a number divisible by 4.
(iii) a number divisible by 5.
(iv) a number divisible by 4 or 5.
(v) a number divisible by 4 and 5.
Solution:
Total number of stones = 100
On which numbers 1 to 100 are marked
∴ Number of possible outcome = 100
(i) A perfect square = which are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 which are 10 inmumbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q24.1
(ii) A number divisible by 4 are 4, 8, 12, 16, …, 96, 100 which one 25 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q24.2
(iii) A number divisible by 5, are 5, 10, 15, 20, ….., 95, 100 which are 20 in numbers,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q24.3
(iv) A number divisible by 4 or 5 are 4, 5, 8, 10,12, 15, 16, 20, 24, 25, 28, 30, 32, 35, 36, 40, 96,95, 100 which are 40 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q24.4
(v) A number divisible by 4 and 5 are 20, 40, 60, 80 and 100 which are 5 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q24.5

Question 25.
A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on the paper only, find the probability that it will fall and land :
(i) inside the circle.
(ii) outside the circle.
Solution:
Diameter of the circle = 20 cm
Length of rectangular paper = 40 cm
and width = 30 cm
Area of rectangle = 40 x 30 = 1200 cm2
∴ Number of possible out come = 1200
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q25.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q25.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q25.3

Question 26.
Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is :
(i) 4 or 5
(ii) 7, 8 or 9.
(iii) between 5 and 8
(iv) more than 10
(v) less than 6
Solution:
Two dice having numbers 1 to 6 are rolled together.
∴ Number of possible outcome = 6 x 6 = 36
(i) If sum is 4 or 5 of numbers on the upper most face,
Their number of favourable outcome = (1,3), (2, 2), (3, 1), (1, 4), (2, 3), (3, 2), (4, 1) which are 7 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q26.1
(ii) If sum of number on the upper faces by 7, 8 or 9, then these can be (1, 6), (2, 6), (3, 6), (4, 3), (5. 2). (6, 1). (4, 3), (2, 5), (3, 4), (4, 4), (5, 3), (5, 4), (6, 2), (6, 3), (4, 5) which are total 15 in numbers.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q26.2
(iii) Sum is between 5 and 8.
i.e. sum is 6 or 7.
These can be.
(1, 5), (2, 5), (1, 6), (5, 1), (5, 2), (6, 1), (3, 3), (3, 4), (4, 3), (2, 4), (4, 2) which are 11 in number.
∴ Number of favourable outcome = 11
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q26.3
(iv) If sum is more than 10, then these can be (5. 6), (6, 5), (6, 6) which are 3 in number
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q26.4
(v) If sum is less than 6, then there can be (1, 2), (2, 1), (1, 3), (3, 1), (4, 1), (1, 4), (2, 3), (3, 2), (2, 4), (4, 2)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q26.5

Question 27.
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting :
(i) exactly two heads.
(ii) at least two heads.
(iii) atmost two heads.
(iv) all tails.
(v) at least one tail.
Solution:
3 coins one tossed together.
∴ Number of possible outcome = 23 = 2 x 2 x 2 = 8
i.e. HHH, HHT, HTH, HTT, TTT, THH, THT, TTH
(i) exactly two heads these can be HHT, THH, HTH = 3 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q27.1
(ii) At least two heads : These can be HHH,HHT, HTH, THH = 4 in numbers
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q27.2
(iii) Atmost two heads : These can be THH, HHT, HTH, HTT, THT, TTH, TTT which is 7 in numbers.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q27.3
(iv) All tails : There can be TTT i. e., only one
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q27.4
(v) At least one tail : There can be HHT, HTH, HTT, TTT, THH, THT, TTH = 7 in number
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q27.5

Question 28.
Two dice are thrown simultaneously. What is the probability that :
(i) 4 will not come up either time ?
(ii) 4 will come up at least once ?
Solution:
Two dice are thrown simultaneous,
and each dice has 1-6 numbers on its faces
∴ Number of possible outcome = 6 x 6 = 36
(i) 4 will not come up either time
∴ Number of favourable outcomes = (6 – 1)² = (5)² = 25
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q28.1
(ii) 4 will come up at least once
∴ These can be (1, 4), (2, 4), (3, 4), (4, 4), (4, 5),(4, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
∴ Number of favourable outcome =11
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q28.2

Question 29.
Cards marked with numbers 1, 2, 3, 4, …,20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is
(i) a prime number
(ii) divisible by 3
(iii) a perfect square ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q29.1

Question 30.
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on :
(i) the same day
(ii) consecutive day
(iii) different days.
Solution:
∵ Office are open for 5 days a week
∴ Number of possible outcomes for 2 employees = 5 x 5 = 25
Let the five days of working be denoted by M, T, W, Th, F for Mondays, Tuesday, Wednesday Thursday and Friday respectively
(i) On Same day.
Favourable outcome will be M;T;T;M;T;W;W;T;W;H; TH.W ; TH ; F and F, TH which are 8 in all
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q30.1
(iii) Absent on different days : Then = 1- P(absent on the same day)
= 1 – \(\frac { 1 }{ 5 }\)= \(\frac { 4 }{ 5 }\) [from (i)]
∵ P(E) + P(not E) = 1

Question 31.
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of white ball; find the number of black balls in the box.
Solution:
In a box, some balls are black while 30 balls are white
Let number of black balls = x
Then number of possible outcome = x + 30
Probability of drawing a black ball = \(\frac { 2 }{ 5 }\) of a white balls.
Now, In case of black ball,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q31.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q31.2
Either x + 30 = 0, then x = -30 which is not possible
or x – 12 = 0, then x = 12
Hence number of black balls = 12

Question 32.
From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed. A card is now drawn at random.
What is the probability that the card drawn is
(i) a face card (King, Jack or Queen)
(ii) an even numbered red card ? (2011)
Solution:
No. of total cards = 52 cards removed of 4 colors = 3, 6, 9, 12 = 4 x 4 = 16
Remain using cards = 52 – 16 = 36
(i) No. of faces cards = 2 x 4 = 8 cards (excluding queen)
∴ Probebility P(E) = \(\frac { 8 }{ 36 }\) = \(\frac { 2 }{ 9 }\)
(ii) An even number red cards = 2, 4, 8, 10 = 4×2 = 8 cards
∴ Probebility P(E) = \(\frac { 8 }{ 36 }\) = \(\frac { 2 }{ 9 }\)

Question 33.
A die has 6 faces marked by the given numbers as shown below:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q33.1
The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than -3.
(iii) the smallest integer.
Solution:
Total outcomes n(S)= 6
(i) a positive integer = (1, 2, 3)
No. of favourables n(E) = 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q33.2

Question 34.
A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is :
(i) a green ball.
(ii) a white or a red ball.
(iii) neither a green ball nor a white ball. (2015)
Solution:
A bag contains 5 white balls, 6 red balls and 9 green balls.
∴ Total number of balls = 5 + 6 + 9 = 20 balls
One ball is drawn at random.
(i) Probability of a green ball = \(\frac { 9 }{ 20 }\)
(ii) Probability of a white or a red ball = \(\frac { 5 + 6 }{ 20 }\) = \(\frac { 11 }{ 20 }\)
(iii) Probability of neither a green ball nor a white ball = \(\frac { 6 }{ 20 }\) = \(\frac { 3 }{ 10 }\) (Only red balls )

Question 35.
A game of numbers has cards marked with 11, 12, 13, , 40. A card is drawn at random. Find the probability that the number on the card drawn is :
(i) A perfect square
(ii) Divisible by 7
Solution:
(i) The perfect squares lying between 11 and 40 and 16, 25 and 36.
So the number of possible outcomes is 3.
Total number of cards from 11 to 40 is 40 – 11 + 1 = 30
Probability that the number on the card drawn is a perfect square
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q35.1
So, the probability that the number on the
card drawn is a perfect square is \(\frac { 1 }{ 10 }\) .
(ii) The numbers lfom 11 to 40 that are divisible by 7 are 14, 21, 28 and 35.
So the number of possible outcomes is 4. Total number of cards from 11 to 40 is 30.
Probability that the number on the card drawn is divisible by 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C Q35.2
So, the probability that the number on the card drawn is divisible by 7 is \(\frac { 2 }{ 15 }\).

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C are helpful to complete your math homework.

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