Category: Class 10

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0; 3x – 9y – 2 = 0
(ii) 2x + y = 5; 3x + 2y = 8
(iii) 3x – 5y = 20; 6x -10y = 40
(iv) x – 3y – 7 = 0; 3x – 3y – 15 = 0
Solution:
(i) For x – 3y – 3 = 0, 3x – 9y – 2 = 0
∴ a₁ = 1, b₁ = – 3, C₁ = – 3, a₂ = 3, b₂ = – 9, C₂ = -2

∴ The given system has no solution.

(ii) 2x + y – 5 = 0, 3x + 2y – 8 = 0
∴ a₁ = 2, b₁ = 1, c₁ = -5, a₂ = 3, b₂ = 2, c₂ = -8

∴ The given system has a unique solution.
Now, using cross multiplication method, we have

(iii) For 3x – 5y – 20 = 0, 6x – 10y – 40 = 0

∴ The given system of linear equation has infinitely many solutions

(iv) For x – 3y – 7 = 0, 3x – 3y – 15 = 0

∴ The given statement has unique solution. Now, using cross multiplication method, we have

Question 2.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?
2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1
Solution:
(i) We have, 2x + 3y = 7 and
(a – b) x + (a + b) y = (3a + b – 2)
⇒ 2x + 3y – 7 = 0
and (a – b) x + (a + b) y -(3a + b – 2) = 0
a₁ = 2, b₁ = 3, c₁ = – 7, a₂ = (a- b) , b₂ = (a + b) , c₂ = -(3a + b – 2)
For infinite number of solutions,

From the first two equations, we get

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9; 3x + 2y = 4
Solution:
Method-1 [Substitution method]:
8x + 5y = 9 … (1)
3x + 2y = 4 … (2)

Question 4.
Form the pair of linear equations in the following problems and find their solutions (if they exist)by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
(i) Let the fixed charges = ₹ x
and charges of food per day = ₹ y
For student A : Number of days = 20
∴ Cost of food for 20 days = ₹ 20y
According to the problem,
x + 20y = 1000
⇒ x + 20y – 1000 = 0 …. (1)
For student B : Number of days = 26
Cost of food for 26 days = ₹ 26y
According to the problem,
x + 26y = 1180
⇒ x + 26y – 1180 = 0 … (2)
Solving these by cross multiplication, we get

Thus x = 400 and y = 30
∴ Fixed charges = ₹ 400
and cost of food per day = ₹ 30

(ii) Let the numerator = x
and the denominator = y

From equations (1) and (2) , we have

(iii) Let the number of correct answers = ₹ x
and the number of wrong answers = ₹ y
Case-I: Marks for all correct answers
= (3 × x) = 3x
Mark for all wrong answers = (1 × y) = y
∴ According to the condition :
3x – y = 40 ⇒ 3x – y – 40 = 0
Case-II: Mark for all correct answers
= (4 × x)=
Marks for all wrong answers = (2 × y)
= 2y
∴ According to the condition :
4x – 2y = 50
⇒ 2x – y = 25 ⇒ 2x – y – 25 = 0 … (2)
From (1)and (2) , we have a₁ = 3, b₁ = -1, c₁ = -40, a₂ = 2, b₂ = -1, c₂ = -25

Now, total number of questions = [Number of correct answers] + [Number of wrong answers]
= 15 + 5 = 20
Thus, required number of questions = 20.

(iv) Let the speed of car-I be x km/hr.
and the speed of car-II be y km/hr.
Case-I:

Distance travelled by car-I = AC
∵ AC = time × speed = 5 × x km, AC = 5x
Distance travelled by car-II, BC = 5y
Since AB = AC – BC,
100 = 5x – 5y
⇒ 5x – 5y – 100 = 0
⇒ x – y – 20 = 0 …. (1)


Thus, speed of car-I = 60 km/hr
Speed of car-II = 40 km/hr

(v) Let the length of the rectangle = x units
and the breadth of the rectangle = y units
∴ Area of rectangle = x × y = xy

Condition-II:
(Length + 3) × (Breadth + 2) = Area + 67
⇒ (x + 3) (y + 2)= xy + 67 ⇒ 2x + 3y + 6 = 67
⇒ 2x + 3y – 61 = 0….(2)
Now, using cross multiplication method in (1)and (2) , where

Thus, length of the rectangle = 17 units and breadth of the rectangle = 9 units.

MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements

MP Board Class 10th Science Chapter 5 Intext Questions

Intext Questions Page No. 81

Question 1.
Did Dobereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.
Answer:
In Newland’s Octaves, the properties of lithium and sodium were found to be the same. This arrangement is also found in Dobereiner triads.

Question 2.
What were the limitations of Dobereiner’s classification?
Answer:
Dobereiner could identify only three ‘triads’ from the elements known at that time. Hence this system of classification into triads was not found to be useful.

Question 3.
What were the limitations of Newlands’ Law of Octaves?
Answer:

The Limitations of Newlands’ Law of Oclaves is as follows:

1. It was found that the Law of Octaves was applicable only upto calcium, as after calcium every eighth element did not possess properties similar to that of the first.
2. It was assumed by Newlands’ that only 56 elements existed in nature and no more elements would be discovered in the future. But later on, several new elements were discovered, whose properties did not fit into the Law of Octaves.
3. In order to fit elements into his Table, Newlands adjusted two elements in the same slot, but also put some unlike elements under the same note.

Intext Questions Page No. 85

Question 1.
Use Mendeleev’s Periodic Table to predict the formulae for the oxides of the following elements: K, C, Al, Si, Ba.
Answer:

1. K is in I group. Its oxide is K₂O
2. C, is in IV group, its oxide is CO₂
3. Al, is in III group, its oxide is Al₂O₃
4. Si, is IV group, its oxide is SiO₂
5. Ba, is in II group, its oxide is BaO

Question 2.
Besides gallium, which other elements have since been discovered that were left by Mendeleev in his Periodic Table? (any two)
Answer:
Scandium and germanium.

Question 3.
What were the criteria used by Mendeleev in creating his Periodic Table?
Answer:
Mendeleev used the relationship between the atomic masses of the elements and their physical and chemical properties. Among chemical properties, he examined the compound formed by elements with oxygen and hydrogen. He found that if the 63 elements known at that time were arranged in the increasing order of their atomic masses, the properties of elements and also formulae of their oxides and hydrides gradually changed from element to element and at a certain interval they suddenly started almost repealing relationship was expressed by Mendeleev’s periodic law. i,e the properties of examinants are the periodic functions of their atomic masses.

Question 4.
Why do you think the noble gases are placed in a separate group?
Answer:
All noble gases are inert elements. Their properties are different from other elements and are the least reactive. Therefore, the noble gases are placed in a separate group.

Intext Questions Page No. 90

Question 1.
How could the Modern Periodic Table remove various anomalies of Mendeleev’s Periodic Table?
Answer:

1. In the Modern periodic Table atomic number of an elements is a more fundamental property than its atomic mass.
2. The anomalous position of hydrogen can be discussed after we see what are the basis on which the position of an elements in the Modern Periodic Table depends.
3. The elements present in any one group have the same number of valence electrons.
4. Atoms of different elements with the same number of occupied shells are placed in the same period.
5. In the Modern Periodic Table, a zig-zag line separated metals from non-metals.

Question 2.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?
Answer:
Calcium (Ca) and Strontium (Sr) is expected to show chemical reactions similar to magnesium (Mg). This is because the number of valence electrons (2) is the same in all of these three elements and since chemical properties are due to valence electrons, they show the same chemical reactions.

Question 3.
Name:

1. Three elements that have a single electron in their outermost shells.
2. Two elements that have two electrons in their outermost shells.
3. Three elements with filled outermost shells.

Answer:

1. Lithium (Li), Sodium (Na), and Potassium (K) has a single electron in their outermost shells.
2. Magnesium (Mg) and Calcium (Ca) have two electrons in their outermost shells.
3. Neon (Ne), Argon (Ar), and Xenon (Xe) have filled outermost shells.

Question 4.
Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?
Answer:

Lithium, sodium and potassium – These three elements have one electron in their outermost orbit.

b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?
Answer:
Both helium (He) and neon (Ne) have filled outermost shells. Helium has a duplet in its K shells, while neon has an octet in its L shells.

Question 5.
In the Modern Periodic Table, which are the metals among the first ten elements?
Answer:
In the modem periodic table, Lithium and Beryllium are the metals among the first 10 elements.

Question 6.
By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristic?
Ga Ge As Se Be
Answer:
Since, ‘Be’ lies to the extreme left-hand side of the periodic table, ‘Be’ is the most metallic among the given elements.

MP Board Class 10th Science Chapter 5 Ncert Textbook Exercises

Question 1.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of the periodic table.
(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c) The atoms lose their electrons more easily.

Question 2.
Element X forms a chloride with the formula XCl₂, which is solid with a high melting point. X would most likely be in the same group of the Periodic table as:
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(b) Mg

Question 3.
Which element has:
(a) two shells, both of which are completely filled with electrons?
(b) Electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?
Answer:
(a) Neon
(b) Magnesium
(c) Silicon
(d) Boron
(e) Carbon

Question 4.
(a) What property do all elements in the same column of the Periodic table as Boron have in common?
(b) What property do all elements in the same column of the Periodic table as Fluorine have in common?
Answer:
(a) Valency equal to 3.
(b) Valency equal to 1.

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.)
N(7) F(9) P(15) Ar(18)
Answer:
(a) The atomic number of this element is 17.
(b) It would be chemically similar to F(9) with configuration as 2, 7.

Question 6.
The position of three elements A, B and C in the Periodic table is shown below:

(a) State whether A is a metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B?
(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
(a) A is a non-metal.
(b) C is less reactive than A because reactivity decreases down the group in halogens.
(c) C should be smaller in size than B as moving across a period, the nuclear charge increases and therefore, electrons come closer to the nucleus.
(d) A will form an anion as it will accept an electron to complete its octet.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?
Answer:
Nitrogen (7) : 2, 5
Phosphorus (15) : 2, 8, 5
Nitrogen is more electronegative because Metallic character decreases across a period and increased down a group.

Question 8.
How does the electronic configuration of an atom relate to its position in the Modern Periodic table?
Answer:
In the modern periodic table, atoms with similar electronic configurations are placed in the same column. In a group, the number of valence electrons remains the same. Elements across a period show an increase in the number of valence electrons.

Question 9.
In the Modern Periodic table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21, and 38. Which of these have physical and chemical properties resembling calcium?
Answer:
The element with atomic number 12 has the same chemical properties as that of calcium. This is because both of them have same number of valence electrons (2).

Question 10.
Compare and contrast the arrangement of elements in Mendeleev’s Periodic table and the Modern Periodic table.
Answer:

Mendeleev s Periodic table:

1. 1. Elements are arranged in the increasing order of their atomic mass.
   2. This table has 8 groups and 6 periods. And each group is subdivided as an A and B.
   3. In this table, Hydrogen has no position.
   4. No position for isotopes, because in Mendeleev period these are not discovered.

Modern Periodic table:

1. 1. Elements are arranged in the increasing order of their atomic number.
   2. It has 18 groups and 7 periods.
   3. Inert gases are placed in separate groups.
   4. In this table, a zigzag line separates Metals from Non-metals.

(or)

Mendeleev’s Periodic table vs Modern Periodic table:

1. Elements are arranged in the increasing order of their atomic masses, while in Modern Periodic table elements are arranged in the increasing order of their atomic numbers.
2. There are a total of 7 groups (columns) and 6 periods (rows) while in Mendeleev’s’ Periodic Table, there are a total of 18 groups (columns) and 7 periods (rows).
3. Elements having similar properties were placed directly under one another, while in Mendeleev’s’ Periodic Table elements having the same number of valence electrons are present in the same group.
4. In Mendeleev’s Periodic Table the position of hydrogen could not be explained, while in Modern Periodic table hydrogen is placed above alkali metals.
5. No distinguishing positions for metals and non-metals in Mendeleev’s Periodic Table while in Modern Periodic Table metals are present at the left-hand side of the periodic table whereas nonmetals are present at the right-hand side.

MP Board Class 10th Science Chapter 5 Additional Questions

MP Board Class 10th Science Chapter 5 Multiple Choice Questions

Question 1.
Which of the following statements is correct about the trends when going down in a group of the periodic table?
(a) Elements become less electropositive in nature.
(b) Element oxides become more acidic.
(c) Valence electrons increases.
(d) Elements lose their electrons more easily.
Answer:
(d) Elements lose their electrons more easily.

Question 2.
Element A forms a chloride with the formula ACl₃, which is a stable compound. A would most likely be the same group of the Periodic Table as –
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(c) Al

Question 3.
Which of the following are coin metals?
(a) Ne, Ca, Na
(b) H₂, N₂, O₂
(c) Li, Na, K
(d) Cu, Au, Ag
Answer:
(d) Cu, Au, Ag

Question 4.
Who gave the triad arrangement of elements?
(a) Mendeleev
(b) Newlands
(c) Dalton
(d) Dobereiner
Answer:
(d) Dobereiner

Question 5.
Newlands periodic table is based on the
(a) Atomic weight
(b) Atomic number
(c) Atomic radius
(d) Atomic volume
Answer:
(a) Atomic weight

Question 6.
Which of the following is not gas in normal atmospheric condition?
(a) Helium (He)
(b) Argon (Ar)
(c) Bromine (Br)
(d) Chlorine (Cl)
Answer:
(c) Bromine (Br)

Question 7.
While moving left to right across a period, the atomic radii –
(a) Remains the same
(b) Approaches zero
(c) Decreases
(d) Increases first then decreases
Answer:
(c) Decreases

Question 8.
Which element is a metalloid?
(a) Carbon
(b) Nitrogen
(c) Oxygen
(d) Silicon
Answer:
(d) Silicon

Question 9.
Moseley’s periodic table is based on
(a) Atomic mass
(b) Mass number
(c) Atomic number
(d) Atomic volume
Answer:
(c) Atomic number

Question 10.
Which of the following is a group of highly electronegative elements?
(a) Cl, Br, I
(b) S, Se, Te
(c) Na, K, Rb
(d) Ca, Sr, Ba
Answer:
(a) Cl, Br, I

Question 11.
Which of the following elements is a non-metal?
(a) Aluminium
(b) Chlorine
(c) Sodium
(d) Silicon
Answer:
(b) Chlorine

Question 12.
As we move down in a group in Modern Periodic Table, the size of elements generally
(a) increases
(b) decreases
(c) remain the same
(d) first, increase then decrease
Answer:
(a) increases

Question 13.
As we move from top to bottom in a group in Modern Periodic Table, the electronegativity of elements
(a) Increases
(b) Decreases
(c) No change
(d) Not certain
Answer:
(b) Decreases

Question 14.
Which group of elements is considered most electropositive?
(a) Group 1
(b) Group 2
(c) Group 17
(d) Group 18
Answer:
(a) Group 1

Question 15.
Group 1 elements are also called as:
(a) Alkali metals
(b) Alkaline earth metals
(c) Halogens
(d) Noble gases
Answer:
(a) Alkali metals

Question 16.
Group 17 elements are also called as:
(a) Alkali Metals
(b) Alkaline Earth Metals
(c) Halogens
(d) Noble Gases
Answer:
(c) Halogens

Question 17.
How many elements were known when Mendeleev started his work?
(a) 100
(b) 215
(c) 65
(d) 80
Answer:
(c) 65

Question 18.
Why Mendeleev left spaces in his Periodic Table?
(a) A mistake
(b) For future elements
(c) For Isotopes
(d) For Isobars
Answer:
(b) For future elements

Question 19.
Why Lanthanoids and Actinoids are placed below in the Periodic Table?
(a) A mistake
(b) Better representation and view
(c) They were found very recently
(d) All of the above
Answer:
(c) They were found very recently

Question 20.
A period may have elements with –
(a) Variable atomic sizes
(b) Variable atomic number
(c) Variable valency
(d) All of the above
Answer:
(d) All of the above

Question 21.
Element A belongs to group 15. The formula of its hydride will:
(a) AH
(b) AH₂
(c) AH₃
(d) A₃H
Answer:
(c) AH₃

Question 22.
An electropositive element, A with 2 valence electron will form which type of oxide?
(a) AO
(b) A₂O
(c) AO₂
(d) AO₃
Answer:
(a) AO

Question 23.
Most electronegative element in our Periodic Table:
(a) Iron
(b) Nitrogen
(c) Carbon
(d) Flourine
Answer:
(d) Flourine

Question 24.
Which of the following elements where not among metals/elements named to fill the gap of Mendeleev’s Periodic Tablespaces?
(a) Cobalt
(b) Scandium
(c) Gallium
(d) Germanium
Answer:
(a) Cobalt

Question 25.
In a group, all elements have similar ………..
(a) Electronic configuration
(b) Valence electron
(c) Electronegativity
(d) All of these
Answer:
(b) Valence electron

Question 26.
Which among the following is a noble gas?
(a) C
(b) N
(c) O
(d) Ne
Answer:
(d) Ne

Question 27.
Which of the following elements has electronic configuration E = 2, 6?
(a) C
(b) N
(c) O
(d) Ne
Answer:
(c) O

Question 28.
Which of the following elements is a metalloid?
(a) B
(b) Al
(c) S
(d) P
Answer:
(a) B

Question 29.
Which of the following is not a halogen?
(a) Br
(b) I
(c) Te
(d) At
Answer:
(c) Te

Question 30.
Which element has a total of two shells, with four valence electrons?
(a) C
(b) N
(c) Br
(d) Co
Answer:
(a) C
(ii) Match column A’s description with column B’s Particulars.

Answers:

1. → 12
2. → 3
3. → 2
4. → 4
5. → 5
6. → 13
7. → 11
8. → 14
9. → 10
10. → 10
11. → 6
12. → 1
13. → 7
14. → 8

MP Board Class 10th Science Chapter 5 Very Short Answer Type Questions

Question 1.
What would be the maximum number of electrons present in the outermost shell of atoms in the first period of Periodic Table?
Answer:
Two

Question 2.
What is Solder?
Answer:
It is an alloy of lead (Pb) and tin (Sn).

Question 3.
What is anode mud?
Answer:
During electrolytic refining, the soluble impurities go into the solution, Whereas, the insoluble impurities settle down at the bottom of anode and are known as anode mud.

Question 4.
Which metal is used with iron oxide to join railway tracks or cracked machine parts?
Answer:
Aluminium.

Question 5.
Give the thermit reaction.
Answer:
Fe₂O₃(s) + 2Al(s) → 2 Fe(l) + Al₂O₃(s) + Heat

Question 6.
Roasting is used for the extraction of which ore?
Answer:
Sulphide ore.

Question 7.
Name the metal lowest inactivity series (relative reactivities of metals).
Answer:
Au or Gold.

Question 8.
Which gas is evolved when a metal reacts with nitric acid?
Answer:
Hydrogen gas.

Question 9.
Name any two metal that does not react with water at all.
Answer:
Lead, copper, gold, silver. (any two)

Question 10.
Complete the following reaction: Metal oxide + water →.
Answer:
Metal hydroxide.

Question 11.
Which material is used to coat electric wires in homes?
Answer:
PVC or Polyvinylchloride.

Question 12.
Name any two metals that are poor conductors of heat.
Answer:
Lead and mercury.

MP Board Class 10th Science Chapter 5 Short Answer Type Questions

Question 1.
What are the limitations of the Modern Periodic Table?
Answer:
The limitations of the Modern Periodic Table:
Position of hydrogen still dicey. It is not fixed till now. Position of lanthanides and actinides has not been given inside the main body of the Periodic Table. It does not reflect the exact distribution of electrons of some of the transition and inner transition elements.

Question 2.
Two elements X and Y have atomic numbers 12 and 16 respectively. Write the electronic configuration for these elements. To which period of the Modern Periodic Table do these two elements belong? What type of bond will be formed between them and why?
Answer:
Electronic configuration of X (Z= 12): 2, 8,2
Electronic configuration of Y (Z = 16): 2, 8,6
Both these elements belong to the third period. An ionic bond is formed between X and Y due to the transfer of two electrons from X to Y.

Question 3.
The present classification of elements is based on which fundamental property of elements?
Answer:
Atomic number.

Question 4.
Li, Na and K are the elements of a Dobereiner’s Triad. If the atomic mass of Li is 7 and that of K is 39, what would be the atomic mass of Na?
Answer:
According to of Dobereiner’s law of triads, the atomic mass of the middle element, in this case, Na, should be the arithmetic mean of Li and K. Thus, Arithmetic mean of Li and K = (7 + 39)/2 = 23.

Question 5.
Define Dobereiner’s law of triads.
Answer:
It states, “When elements are placed in order of the ascending order of atomic masses, groups of three elements having similar properties are obtained. The atomic mass of the middle element of the triad is equal to the mean of the atomic masses of the other two elements of the triad.”

Question 6.
Why did Dobereiner’s system of classification fail?
Answer:
The major drawback of Döbereiner’s classification was that it was valid only for a few groups of elements known during that time. He was able to identify three triads only. Also, more accurate measurements of atomic masses showed that the mid element of the triad did not really have the mean value of the sum of the other two elements of the triad. For elements of very low mass or very high mass, the law did not hold good. For example, Fluorine (F), Chlorine (Cl), Bromine (Br). The atomic mass of Cl is not an arithmetic mean of atomic masses of F and Br.

Question 7.
Explain the position of metalloids in the Modern Periodic Table.
Answer:
In the Modern Periodic Table, a zig-zag line separates metals from non-metals. The borderline elements – boron, silicon, germanium, arsenic, antimony, tellurium and polonium – are intermediate in properties and are called metalloids or semi-metals.

Question 8.
Why silicon is classified as metalloid?
Answer:
Silicon is classified as a semi-metal or metalloid because it exhibits some properties of both metals and non-metals.

Question 9.
State Newlands law of octaves.
Answer;
Elements are arranged in increasing order of their atomic masses such that the properties of the eighth element are the repetition of the properties of the first element (similar to eighth note in an octave of music).

Question 10.
X and Y are the two elements having similar properties which obey Newlands law of octaves. How many elements are there in between X and Y?
Answer:
The law states there are eight elements in an octave (row). A number of elements between X and Y are six.

Question 11.
What are the drawbacks of Newlands law of octaves?
Answer:
Following are the major drawbacks:

1. Worked well with lighter elements (upto calcium. After those elements in the eighth column did not possess properties similar to elements in the first column.
2. Newland assumed only 56 elements existed so far. Later, new elements were discovered which did not fit into octaves table.
3. Newland adjusted few elements in the same slot through their properties were quite different, e.g., Cobalt and nickel are in the same slot and these are placed in the same column as fluorine, chlorine and bromine which have very different properties than these elements. Iron, which resembles cobalt and nickel in properties, has been placed far away from these elements.

MP Board Class 10th Science Chapter 5 Long Answer Type Questions

Question 1.
What are the salient features of the Modern Periodic Table?
Answer:
In a period of the Periodic Table, the number of valence electrons increases as the atomic number increases. As a result, elements change from metal to metalloid to nonmetal to a noble gas. Atomic size is a periodic property. As atomic number increases in a period, the atomic radius decreases. As atomic number increases in a group, atomic radius increases.

Positive ions have smaller atomic radii than the neutral atoms from which they derive. Negative ions have larger atomic radii than their neutral atoms. Positive ions in the same group increase in size down the group. In a group, each element has the same number of valence electrons. As a result, the elements in a group show similar chemical behaviour.

Metallic character decreases from left to right in a period because of the increase in the effective nuclear charge. Non-metallic character increase from left to right in a period because of the increase in effective nuclear charge. Non-metallic character decreases down the group because of increase in the size of the atom.

Question 2.
What periodic trends do we observe in terms of atomic radii or atomic sizes in Modern Periodic table?
Answer:
Following two trends are observed:
1. Within each column (group), atomic radius tends to increase from top to bottom. This trend results primarily from the increase in the number of the outer electrons. As we go down a column, the outer electrons have a greater probability of being farther from the nucleus, causing the atom to increase in size.

2. Within each row (period), the atomic radius tends to decrease from left to right. The major factor influencing this trend is the increase in the nuclear charge as we move across a row. The increasing effective nuclear charge steadily draws the valence electrons closer to the nucleus, causing the atomic radius to decrease.

Question 3.
An element A with atomic number 19 combines separately with NO₃^–and (SO₄)₂^–,(PO₄)₃^–radicals:
(a) Give the electronic configuration of element A.
(b) Write the formulae of the three compounds so formed.
(c) To which group of the periodic table does the element ‘R’ belong?
(d) Does it form covalent or ionic compound? Why?
Answer:
(a) Electronic configuration of A: 2,8, 8, 1.
(b) Compounds formed are A(NO₃), A₂(SO₄) and K₃(PO₄).
(c) A has one valence electron and hence, it belongs to the first group.
(d) It forms the ionic compound.

Question 4.
Describe types of periods, blocks and trends of periodic properties along periods associated with Modern Periodic Table.
Answer:
Periods:
First period (Atomic number 1 and 2): This is the shortest period. It contains only two elements (hydrogen and helium).

Second Period: (Atomic number 3 to 10): It contains eight elements (lithium to neon).

Third period (Atomic number 11 to 18): It contains eight elements (sodium to argon).

Fourth period (Atomic number 19 to 36): Row contains eighteen elements (potassium to krypton). i.e., 8 normal elements and 10 transition elements.

5th period (Atomic number 37 to 54): Contains 18 elements (rubidium to xenon) includes 8 normal elements and 10 transition elements.

Sixth period (Atomic number 55 to 86): The longest period. It contains 32 elements (caesium to radon) has 8 normal elements, 10 transition elements and 14 inner transition elements (lanthanides).

7th period (Atomic number 87 to 118): As like the sixth period, this period also can accommodate 32 elements. Till now 26 elements have been authenticated by IUPAC.

Blocks in Periodic Table:
The periodic table includes “blocks” defined in terms of which type of orbital is being filled via the Aufbau principle. This gives us the s-block, p-block, d-block, and f-block.

Blocks:
The s-, p-, d-, and f-blocks contain elements with outer electrons in the same type of orbital. Another key link between electron arrangement and position in the periodic table is that elements in any one main group have the same number of electrons in their highest energy level. The number of elements discovered so far is 118. The last element authenticated by IUPAC is Cn112 (Copernicium).

Properties of Periods: As you proceed to the left in a period or as you proceed down within a group:

1. The metallic strength increases (Non-Metallic Strength decreases).
2. The atomic radius increases.
3. The ionization potential decreases.
4. The electron affinity decreases.
5. The electronegativity decreases.

MP Board Class 10th Science Chapter 5 Textbook Activities

Class 10 Science Activity 5.1 Page No. 84

1. Looking at its resemblance to alkali metals and the halogen family, try to assign hydrogen a correct position in Mendeleev’s Period Table.
2. To which group and period should hydrogen be assigned?

Observations:

1. No position can be fixed for hydrogen in the Mendeleev’s Periodic Table.
2. Properties of hydrogen fit with alkali metal as it combines with halogens, oxygen and sulphur to form compounds.
3. Properties of hydrogen also fit or are similar to halogen as it exists in the form of diatomic molecules and combines with metals and non-metals forming covalent compounds.

Class 10 Science Activity 5.2 Page No. 85

1. Consider the isotopes of chlorine, Cl-35 and CI-37.
2. Would you place them in different slots because their atomic masses are different?
3. Or would you place them in the same position because their chemical properties are the same?

Observations:
Two isotopes of chlorine are Cl-35 and Cl-37. Both isotopes have the same chemical properties and hence, both isotopes should be placed in the same position.

Class 10 Science Activity 5.3 Page No. 85

1. How were the positions of cobalt and nickel resolved in the Modern Periodic Table?
2. How were the positions of isotopes of various elements decided in the Modern Periodic Table?
3. Is it possible to have an element with atomic number 1, 5 placed between hydrogen and helium?
4. Where do you think should hydrogen be placed in the Modern Periodic Table?

Observations:
The position of Cobalt and Nickel were decided by placing them in the increasing order of atomic number in the Modern Periodic Table. Since isotopes are elements with the similar atomic number they are placed in the same position as its basic elements in the modern periodic table.

Class 10 Science Activity 5.4 Page No. 87

1. Look at group 1 of the Modern Periodic Table, and name the elements present in it. Write down the electronic configuration of the first three elements of group 1.
2. What similarity do you find in their electronic configurations?
3. How many valence electrons are present in these three elements?

Observations:
Elements present in Group 1 are:

Electronic configuration of the first three elements of Group I are as below:

Class 10 Science Activity 5.6 Page No. 87

1. If you look at the Modern Periodic Table, you will find that the elements Li, Be, B, C, N, O, F, and Ne are present in the second period. Write down their electronic configurations.
2. Do these elements also contain the same number of valence electrons?
3. Do they contain the same number of shells?

Observations:
No, these elements contain variable valence electrons as they belong to different groups:

Class 10 Science Activity 5.6 Page No. 88

1. How do you calculate the valency of an element from its electronic configuration?
2. What is the valency of magnesium with atomic number 12 and sulphur with atomic number 16?
3. Similarly, find out the valencies of the first twenty elements.
4. How does the valency vary in a period on going from left to right?
5. How does the valency vary in going down a group?

Observations:

1. Valency of an element can be calculated by the numbers of valence electron present.
2. Valency of Magnesium: 2
3. Valency of Sulphur: 2 Variation of valency while moving left to right in a period.
   1 → 2 → 3 → 4 → 3 → 2 → 1 → 0
4. Variation of valency while going down in a group does not change.

Class 10 Science Activity 5.7 Page No. 88

1. Atomic radii of the elements of the second period are given below:
   
2. Arrange them in decreasing order of their atomic radii.
3. Are the elements now arranged in the pattern of a period in the Periodic Table?
4. Which elements have the largest and the smallest atoms?
5. How does the atomic radius change as you go from left to right in a period?

Observations:
Decreasing order of atomic radii of following elements:

1. O < N < C < B < Be < Li
2. No in pattern.
3. Oxygen is smallest as per given data while Li is largest.
4. Atomic radius reduces while moving right in a group a, nuclear charge increase.

Class 10 Science Activity 5.8 Page No. 89

1. Study the variation in the atomic radii of first group elements given below and arrange them in increasing order.
   
2. Name the elements which have the smallest and the largest atoms.
3. How does the atomic size vary as you go down a group?

1. Sodium (Na) has the smallest atom and calcium (Ca) has the largest atom.
2. Atomic size increases as we go down a group.

Class 10 Science Activity 5.9 Page No. 89

1. Examine elements of the third period and classify them as metals and non-metals.
2. On which side of the Periodic Table do you find the metals?
3. On which side of the Periodic Table do you find the non-metals?

Observations:
Elements of the third period are:

Class 10 Science Activity 5.10 Page No. 89

1. How do you think the tendency to lose electrons change in a group?
2. How will this tendency change in a period?

Observations:
Metallic property reduces while moving right in a period.

Class 10 Science Activity 5.11 Page No. 90

1. How would the tendency to gain electrons change as you go from left to right across a period? How
2. would the tendency to gain electrons change as you go down a group?

Observations:

1. The electrons increases as we go left to right in a period up to 17th group. It decreases in the 18th
2. group. The tendency of gaining the electrons decreases as we go down a group.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) The sides are : 7 an, 24 cm, 25 cm
Here, (7 cm)² = 49 cm²
(24 cm)² = 576 cm²
(25 cm)² = 625 cm²
∵ (49 + 576)cm² = 625 cm²
∴ It is a right triangle.
Hypotenuse = 25 cm.

(ii) The sides are: 3 cm, 8 cm, 6 cm
Here, (3 cm)² = 9 cm²
(8 cm)² = 64 cm²
(6 cm)² = 36 cm²
∵ (9 + 36) ≠ 64 cm²
∴ It is not a right triangle.

(iii) The sides are : 50 cm, 80 cm, 100 cm
Here, (50 cm)² = 2500 cm²
(80 cm)² = 6400 cm²
(100 cm)² = 10000 cm²
∵ (2500 + 6400) cm² ≠ 10000 cm²
∴ It is not a right triangle.

(iv) The sides are : 13 cm, 12 cm, 5 cm
Here, (13 cm)² = 169 cm²
(12 cm)² = 144 cm²
(5 cm)² = 25 cm²
∵ (144 + 25)cm² = 169 cm²
∴ It is a right triangle.
Hypotenuse = 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR.
Solution:
In ∆QMP and ∆QPR,
∠QMP = ∠QPR [Each 90°]
∠Q = ∠Q [Common]
⇒ ∆QMP ~ ∆QPR …. (1) [AA similarity]
Again, in ∆PMR and ∆QPR,
∠PMR = ∠QPR [Each = 90°]
∠R = ∠R [Common]
⇒ ∆PMR ~ ∆QPR …… (2) [AA similarity]
From (1) and (2), we have

Question 3.
In the figure, ABD is a triangle, right angled at A and AC ⊥ BD.

Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
Solution:

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
We have, right ∆ABC such that ∠C = 90° and AC = BC.

∴ By Pythagoras theorem, we have AB² = AC² + BC² = AC² + AC² = 2AC²
[∵ BC = AC (given)]
Thus, AB² = 2AC²

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.
Solution:
We have, an isosceles AABC such that BC = AC.

Also, AB² = 2AC²
∴ AB² = AC² + AC²
But AC =BC
∴ AB² = AC² + BC²
∴ Using the converse of Pythagoras theorem, ∠ACB = 90°
f.e., ∆ABC is a right angled triangle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
In equilateral triangle, altitude bisects the base.
⇒ AD = DB

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Let us have a rhombus ABCD.

∵ Diagonal of a rhombus bisect each other at right angles.
∴ OA = OC and OB = OD
Also, ∠AOB = ∠BOC [Each = 90°]
And ∠COD = ∠DOA [Each = 90°]
In right ∆AOB, we have,
AB² = OA² + OB² …… (1)
[Using Pythagoras theorem]
Similarly, in right ∆BOC,
BC² = OB² + OC² …… (2)
In right ∆COD,
CD² = OC² + OD² …… (3)
In right ∆AOD,
DA² = OD² + OA² ……. (4)
Adding (1), (2), (3) and (4)
AB² + BC² + CD² + DA²
= [OA² + OB²] + [OB² + OC²] + [OC² + OD²] + [OD² + OA²]
= 2OA² + 2OB² + 2 OC² + 2OD² = 2[OA² + OB² + OC² + OD²]
= 2[OA² + OB² + OA² + OB²]

Thus, sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Question 8.
In the figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF²

Solution:
We have a point in the interior of a ∆ABC such that
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
(i) Let us join OA, OB and OC.

In right ∆OAF, by Pythagoras theorem
OA² = OF² + AF² …(1)
Similarly, from right triangle ODB and OEC, we have
OB² = BD² + OD², …(2)
and OC² = CE² + OE² …(3)
Adding (1), (2) and (3), we get OA² + OB² + OC²
= (AF² + OF²) + (BD² + OD²) + (CE² + OE²)
⇒ OA² + OB² + OC²
= AF² + BD² + CE² + (OF² + OD² + OE²)
⇒ OA² + OB² + OC² – (OD² + OE² + OF²)
= AF² + BD² +CE²
⇒ OA² + OB² + OC² – OD² – OE² – OF²
= AF² + BD² + CE²

(ii) In right triangle OBD and triangle OCD, by Pythagoras theorem:
OB² = OD² + BD² and OC² = OD² + CD²
⇒ OB² – OC² = OD² + BD² – OD² – CD²
⇒ OB² – OC² = BD² – CD² ….. (1)
Similarly, we have
OC² – OA² = CE² – AE² …… (2)
and OA² – OB² = AF² – BF² ….. (3)
Adding (1), (2) and (3), we get (OB² – OC²) + (OC² – OA²) + (OA² – OB²) = (BD² – CD²) + (CE² – AE²) + (AF² – BF²)
⇒ 0 = BD² + CE² + AF² – (CD² + AE² + BF²)
⇒ BD² + CE² + AF² = CD² + AE² + BF²
or AF² + BD² + CE² = AE² + BF² + CD²

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Let PQ be the ladder and PR be the wall

⇒ PQ = 10 m, PR = 8 m
Now, in the right ∆PQR, PQ² = PR² + QR²
⇒ 10² = 8² + QR²
[using Pythagoras theorem]
⇒ QR² = 10² – 8² = (10 + 8)(10 – 8)
= 18 × 2 = 36
QR = \(\sqrt{36}\) = 6m
Thus, the distance of the foot of the ladder from the base to the wall is 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB is the wire and BC is the vertical pole. The point A is the stake.

Now, in the right AABC, using Pythagoras Theorem, we have

Thus, the stake is required to be taken at \(6 \sqrt{7}\)m from the base of the pole to make the wire taut.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after \(1 \frac{1}{2}\) hours?
Solution:
Let the point A represent the airport. Plane-I fly towards North,
∴ Distance of the plane-I from the airport after \(1 \frac{1}{2}\) hours = speed × time

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops?
Solution:
Let the two poles AB and CD are such that the distance between their feet AC = 12m.
∵ Height of pole-1, AB = 11 m

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
Solution:
We have a right ∆ABC such that ∠C = 90°.
Also, D and E are points on CA and CB respectively.
rain We have a right ∆ABC such that ∠C = 90°.

Let us join AE and BD.
In right ∆ACB, using Pythagoras theorem
AB² = AC² + BC² …… (1)
In right ∆DCE, using Pythagoras theorem,
DE² = CD² + CE² …… (2)
Adding (1) and (2), we get
AB² + DE² = [AC² + BC²] + [CD² + CE²]
= AC² + BC² + CD² + CE² = [AC² + CE²] + [BC² + CD²] …. (3)
In right ∆ACE,
AC² + CE² = AE² …… (4)
In right ∆BCD,
BC² + CD² = BD² …….. (5)
From (3), (4) and (5), we have AB² + DE² = AE² + BD² or AE² + BD² = AB² + DE²

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB² = 2AC² + BC².

Solution:

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD² = 7AB².
Solution:
We have an equilateral ∆ABC; in which D is a point on BC such that BD = \(\frac{1}{3}\) BC.

Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
We have an equilateral ∆ABC, in which AD ⊥ BC.
Since, an altitude in an equilateral A, bisects the corresponding side.

∴ D is the mid-point

Question 17.
Tick the correct answer and justify : In ∆ABC, AB = \(6 \sqrt{3}\) cm, AC = 12 cm and BC = 6 cm. The angle B is
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution:
(C): We have, AB = \(6 \sqrt{3}\) cm, AC = 12 cm, and BC = 6 cm

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 1.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Here, r = 6 cm and θ = 60°

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Let radius of the circle be r.
∴ 2πr = 22

Question 3.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Length of minute hand = radius of the circle (r) = 14 cm
∵ Angle swept by the minute hand in 60 minutes = 360°
∴ Angle swept by the minute hand in 360°
5 minutes = \(\frac{360^{\circ}}{60}\) × 5 = 30°
Now, area of the sector with r = 14 cm and θ = 30°
= \(\frac{\theta}{360^{\circ}}\) × πr² = \(\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 14 × 14 cm²
= \(\frac{11 \times 14}{3} \mathrm{cm}^{2}=\frac{154}{3} \mathrm{cm}^{2}\)
Thus, the required area swept by the minute hand in 5 minutes = \(\frac{154}{3}\) cm²

Question 4.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major segment.(Use π = 3.14)
Solution:
Length of the radius (r) = 10 cm
Sector angle (θ) = 90°

= [314 – 78.5] cm² = 235.5 cm²

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord
Solution:
Here, radius(r) = 21 cm and θ = 60°
(i) Circumference of the circle = 2πr
= 2 × \(\frac{22}{7}\) × 21 cm = 2 × 22 × 3 cm = 132 cm

∴ Length of the arc APB

Question 6.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and \(\sqrt{3}=\) = 1.73)
Solution:
Here, radius (r) = 15 cm and
Sector angle (θ) = 60°
∴ Area of the sector
\(\frac{\theta}{360^{\circ}}\) × πr² = \(\frac{60^{\circ}}{360^{\circ}} \times \frac{314}{100}\) × 15 × 15 cm²
= \(\frac{11775}{100}\) cm² = 117.75 cm²
Since ∠O = 60° and OA = OB = 15 cm
∴ AOB is an equilateral triangle.

⇒ AB = 15 cm and ∠A = 60°
Draw OM ⊥ AB,
In ∆AMO

Now area of the minor segment = (Area of minor sector) – (ar ∆AOB)
= (117.75 – 97.3125) cm² = 20.4375 cm²
Area of the major segment = [Area of the circle] – [Area of the minor segment]
= πr² – 20.4375 cm²
= [\(\frac{314}{100}\) × 15²] – 20.4375 cm²
= (706.5 – 20.4375) cm² = 686.0625 cm²

Question 7.
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and \(\sqrt{3}=\) = 1.73 )
Solution:
Here θ = 120° and r = 12 cm

Draw OM ⊥ AB
In ∆AOB, ∠O = 120°
By angle sum property,
∠A + ∠B + ∠O = 180°
⇒ ∠A + ∠B = 180° – 120° = 60°
∵ OB = OA = 12 cm
⇒ ∠A = ∠B = 30°

= 36 × 1.73 cm² = 62.28 cm²
∴ Area of the minor segment = [Area of sector] – [Area of ∆AOB]
= [150.72 cm²] – [62.28 cm²] = 88.44 cm²

Question 8.
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10m long instead of 5 m. (Use π = 3.14)

Solution:
Here, length of the rope = 5 m
∴ Radius of the circular portion grazed by the horse(r) = 5 m
(i) Area of the circular portion grazed

(ii) When length of the rope is increased to 10 m, then, r = 10 m
Area of the new circular portion grazed
= \(\frac{\theta}{360^{\circ}}\) × πr² = \(\frac{90^{\circ}}{360^{\circ}} \times \frac{314}{100}\) × (10)² m²
= \(\frac{1}{4}\) × 314 m² = 78.5 m²
∴ Increase in the grazing area = (78.5 – 19.625) m² = 58.875 m²

Question 9.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Solution:
Diameter of the circle = 35 mm
∴ Radius (r) = \(\frac{35}{2}\) mm
(i) Circumference = 2πr
= 2 × \(\frac{22}{7} \times \frac{35}{2}\) mm = 22 × 5 mm = 110 mm
Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors = 35 mm
∴ Length of 5 pieces = 5 × 35 mm = 175 mm
∴ Total length of the silver wire = (110 + 175) mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,

Question 10.
An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:
Here, radius (r) = 45 cm
Since circle is divided into 8 equal parts,
∴ Sector angle corresponding to each part

Question 11.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution:
Here, radius (r) = 25 cm
Sector angle (θ) = 115°
∴ Total area cleaned at each sweep of the blades

Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Solution:
Here, radius (r) = 16.5 km and sector angle (θ) = 80°
∴ Area of the sea surface over which the ships are warned

Question 13.
A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use \(\sqrt{3}=\) = 1.7)

Solution:
Here, r = 28 cm
Since, the circle is divided into six equal sectors.

Now, area of 1 design = Area of segment APB = Area of sector ABO – Area of ∆AOB ………..(2)
In ∆AOB, ∠AOB = 60°, OA = OB = 28 cm
∴ ∠OAB = 60° and ∠OBA = 60°
⇒ ∆AOB is an equilateral triangle.
⇒ AB = AO = BO ⇒ AB = 28 cm
Draw OM ⊥ AB
∴ In right ∆AOM, we have

= 14 × 14 × 1.7 cm² = 333.2 cm² ……………(3)
Now, from (1), (2) and (3), we have
Area of segment APB
= 410.67 cm² – 333.2 cm² = 77.47 cm²
⇒ Area of 1 design = 77.47 cm²
∴ Area of the 6 equal designs = 6 × (77.47) cm² = 464.82 cm²
Hence, the cost of making the design at the rate of ₹ 0.35 per cm²
= ₹ 0.35 × 464.82 = ₹ 162.68.

Question 14.
Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is

Solution:
(D) Here, radius = R
Angle of a sector (θ) = p
∴ Area of the sector

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic
progression, and why’
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and 8 for each
additional km.
(ii) The amount of air present In a cylinder when a vacuum pump removes 1/4 of the
air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs 150 for
the first metre and rises by 50 for each subsequent metre.
(iv) The amount of money in the account every year, when 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Let us consider, the first term
(T₁) Fare for the first 1 km = ₹ 15
Since, the taxi fare beyond the first 1 km is ₹ 8 for each additional km.
∴ Fare for 2 km = ₹ 15 + 1 × ₹ 8
T₂ = a + 8 [where a = 15]
Fare for 3 km = ₹ 15 + 2 × ₹ 8
⇒ T₃ = a + 16
Fare for 4 km= ₹ 15 + 3 × ₹ 8
⇒ T₄ = a + 24
Fare for 5 km = ₹ 15 + 4 × ₹ 8
⇒ T₅ = a + 32
Fare for n km = ₹ 15 + (n – 1)8
⇒ T_n = a + (n – 1)8
We see that above terms form an AP with common difference 8.

(ii) Let the amount of air present in the cylinder be x

The above terms are not in A.P.

(iii) Here, the cost of digging for first 1 metre = ₹ 150
The cost of digging for first 2 metres = ₹ 150 + ₹ 50 = ₹ 200
The cost of digging for first 3 metres = ₹ 150 + (₹ 50) × 2 = ₹ 250
The cost of digging for first 4 metres = ₹ 150 + (₹ 50) × 3 = ₹ 300
∴ The terms are: 150, 200, 250, 300,…..
Since, 200 – 150 = 50 and 250 – 200 = 50
(200 – 150) (250 – 200) = 50
∴ The above terms form an AP with common difference 50.

(iv)

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given
as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = 1/2
(y) a = -1.25, d = -0.25
Solution:
(i) ∵ T_n = a + (n – 1)d
∴ For a = 10 and d = 10, we have:
T₁ =10 + (1 – 1) × 10 = 10 + 0 = 10
T₂ = 10 + (2 – 1) × 10 = 10 + 10 – 20
T₃ = 10 + ( 3 – 1) × 10 = 10 + 20 = 30
T₄ = 10 + (4 – 1) × 10 = 10 + 30 = 40
Thus, the first four terms are:
10, 20, 30, 40

(ii) ∵ T_n = a + (n – 1)d
∴ For a = -2 and d = 0,we have:
T₁ = -2 + (1 – 1) × 0 = -2 + 0 = -2
T₂ = -2 + (2 – 1) × 0 = -2 + 0 = -2
T₃ = -2 + (3 – 1) × 0= -2 + 0 = -2
T₄ = -2 + (4 – 1) × 0 = -2 + 0 = -2
∴ Thus, the first four terms are: -2, -2, -2, -2

(iii) ∵ T_n = a + (n – 1)d
For a = 4 and d = -3, we have:
T₁ = 4 + (1 – 1) × (-3) = 4 + 0 = 4
T₂ = 4 + (2 – 1) × (-3) = 4 + (-3) = 1
T₃ = 4 + (3 – 1) × (-3) = 4 + (-6) = -2
T₄ = 4 + (4 – 1) × (-3) = 4 + (-9) = -5
Thus, the first four terms are:
4, 1, -2, -5

(iv)

(v) ∵ T_n = a + (n – 1)d
∴ For a = -1.25 and d = -0.25, we have
T₁ = -1.25 + (1 – 1) × (-0.25) = -1.25 + 0
= -1.25
T₂ = -1.25 + (2 – 1) × (-0.25) = -1.25 + (-0.25) = -1.50
T₃ = -1.25 + (3 – 1) × (-0.25) = -1.25 + (-0.50) = -1.75
T₄ = -1.25 + (4 – 1) × (-0.25) = -1.25 + (-0.75) = -2.0

Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3,…..
(ii) -5, -1, 3, 7,….
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots .\)
(iv) 0.6, 1.7, 2.8, 3.9, …..
Solution:
(i) We have : 3, 1, -1, -3, …
⇒ T₁ ⇒ 3 = a = 3
T₂ = 1, T₃= -1, T₄= -3
∴ T₂ – T₁ = 1 – 3 = -2
T₄ = T₃ = -3 – (-1) = -3 + 1 = -2 ⇒ d = -2

(ii) We have : -5, -1, 3, 7,….
⇒ T₁ = -5 ⇒ a = -5, T₂ = -1, T₃ = 3, T₄ = 7
∴ T₂ – T₁ = -1 – (-5) = -1 + 5 = 4
and T₄ – T₃ = 7 – 3 = 4 ⇒ d = 4
Thus a = -5, d = 4

(iii)

Question 4.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16,
(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ….
(iii) -1.2, -3.2, -5.2, -7.2, ……
(iv) -10, -6, -2, 2, ….
(v) \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots\)
(vi) 0.2, 0.22, 0.222, 0.2222, ….
(vii) 0, -4, -8, -12,
(ix) 1, 3, 9, 27, …..
(x) o, 2a, 3a, 4a,…
(xi) a, a₂, a₃, a₄,….
(xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\)
(xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots .\)
(xiv) 1², 3², 5², 7², ….
(xv) 1², 5², 7², 73, …..
Solution:
(i)

∴ The given numbers do not form an AP

(ii)

(iii) We have : -1.2, -3.2, -5.2, -7.2,
∴ T₁ = -1.2, T₂ = -3.2, T₃ = -5.2, T₄ = -7.2
T₂ – T₁ = -3.2 + 1.2 = -2
T₃ – T₂ = -5.2 + 3.2 = -2
T₄ – T₃ = -7.2 + 5.2 = -2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = -2 d = -2
∴ The given numbers form an AP such that d = -2.
Now, T₅ = T₄ + (-2) = -7.2 + (-2) = -9.2,
T₆ = T₅ + (-2) = -9.2 + (-2) = -11.2 and
T₇ = T₆ + (-2) = -11.2 + (-2) = -13.2
Thus, d = -2 and T₅ = -9.2, T₅ = -11.2 and T₆ = -13.2

(iv) We have : -10, -6, -2, 2,
T₁ = -10, T₂ = -6, T₃ = -2, T₄ = 2
T₂ – T₁ = -6 + 10 = 4
T₃ – T₂ = -2 + 6 = 4
T₄ – T₃ = 2 + 2 = 4
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 4 ⇒ d = 4
∴ The given numbers form an AP Now, T₅ = T₄ + 4 = 2 + 4 = 6,
T₆ = T₅ + 4 = 6 + 4 = 10,
T₇ = T₆ + 4 = 10 + 4 = 14
Thus, d = 4 and T₅ = 6, T₆ = 10, T₇ = 14

(v) We have :
3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),….

(vi) We have : 0.2, 0.22, 0.222, 0.2222,….

∴ The given numbers do not form an AP

(vii) We have : 0, – 4, – 8, – 12,
∴ T₁ = 0, T₂ = -4, T₃ = -8, T₄ = -12 T₂ – T₁ = -4 – 0 = -4
T₃ – T₂ = – 8 + 4 = – 4
T₄ – T₃ = -12 + 8 = -4
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = -4 ⇒ d = -4
∴ The given numbers form an AP
Now, T₅ = T₄ + (- 4) = -12 + (- 4) = -16
T₆ = T₅ + (-4) = -16 + (-4)= -20
T₇ = T₆ + (- 4) = -20 + (- 4) = -24
Thus, d = – 4 and T₅ = -16, T₆ = -20,
T₇ = -24.

(viii)

(ix) We have, 1, 3, 9, 27,….

(x) We have : a, 2a, 3a, 4a,

(xi)

(xii)

(xiii)

(xiv)
We have : 1², 3₂, 5₂, 7₂,….

(xv) We have : 1₂ , 5₂, 7₂, 7₂,….

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.4

Question 1.
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
We have ar(∆ABC) = 64 cm²
ar(∆DEF) = 121 cm² and EF = 15.4 cm[Given]
∵ ∆ABC ~ ∆DEF
∴ \(\frac { { ar }(\Delta ABC) }{ { ar }(\Delta DEF) } =\left( \frac { BC }{ EF } \right) ^{ 2 }\)
[Ratios of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides]

Thus, BC = 11.2 cm

Question 2.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution:
In trapezium ABCD, AB || DC. Diagonals AC and BD intersect at O.
In ∆AOB and ∆COD,

∠AOB = ∠COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate angles]
∴ Using AA criterion of similarity, we have
∆AOB ~ ∆COD

i. e., ar(∆AOB) : ar(∆COD) = 4 : 1

Question 3.
In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \(\frac { { ar }(\Delta ABC) }{ { ar }(\Delta DBC) } =\frac { AO }{ DO } \)

Solution:
We have, ∆ABC and ∆DBC are on the same base BC. Also BC and AD intersect at O.
Let us draw AE⊥BC and DF⊥BC.
In ∆AOE and ∆DOF,
∠AEO = ∠DFO = 90° ……….. (1)
Also, ∠AOE = ∠DOF …………… (2)
[Vertically Opposite Angles]

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
We have ∆ABC and ∆DEF, such that ∆ABC ~ ∆DEF and ar(∆ABC) = ar(∆DEF).
Since the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.


i.e., the corresponding sides of ∆ABC and ∆DEF are equal.
⇒ ∆ABC ≅ ∆DEF [By SSS congruency]

Question 5.
D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution:
We have a ∆ABC in which D, E and F are mid points of AB, BC and CA respectively. D, E and F are joined to form ∆DEF.

Now, in ∆ABC, D and F are the mid-points of sides AB and AC.
∴ \(\frac{A D}{D B}=\frac{A F}{F C}\) = 1
∴ By the converse of the basic proportion¬ality theorem, we have,
DF||BC ⇒ DF||BE
Similarly; EF||AB ⇒ EF||BD
Since, DF||BE and DB||EF
∴ Quadrilateral BEFD is a parallelogram.
⇒ FE = BD = \(\frac{1}{2}\)AB …………. (1)
Similarly, quadrilateral ECFD is a parallelogram.
⇒ DF = EC = \(\frac{1}{2}\)BC ………… (2)
and DE = FC = \(\frac{1}{2}\)AC ………….. (3)
Now, in ∆ABC and ∆DEF

Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
We have two triangles ABC and DEF such that ∆ABC ~ ∆DEF

AM and DN are medians corresponding to BC and EF respectively.
∵ ∆ABC ~ ∆DEF
∴ The ratio of their areas is equal to the square of the ratio of their corresponding sides.

Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
We have a square ABCD, whose diagonal is AC. Equilateral ∆BQC is described on the side BC and another equilateral ∆APC is described on the diagonal AC.

∵ All equilateral triangles are similar.
∴ ∆APC ~ ∆BQC
∴ The ratio of their areas is equal to the square of the ratio of their corresponding sides.

Tick the correct answer and justify:
Question 8.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Solution:
(C) : We have an equilateral ∆ABC and D is the mid point of BC. DE is drawn such that BDE is also an equilateral traingle. Since all equilateral triangles are similar,

∴ ∆ABC ~ ∆BDE
⇒ The ratio of their areas is equal to the square of the ratio of their corresponding sides.
∴ \(\frac { { ar }(\Delta ABC) }{ { ar }(\Delta BDE) } =\left( \frac { AB }{ BD } \right) ^{ 2 }\) …………. (1)
∵ AB = AC = BCfsides of equilateral AABC]
and BD = \(\frac{1}{2}\)BC [∵ D is the mid point of BC]
⇒ BC = 2BD = AB …………… (2) [∵ AB = BC]
From (1) and (2), we have

⇒ ar(∆ABC) : ar(∆BDE) = 4 : 1

Question 9.
Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Solution:
(D) : We have two similar triangles such that the ratio of their corresponding sides is 4 : 9.
∴ The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
∴ \(\frac { { ar }(\Delta -I) }{ { ar }(\Delta -II) } =\left( \frac { 4 }{ 9 } \right) ^{ 2 }=\frac { 16 }{ 81 } \)
⇒ ar(∆-I): ar(∆-II) = 16 : 81

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Solution:
(i) we have p(x) = x² – 2x – 8
= x² + 2x + – 4x – 8
= x(x + 2) – 4(x + 2)
= (x – 4)(x + 2)
For p(x) = 0, we must have (x – 4)(x + 2) = 0
Either x – 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = -2
∴ The zeroes of x² – 2x – 8 are 4 and -2
Now, sum of zeroes


Thus, the relationship between the zeroes and the coefficients in the polynomial x² – 2x – 8 is verified.

(ii) We have p(s) = 4s² – 4s + 1
= 4s² – 2s – 2s + 1
= 2s(2s – 1) – l(2s – 1)
= (2s – 1)(2s – 1)

Thus, the relationship between the zeroes and coefficients in the polynomial 4s² – 4s + 1 is verified.

(iii) We have p(x) = 6x² – 3 – 7x
= 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
For p(x) = 0, we have,

Thus, the relationship between the zeroes and coefficients in the polynomial 6x² – 3 – 7x is verified.

(iv) We have, p(u) = 4u² + 8u = 4u(u + 2)
For p(u) = 0,
we have
Either 4u = 0 ⇒ u = 0
or u + 2 = 0 ⇒ u = -2
∴ The zeroes of 4u² + 8u are 0 and – 2.
Now, 4u² + 8u can be written as 4u² + 8u + 0.

Thus, the relationship between the zeroes and coefficients in the polynomial 4u² – 4s + 1 is verified.

(v) We have, p(t) = t² – 15 = (t)² – \((\sqrt{15})^{2}\)

Thus, the relationship between zeroes and the coefficients in the polynomial t² – 15 is verified.

(vi) We have, p(x) = 3x² – x – 4
= 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1)(3x – 4)
For p(x) = 0 we have,
Either (x + 1) = 0 ⇒ x = -1

Thus, the relationship between the zeroes and coefficients in the polynomial 3x² – x – 4 is verified

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac{1}{4}\), -1
(ii) \(\sqrt{2}, \frac{1}{3}\)
(iii) 0, \(\sqrt{5}\)
(iv) 1, 1
(v) \(-\frac{1}{4}, \frac{1}{4}\)
(vi) 4,1
Solution:
(i) Since, sum of zeroes, \((\alpha+\beta)=\frac{1}{4}\)
Product of zeroes, αβ = -1
∴ The required quadratic polynomial is

have samezeroes, therefore (4x² – x – 4) is the required quadratic polynomial.

(ii) Since, sum of zeroes, \((\alpha+\beta)=\sqrt{2}\)
product of zeroes, αβ = \(\frac{1}{3}\)
∴ The required quadratic polynomial is

(iii) Since, sum of zeroes, (α + β) = 0
Product of zeroes, αβ = \(\sqrt{5}\)
∴ The required quadratic polynomial is

(iv) Since, sum of zeroes, (α + β) = 1
Product of zeroes, αβ = 1
∴ The required quadratic polynomial is

(v) Since, sum of zeroes, (α + β) = \(-\frac{1}{4}\)
Product of zeroes, αβ = \(\)[\frac{1}{4}/latex]
∴ The required quadratic polynomial is

same zeroes, therefore, the required quadratic polynomial is (4x² + x + 1)

(vi) Since, sum of zeroes, (α + β) = 4 and
product of zeroes, αβ = 1
∴ The required quadratic polynomial is
x² – (α + β)x + αβ = x² – 4x + 1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.6 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6

Question 1.
In the figure, PS is the bisector of ∠QPR of ∆PQR. Prove that \(\frac{Q S}{S R}=\frac{P Q}{P R}\).

Solution:
We have, ∆PQR in which PS is the bisector of ∠QPR.
∴ ∠QPS = ∠RPS
Let us draw RT || PS to meet QP produced at T, such that
∠1 = ∠RPS [Alternate angles]
Also, ∠3 = ∠QPS [Corresponding angles]

But ∠RPS = ∠QPS [Given]
∴ ∠1 = ∠3
∴ PT = PR
Now, in ∆QRT, PS || RT [By construction]
∴ Using the Basic Proportionality Theorem, we have

Question 2.
In the figure, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that
(i) DM² = DN.MC
(ii) DN² = DM.AN

Solution:

Question 3.
In the figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² – AB² + BC² + 2BC.BD

Solution:
∆ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced.
∵ In ∆ADB, ∠D = 90°
∴ Using Pythgoras Theorem, we have
AB² = AD² + DB² ….. (1)
In right ∆ADC, ∠D = 90°
∴ Using Pythagoras Theorem, we have
AC² = AD² + DC²
= AD² + [BD + BC]²
= AD² + [BD² + BC² + 2BD.BC]
⇒ AC² = [AD² + DB²] + BC² + 2BC – BD
⇒ AC² = AB² + BC² + 2BC – BD [From (1)] Thus, AC² = AB² + BC² + 2 BC.BD

Question 4.
In the figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² – 2BC.BD.

Solution:
We have ∆ABC in which ∠ABC < 90°
and AD ⊥ BC
In right ∆ADB, ∠D = 90°
Using Pythagoras Theorem, we have
AB² = AD² + BD² …… (1)
Also in right ∆ADC, ∠D = 90°
Using Pythagoras Theorem, we have AC² = AD² + DC²
= AD² + [BC – BD]² = AD² + [BC² + BD² – 2BC.BD]
= [AD² + BD²] + BC² – 2BC.BD = AB² + BC² – 2BC.BD [From (1)]
Thus, AC² = AB² + BC² – 2BC.BD, which is the required relation.

Question 5.
In the figure, AD is a median of triangle ABC and AM ⊥ BC. Prove that


Solution:

Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
We have a parallelogram ABCD.

AC and BD are the diagonals of parallelogram ABCD.
∵ Diagonals of a parallelogram bisect each other.
∴ O is the mid-point of AC and BD.
Now, in ∆ABC, BO is a median.

Question 7.
In the figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC – ∆DPB
(ii) AP.PB = CP.DP

Solution:
We have two chords AB and CD of a circle. AB and CD intersect at P.
(i) In ∆APC and ∆DPB,
∴ ∠APC = ∠DPB ….. (1)
[Vertically opp. angles]
∠CAP = ∠BDP …… (2)
[Angles in the same segment]
From (1) and (2) and using AA similarity, we have
∆APC ~ ∆DPB

(ii) Since, ∆APC ~ ∆DPB [As proved above]
∴ Their corresponding sides are proportional,

⇒ AP.BP = CP.DP, which is the required relation.

Question 8.
In the figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) ∆PAC – ∆PDB
(ii) PA.PB = PC.PD

Solution:
We have two chords AB and CD when produced meet outside the circle at P.
(i) Since in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle,

Question 9.
In the figure, D is a point on side BC of ∆ABC such that \(\frac{B D}{C D}=\frac{A B}{A C}\). Prove that AD is the bisector of ∠BAC.

Solution:
Let us produce BA to E such that AE = AC
Join EC.

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds ?

Solution:
Let us find the length of the string that Nazima has out.
In right ∆OAB, OB² = OA² + AB²
∴ OB² = (2.4)² + (1.8)²
⇒ OB² = 5.76 + 3.24 = 9.00
\(\Rightarrow \quad O B=\sqrt{9.00}=3 \mathrm{m}\)
i.e., Length of string she has out = 3 m

Since, the string is pulled in at the rate of 5 cm/sec,
∴ Length of the string pulled in 12 seconds

In the ∆PBC, let PB be the required horizontal distance of fly.
Since, PB² = PC² – BC² [By Pythagoras theorem]
∴ PB² = (2.4)² – (1.8)² = 5.76 – 3.24 = 2.52

Thus, the horizontal distance of the fly from Nazima after 12 seconds
= (1.59 + 1.2) m (approximately)
= 2.79 m (approximately)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
Mode:
Here, the highest frequency is 23.
The frequency 23 corresponds to the class interval 35 – 45.
∴ The modal class is 35 – 45
Now, class size, h = 10
Lower limit, l = 35
Frequency of the modal class (f₁) = 23
Frequency of the class preceding the modal class (f₀) = 21
Frequency of the class succeeding the modal class (f₂) = 14
∴ Mode = l + \(\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]\) × h

Mean: Let assumed mean, a = 40
Class size, h = 10
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-40}{10}\)

∴ Required mean = 35.38 years.
Interpretation:
The maximum number of patients admitted in the hospital are of age 36.8 years while the average age of patients is 35.37 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.
Solution:
Here, the highest frequency = 61.
∵ The frequency 61 corresponds to class 60 – 80
∴ The modal class is 60 – 80
∴ We have l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38

Thus, the modal lifetimes of the components is 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:
Mode:
∵ The maximum number of families i.e., 40 have their total monthly expenditure is in interval 1500 – 2000.
∴ The modal class is 1500 – 2000 and l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33

Thus, the required modal monthly expenditure of the families is ₹ 1847.83.
Mean: Let assumed mean, a = 3250
and class size, h = 500
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-3250}{500}\)
∴ We have the following table;

∴ \(\overline{x}\) = a + h × [\(\frac{1}{N}\) Σf_iu_i] = 3250 + 500 × \(\left[\frac{-235}{200}\right]\)
= 3250 – \(\frac{1175}{2}\) = 3250 – 587.50 = 2662.5
Thus, the mean monthly expenditure = ₹ 2662.50

Question 4.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
Mode : Since greatest frequency i.e., 10 corresponds to class 30 – 35.
∴ Modal class = 30 – 35 and h = 5, l = 30, f₁ = 10, f₀ = 9, f₂ = 3
∴ Mode = l + \(\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]\) × h
= 30 + \(\left[\frac{10-9}{20-9-3}\right]\) × 5
= 30 + \(\left[\frac{10-9}{20-9-3}\right]\) × 5 = 30 + 0.625 = 30.6 (approx)
Mean: Let the assumed mean, a = 37.5 and class size, h = 5
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-37.5}{5}\)
∴ We have the following table:


Interpretation:
The maximum teacher-student ratio is 30.6 while average teacher-student ratio is 29.2

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data
Solution:
∵ The class 4000 – 5000 has the highest frequency i.e., 18
∴ Modal class = 4000 – 5000.
Also, h = 1000, l = 4000, f₁ = 18, f₀ = 4, f₂ = 9

Thus, the required mode is 4608.7 runs.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Solution:
∵ The class 40 – 50 has the maximum frequency i.e., 20
∴ Modal class = 40 – 50
∴ f₁ = 20, f₀ = 12, f₁ = 11 and l = 40. Also, h = 10

Thus, the required mode is 44.7 cars.

MP Board Class 10th Science Solutions Chapter 12 Electricity

MP Board Class 10th Science Chapter 12 Intext Questions

Class 10th Science Chapter 12 Intext Questions Page No. 200

Question 1.
What does an electric circuit mean?
Answer:
A continuous and closed path of an electric current is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
The SI unit of electric current is ampere (A).
When I coulomb of electric charge flows through any cross. Section of a conductor in I second, the electric current flowing through it is said to be 1 ampere.
∵ 1 ampere = 1C/1s

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
The SI unit of electric charge is column, which is equivalent to the charge contained in nearly 6 × 10¹⁸ electrons.

Class 10th Science Chapter 12 Intext Questions Page No. 202

Question 1.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
A source of electricity such as battery, cell, power supply etc. can maintain a potential difference.

Question 2.
What is meant by saying that the potential difference between two points is 1 V?
Answer:
One Volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

1 V = 1 Jc^-1

Question 3.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
W = VQ
= 6 × 1 = 6 joules
Hence 6 joules of energy is given to each coulomb of charge passing through a 6 V battery.

Class 10th Science Chapter 12 Intext Questions page No. 209

Question 1.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends upon main four factors which are given below:

1. Length of the conductor.
2. Cross-sectional area of the conductor.
3. Natural of material of the conductor.
4. Temperature of the conductor.

Question 2.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
Thicker the wire, lower is the resistance of the wire. Hence, current flows more easily through a thick wire. It is because the resistance of a conductor is inversely proportional to its area of cross-section.

Question 3.
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
As per ohm’s law, V = IR
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\) —— (i)
Potential difference is half

∴ Current flowing is also half of ts former value.

Question 4.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
a) Resistivity of iron = 10.0 × 10^-8
Resistivity of Mercury = 94.0 × 10^-8
Resistivity of an alloy is greater than iron. By this we conclude that, Iron is good conductor of heat comparing to Mercury.
b) Resistivity of silver is less, hence it is a good conductor of heat.

Question 5.
Use the data in Table 12.1 to answer the following:
Electrical resistivity of some substances at 20°C

(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer:
(a) Resistivity of iron = 10.0 × 10^-8 Ωm
Resistivity of mercury = 94.0 × 10^-8 Ωm
Hence, resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.
(b) It can be observed from table that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.

Class 10th Science Chapter 12 Intext Questions page No. 213

Question 1.
Draw a schematic diagram of a circuit consisting of a battery of ‘three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:
Schematic diagram of a circuit:

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

Hence, reading in the ammeter would be 0.24 A and reading in voltmeter would be 2.88 V.

Class 10th Science Chapter 12 Intext Questions Page No. 216

Question 1.
Judge the equivalent resistance when the following are connected in parallel:
(a) 1Ω, and 106 Ω
(b) 1 Ω, and 103 Ω and 106 Ω
Answer:
(a) 1 ohm and 10⁶ times ohm
R₁ = 1 ohm
R₂ = 10⁶ times = 1000000 ohm
Total resistance (parallel)
1/R = 1/R₁ + 1/R₂
= 1/1 + 1/1000000
= 1000000 + 1 /1000000
= 1000000/ 1000000
1/R = 1/1 ohm (approx.)
∴ Resistance = 1 ohm (approx.)

(b) Given R₁ = 1 ohm
R₂ = 10³ ohm
R₃ = 10⁶ ohm

Question 2.
An electric lamp of 100Ω, a toaster of resistance 50Ω and a water filter of resistance 500Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:
Resistance of Electric 1 amp R₁ = 100Ω
Resistance of Tosser, R₂ = 50Ω
Resistance of water filter, R₃ = 50Ω
Potential difference, V = 220V
When these are connected in parallel,


7.04A of electricity is obtained by three appliances
Resistance of an electric iron connected to the same source that takes as much current as all three appliances

∴ Resistance of iron box = 31.25Ω
Electricity flowing through this = 7.04A.

Question 3.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
(i) In parallel circuit, if one electrical appliance stops working due to some defect, then all other appliances keep working normally. In series circuit, if one electrical appliance stops working due to some defect, then all other appliances also stop working.

(ii) In parallel circuits, each electrical appliance gets the same voltage as that of the power supply line. In series circuit, appliances do not get the same voltage, as that of the power supply line.

(iii) In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high. In the series connection, the overall resistance of the circuit increases too much due to which the current from the power supply is low.

Question 4.
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω?
Answer:
Given,
R₁ = 2 Ω
R₂ = 3 Ω
R₃ = 6 Ω
(a) When R₂ and R₃ are connected in parallel with R₁ in series we receive

(b) When R₁, R₂, R₃ are connected in parallel we receive

Question 5.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(a) For highest resistance according to question resistances must be connected in series:

4 Ω, 8 Ω, 12 Ω, 24 Ω
R₁ = 4 Ω
R₂ = 8 Ω
R₃ = 12 Ω
R₄ = 24 Ω
Total resistance in series = R₁ + R₂ + R₃ + R₄
= 4 + 8 + 12 + 24
= 48 Ω
Now, resistance is maximum when connected in series.

(b) For lowest resistance, resistances must be connected in parallel.

Resistance is lowest when connected in parallel.

Class 10th Science Chapter 12 Intext Questions Page No. 218

Question 1.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
In cord of an electric heater, as current flows these become hot arid glows but in case of electric heater this will not happen.

Question 2.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
Given, Charge = 96000 coulomb
Time = 1 hour = 3600 seconds
Potential difference = 50 V Ω
As we know, I = Q/T
I = 96000/36000 = 80/3 amp.
V = I × R
50 = 80/3 × R
So, R = 15/8 Ω
Now, Heat = I²RT
= (80/3)² × 15/8 × 3600
800 × 6000
= 4800000 joules or 4.8 × 10⁶ J

Question 3.
An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
As per Joule’s law
H = VIt
H = IR = 5A × 20Ω = 100V
I = 5A, t = 30 sec.
∴ H = 100 × 5 × 30 J
= 15000 J = 1.5 × 10⁴ J.

Class 10th Science Chapter 12 Intext Questions Page No. 220

Question 1.
What determines the rate at which energy is delivered by a current?
Answer:
The rate at which electric energy is dissipated or consumed in an electric circuit is termed as electric power.
P = VI.

Question 2.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
P = VI
V = 220V, and I = 5A.
Power P = 220 × 5 = 1100 W
Power of the motor = P × t
P = 1100 W.
t = 2 Hr 2 × 60 × 60 W
= 7200 S
∴ Energy consumed, E = 1100 × 7200 J
= 7920000
= 7.92 × 10⁶J.

MP Board Class 10th Science Chapter 12 NCERT Textbook Exercises

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio \(\frac { R }{ R` } \) is –
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) \(\frac { { V }^{ 2 } }{ R } \)
Answer:
(b) IR²

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W

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

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
Voltmeter is connected in parallel in the circuit to measure the potential difference between two points.

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:
Given, Diameter = 0.5 mm = 0.0005 m

So, the resistance is one-fourth if the diameter is doubled.

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

Plot a graph between V and I and calculate the resistance of that resistor are given below:
Answer:

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
Given, V = 12v
I = 2.5 mA = 2.5 × 10^-3 A
As we know, R = \(\frac { V }{ 1 } \)
R = \(\frac{12 \mathrm{V}}{2.5 \times 10^{-3} \mathrm{A}}\) = 48000 Ω
So, R = 4.8 kΩ

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω, and 12 Ω respectively. How much current would flow through the 12 Ω resistor.
Answer:
R₁ = 0.2 Ω R₂ = 0.3 Ω R₃ = 0.4 Ω R₄ = 0.5 Ω R₅ = 12 Ω

Given, V = 9 V
R5 = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
= 13.4 Ω
According to formula I = \(\frac { V }{ R } \)
= \(\frac{9 \mathrm{V}}{13.4 \Omega}\)
= 0.67 A
0.67 A current would flow through the 12 Ω resistor.

Question 10.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Given
I = 5 A V = 220 V
Now, Let the number of 176 Ω resistors be n.

So, four resistors of 176 Ω are required to carry 5 A on a 220 V line.

Question 11.
Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of (i) 9Ω, (ii) 4Ω.
Answer:
If resistors are connected in series 6Ω + 6Ω + 6Ω =18Ω This ia not correct
When they are connected in parallel
\(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=3\) This is also wrong,
i) When they are connected in parallel

Two 6Ω resistors are connected in parallel

If 3rd resistor of 6Ω and 3Ω are connected in series, it becomes 6Ω + 3Ω = 9Ω.
ii) When they are connected in series

Resistance = 6Ω + 6Ω
= 12Ω
If 3rd resistor 6Ω is connected to 12Ω in parallel

Total resistance = 4Ω.

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Here V = 220V
I – 5A

∴ 110 lamps can be connected in parallel with each other across the two wires.

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:
(i) If coils are connected separately V = 220 V
Resistance R₁ = 240
As per ohm’s law, V = IR

∴ If coils are connected separately 9.16A electricity flows in the coil.
(ii) If coils are connected in series
Resistance R₂ = 24Ω + 24Ω = 48Ω
As per ohm’s law V = IR

∴ If coils are connected in series 4.58A electricity flows.

Question 14.
Compare the power used in the 20 resistors in each of the following circuits:
(i) a 6 V battery in series with 1Ω and 2Ω resistors, and
(ii) a 4 V battery in parallel with 12Ω and 2Ω resistors.
Answer:
(i) Potential Difference V = 6V
If 1Ω and 2Ω resistors are connected in series, then Resistance
R = 1 + 2 = 3W.
As per ohm’s law
\(\mathrm{I}=\frac{6}{3}=2 \mathrm{A}\)
P(I²)R = (2)² × 2 = 8W.
(ii) Potential difference V = 4V
If 12Ω and 2Ω resistors are connected in parallel, voltage is equal
Voltage of resistance 2 W is 4 volts

∴ Power of 2Ω is 8 W.

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Both lamps are connected in parallel potential difference = 220 V
Power = V × I.

Question 16.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
250W TV set in used in 1 Hr, then its energy
= 250 × 3600 = 9 × 10⁵
energy of Toaster = 1200 × 600
If it is used in 10 minutes, then its power = 1200 × 600 × 7.2 × 10⁵ J
∴ Energy of 250 W TV set is used in 1 Hr is greater than 1200 W toaster used in 10 minutes.

Question 17.
An electric heater of resistance 8Ω draws 15 A from the service mains 2 hours.
Calculate the rate at which heat is developed in the heater.
Answer:
P = I²R
R = 8Ω, I = 15A
P= (15)² × 8 = 1800 J/s.
∴ Rate at which heat is developed in the heater
= 1800 J/s.

Question 18.
Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why arc the conductors of electric heating devices, such as bread-toasters and electric irons, majle of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
(a) Tungsten used almost exclusively for filament of electric lamps because its melting point is very high and resistivity of tungsten is low, it has special property of glowing on heating, it does not oxidized.

(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is higher than metal so alloy produces large amount of heat. They do not get oxidized. Melting point is also high.

(c) As in series, circuit voltage is divided. Components of a series circuit receives only small voltage so the amount of current decreases and the device turn hot and do not work properly. Hence, series arrangement is avoided being used in domestic circuits.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A) i.e., when area of cross section decreases the resistance increases or vice versa.

(e) Copper and aluminium are usually used for electricity transmission because they are good conductors of electricity and also have low resistivity and do not get heated.

MP Board Class 10th Science Chapter 12 Additional Important Questions

MP Board Class 10th Science Chapter 12 Multiple Choice Questions

Question 1.
SI unit of an electric current is equal to
(a) 6 × 10⁸ C
(b) 9.8 × 10¹⁰ C
(c) 1.6 × 10^-19 C
(d) None of these
Answer:
(d) None of these

Question 2.
Which materials have changing resistivity?
(a) Conductors
(b) Semiconductors
(c) Non-metals
(d) Super metals
Answer:
(b) Semiconductors

Question 3.
If wire in a circuit is doubled its resistivity will be?
(a) Two times
(b) Ten times
(c) Zero
(d) half
Answer:
(d) half

Question 4.
A wire used in circuit must be
(a) Highly resistant
(b) Non resistance
(c) Have high melting point
(d) None of these
Answer:
(b) Non resistance

Question 5.
A heating w ire must have
(a) High melting point
(b) High conductivity to heat
(c) Both
(d) None
Answer:
(b) High conductivity to heat

Question 6.
SI unit of resistance is:
(a) Coulomb
(b) Ampere
(c) Ohm
(d) Ohm – meter
Answer:
(c) Ohm

Question 7.
Symbol for use of voltmeter in circuit is:

Answer:

Question 8.
Symbol for use of battery in circuit is:

Answer:

Question 9.
Relation of potential difference and current is represented by:
(a) Joule’s law
(b) Ohm’s law
(c) Dalton’s law
(d) None
Answer:
(b) Ohm’s law

Question 10.
Formula which represents the relation among potential difference and current following in a circuit is:
(a) V = IR
(b) V = I/R
(c) l = Q/t
(d) R = p \(\frac { l }{ A } \)
Answer:
(a) V = IR

Question 11.
Kilo watt is a unit for:
(a) Power
(b) Electrical energy commercial
(c) Resistance
(d) Current flow in circuit
Answer:
(a) Power

Question 12.
Which one has less resistance?
(a) 10 watt bulb
(b) 15 watt LED
(c) 60 watt bulb
(d) 60 watt LED.
Answer:
(a) 10 watt bulb

Question 13.
In which type of combinations resistance is more?
(a) Series Combination
(b) Parallel combination
(c) both
(d) Single resistance unit.
Answer:
(a) Series Combination

Question 14.
Joule’s law formulation is
(a) H = IRT
(b) H = I2RT
(c) P = VI
(d) RT = PV
Answer:
(d) RT = PV

Question 15.
Hardest metal which can resist maximum heat is:
(a) Lead
(b) Tungsten
(c) Silver
(d) Carbon
Answer:
(b) Tungsten

Question 16.
If current passing a circuit is 10 A and wire is connected to 100 volt line resistant in the wire will be:
(a) 10 A
(b) 10 Ω
(c) 100 Ω
(d) O Ω
Answer:
(b) 10 Ω

Question 17.
An insulator has a resistivity equals to:
(a) 10^-8 Ω m
(b) 10 Ω m
(c) 5 Ω m
(d) 10^-18 Ω
Answer:
(a) 10^-8 Ω m

Question 18.
Calculate current-(I) in circuit given below:
(a) 0.6 A
(b) 20 A
(c) 50 A
(d) 10 A

Answer:
(a) 0.6 A

Question 19.
If resistors are arranged in parallel with value 1 Ω, 2 Ω and 3 Ω Total resistant will be:
(a) more than 3 Ω
(b) less than 3 Ω
(c) less than 1 Ω
(d) equal to 1 Ω
Answer:
(b) less than 3 Ω

Question 20.
The changing 1 kWh to joule will be equal to
(a) 3.6 × 10⁶ J
(b) 3.6 × 10^-6 J
(c) 1.6 × 10⁶ J
(d) 1 × 10⁶ J
Answer:
(a) 3.6 × 106 J

MP Board Class 10th Science Chapter 12 Very Short Answer Type Questions

Question 1.
What kind of circuit is required for flow of current ?
Answer:
Closed.

Question 2.
Draw a schematic diagram of a simple electric with battery, bulb, ammeter and key.
Answer:

Question 3.
Express in formula form:

1. net charge of a circuit
2. time taken for production of a unit amount of current.

Answer:

1. Q = It.
2. t = Q/I.

Question 4.
What is the value of an electric charge for an electron?
Answer:
-1.6 × 10^-19 C.

Question 5.
Formulate relationship between potential difference and current.
Answer:
V = IR.

Question 6.
Draw the graph verifying ohm’s law.
Answer:

Question 7
Draw circuit with three resistor in series combination.
Answer:

Question 8.
Represent with diagram:
(i) A cell
(ii) A bulb.
Answer:

Question 9.
Which material has highest conductivity and least resistance?
Answer:
Super conductors.

Question 10.
Which quality of electrical energy is inversely proportional to electrical power?
Answer:
Resistance.

Question 11.
Calculate resistance between terminal P & Q of following diagram.
Answer:

First R₁ and R₂ are in series.
So, R’ = R₁ + R₂ = 1 Ω + 1 Ω = 2 Ω.
then, R₃ and R₄ are in series
R” = R₃ + R₄ = 1 Ω + 1 Ω = 2 Ω
Now both R’ and R” are in parallel.
\(\frac { 1 }{ R } \) = \(\frac{1}{R^{\prime}}+\frac{1}{R^{\prime \prime}}+\frac{1}{2}+\frac{1}{2}=1\)
So, R = 1 Ω

Question 12.
If a circuit has open key will current flow from it?
Answer:
No.

Question 13.
If ammeter is not placed in circuit will current flow?
Answer:
Yes.

Question 14.
Which gas is filled in electric bulb?
Answer:
Argon.

Question 15.
Write the formula representing relationship between heat produced and current flowing in a circuit.
Answer:
H = I²RT.

Question 16.
Write the formula to calculate power generated by a circuit a with current M.
Answer:
Since, P = VI
Putting value of I, P = VM (I = m, Given)

Question 17.
Convert 5 kWh in joule.
Answer:
1 kWh = 3.6 × 10⁶ Joule
5 kWh = 5 × 3.6 × 10⁶ J.
= 16 × 10⁶ J
or 1.6 × 10⁷ J

Question 18
One volt ampere is equal to how many watt power?
Answer:
1 Watt Power.

MP Board Class 10th Science Chapter 12 Short Answer Type Questions

Question 1.
What is electric current?
Answer:
Total charge which pass through a particular area in unit time is called electric current of that particular conductor.

Question 2.
What is potential difference?
Answer:
Movement of electron need a pressure difference among conductors. Hence, battery or current flow suppliers are added to circuit. This difference in potential is called potential difference.

Question 3.
Define Ohm’s law.
Answer:
Ohm’s law states that current flowing in a circuit is directly proportional to its potential difference. It is represented by formula,
V α I
or V = RI
or V = IR
Here, R is a constant known as resistance.

Question 4.
Why resistance is applied to a circuit?
Answer:
To control the current flow in desired way resistance is applied. Example- Regulator of fan give desired speed of wind flow by controlling flow of current which is being converted to physical energy by fan.

Question 5.
Write down the factors at which the resistance of the conductor depends.
Answer:

1. Length of wire: Resistance is directly proportional to length R ∝ L.
2. Area cross section: Resistance is inversely proportional to area cross section of a wire R ∝ 1/A.
3. Nature of material.
4. Temperature of the conductor.

Question 6.
Arrange the resistivity in decreasing order of following material,
Iron, Silver, Tungsten, Manganin and Glass
Answer:
Glass > Manganin > Iron > Tungsten > Silver.

Question 7.
Give reason for the following:

1. Electric bulbs are usually filled with chemically inactive gases like nitrogen and argon.
2. Copper and aluminium wires are usually employed for electricity transmission.
3. Fuse wire is placed in series with the device.

Answer:

1. Electric bulbs are usually filled with chemically inactive gases like nitrogen and argon because these gases do not react with the hot tungsten filament and hence, prolong the life of filament of the electric bulb.
2. Copper and aluminium wires are usually employed for electricity transmission because copper and aluminium have low resistivity and thus, they are very good conductors of electricity.
3. Fuse wire is placed in series with the device because when large current passes through the circuit the fuse wire gets heated up and melts and whole circuit breaks and the device is protected from the damage.

Question 8.
If R₁ = 10 Ω, R₂ = 20 Ω and R₃ = 30 Ω, calculate the effective resistance when they are connected in series to a battery of 6 V. Also find the current flowing in the circuit.
(a) Current flows through different resistances, when these are joined in series, as shown in the below figure:

(b) R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 30 Ω
Effective resistance, R = R₁ + R₂ + R₃
R 10 + 20 + 30 = 60 Ω
Potential difference, V = 6V,
Current, I = ?
According to Ohm’s law
V = IR

Question 9.
Find the resistance between points A and B in the circuit diagram given below

Answer:
R₂, R₃, R₄ are in series and have resultant resistance R
R’ = R₂ + R₃ + R₄
= 2 + 2 + 2 = 6 Ω
R’ is in parallel combination with R₁
∴ Resultant resistance of the circuit (R’)
⇒ \(\frac { 1 }{ R } \) = \(\frac{1}{R_{1}}+\frac{1}{R_{1}}\) = \(\frac { 1 }{ 6 } \) + \(\frac { 1 }{ 6 } \) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)
∴ Resistance, R = 3 Ω.

MP Board Class 10th Science Chapter 12 Long Answer Type Questions

Question 1.
An electric iron consumes energy at a rate of 840 W when heating is at the maximum and 360 W when the heating is at the minimum. The voltage at which it is running is 220 V. What are the current and resistance in each case?
Solution:
At maximum heating:
The consumption of energy (electric) is given at the rate of 840 W at voltage 220 V.
P = 840 Ω, V = 220 V
Then current, I₁ = ?
∴ P = V × I₁

Question 2.
Three resistors of 5 Ω,10 Ω and 15 Ω are connected in series and the combination is connected to battery of 30 V. Ammeter and Voltmeter are connected in the circuit. Draw a circuit diagram to connect devices all the in correct order. What is the current flowing and potential difference across 10 Ω resistance?
Answer:
Current flowing I = ?
V₂ = ?
Total resistance, R = R₁ + R₂ + R₃
= 5 + 10 + 15 = 30 Ω
Total potential difference, V = 30 volts
According to Ohm’s law V = IR

Potential difference across 10 Ω is 10 volts.

Question 3.
Nichrome wire of length ‘l’ and radius ‘r’ has resistance of 10 Ω. How would the resistance of the wire change when:
(a) (i) Only length of the wire is doubled?
(ii) Only diameter of the wire is doubled? Justify your answer.
(b) Why element of electrical heating devices are made up of alloys?
Solution:
(a) Resistance, R ∝ 1

(i) Resistance is directly proportional to the length of the conductor. If length of nichrome wire (I) is doubled its resistance also gets doubled.
∴ R’ new resistance = 20 Ω

(ii) The resistance of the wire is inversely proportional to the square of its diameter. If the diameter of the wire is doubled, its resistance becomes one-fourth.
∴ R’, new resistance = \(\frac { 10 }{ 4 } \) Ω = 2.5 Ω.

(b) The heating elements of electrical heating appliances are made up of nichrome alloy because,

• Nichrome has very high resistivity due to which it produces a lot of heat on passing current.
• Nichrome does not undergo oxidation easily even at high temperature, it can be kept red hot without burning.

Question 4.
The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10-8 ohm meter, find the length of the wire.
Answer:
Resistance of a wire, R = 10 Ω Radius, r = 0.01 × 10^-2 m
Resistivity, p = 50 × 10^-8 ohm meter Length of the wire, l = ?
Area of cross section, A = πr² = 3.14 × (0.01 × 1o-2)² m²
= 3.14 × 0.01 × 0.01 × 10^-4 m²
= 3.14 × 10^-8 m²

Question 5.
(a) Calculate the resistance of the wire using the graph.
(b) How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?
(c) Define electric power. Derive relation between power, potential difference and resistance.

Answer:
(a) Resistance of wire = Slope of the graph
According to Ohm’s law,
V = IR or R = \(\frac { V }{ I } \)

(b) Resistance, R¹ = 176 Ω, No. of resistors = n,
Current, I = 5A
Potential difference, V = 220 volts, Resultant resistance = R According to Ohm s law,
V = IR, R = \(\frac { V }{ 1 } \) = \(\frac { 220 }{ 5 } \) = 44 Ω

Thus, 4 resistors of 176 Ω in parallel combination are required to carry 5 A on a 220 V line.

(c) Electric power is defined as the electrical work done per unit-time.

The work done, W by current, I when it flows for time t under potential difference V is given by

MP Board Class 10th Science Chapter 12 NCERT Textbook Activities

Class 10 Science Activity 12.1 Pages No. 203.204

• Set up circuit as shown in figure, consisting of a mchrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)
• First use only cell as the source in the circuit. Note the reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in table given.
• Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
• Repeat the above steps using three cells and then four cells in the circuit separately.
• Calculate the ratio of V to 1 for each pair of potential difference V and current I.
  
• Plot a graph between V and 1, and observe the nature of the graph.

Observations:

Approximately same value for \(\frac { V }{ I } \) will be obtained in each case. The straight line is obtained in the V – I graph. Thus, \(\frac { V }{ I } \) is a constant ratio.

Class 10 Science Activity 12.2 Page No. 205

• Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0-5 A range), a plug key and some connecting wires.
• Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit as shown in Fig. 12.1.
  

• Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current the circuit.
• Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
• Now repeat the above step with the 10 W bulb in the gap XY.
• Are the ammeter readings differ for different components in the gap XY? What do the above observations indicate?
• You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

Observations:

• The current is different for different components as certain components offer an easy path for the flow of electric current while the others resist the flow.

Class 10 Science Activity 12.3 Page No. 206

• Complete an electric circuit consisting of cell, an ammeter, a nichrome wire of length l [say, marked (1)] ana a plug key, as shown in Fig. 12.2.
  
  Electric circuit to study the factors on which the resistance of conducting wires depends.

• Figure 12.2. Electric circuit ti study the factors on which the resistance of conducting wires depends
• Now, plug the key. Note the current in the ammeter.
• Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 2l [marked (2) in the Fig. 12.2],
• Note the ammeter reading.
• Now replace the wire by a thicker nichrome wire of the same length / [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
• Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. 12.2] in the circuit. Let the w ire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
• Notice the difference in the current in all cases.
• Does the current depend in the length of the conductor?
• Does the current depend on the area of cross-section of the wire used?

Observations:

• The length of the conductor affects the current flow. Ammeter reading decreases to one-half when the length of the wire is doubled.
• The cross-section of the wire also affects the current flow. The length and cross-section changes affect the resistance also.

Class 10 Science Activity 12.4 Page No. 210

• Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key. as shown in Fig. 12.3. You may use the resistors of values like 1Ω, 2Ω, 3Ω etc, and a battery of 6 V for performing this Activity.
• Plug the key. Note the ammeter reading.
• Change the position if ammeter to anywhere in between the resistors. Note the ammeter reading each time.
• Do you find any change in the value of current through the ammeter?

Observations:

• The value of the current in the ammeter remains unchanged. This shows that it is independent of its position in the electric current. This shows that the current is the same in every part of a series combination.

Class 10 Science Activity 12.5 Page No. 211

• In Activity 12.4, insert a voltmeter across the ands X and Y of the series combination of three resistors, as shown in Fig. 12.4.
• Plug the key in the circuit and note series combination of resistors. Let it be V. Now measure the potential difference across the two terminals of the battery. Compare the two values.
• Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ands X and P the first resistor, as shown in Fig. 12.4.
  
• Plug the key and measure the potential difference across the first resistor. Lei it be V₁.
• Similarly, measure the potential difference across the first resistor, separately. Let these values be V₃, and V₂, respectively.
• Deduce a relationship between V, V₁, V₂ and V₃.

Observations:

• The potential difference (V) is equal to the sum of potential differences V₁, V₂, and V₃.
• Total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors,
V = V₁ + V₂ + V₃

Class 10 Science Activity 12.6 Page No. 211

• Make a parallel combination. X Y of three resistors having resistances R₁, R₂, and R₃, respectively Connect it with a battery, a plug key and an ammeter, as shown in Fig. 12.5. Also connect a voltmeter in parallel with the combination of resistors
  

• Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor (see Fig. 12.6).
• Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor Rl, as shown in Fig. 12.6. Note the ammeter reading, 11.
  

Similarly, measure the currents through R₂ and R₃ Let these be I₂, and I₃, respectively. What is the relationship between I, I₁, I₂ and I₃?

Observations:

• The total current I, is equal to the sum of the separate currents through each branch of the combination,
  I = I₁ + I₂ + I₃

MP Board Class 10th Science Solutions