NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
There can be infinitely many tangents to a circle.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………… point(s).
(ii) A line intersecting a circle in two points is called a ………… .
(iii) A circle can have parallel tangents at the most ………… .
(iv) The common point of a tangent to a circle and the circle is called ……….. .
Solution:
(i) One
(ii) Secant
(iii) Two
(iv) Point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) \( \sqrt{199} \) cm
Solution:

Radius of the circle = 5 cm
OQ = 12 cm
∠OPQ = 90°
[The tangent to a circle is perpendicular to the radius through the point of contact]
PQ² = OQ² – OP² [By Pythagoras theorem]
PQ² = 12² – 5² = 144 – 25 = 199
PQ = \( \sqrt{199} \) cm.
Hence correct option is (d).

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:

A line m is parallel to the line n and a line l which is secant is parallel to the given line.

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NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1

Ex 12.1 Class 10 Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution:
Given: radius of 1^st circle (R₁) = 19 cm
∴ Circumference of 1^st circle = 2πR₁ = 2π(19) cm
Radius of 2^nd circle (R₂) = 9 cm
∴ Circumference of 2^nd circle = 2πR₂ = 2π(9) cm
Let radius of 3^rd circle be R₃
Circumference of 3^rd circle = 2πR₃
According to question,
2πR₁ + 2πR₂ = 2πR₃
⇒ 2π(R₁ + R₂) = 2πR₃
⇒ R₁ + R₂ = R₃
⇒ 19 + 9 = R₃
⇒ R₃ = 28 cm

Class 10 Maths Chapter 12.1 Question 2.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution:
Given: radius of 1^st circle (R₁) = 8 cm
Area of 1^st circle = πR₁² = π(8)²cm²
Radius of 2^nd circle (R₂) = 6 cm
Area of 2^nd circle = πR₂² = π(6)² cm²
Let radius of 3^rd circle be R₃
Area of 3^rd circle = πR₃²
According to question,
πR,² + πR₂² – πR₃²
⇒ R₁² + R₂² = R₃²  ⇒ (8)² + i6)² – R₃²
⇒ 64 + 36 = R₃² ⇒ R₃=  \( \sqrt{100} \) = 10 cm

Areas Related To Circles Exercise 12.1 Question 3.
The figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue,
Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:
Diameter of the region representing gold score is 21 cm
⇒ Radius of the region representing gold region = \(\frac { 21 }{ 2 }\) cm

Exercise 12.1 Class 10 Question 4.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution:
Given: diameter of the wheels of the car = 80 cm
⇒ Radius of the wheel of the car = \(\frac { 80 }{ 2 }\) = 40 cm
Circumference of the wheel = 2πr = 2 x \(\frac { 22 }{ 2 }\) x 40 cm
Speed of the car = 66 km/h
Distance covered in 10 minutes =\(\frac { 66 x 10 }{ 60 }\) = 11 km
= 11 x 1000 x 100 cm = 11,00,000 cm

Ex 12.1 Class 10 Maths Solution Question 5.
Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(a) 2 units
(b) n units
(c) 4 units
(d) 7 units
Solution:
Let radius of the circle = r units
Perimeter of the circle = 2πr
Area of the circle = πr²
According to question,
Perimeter of the circle = Area of the circle
⇒ 2πr = πr²
⇒ r = 2 units
Hence, option (a) is correct.

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

In each of the following, give the justification of the construction also:

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Solution:
Steps of Construction:
1. Draw a line segment AB = 7.6 cm.
2. Draw an acute angle BAX on base AB. Mark the ray as AX.
3. Locate 13 points A₁, A₂, A₃, …… , A₁₃ on the ray AX so that AA₁ = A₁A₂ = ……… = A₁₂A₁₃
4. Join A₁₃ with B and at A₅ draw a line ∥ to BA₁₃, i.e. A₅C. The line intersects AB at C.
5. On measure AC = 2.9 cm and BC = 4.7 cm.

Justification:
In ∆AA₅C and ∆AA₁₃B,

∴ AC : BC = 5 : 8

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction:
1. Construct a ΔABC with AB = 4 cm, BC = 6 cm and AC = 5 cm.
2. Draw an acute angle CBX on the base BC at point B. Mark the ray as BX.
3. Mark the ray BX with B₁, B₂, B₃ such that
BB₁ = B₁B₂ = B₂B₃
4. Join B₃ to C.
5. Draw B₂C’ ∥ B3C, where C’ is a point on BC.
6. Draw C’A’ ∥ AC, where A’ is a point on BA.
7. ΔA’BC’ is the required triangle.

Justification: In ∆A’BC and ∆ABC,

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac { 7 }{ 5 }\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction:
1. Draw a ΔABC with AB = 5 cm, BC = 7 cm and AC = 6 cm.
2. Draw an acute angle CBX below BC at point B.
3. Mark the ray BX as B₁, B₂, B₃, B₄, B₅, B₆ and B₇ such that BB₁= B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
4. Join B₅ to C.
5. Draw B₇C’ parallel to B₅C, where C’ is a point on extended line BC.
6. Draw A’C’ ∥ AC, where A’ is a point on extended line BA.
A’BC’ is the required triangle.

Justification: In ∆ABC and ∆A’BC’,

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac { 1 }{ 2 }\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction:
1. Draw base AB = 8 cm.
2. Draw perpendicular bisector of AB. Mark CD = 4 cm, on ⊥ bisector where D is mid-point on AB.
3. Draw an acute angle BAX, below AB at point A.
4. Mark the ray AX with A₁, A₂, A₃ such that AA₁ =A₁A₂ = A₂A₃
5. Join A₂ to B. Draw A₃B’ ∥ A₂ B, where B’ is a point on extended line AB.
6. At B’, draw B’C’ 11 BC, where C’ is a point on extended line AC.
7. ∆AB’C’ is the required triangle.

Justification: In ∆ABC and ∆A’BC’,

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac { 3 }{ 4 }\) of the corresponding sides of the triangle ABC.
Solution:
Steps of Construction:
1. Draw a line segment BC = 6 cm and at point B draw an ∠ABC = 60°.
2. Cut AB = 5 cm. Join AC. We obtain a ΔABC.
3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.
4. Locate 4 points A₁, A₂, A₃ and A₄ on the ray BX so that BA₁ = A₁A₂ = A₂A₃ = A₃A₄.
5. Join A₄ to C.
6. At A₃, draw A₃C’ ∥ A₄C, where C’ is a point on the line segment BC.
7. At C’, draw C’A’ ∥ CA, where A’ is a point on the line segment BA.
∴ ∆A’BC’ is the required triangle.

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac { 4 }{ 3 }\) times the corresponding sides of ∆ABC.
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180°
⇒ 105° + 45° + ∠C = 180°
⇒ 150° + ∠C = 180°
⇒ ∠C = 30°
Steps of Construction:
1. Draw a line segment BC = 7 cm. At point B, draw an ∠B = 45° and at point C, draw an ∠C = 30° and get ΔABC.
2. Draw an acute ∠CBX on the base BC at point B. Mark the ray BX with B₁, B₂, B₃, B₄, such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
3. Join B₃ to C.
4. Draw B₄C’ ∥ B₃C, where C’ is point on extended line segment BC.
5. At C’, draw C’A’ ∥ AC, where A’ is a point on extended line segment BA.
6. ∆A’BC’ is the required triangle.

Justification: In ∆ABC and ∆A’BC’,

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac { 5 }{ 3F }\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction:
1. Draw a right angled ∆ABC with AB = 4 cm, AC = 3 cm and ∠A = 90°.
2. Make an acute angle BAX on the base AB at point A.
3. Mark the ray AX with A₁, A₂, A₃, A₄, A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
4. Join A₃B. At A₅, draw A₅B’ ∥ A₃B, where B’ is a point on extended line segment AB.
5. At B’, draw B’C’ ∥ BC, where C’ is a point on extended line segment AC.
6. ∆AB’C’ is the required triangle.

Justification:
In ∆ABC and ∆AB’C’,

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NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see figure).

Solution:
Given: length of the rope (AC) = 20 m, ∠ACB = 30°
Let height of the pole (AB) = h metres


Hence, height of the pole = 10 m

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
Let DB is a tree and AD is the broken part of it which touches the ground at C.

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Let l₁ is the length of slide for children below the age of 5 years and l₂ is the length of the slide for elder children

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Solution:
Let h be the height of the tower
In ∆ABC,

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:

Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution:
Let the height of the pedestal AB = h m
Given: height of the statue = 1.6 m, ∠ACB = 45° and ∠DCB = 60°

Question 9.
The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
Given: height of the tower AB = 50 m,
∠ACB = 60°, ∠DBC = 30°
Let the height of the building CD = x m
In ∆ABC,

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.
Solution:

Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.

Solution:
Let the height of the tower AB = h m and BC be the width of the canal.

Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Let height of the tower AB = (h + 7) m
Given: CD = 7 m (height of the building),

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:
Given: height of the lighthouse = 75 m
Let C and D are the positions of two ships.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.

Solution:
Let the first position of the balloon is A and after sometime it will reach to the point D.
The vertical height ED = AB = (88.2 – 1.2) m = 87 m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Let the height of the tower AB = h m
Given: ∠XAD = ∠ADB = 30°
and ∠XAC = ∠ACB = 60°
Let the speed of the car = x m/sec
Distance CD = 6 x x = 6x m
Let the time taken from C to B = t sec.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1

Question 1.
Fill in the blanks by using the correct word given in brackets.
(i) All circles are ……………. . (congruent/similar)
(ii) All squares are …………… . (similar/congruent)
(iii) All …………….. triangles are similar. (isosceles/equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are …………… and
(b) their corresponding sides are …………… (equal/proportional)
Solution:
Fill in the blanks.
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.

Question 2.
Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
Solution:
(i) Examples of similar figures:

• Square
• Regular hexagons

(ii) Examples of non-similar figures:

• Two triangles of different angles.
• Two quadrilaterals of different angles.

Question 3.
State whether the following quadrilaterals are similar or not.

Solution:
No, the sides of quadrilateral PQRS and ABCD are proportional but their corresponding angles are not equal.

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Exercise 2.1

Question 1.
The graphs of y = p(x) are given in figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:
The number of zeroes of p(x) in each graph given; are

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

NCERT Solutions For Class 10 Maths Chapter 4 Question 1.
Check whether the following are quadratic equations:
(i) (x+ 1)²=2(x-3)
(ii) x – 2x = (- 2) (3-x)
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
(iv) (x – 3) (2x + 1) = x (x + 5)
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
(vi) x² + 3x + 1 = (x – 2)²
(vii) (x + 2)³ = 2x(x² – 1)
(viii) x³ -4x² -x + 1 = (x-2)³
Solution:
(i) (x+ 1)²=2(x-3)
⇒ x² + 2x +1 = 2x – 6
⇒ x² + 2x – 2x+1 + 6 = 0
⇒ x² + 7 = 0
⇒ x² + 0x + 1 = 0
Which is of the form
ax² + bx + c = 0
Hence, the given equation is a quadratic equation.

(ii) x² – 2x = (- 2) (3 -x)
⇒ x² -2x = -6 + 2x
⇒ x² -4x + 6 = 0
Which is of the form
ax² + bx + c = 0
Hence, the given equation is a quadratic equation.

(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x² + x-2x-2 = x2 + 3x-x -3
⇒ x² + x – 2x – 2 = x² – 3x + x + 3 = 0
⇒ – 3x + 1 = 0 ⇒ 3x – 1 = 0
Since degree of equation is 1, hence, given equation is not a quadratic equation.

(iv) (x-3) (2x+ 1) = x (x + 5)
⇒ 2x² + x – 6x – 3 = x² + 5x
2x² + x – 6x – 3-x2 – 5x = 0
⇒ x² – 10x -3 = 0
Which is of the form
ax² + bx + c 0
Hence, the given equation is a quadratic equation.

(v) (2x-1)(x-3) = (x + 5)(x-1)
⇒ 2x² – 6x-x + 3 = x² -x + 5x – 5
2x² – 6x-x + 3 = x² + x – 5x + 5 = 0
⇒ x² – 11x + 8 = 0
Which is of the form
ax² + bx + c = 0
Hence, the given equation is a quadratic equation.

(vi) x² + 3x + 1 = (x-2)²
⇒ x² + 3x + 1 = x² + 4 – 4x
⇒ x² + 3x + 1 = x²– 4 + 4c = 0
⇒ 7x – 3 = 0
Since degree of equation is 1, hence, the given equation is not a quadratic equation,

(vii) (x + 2)³ = 2x (x² – 1)
⇒ x³ + 8 + 3.x.2 (x + 2) = 2x³ – 2x
⇒ x³ + 8 + 6x² + 12x = 2x³ – 2x
⇒ x³ – 6x² – 14x – 8 = 0
Which is not of the form
ax² + bx + c = 0
Hence, the given equation is not a quadratic equation.

(viii) x³ – 4x² – x+1 = (x-2)³
⇒ x³ – 4x² – x + 1 = x³-8 + 3x(-2)(x – 2)
⇒ x³ – 4x² -x + 1 = x³ – 6x² + 12x – 8
⇒ 2x² – 13x + 9 = 0
Which is of the form
ax² + bx + c = 0
Hence, the given equation is a quadratic equation.

Exercise 4.1 Maths Class 10 Solutions Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
(i) Let breadth of the rectangular plot = x m
Then, length of the plot = (2x + 1)m
Area of a rectangular plot = l x b ,
⇒ 528 (2x + 1)x
⇒ 528 = 2x² +x
⇒ 2x² + x – 528 = 0
Which is the required quadratic equation.

(ii) Let the two consecutive integers be x and x + 1
Then, x(x+l) = 306
⇒ x² +x-306 = 0
Which is the required quadratic equation.

(iii) Let the present age of Rohan = x years
Rohan’s mother’s present age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
After 3 years, Rohan’s mother’s age = (x + 26 + 3) years
According to question,
(x + 3) (x + 29) = 360
⇒ x² + 29x + 3x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
Which is the required quadratic equation.

(iv) Let speed of the train = x km/h

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

Ex 1.1 Class 10 Question 1.
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255.
Solutions:
(i) Given numbers are 135 and 225.
On applying Euclid’s division algorithm, we have
225 = 135 x 1 + 90
Since the remainder 90 ≠ 0, so again we apply Euclid’s division algorithm to 135 and 90, to get
135 = 90 x 1 + 45
Since the remainder 45 ≠ 0, so again we apply Euclid’s division algorithm to 90 and 45, to get
90 = 45 x 2 + 0
The remainder has now become zero, so we stop.
∵ At the last stage, the divisor is 45
∴ The HCF of 135 and 225 is 45.

(ii) Given numbers are 196 and 38220
On applying Euclid’s division algorithm, we have
38220 = 196 x 195 + 0
Since we get the remainder zero in the first step, so we stop.
∵ At the above stage, the divisor is 196
∴ The HCF of 196 and 38220 is 196.

(iii) Given numbers are 867 and 255
On applying Euclid’s division algorithm, we have
867 = 255 x 3 + 102
Since the remainder 102 ≠ 0, so again we apply Euclid’s division algorithm to 255 and 102. to get
255 = 102 x 2 + 51
Since the remainder 51 ≠ 0, so again we apply Euclid’s division algorithm to 102 and 51, to get
102 = 51 x 2 + 0
We find the remainder is 0 and the divisor is 51
∴ The HCF of 867 and 255 is 51.

Exercise 1.1 Class 10 Maths NCERT Solutions Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solutions:
Let ‘a’ be any positive integer and b = 6.
∴ By Euclid’s division algorithm, we have
a = bq + r, 0 ≤ r ≤ b
a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4,5
Now, ‘a’ may be of the form of 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5
If ‘a’ is of the form 6q, 6q + 2, 6q + 4 then ‘a’ is an even.

In above we can see clearly that the numbers of the form 6q, 6q + 2, 6q + 4 are having the factor 2.
∴ The numbers of the form 6q, 6q + 2, 6q + 4 are even.
If ‘a’ is of the form 6q +1,6q +3, 6q + 5 then ‘a’ is an odd.
As if

∵ We know that the number of the form 2k + 1 is odd.
∴ The numbers of the form 6q + 1, 6q + 3, 6q + 5 are odd.

Class 10 Maths Chapter 1 Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solutions:
Maximum number of columns = HCF of (616, 32)
For finding the HCF we should apply Euclid’s division algorithm
Given numbers are 616 and 32
On applying Euclid’s division algorithm, we have
616 = 32 x 19 + 8
Since the remainder 8 ≠ 0, so again we apply Euclid’s division algorithm to 32 and 8, to get
32 = 8 x 4 + 0

The remainder has now become zero, so we stop,
∵ At the last stage, the divisor is 8
∴ The HCF of 616 and 32 is 8.
Therefore, the maximum number of columns in which an army contingent of 616 members can march behind an army band of 32 members in a parade is 8.

Class 10 Maths Ex 1.1 Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solutions:
Let ‘a’ be any positive integer and b = 3.
∴ By Euclid’s division algorithm, we have a = 3q + r, 0 ≤ r < b
a = 3q + r, 0 ≤ r < 3 [ ∵ b = 3] where q ≥ 0 and r = 0,1, 2
∴ a = 3q or 3q + 1 or 3q + 2
Now

Thus, the square of any positive integer is either of the form 3m or 3m + 1.

Maths 1.1 Class 10 Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solutions:
Let ‘a’ be any positive integer and b = 3.
∴ By Euclid’s division algorithm, we have a = bq + r,0 ≤ r ≤ b
a = 3q + r,0 ≤ r < 3 [ ∵ b = 3] where q ≥ 0 and r = 0. 1, 2
∴ a = 3q or 3q + 1 or 3q + 2
Now

Thus, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

We hope the NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Social Science Geography Chapter 2 Physical Features of India

Question 1.
Choose the right answer from the four alternatives given below:

(i) A landmass bounded by sea on three sides is referred to as
(a) Coast
(b) Island
(c) Peninsula
(d) None of the above
Ans. (c) Peninsula

(ii) Mountain ranges in the eastern part of India forming its boundary with Myanmar are collectivity called
(a) Himachal
(b) Uttarakhand
(c) Purvachal
(d) None of the above
Ans. (c) Purvachal

(iii) The western coastal strip, south of Goa is referred to as
(a) Coromandel
(b) Konkan
(c) Kannad
(d) Northern Circar
Ans. (c) Kannad

(iv) The highest peak in the Eastern Ghats is
(a) Anai Mudi
(b) Kanchenjunga
(c) Mahendragiri
(d) Khasi
Ans. (c) Mahendragiri

Get free NCERT Solutions for Class 9 Social Science Geography Chapter 2 Physical Features of India solved by experts.

Question 2.
Answer the following questions briefly.

(i) What are tectonic plates?
Ans. Large fragments of the earth’s crust torn due to the rising convectional currents are called tectonic plates.

(ii) Which continents of today were part of the Gondwana land?
Ans. The continents are South America, South Africa, Australia and Antarctica.

(iii) What is the bhabar?
Ans. Bhabar is a narrow belt of the Ganga plain covered with pebbles lies along the foothills of the Shiwaliks.

(iv) Name the three major divisions of the Himalayas from north to south.
Ans. (a) The Greater Himalayas or Himadri (Inner Himalayas)
(b) Himachal or Lesser Himalayas (Middle Himalayas)
(c) The Shiwaliks (Outer Himalayas)

(v) Which plateau lies between the Aravali and the Vindhyan ranges?
Ans. Malwa plateau lies between the Aravali and the Vindhyan range.

(vi) Name the island group of India having coral origin.
Ans. Lakshadweep islands is the island group having coral origin.

Question 3.
Distinguish between:

(i) Converging and diverging tectonic plates
Ans:

(ii) Bhangar and Khadar
Ans:

(iii) Western Ghats and Eastern Ghats
Ans:

Question 4.
Describe how the Himalayas were formed.
Answer:
(a) The oldest landmass of India (i.e. Peninsular part) was a part of Gondwana land, which included India, Australia, South Africa and South America as one single landmass.
(b) Convectional currents split the earth’s crust into’a number of fragments, thereby leading to the drifting of the Indo – Australian plate towards the north after being separated from Gondwana land.
(c) The northward drift resulted in the collision of this plate with much larger Eurasian plate.
(d) Due to this collision, the sedimentary rocks which were accumulated in the geosyncline called the Tethys, were folded to form the mountain systems of west Asia and Himalayas.
(e) The Himalayas represent a youthful topography with high peaks, deep valleys and fast-flowing rivers.

Question 5.
Which are the major physiographic divisions of India? Contrast the relief of the Himalayan region with that of the Peninsular plateau.
Answer:
The major physiographic divisions of India are:

• The Himalayan Mountains
• The Northern Plains of India
• The Peninsular Plateau
• The Indian Desert
• The Coastal Plains
• The Islands

Question 6.
Give an account of the Northern Plains of India.
Answer:

1. (a) The Northern Plains of India are alluvial plains of India. They are formed by the sediments brought from the mountains and deposited by the rivers in the depression formed after the uplift of the Himalayas namely the Indus, Ganga and Brahmaputra and their tributaries.
2. (b) The plains spreads over an area of 7 lakh sq. km. The length of this plain is 2,400 km long and 240 km broad.
3. (c) It is densely populated and intensely cultivated area.
4. (d) With adequate water supply and favourable climate, it is agriculturally a very productive part of India.
5. (e) The Northern Plains of India are divided into three divisions.
   • Punjab Plains covers the western part of the Northern plains. They are formed by Indus river and its tributaries.
   • Ganga Plains extends between Ghaggar and the Teesta river, spread over the states of Haryana, Delhi, UP, Bihar, Jharkhand and West Bengal.
   • The Brahmaputra Plain lies to the east of the Ganga plains. It covers the area of Assam.

Question 7.
Write short notes on the following:

(a) The Indian Desert
Ans. (a) The Indian desert lies to the west of the Aravali hills. It is an uneven sandy plain covered with sand dunes.
(b) Barchans (crescent-shaped sand dunes) cover a larger part of the desert. Near the Indo-Pakistan border, longitudinal sand dunes are more common.
(c) It has arid climate with scarce vegetation and rainfall below 150 mm per year.
(d) Rivers/streams appear only during the rainy season and soon afterward disappear in the sand. They do not have enough water to reach the sea. JRiver Luni is the only large river in this area.

(b) The Central Highlands
Ans. (a) The Part of the Peninsular plateau lying north of the Narmada river is called Central Highlands. These highlands are made up of hard igneous and metamorphic rocks.
(b) It is bordered by Aravali range to the north-west. The Central Highlands include the Malwa plateau to the west and Chotanagpur plateau to the east.
(c) The Central Highlands are wider in the west and become narrow eastwards. The eastward extension of the Malwa plateau is locally called Bundelkhand and Baghelkhand. Chotanagpur plateau in the east is drained by Damodar river, a southern tributary of Ganga river.

(c) The Island Groups of India
Ans. India has two groups of islands namely:
(a) Lakshadweep Islands
(b) Andaman and Nicobar islands group

Lakshadweep Islands:
(a) These island groups are located in the Arabian Sea i.e., west of Malabar coast of Kerala.
(b) These islands are of coral origin formed by deposition of the dead remains,
(c) The total area of islands are 32 sq km. In Lakshadweep, administrative headquarters is in Kavaratti islands.
(d) They have a wide diversity of flora and fauna. The Pitti island, an uninhabited island has a bird sanctuaiy.

Andaman and Nicobar Island:
(a) These islands are located in the Bay of Bengal are the raised portion of the submerged mountain ranges projecting out of the sea water.
(b) Large in size and are more numerous. Some of them are of volcanic origin e.g. Barren island the only active volcano,
(c) These islands are of strategic importance as it lies very close to south-east Asia,
(d) The capital city is Port Blair,
(e) Experience equatorial climate and has thick forest cover.

Question 8.
On an outline map of India show the following.
(i) Mountain and hill ranges – the Karakoram, the Zaskar, the Patkai Bum, the Jaintia, the Vindhya range, the Aravali and the Cardamom hills.
(ii) Peaks – K₂, Kanchenjunga, Nanga Parbat and Anai Mudi.
(iii) Plateaus – Chotanagpur and Malwa
(iv) The Indian Desert, Western Ghats, Lakshadweep Islands.
Answer:

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Haloalkanes and Haloarenes NCERT Solutions will help you to score more marks in your CBSE board Examination.

NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

Haloalkanes and Haloarenes is an important chapter from examination perspective. This chapter defines haloalkanes and haloarenes according to the IUPAC system of nomenclature. The reactions involved in the preparation of haloalkanes and haloarenes are also explained here. These are very important for the examination.

The chapter also explains the correlation between haloalkanes and haloarenes. NCERT Solutions for Class 12 Chapter 10 acts as a guide for the students preparing this chapter. The students can refer to these for better practice.

NCERT IN-TEXT QUESTIONS

Question 1.
Write the structures of the following compounds : (C.B.S.E. Delhi 2010)
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert. butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec butyl-2-methylbenzene. (C.B.S.E. Sample paper 2011)
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
H₂SO₄ is an oxidising agent. It oxidises HI produced during the reaction to I₂ and thus prevents the reaction between an alcohol and HI to form alkyl iodide. To prevent this, a non¬oxidising acid like H₃PO₃ is used.

Question 3.
Write the structures of different dihalogen derivatives of propane.
Answer:
Propane (CH₃CH₂CH₃) has two primary and one secondary hydrogen atoms present. Four isomeric dihalogen derivatives are possible. Let the halogen X be Br.

Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one which on photochemical chlorination yields
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
Answer:
The molecular formula C₅H₁₂ represents three structural isomers which are chain isomers.

(i) The isomer is symmetrical with four primary (1°) carbon atoms and one quaternary (4°) carbon atom. Since all the hydrogen atoms are equivalent, it will yield only one monochloride upon photochlorination i.e., chlorination carried in the presence of ultra-violet light.

(ii) In the straight chain isomer pentane, there are three groups of equivalent hydrogen atoms. As a result, three isomeric monochlorides are possible.

(iii) The branched chain isomer has four types of equivalent hydrogen atoms present. It will give four isomeric monochlorides upon chlorination.

Question 5.
Draw the structures of the major monohaloproducts in each of the following reactions:

Answer:

The reaction is carried in the presence of dry acetone upon heating. It is called Finkelstein reaction. In this reaction, I^– ion being a stronger nucleophile displaces Br^– ion. NaBr formed is insoluble in dry acetone whereas Nal dissolves. This shifts the equilibrium in the forward direction.

Under the reaction conditions allylic halogenation will take place. Addition of bromine can be possible in case the reaction is carried at room temperature.

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i)Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii)1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
(i) Chloromethane < Bromomethane <
Dibromomethane < Bromoform
The reason is:
(a)for same alkyl group, B.Pt increases with size of halogen atom.
(b)B.Pt increases as number of halogen atoms increase.
(ii)Isopropyl chloride < 1 – Chloropropane < 1 – Chlorobutane
Reason :
(a)For same halogen, B.Pt. increases as size of alkyl group increases.
(b)B.Pt. decreases as branching increases.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by S_N² mechanism ? Explain your answer.

Answer:
If the leaving group is the same in different isomers of a particular molecular formula, the reactivity of the isomers towards S_N² mechanism decreases with the increase in steric hindrance. In the light of above, the reactivity order in different cases is :

(i) CH₃CH₂CH₂CH₂Br is a primary alkyl halide (1°). It is more reactive than the other isomer which is a secondary (2°) alkyl halide because less steric hindrance is caused by primary alkyl group as compared to secondary alkyl group.
(ii)

is a secondary alkyl halide (2°). It is more reactive than the other isomer which is a tertiary alkyl halide (3°). The explanation is the same.
(iii) Here both the isomers are primary alkyl halides (1°). However, the isomer with CH₃ group at C₂ atom exerts more steric hindrance to the attacking nucleophile at C₁ atom as compared to the other isomer in which a CH₃ group is attached to C₃ atom. It is, therefore, less reactive.

Question 8.
In the following pairs of halogen compounds, which compound undergoes reaction faster ? (C.B.S.E. Delhi 2008, Outside Delhi 2010, 2013)

Answer:
The reactivity of a particular halogen compound towards S_N¹ reaction depends upon the stability of the carbocation formed as a result of ionisation. This is a slow step and is called rate determining step. The order of relative stabilities of different carbocations is in the order : tertiary > secondary > primary. In the light of this, the order of reactivity in the two cases is explained.

1. The isomer (a) is a tertiary alkyl chloride while the other isomer (b) is a secondary alkyl chloride. The isomer (a) is more reactive towards S i reaction since the tertiary carbocation formed in this case is more stable than the secondary carbocation which is likely to be formed in the other case.
2.  The isomer (a) is a secondary alkyl chloride while the other isomer (b) is primary in nature. The secondary alkyl chloride (a) is expected to react faster since the secondary carbocation formed is more stable than the primary carbocation which is likely to be formed in the other case.

Question 9.
Identify A, B, C, D, E, R and R’ in the following :

Answer:

NCERT EXERCISE

Question 1.
Name the following compounds according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary) vinyl or aryl halides.
(i) (CH₃)₂CHCH(C1)CH₃ (C.B.S.E. Delhi 2013)
(ii) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl
(iii) CH₃CH₂C(CH₃)₂CH₂I
(iv) CH3C(C1)(C₂H5)CH₂CH₃
(v) CH₃.C(C₂H₅)₂CH₂Br
(vi) CH₃CH=C(C1)CH₂CH(CH₃)₂
(vii) CH₂=CH-CH₂-Br
(viii) CH₃CH=CHC(Br)(CH₃)₂
(ix) m-C1CH₂C₆H₄CH₂C(CH₃)3
(x) o-BrC₆H₄CH(CH₃)CH₂CH₃
(xi) (CH₃)₃CCH₂CH(Br)C5H₅
(xii) p-ClC₆H₄CH₂CH(CH₃)₂
Answer:

Question 2.
Give the IUPAC names of the following compounds :

Answer:

Question 3.
Write the structures of the following compounds :
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane.
(iii) 2-(2-Chlorophenyi)-1-iodooctane
(iv) 4-tert. butyl -3-iodooctane
(v) 1, 4-Dibromobut-2-ene
(vi) 1-Bromo-4-sec.butyl-2-methylbenzene.
(vii) p-Bromochlorobenzene
(viii) Perfluorobenzene
Answer:

Question 4.
Which one of the following has highest dipole moment?
(a) CH₂Cl₂
(b) CHCl₃
(c) CCl₄
Answer:

CCl₄ is a symmetrical molecule. Therefore, the dipole moments of all four C-Cl bonds cancel each other. Hence its resultant dipole moment is zero.

As shown in the above figure, in CHCl₃, the resultant dipole moments of two C-Cl bonds is opposed by the resultant dipole moments of one C-H and one C-Cl bond. Since the resultant of one C-H and one C-Cl bond is smaller than the resultant of the two C-Cl bonds dipole moments, the opposition is to a small extent. As a result CHCl₃ has a small net dipole moment.

On the other hand, in case of CH₂,Cl₂ the resultant of the dipole moments of two C-Cl bonds is strengthened by the resultant of the dipole moments of two C-H bonds. As a result, CH₂Cl₂ has a higher dipole moment. Hence CH₂Cl₂ has the highest dipole moments among the three compounds.
Hence, the given compounds can be arranged in the increasing order of their dipole moments as
CCl₄ < CHCl₃ < CH₂Cl₂

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monobromo compound in bright sunlight. Identify the hydrocarbon.
Answer:
A hydrocarbon with the molecular formula, C₅H₁₀ belongs to the group with a general molecular form C_nH_2n. therefore, it may either be an alkene or a cycloalkane since hydrocarbon does not react with chlorine in the dark, it cannot be alkene. Further, the hydrocarbon gives a single monochloro compound, C₅H₉Cl by reacting with chlorine in might sunshine since the formed compound is monochloro one all the C-H bonds should be equivalent. Hence the compound should be a cycloalkane. Hence the compound is C₅H₁₀ (cyclopentane).

Question 6.
Write the isomers of the compound having the formula C₄H₉Br. (Haryana Board 2013)
Answer:
The compound has the following structural isomers.

2-Bromobutane has a chiral carbon and it is expected to exhibit optical isomerism.

Question 7.
Write equations for the preparation of 1-Iodobutane from :
(a) Butan-1- ol
(b) 1-Chlorobutane
(c) But-1-ene.
Answer:

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
Nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide ion exists as a hybrid of the following two structures. It can attack either

through carbon to form cyanides (or nitriles) or through nitrogen to form isocyanides (or carbyl amines). For more details, consult section 11.7.

Question 9.
Which compound in the following pairs will react faster in the Sn² reaction?
(1) CH₃Br or CH₃I
(1) (CH₃)₃CCl or CH₃Cl (C.B.S.E. 2008)
Answer:
1. In the S_N2 mechanism the reactivity of halides for the same alkyl group increase in order. This happens because as the size increases the halide ion becomes a better leaving group.
R-F << R-Cl < R – Br < R-I
Therefore, CH₃I will react faster than CH₃Br in S_N2 reaction with image 17.

2. The S_N2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in the case (CH₃)₃ CCl, the attack of the nucleophile at the carbon atom is hindered by the presence of the bulky substituents on that carbon atom bearing-the leaving the group in CH₃Cl. Hence CH₃Cl reacts faster than (CH₃)₃ CCl in S_N2 reaction with

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of following alkyl halides with sodium ethoxide in ethanol.
(i) 1-Bromo-l-metbylcyclohexane
(ii) 2-Chloro-2-methyl butane
(iii) 3-Bromo-2, 2, 3-trimethylpentane.
Answer:
(i) 1-Bromo-l-methylcyclohexane has two β-hydrogen atoms. This will give a mixture of two alkenes as a result of dehydrohalogenation. Since alkene (B) is more substituted according to SaytzefFs rule, it is more stable and will be the major product. The same rule applies to the other alkyl halides also.

(ii) The compound has two sets of β-hydrogen atoms. Therefore, two elimination products are formed. However, a more substituted alkene is formed in greater proportion as compared to a less substituted alkene.

The explanation is similar. More substituted alkene is formed in preference to less substituted alkene.

Question 11.
How will you bring about the following conversions?   (Haryana Board 2011)
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethane
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromometbane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chiorobutane to n-octane
(x) Benzene to biphenyl
Answer:

Question 12.
Explain why:
(i) Dipole moment of chlorobenzene is lower than that of cyclohexyl chloride (C.B.S.E 2016)
(ii) Alkyl halides though polar, are immiscible with water.
(iii) Grignard reagents should be prepared under anhydrous conditions.
Answer:

The polarity of C- Cl bond in chlorobenzene is less than that of same bond in cyclohexyl chloride because of carbon atom involved in chlorobenzene is more electronegative (greater s-character) as compared to the carbon atom in case of cyclohexyl chloride (lesser s-character). Therefore, the dipole moment of chlorobenzene is less with respect to cyclohexyl chloride.

(ii) In water, H₂O molecules are linked to each other by intermolecular hydrogen bonding. Although alkyl halides also contain polar C – X bonds, they cannot break the hydrogen bonding in H₂0 molecules. This means that there is hardly any scope for the association between molecules of alkyl halides and water. They, therefore, exist as separate layers and are immiscible with each other. For more details, consult section 11.6.

(iii) Grignard reagents (R – Mg – X) should be prepared under anhydrous conditions because these are readily decomposed by water to form alkanes.

That is why ether used as solvent in the preparation of Grignard reagent is completely anhydrous in nature.

Question 13.
Give the uses of freon-12, D.D.T., carbon tetrachloride and iodoform?
Answer:

1. Freons are the trade names for the commercially used fluoro chloromethanes with the formula CF_xCl_y (x + y = 4). A few examples are:
   CF₄ (Freon-14), CF₃C1 (Freon-13), CF₂Cl₂ (Freon-12), CFCl₃ (Freon-11)
   Out of the various freons mentioned, Freon- 12 is the most common refrigerant. It is prepared by passing hydrogen fluoride
   through carbon tetrachionde in the presence of antimony trichioride catalyst.
   
   in addition to their use as refrigerants in place of highly toxic liquid sulphur dioxide (SO₂) and ammonia (NH₃), large amount of CFCs are also used in the manufacture of disposable foam products such as cups and plates, as aerosol propellants in spray cans and as solvents to clean freshly soldered electronic circuit boards.
2. D.D.T. is the abbreviated form of p, p’-dichlorodiphenyltrichloroethane and its actual IUPAC naine has been given above. It is
   prepared by heating chiorobenzene with chlorai (trichioroacetaldehyde) in the presence of conc. H₂S0₄
   
3. Carbon tetrachloride (CC14) is also a colourless oily liquid just like chloroform. It is completely immiscible with water but
   dissolves in organic solvents.
   Carbon tetrachloride is a very useful solvent for oils, fats, resins etc. Ills used as a cleansing agent both in industry and in home because it can easily dissolve grease and other organic matter. But it mainly finds application for the manufacture of refrigerants, propellants for aerosol cans and some pharmaceuticals.
4. lodoform is a yellow crystalline solid with a characteristic unpleasant smell. It is insoluble in water but dissolves in alcohol, ether and other organic solvents.
   lodoform can be prepared in the laboratory by treating ethyl alcohol or acetone with sodium hydroxide and iodine. The reaction is known as haloform or iodoform reaction.
5. Physiological effects: lodoform is used as an antiseptic, particularly for dressing wounds. Actually, on coming in contact with skin (organic mater) it decomposes and slowly loses iodine which accounts for the antiseptic properties of iodoform.

Question 14.
Write the structures of the major products in each of the following reactions :

Answer:

Question 15.
Explain the following reaction :    (C.B.S.E. Delhi 2009 Comptt.)

Answer:
KCN is a resonance hybrid of two contributing structures :

This shows that the cyanide ion is an ambident nucleophile and the nucleophile attack is possible either through carbon atom or nitrogen atom resulting in cyanides and isocyanides respectively. In this case, in the presence of polar solvent, KCN readily ionises to furnish ions. The nucleophile attack takes place predominantly through a carbon atom and not through nitrogen atom as C- C bond is more stable than C -N bond.

Question 16.
Arrange the compounds of each set in order of decreasing reactivity towards (S_N²) displacement:
(a) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(b) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
(c) 1-Bromobutane, l-Bromo-2, 2-dimethylpropane, l-Bromo-2-methylbutane, l-Bromo-3-methylbutane. (C.B.S.E. Outside Delhi 2011)
Answer:
The reactivity of a particular haloalkane towards S_N² reaction is inversely proportional to the steric hindrance around the carbon atom involved in C – X bond. More the steric hindrance, lesser will be the reactivity. In the light of this, the decreasing order of reactivity in all the three cases is as follows :

Question 17.
Out of  C₆H₅CH₂Cl and C₆H₅CH(C1)C₆H₅ which is more easily hydrolysed by aqueous KOH?
Answer:
The compound C₆H₅CH₂Cl is a primary aralkyl halide while C₆H₅CH(Cl)C₆H₅ is secondary in nature. The hydrolysis of both these compounds with aqueous KOH (polar) is likely to proceed by S_N¹ mechanism due to the following reasons.
(a) The carbocations formed in both the cases as a result of ionisation are resonance stabilised due to the presence of phenyl groups at the a-position(s).
(b) As water is a polar solvent, it is expected to favour ionisation of the two halogen-substituted compounds leading to S_N¹ mechanism.
The carbocations that are formed as a result of ionisation in the slow steps are shown :

The ease of hydrolysis depends upon the relative stability of the carbocation/s that are formed in two cases. The secondary carbocation is more stable since the positive charge on the carbocation is delocalised on two phenyl groups that are present at the a-positions. On the other hand, there is only one phenyl group in primary carbocation available for charge delocalisation.
Thus, we may conclude that C₆H₅CHClC₆H₅ is more easily hydrolysed by aqueous KOH as compared to C₆H₅CH₂Cl.

Question 18.
p-Dichlorobenzene has higher m.p. and lower solubility than those of o-and m-isomers. Discuss.
Answer:
p-dichlorobenzene has a higher melting point than its o-isomer due to the symmetry of the p-isomer that fits in the crystal lattice better than the o- or m- isomer. Therefore, it has stronger intermolecular forces of attraction than o- and m- isomers, and thus greater energy are required to break crystal lattice to melt or dissolve the p-isomer than the corresponding o- and m- isomers. In other words, the melting point of the p-isomer is higher and its solubility is lower than corresponding m- and o- isomers.

Question 19.
How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenyl ethanoic acid
(vii) Ethanol to propane nitrite
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4 – dimethyl hexane
(x) 2-Methylpropene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-l-ene to n-butyl iodide
(xiii) 2-Chloropropane to propan-l-ol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide.
Answer:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
If aqueous solution, KOH is almost completely ionized to give OH^– ions which being a strong nucleophile brings about a substitution reaction on alkyl halides to form alcohols. Further in the aqueous solution, OH^– ions are highly solvated (hydrated). This solvation reduces the basic character of OH^– ions which, therefore, fails to abstract a hydrogen from the P-carbon of the alkyl chloride to form alkenes. In contrast, an alcoholic solution of KOH contains alkoxide (RO^–) ion which being a much stronger base than OH^– ions preferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.

Question 21.
Primary alkyl halide (a) C₄H₉Br was reacted with alcoholic KOH to give compound (b). Compound (b) was reacted with HBr to give (c) which was an isomer of (a). When (a) was reacted with sodium metal, it gave a compound (d) C₈H₁₈, that was different than the compound when n-butyl bromide was reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer:
The two primary alkyl bromides are possible from the molecular formula (a) C₄H₉Br. These are:

According to the available information, the isomer (I) does not represent the correct compound because this on reacting with sodium metal (Wurtz reaction) will give n-octane. (C₈H₁₈) which is not actually formed

Question 22.
What happens when
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) bromobenzene is treated with Mg in the presence of dry ether,
(iii) chlorobenzene is subjected to hydrolysis,
(iv) ethyl chloride is treated with (aq.) KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether,
(vi) methyl chloride is treated with KCN?
Answer:
(i) But-l-ene is formed as the product as a result of dehydrohalogenation.

(ii) Phenyl magnesium bromide (Grignard reagent) is formed as a result of the reaction.

(iii) Chlorobenzene will not get hydrolysed on boiling with NaOH. No product will be formed.
(iv) Ethyl alcohol is formed as the product

(v) Ethane is formed as a result of Wurtz reaction

(vi) Methyl cyanide is formed.

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