Category: Class 10

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

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

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Solution:
Here, r = 1 cm and h = 1 cm.
Volume of the conical part = \(\frac{1}{3}\) πr²h and volume of the hemispherical part = \(\frac{2}{3}\) πr³h

∴ Volume of the solid shape
= \(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr³h = \(\frac{1}{3}\) πr²h[h + 2r]
= \(\frac{1}{3}\) π(1)² [1 + 2(1)] cm³
= (\(\frac{1}{3}\) π × 1 × 3) cm³ = π cm³

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Here, diameter = 3 cm
⇒ Radius (r) = \(\frac{3}{2}\) cm
Total height = 12 cm
Height of a cone (h₁) = 2 cm

∴ Height of both cones = 2 × 2 = 4 cm
⇒ Height of the cylinder (h₂) = (12 – 4) cm = 8 cm.
Now, volume of the cylindrical part = πr²h₂
Volume of both conical parts = 2[\(\frac{1}{3}\) πr²h₁]
∴ Volume of the whole model

Question 3.
A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

Solution:
Since a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.

Diameter = 2.8 cm
⇒ Radius (r) = 1.4 cm
∴ Length of the cylindrical part (h)
= 5 cm – (1.4 + 1.4) cm = 5 cm – 2.8 cm = 2.2 cm
Now, volume of the cylindrical part = πr²h and volume of both the hemispherical ends
= 2(\(\frac{2}{3}\) πr³) = \(\frac{4}{3}\) πr³
∴ Volume of a gulab jamun

Volume of 45 gulab jamuns

Question 4.
A pen stand made up of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).

Solution:
Dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm.
∴ Volume of the cuboid = 15 × 10 × \(\frac{35}{10}\) cm³
= 525 cm³
Since, each depression is conical in shape with base radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of each depression
= \(\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times\left(\frac{5}{10}\right)^{2} \times \frac{14}{10} \mathrm{cm}^{3}=\frac{11}{30} \mathrm{cm}^{3}\)
Since, there are 4 depressions
∴ Total volume of 4 depressions = \(\frac{44}{30}\) cm³
Now, volume of the wood in entire stand = [Volume of the wooden cuboid] – [Volume of 4 depressions]
= 525cm³ – \(\frac{44}{30}\) cm³
= \(\frac{15750-44}{30} \mathrm{cm}^{3}=\frac{15706}{30} \mathrm{cm}^{3}\)
= 523.53 cm³

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Height of the conical vessel (h) = 8 cm
Base radius (R) = 5 cm
Volume of water in conical vessel = \(\frac{1}{3}\) πR²h

Thus, the required number of lead shots = 100

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)
Solution:
Height of the big cylinder (h) = 220 cm
Base radius (r) = \(\frac{24}{2}\) cm = 12cm
∴ Volume of the big cylinder
= πr²h = (π(12)² × 220) cm³
Also, height of smaller cylinder (h₁) = 60 cm

Base radius (r₁) = 8 cm
∴ Volume of the smaller cylinder πr₁²h₁
= (π(8)² × 60) cm³
∴ Volume of iron pole = [Volume of big cylinder] + [Volume of the smaller cylinder]
= (π × 220 × 12² × π × 60 × 8²) cm³
= 3.14[220 × 12 × 12 + 60 × 8 × 8] cm³

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Height of the conical part = 120 cm
Base radius of the conical part = 60 cm
Volume of the conical part
= \(\frac{1}{3}\) πr²h = [\(\frac{1}{3} \times \frac{22}{7}\) × (60)² × 120] cm³
Radius of the hemispherical part = 60 cm.

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Volume of the cylindrical part = πr²h
= (3.14 × 1² × 8) cm³ = \(\frac{314}{100}\) × 8 cm³

∴ Total volume of the glass vessel = [Volume of cylindrical part] + [Volume of spherical part]

⇒ Volume of water in the vessel = 346.51 cm³
Since the child finds the volume as 345 cm³
∴ The child’s answer is not correct.
The correct answer is 346.51 cm³.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos48° – sin42°
(iv) cosec31°- sec59°
Solution:

(iii) cos 48° – sin 42°
cos 48° = cos (90° – 42°) = sin 42°
[∵ cos (90° – A) = sin A]
∴ cos 48° – sin 42° = sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
cosec 31° = cosec (90° – 59°) = sec 59° [ ∵ cosec (90° – A) = sec A]
∴ cosec 31° – sec 59° = sec 59° – sec 59° = 0

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution;
(i) L.H.S. = tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan 23° tan 42° tan (90° – 23°)
= cot 42° tan 23° tan 42° cot 23° [ ∵ tan (90° – A) = cot A]

(ii) L.H.S. = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin38° sin(90° – 38°)
= cos 38° sin 38° – sin 38° cos 38° [ ∵ sin(90° – A) = cosA and cos(90° – A)= sinA]
= 0 = R.H.S.
⇒ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot(A – 18°)
cot(90 – 2A) = cot (A – 18)
90 – 2A = A – 18
90 + 18 = A + 2A
3A= 108
A = \(\frac {108}{3}\)
A = 36°.

Question 4.
If tan A = cotB, prove that A + B = 90°.
Solution:
tan A = cot B and cot B = tan (90° – B) [∵ tan (90° – θ) = cot θ]
∴ A = 90° – B ⇒ A + B = 90°

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A= cosec (A – 20)
cosec (90 – 4A) = cosec (A – 20)
90 – 4A = A – 20
90 + 20 =A + 4A
110 = 5A
A = \(\frac{110}{5}\)
A = 22°.

Question 6.
If A, 6 and C are interior angles of a triangle ABC, then show that \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)
Solution:
Since, sum of the angles of ∆ABC is 180° i.e.,
A + B + C = 180°
∴ B + C = 180° – A
Dividing both sides by 2, we get

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75° = sin (90 – 23) + cos (90 – 15)
[sin(90 – θ) = cosθ
cos(90 – θ) = sinθ]
= cos 23 + sin 15

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

Question 1.
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Solution:
Let the required point be P(x, y).
Here the end points are (-1, 7) and (4, – 3)
Ratio = 2 : 3 = m₁ : m₂
∴ x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)
= \(\frac{(2 \times 4)+3 \times(-1)}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)
And y = \(\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\)
= \(\frac{2 \times(-3)+(3 \times 7)}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Thus, the required point is (1, 3)

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Solution:
Let the given points be A(4, -1) and B(-2, -3)

Let the points P and Q trisects AB.
i.e., AP = PQ = QB
i.e., P divides AB in the ratio of 1 : 2 and Q divides AB in the ratio of 2 : 1
Let the coordinates of P be (x, y)

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the figure. Niharika runs 1/4^th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5^th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag ?.

Solution:
Let us consider ‘A’ as origin, then
AB is the x-axis.
AD is the y-axis.
Now, the position of green flag-post is (2, \(\frac{100}{4}\)) or (2, 25)
And the position of the red flag-post is (8, \(\frac{100}{5}\)) or (8, 20)
Distance between both the flags

Let the mid-point of the line segment joining the two flags be M(x, y).

or x = 5 and y = 22.5
Thus, the blue flag is on the 5th line at a distance 22-5 m above AB.

Question 4.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Let the given points are A(-3, 10) and B(6, -8).
Let the point P(-1, 6) divides AB in the ratio m₁ : m₂.
∴ Using the section formula, we have

⇒ -1(m₁ + m₂) = 6m₁ – 3m₂
and 6(m₁ + m₂) = – 8m₁ + 10m₂
⇒ -m₁ – m₂ – 6m₁ + 3m₂ = 0
and 6m₁ + 6m₂ + 8m₁ – 10m₂ = 0
⇒ -7m₁ + 2m₂ = 0 and 14m₁ – 4m₂ = 0 or 7m₁ – 1m₂ = 0
⇒ \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\) and \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\)
⇒ 2m₂ = 7m₁ and 7m₁ = 2m₂
⇒ m₁ : m₂ = 2 : 7 and m₁ : m₂ = 2 : 7
Thus, the required ratio is 2 : 7.

Question 5.
Find the ratio in which the line segment joining 4(1, -5) and B(-4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Solution:
The given points are A( 1, -5) and B(-4, 5). Let the required ratio = k:l and the required point be P(x, y).
Since the point P lies on x-axis,
∴ Its y-coordinate is 0.

Question 6.
If (1, 2), (4, y), (x, 6) and (3,5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let the given points are A( 1, 2), B(4, y), C(x, 6) and D(3, 5)

Since, the diagonals of a parallelogram bisect each other.
∴ The coordinates of P are :

∴ The required values of x and y are 6 and 3 respectively.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and Bis (1,4).
Solution:
Let centre of the circle is 0(2, -3) and the end points of the diameter be A(x, y) and B(1, 4)

Since, the centre of a circle bisects the diameter.
∴ 2 = \(\frac{x+1}{2}\) ⇒ x + 1 = 4 or x = 3
And -3 = \(\frac{y+4}{2}\) ⇒ y + 4 = -6 or y = -10
Thus, the coordinates of A are (3, -10)

Question 8.
If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = \(\frac{3}{7}\) AB and Plies on the line segment AB.
Solution:
Here, the given points are A(-2, -2) and B(2, -4)
Let the coordinates of P are (x, y)
Since, the point P divides AB such that
AP = \(\frac{3}{7}\)AB or \(\frac{A P}{A B}\) = \(\frac{3}{7}\)
⇒ AB = AP + BP
⇒ \(\frac{A P+B P}{A P}\) = \(\frac{7}{3}\)

Question 9.
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution:
Here, the given points are A(-2, 2) and B(2, 8)
Let P₁, P₂ and P₃ divide AB in four equal parts.

Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
[Hint: Area of a rhombus = \(\frac{1}{2}\) (product of its diagonals)]
Solution:
Let the vertices of the given rhombus are A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x² – 3x – 10 = 0
(ii) 2x² + x – 6 = 0
(iii) \(\sqrt{2} x^{2}+7 x+5 \sqrt{2}\) = 0
(iv) 2x² – x + \(\frac{1}{8}\) = 0
(v) 100x² – 20x + 1 =0
Solution:
(i) We have, x² – 3x – 10 = 0
⇒ x² – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0 (x – 5)(x + 2) = 0
Either x – 5 = 0 or x + 2 = 0
x = 5 or x = – 2
Thus, the required roots are 5 and -2.

(ii) We have, 2x² + x – 6 = 0
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2)(2x – 3) = 0
Either x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or x = \(\frac{3}{2}\)
Thus, the required roots are -2 and \(\frac{3}{2}\)

(iii)

(iv) We have, 2x² – x + \(\frac{1}{8}\) = 0
⇒ 16x² – 8x + 1 = 0
⇒ 16x² – 4x – 4x + 1 = 0
⇒ 4x(4x – 1) -1(4x – 1) = 0
⇒ 4x – 1 = 0
⇒ x = \(\frac{1}{4}\)

(v) We have 100x² – 20x + 1 = 0
⇒ 100x² – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) -1(10x – 1) = 0
⇒ (10x – 1)(10x – 1) = 0
⇒ (10x – 1) = 0
⇒ x = \(\frac{1}{10}\)
Thus the required roots are \(\frac{1}{10}, \frac{1}{10}\)

Question 2.
Solve the problems:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Solution:
(i) Let John had x marbles and Jivanti had (45 – x) marbles.
When both of them lost 5 marbles then equation becomes (x – 5) × (45 – x – 5) = 124
⇒ (x – 5) × (40 – x) = 124
⇒ x² – 45x + 324 = 0
⇒ x² – 9x – 36x + 324 = 0
⇒ x(x – 9) – 36(x – 9) = 0
⇒ (x – 9)(x – 36) = 0
Either x – 9 = 0 or x – 36 = 0
Thus, x = 9 or x = 36
∴ If John had 9 marbles, then Jivanti had 45 – 9 = 36 marbles.
If John had 36 marbles, then Jivanti had 45 – 36 = 9 marbles.

(ii) Let the number of toys produced in a day be x.
Then cost of 1 toy = \(\frac{750}{x}\)
⇒ \(\frac{750}{x}\) = 55 – x
⇒ 750 = 55x – x²
⇒ x² – 55x + 750 = 0
⇒ x² – 30x – 25x + 750 = 0
⇒ x(x – 30) – 25(x – 30) = 0
⇒ (x – 30)(x – 25) = 0
Either x – 30 = 0 or x – 25 = 0
x = 30 or x = 25

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one of the numbers be x.
∴ Other number = 27 – x
According to the condition,
x(27 – x) = 182
⇒ 21x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 13x – 14x +182 = 0
⇒ x(x – 13) – 14(x – 13) = 0
⇒ (x – 13)(x – 14) = 0
Either x – 13 = 0 or x -14 = 0
⇒ x = 13 or x = 14
Thus, the required numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the two consecutive positive integers be x and (x + 1).
Since, the sum of the square of the numbers is 365.
∴ x² + (x + 1)² = 365
⇒ x² + (x² + 2x + 1) = 365
⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x + 1 – 365 = 0
⇒ 2x² + 2x – 364 = 0
⇒ x² + x – 182 = 0
⇒ x² + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x -13) = 0
Either x + 14 = 0 or x – 13 = 0
⇒ x = -14 or x = 13
Since, x has to be a positive integer
⇒ x = -14 is rejected.
∴ x = 13 ⇒ x + 1 = 13 + 1 = 14
Thus, the required consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of the given right triangle be x cm.
∴ Its height = (x – 7) cm

Squaring both sides, we get
169 = x² + (x – 7)²
⇒ 169 = x² + x² – 14x + 49
⇒ 2x² – 14x + 49 – 169 = 0
⇒ 2x² – 14x – 120 = 0
⇒ x² – 7x- 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
Either x – 12 = 0 or x + 5 = 0
⇒ x = 12 or x = -5
But the sides of a triangle can never be negative
⇒ x = -5 is rejected.
∴ x = 12
∴ Length of base = 12 cm
⇒ Length of altitude = (12 – 7)cm = 5 cm
Thus, the required base = 12 cm and altitude = 5 cm

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced in a day = x
∴ Cost of production of each article = ₹ (2x + 3)
According to the condition,
Total cost = ₹ 90
⇒ x × (2x + 3) = 90
⇒ 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² – 12x + 15x – 90 = 0
⇒ 2x(x – 6) + 15(x – 6) = 0
⇒ (x – 6)(2x + 15) = 0
Either x – 6 = 0 or 2x + 15 = 0
⇒ x= 6 or x = \(\frac{-15}{2}\)
But the number of articles produced can never be negative.
⇒ x = \(\frac{-15}{2}\) is rejected
∴ Cost of production of each article = ₹ (2 × 6 + 3) = ₹ 15
Thus, the required number of articles produced is 6 and the cost of each article is ₹ 15.

MP Board Class 10th Science Solutions Chapter 10 Light Reflection and Refraction

MP Board Class 10th Science Chapter 10 Intext Questions

Class 10th Science Chapter 10 Intext Questions Page No. 168

Question 1.
Define the principal focus of a concave mirror.
Answer:
The number of rays parallel to the principal axis are falling on a concave mirror which meat at a point is called principal focus of the concave mirror.

(or)

Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of concave mirror.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
R = 2f Here R = 20 cm
20 = 2f
∴ \(f=\frac { 20 }{ 2 } =10\)
∴ Focal length = 10 cm.

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror can give an erect and enlarged image of an object when object is placed between the pole and principal focus.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
A convex mirror when fitted at rear-view position of vehicles, it gives a wider field of view, with which driver can see most of the traffic behind him. Convex mirrors give a virtual, erect and diminished image of the objects in front of it. So, we prefer a convex mirror as a rear-view mirror in vehicles.

Class 10th Science Chapter 10 Intext Questions Page No. 171

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Radius of curvature, R = 32 cm
Radius of curvature = 2f
\(R=2f=\frac { R }{ 2 } =\frac { 32 }{ 2 } =16\)
∴ Convex mirror focal length is = 16cm

Question 2.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:

Let the height of the object = h₀ = h
Then, height of the image, h₁ = -3h (Image formed is real)
= \(\frac { -3h }{ h } \) = \(\frac { -v }{ u } \)
Object-distance, u = – 10 cm
v = 3 × (- 10)
= – 30 cm
Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm in front of the given concave mirror.

Class 10th Science Chapter 10 Intext Questions Page No. 176

Question 1.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
Lightray bend towards normal. Because when a ray of light enters from rearer medium to denser medium, it changes its direction in the second medium.

Question 2.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 ms-1.
Answer:
Refractive index of a medium:
µm = Speed of light in vacuum/Speed of light in the medium
Speed of light in vacuum, c = 3 × 10⁸ ms^-1
Refractive index of glass, µ_g = 1.50
Speed of light in the glass,
v = Speed of light in vacuum / Refractive index of glass
= c/µ_g
= 3 × 10⁸/1.50
= 2 × 10⁸ ms^-1.

Question 3.
Find out, from Table the medium having highest optical density. Also find the medium w ith lowest optical density.
Table:

Answer:
Highest optical density = Diamond.
Lowest optical density = Air.

Optical density of a medium is proportional to the refractive index. Hence, medium with highest refractive index will have the highest optical density and vice-versa. It can be observed from table that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in table.

Answer:
Light travel faster in water when compared to kerosene and turpentine, since the refractive index of water is lower than kerosene and turpentine. The speed of light is inversely proportional to the refractive index.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
It means Ratio of velocity of light in air and velocity of air in diamond is 2.42.

Class 10th Science Chapter 10 Intext Questions Page No. 184

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of lens whose focal length is 1 metre 1 D = 1 m^-1

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
Image of Needle is real and inverted means this is real image it is 2f
Image is at a distance of 50 cm
Hence needle is kept 50 cm in front of convex lens.
Distance of object, u = – 50 cm.
Distance of image v = 50 cm
Focal length f = ?
As per lens formula.

f = 25 cm = 0.25 m
Power of the lens

Power of the lens P = + 4D.

Question 3.
Find the power of a concave lens of focal length 2 m.
Answer:
Focal length of concave lens, f = 2 m
Power of lens, P = \(\frac { 1 }{ f } \) = \(\frac { 1 }{ (-2) } \) = -0.5D

MP Board Class 10th Science Chapter 10 NCERT Textbook Exercises

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) A lens allows light to pass through it, but clay does not have that property.

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
(b) When an object is placed at the centre of curvature in front of a convex lens, its image is formed at the centre of curvature on the other side of the lens.

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be:
(а) both concave
(b) both convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex, but the lens is concave
Answer:
(а) both concave

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

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
Answer:
(c)

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer:
Range of the distance of the object = 0 cm to 15 cm.
Nature of the image = virtual, erect and larger than the object.

Question 8.
Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Concave mirror: Concave mirrors can produce powerful parallel ’ beam of light when the light source is placed at their principal focus. Hence, we can visualize ways easily in little light.

(b) Convex mirror: A convex mirror when fitted at rear view position of vehicles, it gives a wider field of view, with which driver can see most of the traffic behind him.

(c) Concave mirror: They are converging mirrors. This is because it concentrates the parallel rays of sun at principal focus and increase intensity of light falling on it.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
Yes, the convex lens will form complete image of the object, even if its one half is covered with black paper. Following two cases can better explain it:

Case I : When the upper half of the lens is covered.
In this case, a ray of light coming from the object is being refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Case II : If the lower half of the lens is covered.
In this case, a ray of light coming from the object is being refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Question 10.
An object 5 cm in length.is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Converging lens means a convex lens. As the distances given in the question are large, so we choose a scale of 1 : 5, i.e., 1 cm represents 5 cm. Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. Now, we draw the ray diagram as follows;

1. Draw a horizontal line to represent ‘die principal axis of the convex lens.
2. Centre line is shown by DE.
3. Mark two foci F and F’ on two sides of the lens, each at a distance of 2 cm from the lens.
4. Draw an arrow AB of height 1 cm on the left side of lens at a distance of 5 cm from the lens.
5. Draw a line AD parallel to principal axis and then allow it to pass straight through the focus (F’) on the right side of the lens.
   
6. Draw a line from A to C (centre of the lens), which goes straight without deviation.
7. Let the two lines starting from A meet at A’.
8. Draw A’B’, perpendicular to the principal axis from A’.
9. Now A’B represents the real, but inverted image of the object AB.
10. Then, measure CB’ and A’B’. It is found that CB’ = 3.3 cm and A’B’ = 0.7 cm.
11. Thus the final position, nature and size of the image A’B’ are
    • (a) Position of image A’B’ = 3.3 cm × 5 = 16.5 cm from the lens on opposite side.
    • (b) Nature of image A’B’: Real and inverted.
    • (c) Height of image A’B’: 0.7 × 5 = 3.5 cm, i.e., image is smaller than the object.

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

Answer:
Focal length of concave lens f = – 15 cm
Image distance, v = – 10 cm
According to the lens formula,
\(\frac { 1 }{ v } \) – \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)

On solving we get, u = – 30 cm
The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the above ray diagram.

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Focal length of convex mirror,
f = +15 cm
Object distance, u = -10 cm
As per lens formula

Magnification \(=\frac{v}{u}=\frac{-6}{-10}=0.6\)
Virtual image is formed at the distance of 6 cm and it is erect.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
The positive sign means image formed by a plane mirror is virtual and erect. Since, the magnification is I, it means that the size of the image is equal to the size of the object.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Object distance, u = – 20 cm
Object height, h = 5 cm
Radius of curvature, R = 30 cm
Radius of curvature = 2 × focal length
R = 2f ⇒ f = 15 cm
According to the mirror formula,
\(\frac { 1 }{ v } \) + \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
The positive value indicates that the image is formed behind the mirror.

The positive value of image height indicates image is virtual, erect and smaller in size.

Question 15
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer:
Object-distance, u = – 27 cm
Object-height, h = 7 cm
Focal length, f = – 18 cm
According to the mirror formula,
\(\frac { 1 }{ v } \) + \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
Putting values, \(\frac { 1 }{ v } \) + \(\frac { 1 }{ (-27) } \) = \(\frac { 1 }{ (-18) } \)
So, v = – 54 cm
The screen should be placed at a distance of 54 cm in front of the given mirror
and \(\frac{h_{2}}{h_{1}}=\frac{-v}{u}\)
h₂ = -14 cm
The negative value of image indicates that the image is inverted.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Given, P = -2D
Power of lens. p = \(\frac { 1 }{ f } \)
and, f = – \(\frac { 1 }{ 2 } \)
\(\frac{h_{2}}{h_{1}}=\frac{-v}{u}\)
= – 0.5 m
A concave lens, because it has a negative value of focal length.

Question 17.
A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Given, P = 1.5.D
Power of lens, P = \(\frac { 1 }{ f } \)
and, focal length f = \(\frac { 1 }{ 1.5 } \)
= \(\frac { 10 }{ 15 } \) = 0.66 m
A convex lens, because it has a positive focal length. Lens is converging.

MP Board Class 10th Science Chapter 10 Additional Important Questions

MP Board Class 10th Science Chapter 10 Multiple Choice Questions

Question 1.
The image formed by a convex lens is virtual, erect and larger than the object. The position of the object must be:
(a) Between the lens and its focus
(b) At the focus
(c) At twice the focal length
(d) At infinity
Answer:
(a) Between the lens and its focus

Question 2.
A real image formed by a convex lens is always:
(a) On the same side of the lens as the object
(b) Erect
(c) Inverted
(d) Smaller than the object
Answer:
(c) Inverted

Question 3.
If an object is moved towards a convex lens, the size of its image:
(a) Decreases
(b) Increases
(c) First decreases and then increases
(d) Remains the same
Ans.
(b) Increases

Question 4.
An object is placed at a distance of 30 cm from a concave mirror of focal length 15 cm. The image will be:
(a) Real and of same size
(b) Real and magnified
(c) Real and diminished
(d) virtual and magnified
Answer:
(a) Real and of same size

Question 5.
A concave mirror always forms real and inverted image except when the object is placed:
(a) At infinity
(b) Between F and C
(c) At F
(d) Between F and pole of the mirror
Answer:
(b) Between F and C

Question 6.
The mirror which has a wide field of view must be:
(a) Concave
(b) Convex
(c) Plane
(d) None of these
Ans.
(b) Convex

Question 7.
The image formed by a concave mirror:
(a) Is always real
(b) Is always virtual
(c) Can be both real and virtual
(d) None of these
Ans.
(c) Can be both real and virtual

Question 8.
An object is placed 20 cm from a convex lens of focal length 10 cm. The image must be:
(a) Real and diminished
(b) Real and of same size
(c) Real and enlarged
(d) Virtual and enlarged
Answer:
(b) Real and of same size

Question 9.
The ratio of the focal length of spherical mirror to its radius of curvature is:
(a) 0.5
(b) 1
(c) 2
(d) 3
Answer:
(a) 0.5

Question 10.
A real and inverted image of the same size is formed by a concave mirror when the object is placed:
(a) Between the mirror and its focus.
(b) Between the focus and the centre of curvature.
(c) At the centre of curvature.
(d) Beyond the centre of curvature.
Ans.
(c) At the centre of curvature.

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

Question 1.
What is a mirror? Mention the different types of mirrors commonly used.
Answer:
Mirror: A highly polished surface which is smooth enough to reflect a good fraction of light incident on it is called a mirror. The mirror may be a highly polished metal surface or an ordinary glass plate coated with a thin silver layer.

Question 2.
What is the number of images of an object held between two plane parallel mirrors?
Answer:
Infinity.

Question 3.
Does the refractive index for a given pair of media depend on the angle of incidence?
Answer:
No, it is independent of the angle of incidence.

Question 4.
The refractive index of water with respect to air is \(\frac { 4 }{ 3 } \). What is the refractive index of air with respect to water?
Answer:
Refractive index of air with respect to water = \(\frac { 3 }{ 4 } \)

Question 5.
Can absolute refractive index of a medium exceed unity?
Answer:
No, because speed of light is maximum in vacuum.

Question 6.
Why does a ray of light bend when it travels from one medium to another?
Answer:
The bending of light or refraction occurs due to the change in the speed of light as it passes from one medium to another due to change in the density of the medium.

Question 7.
What happens when a ray of light strikes the surface of separation between the two media at right angle?
Ans.
The ray of light passes undeflected from one medium to another.
Here, ∠i = ∠r = 0°

Question 8.
What do you mean by a magnification less than unity?
Answer:
It means that the size of the image is smaller than the size of the object.

Question 9.
Which spherical mirror has

1. a real focus and
2. a virtual focus?

Answer:

1. A concave mirror has a real focus.
2. A convex mirror has a virtual focus.

Question 10.
State the position of the object for which a concave mirror produces virtual magnified image.
Answer:
The object should be placed between F and P of the concave mirror.

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

Question 1.
Name the type of mirror(s) that should be used to obtain:
(i) a magnified and virtual image.
(ii) a diminished and virtual image of an object.
Draw labelled diagrams to show the formation of the required image in each of the above two cases. Which of these mirrors could also form a magnified and real image of the object? State the position of object for which this could happen.
Answer:
(i) Concave mirror.

Fig. 10.3 : Concave mirror with the object between F and P.

(ii) Convex mirror.

Fig. 10.4: Convex mirror with the object between pole and infinity.

Question 2.
Explain the uses of concave and convex mirrors.
Answer:
Uses of concave mirrors:
1. Shaving mirror : A concave mirror is used as a shaving or make-up mirror because it forms erect and enlarged image of the face when it is held closer to the face.

2. As head mirror : E.N.T. specialists use a concave mirror on their forehead. The light from a lamp after reflection from the mirror is focussed into the throat, ear or nose of the patient making the affected part more visible.

3. In ophthalmoscope : It consists of a concave mirror with a small hole at its centre. The doctor looks through the hole from behind the mirror while a beam of light from a lamp reflected from it, is directed into the pupil of patient’s eye which makes the retina visible.

4. In headlights : Concave mirrors are used as reflectors in headlights of motor vehicles railway engines, torch lights etc. The source is placed at the focus of the concave mirror. The light rays after reflection travel over a. large distance as a parallel beam of high intensity.

5. In astronomical telescopes : A concave mirror of large diameter (5 m or more) is used as objective in an astronomical telescope. It collects light from the sky and makes visible even those faint stars which cannot be seen with naked eye.

6. In solar furnaces : Large concave mirrors are used to concentrate sunlight to produce heat in solar furnace.

Uses of convex mirrors : Drivers use convex mirror as a rear-view mirror in automobiles because of the following two reasons:

1. A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
2. A convex mirror has a wider field of view than a plane mirror of the same size as shown in Fig. 10.5.

Fig. 10.5: Field of view of (a) a plane mirror, (b) a convex mirror.

Thus, convex mirrors enable the driver to view much larger traffic behind him than would be possible with a plane mirror. The main disadvantage of a convex mirror is that it does not give the correct distance and the speed of the vehicle approaching from behind.

Question 3.
State the characteristics of the image formed by a convex mirror. What is the value of angle of incidence and angle of reflection when a ray of light retraces its path after reflection from a convex mirror? Illustrate with the help of a ray diagram.
Answer:
Properties of the image formed by a convex mirror:

(a) The image is always virtual and erect.
(b) The image is highly diminished or point sized.
(c) It is always formed between F and P.
(d) As the object is moved towards the pole of a convex mirror, image also moves towards its pole and gradually increases in size till its size becomes almost equal to that of the object.

When array of light retraces its path, ∠i = ∠r = 0°.

Fig. 10.6: A ray directed towards C is reflected back along same path after reflection from a convex mirror.

Question 4.
State the new Cartesian sign convention followed for reflection of light by spherical mirrors.
Answer:
According to this convention:

1. The object is on the left of the mirror. So all the ray diagrams are drawn with the incident light travelling from left to right.
2. All the distances parallel to the principal axis are measured from the pole of the mirror.
3. All distances measured in the direction of incident light are taken as positive.
4. All distances measured in the opposite direction of incident light are taken as negative.
5. Heights measured upwards and perpendicular to the principal axis are taken positive.
6. Heights measured downwards and perpendicular to the principal axis are taken negative.

Fig. 10.7: New Cartesian sign convention for reflection of light by spherical mirrors.

Question 5.
State the type of mirror preferred as
(i) rear view mirror in vehicles
(ii) shaving mirror. Justify your answer giving two reasons in each case.
Answer:
(i) A convex mirror is preferred as a rear-view mirror because:

(a) It always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
(b) It has wider field of view.

(ii) A concave mirror is preferred as a shaving mirror because when it is held closer to the face, it forms:

(a) an enlarged image of the face.
(b) an erect image of the face.

Question 6.
State the laws of refraction of light.
Answer:
Laws of refraction of light: The refraction of light obeys the following two laws:
1st law: The incident ray, the refracted ray and normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

2nd law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media.
Mathematically,

The ratio µ21 is called refractive index of the second medium with respect to the first medium. The second law of refraction is also called Snell’s law of refraction.

Question 7.
What is the physical significance of refractive index?
Answer:
The refractive index of any medium gives the ratio of the speed of light in vacuum to the speed of light in that medium. For example, the refractive index of water, µ_w = 1.33. This means that the ratio of the speed of light in vacuum or air to the speed of light in water is 1.33.

Question 8.
What do you mean by optically denser and optically rarer media? How is the speed of light related to optical density?
Answer:
The optical density of a medium represents its ability to refract light. A medium having larger refractive index is called optically denser medium than the other. The other medium having lower refractive index is called optically rarer medium.

The speed of light is higher in a rarer medium than a denser medium. Thus, a ray of light travelling from a rarer medium to a denser medium slows down and bends towards the normal.

When it travels from a denser medium to a rarer medium, it speeds up and bends away from the normal.

Table:
Refractive indices of some material media (with respect to vacuum)

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

Question 1.
With the help of a ray diagram, state and explain the laws of reflection of light at a plane mirror. Mark the angles of incidence and reflection clearly on the diagram.
Answer:
As shown in Fig. 10.8, when a ray of light is incident on a mirror, it gets reflected in accordance with the following laws of reflection.
1st law: The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
2nd law: The angle of incidence (i) is equal to the angle of reflection (r) i.e.∠i = ∠r

Fig.10.8: Reflection in a plane mirror.

Question 2.
What is lateral inversion of an image? What is the cause of lateral inversion?
Answer:
Lateral inversion: If we stand before a plane mirror and move our right hand, our image appears to move its left hand. In fact, our entire image is reversed sideways. This sideways reversal of the image is known as lateral inversion.

Cause of lateral inversion: Lateral inversion is due to the fact that in a plane mirror the image is as far behind the mirror as the object is in front of it, and that the front of the image and the front of the object face each other. The laterally inverted image of the word PAPYRUS is as shown in Fig. 10.9. The images of symmetrical letters like A, H, I, M, O, T, U, V, W, X, Y, 8 are not affected by lateral inversion.

Fig.10.9: Lateral inversion before a mirror.

Question 3.
Define the following terms in connection with spherical mirrors:
(i) Angular aperture
(ii) Centre of curvature
(iii) Radius of curvature
(iv) Principal axis
(v) Linear aperture
(vi) Pole
(vii) Principal force
(viii) Focal length
(ix) Principle focus Focal plane.
Answer:
Definition in connection with spherical mirrors: In Fig 10.10, let APB be a principal section of a spherical mirror, i.e., the section cut by a plane passing through pole and centre of curvature of the mirror
(i) Angular aperture : It is the angle ACB subtended by the boundary of the spherical mirror at its centre of curvature.
(ii) Centre of curvature : It is the centre C of the sphere of which the mirror forms a part.
(iii) Radius of curvature : It is the radius R (= AC or BC) of the sphere of which the mirror forms a part.
(iv) Principal axis : The line passing through the pole and the centre of curvature of mirror is called its principal axis.
(v) Linear aperture : It is the diameter AB of the circular boundary of the spherical mirror.
(vi) Pole: It is the middle point P of the spherical mirror.

Fig. 10.10: Characteristics of a concave mirror.

Question 4.
Deduce a relation between focal length (f) and radius of curvature (R) for a concave mirror.
Answer:
Relation between f and R for a concave mirror: As shown in Fig. 10.11, consider a ray AB parallel to the principal axis and incident at the point B of a concave mirror. After reflection from the mirror, this ray passes through its focus F, obeying the laws of reflection. If C is the centre of curvature, then CP = R, is the radius of curvature and CB is normal to the mirror at point B.

Fig. 10.11: Relation betweenTand R for a concave mirror.

According to the law of reflection, ∠i = ∠r
As AB is parallel to CP, so ∠a = ∠i (Alternate angles)
∠a = ∠r
Thus, ∆ BCF is isosceles,
Hence, CF = FB.
If the aperture (or size) of the mirror is small, then B lies close to P, so that,
FB = FP
FP = CF = \(\frac { 1 }{ 2 } \) CP
or f = \(\frac { R }{ 2 } \)
or Focal lenght = \(\frac { 1 }{ 2 } \) × Radius of curvature
Thus, the principal focus of a spherical mirror lies midway between the pole and the centre of curvature.

Question 5.
What happens to the size of the image formed by a convex mirror, when an object is gradually moved towards the mirror?
Answer:
When the object is at position A1B1, its virtual image is at a1b1 When the object is at position A2B2, its virtual image is at a2b2. So, when an object is gradually moved towards the pole of a convex mirror, its image also moves towards its pole and gradually increases in size till it has a size almost equal to that of the object. However, the image is always formed between F and P.

MP Board Class 10th Science Chapter 10 NCERT Text Book Activities

Class 10 Science Activity 10.1 Page No. 161

• Take a large shining spoon. Try to view your face in its curved surface.
• Do you get the image? Is it smaller or larger?
• Move the spoon slowly away from your face. Observe the image. How docs it change?
• Reverse the spoon and repeat the Activity. How does the image look like now?
• Compare the characteristics of the image on the tw o surfaces.

Observations:

• The image is formed that is smaller in size.
• On moving the spoon away, the size of the image gradually decreases with increasing field of view.
• On reversing the spoon, spoon w hen dose forms elect and magnified image on the inner curved ..urface. As we move away, image gets inverted and gradually decrease in Its size.
• The image on outer surface of spoon is erect and gradually decreases in size on moving away. The image of spoon on inner surface is erect and gets inverted on moving away. The size also decreases when moved away.

Class 10 Science Activity 10.2 Page No. 162

Caution: Do not look at the Sun directly or even into a mirror reflecting sunlight. It may damage your eyes.

• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
• Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?

Observations:

• The paper starts burning when the mirror and paper are held in the same position for a few minutes as light rays from the sun, sharply focuses on this point and due to which heat concentrates at a point resulting in burning because of intense heating.

Class 10 Science Activity 10.3 Pages No. 163,164

You have already learnt a way of determining the focal length of a concave mirror. In activity 10.2, you have seen that the sharp bright spot of light you got on the paper is, in fact, the image of the Sun. It was a tiny, real, inverted image. You got the approximate focal length of the concave mirror by measuring the distance of the image from the mirror:

• Take a concave mirror. Find out its approximate focal length in the w ay described above. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
• Mark a line on a table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
• Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines it equal to the focal length of the minor. These lines will now correspond to the positions of the points P, F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus F lies midway between the pole P and the centre of curvature C.
• Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
• Observe the image carefully. Note down its nature, position and relative size with respect to the object size.
• Repeat the activity by placing the candle-(a) just beyond C, (b) at C, (c) between F and C. (d) at F, and (c) between P and F.
• In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case, Then, look for its virtual image in the mirror itself.
• Note down and tabulate your observations.

Observations:

Class 10 Science Activity 10.4 Page No. 166

• Draw neat ray diagrams for each position of the object shown in previous activity 10.3 observations.
• You may take any two of the rays mentioned in the previous section for locating the image.
• Compare your diagram with those given in Fig. 10.1.
• Describe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.

Observations:

Fig. 10.1: Ray diagrams for the image formation by a concave mirror.

Class 10 Science Activity 10.5 Page No. 167

• Take a convex mirror. Hold it in one hand.
• I Iold a pencil in the upright position in the other hand.
• Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
• Move the pencil away from the mirror slowly. Does the image become smaller or larger?
• Repeat this Activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the min or?

Observations:

• The image is erect and diminished. The image becomes smaller or moving pencil away.
• The image moves closer to the focus as the object is moved away from the mirror.

Class 10 Science Activity 10.6 Page No 167

• Observe the image of a distant object, say a distant tree, in a plane mirror.
• Could you see a full-length image?
• Try with plane minors of different sizes. Did you see the entire object in the image?
• Repeat this Activity with a concave mirror. Did the mirror show full length image of the object?
• Now try using a convex mirror. Did you succeed? Explain your observations with reason.

Observations:

• No, full – length image of a distant object is not seen in a plane mirror.
• The entire images of the objects were not seen.
• No, the mirror do not show full length image of the object.
• Yes, with the convex mirror we can see full length image of distant object with under field of view’ this is because the convex mirror are used as rear or side view mirrors in vehicles. The image formed is diminished, erect and virtual.

Class 10 Science Activity 10.7 Page No. 172

• Place a coin at the bottom of a bucket filled with water.
• With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
• Repeat the Activity. Why did you not succeed in doing it in one go?
• Ask your friends to do this. Compare your experience with theirs.

Observations:

• No, we cannot succeed in picking up the coin.
• This happens due to refraction, the coin appears to he at some other place from where it is actually present. The light rays coming out from water tends to bend creating this problem.

Class 10 Science Activity 10.8 Page No. 172

• Place a large shallow bowl on a Table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.
• Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?

Observations:

• Yes, on pouring water, it again becomes visible and little raised due to refraction.

Class 10 Science Activity 10.9 Page No. 172

• Draw a thick straight line in ink, over a sheet of white paper placed on a Table.
• Place a glass slab over the line in such a way that one of its edges make an angle with the line.
• Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges ?
• Next, place the glass slab such that it is normal to the line. What do you observe now’ Does the part of the line under the glass slab appear bent?
• Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?

Observations:

• The line under the glass slab appear bent at the edges due to refraction.
• No, now it does not appear bend as a ray of light perpendicular to the plane of a refracting medium does not change angle during refraction.
• Yes, this is also due to refraction that apparent position of image of object seems nearer than its actual position.

Class 10 Science Activity 10.10 Page No. 173

• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
• Join O and O’. Also produce EF up to P, as shown by a dotted line in Fig. 10.2.

Observations:

Fig. 10.2: Refraction of light through a rectangular glass slab.

Class 10 Science Activity 10.11 Page No. 177

Caution: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do so.

• Hold a convex lens in your hand. Direct it towards the Sun.
• Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.
• Hold the paper and the lens in the same position for a while. Keep observing the paper. What happened? Why? Recall your experience in Activity 10.2.

Observations:

• This is due to sharp focusing of all rays at a single point after passing through the lens. The concentration of light rays increases at a point resulting in burning of paper.

Class 10 Science Activity 10.12 Page No. 178

• Take a convex lens. Find its approximate focal length in a way described in Activity 10.11.
• Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.
• Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
• The two lines on cither side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F₂ and 2F₂, respectively.
• Place a burning candle, far beyond 2F, to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
• Note down the nature, position and relative size of the image.
• Repeat this Activity by placing object just behind 2F₁ between F₁ and 2F₁ at F₁, between F₁ and O. Note down and tabulate your observations.

Observations:

Class 10 Science Activity 10.13 Page No. 179

• Take a concave lens. Place it on a lens stand.
• Place a burning candle on one side of the lens.
• Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not, observe the image directly through the lens.
• Note down the nature, relative size and approximate position of the image.
• Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

Observations:

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.
Which term of the AP : 121,117,113, ……., is its first negative term? [Hint: Find n for a_n < 0]
Solution:
We have the A.P. having a = 121 and d = 117 – 121 = -4

Thus, the first negative term is 32nd term.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP
Solution:

Question 3.
A ladder has rungs 25 cm apart (see figure).The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \(2 \frac{1}{2}\) m apart, what is the length of the wood required for the rungs? 250
[Hint: Number of rungs = \(\frac{250}{25}+1\)]

Solution:

Length of the 1^st rung (bottom rung) = 45 cm
Length of the 11^th rung (top rung) = 25 cm
Let the length of each successive rung decreases by x cm.
∴ Total length of the rungs = 45 cm + (45 – x) cm + (45 – 2x) cm + ….. + 25 cm
Here, the number 45, (45 – x), (45 – 2x), …., 25 are in an AP such that
First term (a) = 45
and last term (l) = 25
Number of terms, n = 11

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. [Hint: S_x-1 = S₄₉ – S_x]
Solution:
We have the following consecutive numbers on the houses of a row; 1, 2, 3, 4, 5, ….. , 49.
These numbers are in AP, such that a = 1, d = 2 – 1 = 1, n = 49
Let one of the houses be numbered as x.
∴ Number of houses preceding it = x – 1
Number of houses following it = 49 – x
Now, the sum of the house-numbers preceding

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see fig.) Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the 1st step \(\frac{1}{4} \times \frac{1}{2} \times 50 m^{3}\)]

Solution:


\(d=\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\)
Here, total number of steps n = 15
Total volume of concrete required to build 15 steps is given by the sum of their individual volumes.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions 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 :
I. Draw a line segment AB = 7.6 cm.
II. Draw a ray AX making an acute angle with AB.
III. Mark 13 = (8 + 5) equal points on AX, such that AX₁ = X₁X₂ = ……….. X₁₂X₁₃.
IV. Join points X₁₃ and B.
V. From point X₅, draw X₅C || X₁₃B, which meets AB at C.
Thus, C divides AB in the ratio 5 : 8 On measuring the two parts, we get AC = 2.9 cm and CB = 4.7 cm.

Justification:
In ∆ABX₁₃ and ∆ACX₅, we have
CX₅ || BX₁₃
∴ \(\frac{A C}{C B}=\frac{A X_{5}}{X_{5} X_{13}}=\frac{5}{8}\) [By Thales theorem]
⇒ AC : CB = 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 :
I. Draw a ∆ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.
II. Draw a ray BX making an acute angle ∠CBX.
III. Mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X₁, X₂, X₃ on BX₁ such that BXj = X₁X₂ = X₂X₃.

IV. Join X₃C.
V. Draw a line through X₂ such that it is parallel to X₃C and meets BC at C’.
VI. Draw a line through C parallel to CA which intersect BA at A’.
Thus, ∆A’BC’ is the required similar triangle.
Justification :
By construction, we have X₃C || X₂C’

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 :
I. Construct a ∆ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.

II. Draw a ray BX such that ∠CBX is an acute angle.
III. Mark 7 points of X₁, X₂, X₃, X₄, X₅, X₆ and X₇ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ – X₄X₅ = X₅X₆ = X₆X₇
IV Join X₅ to C.
V. Draw a line through X₇ intersecting BC (produced) at C’ such that X₅C || X₇C’
VI. Draw a line through C’ parallel to CA to intersect BA (produced) at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ Using AA similarity, ∆ABC ~ ∆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 :
I. Draw BC = 8 cm
II. Draw the perpendicular bisector of BC which intersects BC at D.

III. Mark a point A on the above perpendicular such that DA = 4 cm.
IV. Join AB and AC.
Thus, ∆ABC is the required isosceles triangle.
V. Now, draw a ray BX such that ∠CBX is an acute angle.
VI. On BX, mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X₁, X₂ and X₃ such that BX₁ = X₁X₂ = X₂X₃
VII. Join X₂C.
VIII. Draw a line through X₃ parallel to X₂C and intersecting BC (extended) to C’.
IX. Draw a line through C’ parallel to CA intersecting BA (extended) to A’, thus, ∆A’BC’ is the required triangle.
Justification:
We have C’A’ || CA [By construction]
∴ Using AA similarity, ∆A’BC’ ~ ∆ABC

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 :
I. Construct a ∆ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
II. Draw a ray BX such that ∠CBX is an acute angle.

Mark four points [greater of 3 and 4 in \(\frac{3}{4}\)] X₁, X₂, X₃, X₄ on BX such that 4
BX₁ = X₁X₂ = X₂X₃ = X₃X₄
IV. Join X₄C and draw a line through X₃ parallel to X₄C to intersect BC at C’.
V. Also draw another line through C’ and parallel to CA to intersect BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
In ∆BX₄C we have
X₄C || X₃C’ [By construction]

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:
Steps of Construction :
I. Construct a AABC such that BC = 7 cm, ∠B = 45°, ∠A = 105° and ∠C = 30°
II. Draw a ray BX making an acute angle ∠CBX with BC.

III. On BX, mark four points [greater of 4 and 3 in \(\frac{4}{3}\) ] X₁, X₂, X₃ and X₄ such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄.
IV. Join X₃C.
V. Draw a line through X₄ parallel to X₃C intersecting BC(extended) at C’.
VI. Draw a line through C parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle. Justification:
By construction, we have
C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]

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 \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
I. Construct the right triangle ABC such that ∠B = 90°, BC = 4 cm and BA = 3 cm.
II. Draw a ray BX such that an acute angle ∠CBX is formed.
III. Mark 5 points X₁, X₂, X₃, X₄ and X₅ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅.
IV. Join X₃C.
V. Draw a line through X₅ parallel to X₃C, intersecting the extended line segment BC at C’.
VI. Draw another line through C’ parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions 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 by the rope with the ground level is 30° (see figure).

Solution:
In right ∆ABC,

Thus, the required height of the pole is 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 the tree is broken at A and its top is touching the ground at B.
Now, in right ∆AOB, we have
\(\frac{A O}{O B}\) = tan 30°

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 angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
In the figure, DE is the slide for younger children, whereas AC is the slide for elder children.

In right ∆ABC, AB = 3m

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:
In right ∆ABC, AB = height of the tower and point C is 30 m away from the foot of the tower,

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:
Let OB = Length of the string
AB = 60 m = Height of the kite.
In the right ∆AOB,
∴ \(\frac{O B}{A B}\) = cosec 60° = \(\frac{2}{\sqrt{3}}\)

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:
Here, OA is the building.

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:
Let the height of the building be BC
∴ BC = 20 m and height of the tower be CD.
Let the point A be at a distance y metres from the foot of the building.
Now, in right ∆ABC,

Thus, the height of the tower is 14.64 m.

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:
In the figure, DC represents the statue and BC represents the pedestal.
Now, in right ∆ABC, we have
\(\frac{A B}{B C}\) = cot 45° = 1
⇒ \(\frac{A B}{h}\) = 1 ⇒ AB = h metres.
Now in right ∆ABD,

Question 9.
The angle of elevation of the top of a building from the foot of the 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:
In the figure, let height of the building = AB = h m

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 distances of the point from the poles.
Solution:
Let AB = h metres = CD

and AP = x m
∴ CP = (80 – x) m
Now, in right ∆APB,
we have

Thus, the required point is 20 m away from the first pole and 60 m away from the second pole.
Height of each pole = 34.64 m.

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 tower is 30° (see figure). Find the height of the tower and the width of the canal.

Solution:
Let the TV tower be AB = h m.
Let the point ‘C be such that BC = x m and CD = 20 m.
Now, in right ∆ABC, we have

Thus, the height of the tower = 17.32 m.
Also width of the canal = 10 m.

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:
In the figure, let AB be the height of the building.
∴ AB = 7 metres.
Let BC = X m = AE

Let CD be the height of the cable tower and DE = h m
∴ In right ∆DAE, we have

Question 13.
As observed from the top of a 75 m high light house 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 light house, find the distance between the two ships.
Solution:
In the figure, let AB represent the light house.
∴ AB = 75 m.
Let the positions of two ships be C and D such that angle of depression from A are 45° and 30° respectively.
Now, in right ∆ABC,
we have

Thus, the required distance between the ships is 54.9 m.

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 some time, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.

Solution:
In the figure, let C be the position of the observer (the girl).
A and P are two positions of the balloon.
CD is the horizontal line from the eyes of the observer (girl).
Here PD = AB = 88.2 m – 1.2 m = 87 m
In right ∆ABC, we have
\(\frac{A B}{B C}\) = tan 60°
⇒ \(\frac{87}{B C}=\sqrt{3}\) ⇒ BC = \(\frac{87}{\sqrt{3}}\) m
In right ∆PDC, we have

Thus, the required distance between the two positions of the balloon = \(58 \sqrt{3}\) 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:
In the figure, let AB be the height of the tower and C, D be the two positions of the car.

⇒ AC = \(\sqrt{3} \times \sqrt{3}\) × AD = 3 AD
Now, CD = AC – AD = BAD – AD = 2AD
Since the distance 2 AD is covered in 6 seconds,
∴ The distance AD will be covered in \(\frac{6}{2}\) i.e., 3 seconds,
Thus, the time taken by the car to reach the tower from D is 3 seconds.

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:
Let the height of the tower be represented by AB in the figure.

⇒ h = ± 6 m
∴ h = 6 m [∵ Height is can be only positive]
Thus, the height of the tower is 6 m.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² – 3x + 5 = 0
(ii) 3x² – 4\(\sqrt{3} x\) + 4 = 0
(iii) 2x² – 6x + 3 = 0
Solution:
(i) Comparing the given quadratic equation with ax² + bx + c = 0, we get a = 2, b = -3, c = 5
b² – 4 ac = (-3)² – 4(2)(5) = 9 – 40 = -31 < 0
Since b² – 4ac is negative.
∴ The given quadratic equation has no real roots.

(ii) Comparing the given quadratic equation with ax² + bx + c = 0, we get

Since, b² – 4ac is zero.
∴ The given quadratic equation has two real roots which are equal. Hence, the roots are

(iii) Comparing the given quadratic equation with ax² + bx + c = 0, we get a = 2,b = -6,c = 3
∴ b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24 = 12 > 0
Since, b² – 4ac is positive.
∴ The given quadratic equation has two real and distinct roots, which are given by

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(k – 2) + 6 = 0
Solution:
(i) 2x² + kx + 3 = 0
Here, a = 2, b = k, c = 3
b² – 4ac = (k)² – 4(2) (3)
= k² – 24
∵ For a quadratic equation to have equal root, b² – 4ac = 0

(ii) kx(k – 2) + 6 = 0
Comparing kx² – 2kx + 6 = 0 with ax² + bx + c, we get
a = k, b = -2k, c = 6
∴  b² – 4ac = (-2k)² – 4(k)(6) = 4k² – 24k
Since, the roots are equal
∴  b² – 4ac ⇒ 4k² – 24k = 0.
⇒ 4k(k – 6) = 0
⇒ 4k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
But k cannot be 0, otherwise, the given equation is not quadratic. Thus, the required value of k is 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m² ? If so, find its length and breadth.
Solution:
Let the breadth be x m.
∴ Length = 2x m
Now, Area = Length × Breadth = 2x × x m² = 2x² m²
According to the given condition,
2x² = 800

Therefore, x = 20 or x = -20
But x = -20 is not possible
[ ∵ Breadth cannot be negative]
∴ x = 20 ⇒ 2x = 2 × 20 = 40
Thus, length = 40 m and breadth = 20 m

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the age of one of the friends be ‘x’.
Age of another friend is (20 – x).
4 years back age of 1st friend is (x – 4)
4 years back age of 2nd friend = (20 – x – 4) = (16 – x)
Product of their ages is 48.
∴ (x – 4) (16 – x) = 48
16x – x² – 64 + 4x = 48
-x² + 20x – 64 = 48
-x² + 20x – 64 – 48 = 0
-x² + 20x – 112 = 0
x² – 20x + 112 = 0
Here, a = 1, b = -20, c = 112
b² – 4ac = (-20)² – 4( 1)( 112) = 400 – 448 = – 48
Here, b² – 4ac = -48 < 0.
∴ It has no real roots.
∴ This situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Solution:
Let the breadth of the rectangle be x m.
Since, the perimeter of the rectangle = 80 m
∴ 2(Length + breadth) = 80
2(Length + x) = 80

⇒ Length = (40 – x) m
∴ Area = (40 – x) × x m² = (40x – x²) m²
According to the given condition,
Area of the rectangle = 400 m²
⇒ 40x – x² = 400
⇒ – x² + 40x – 400 = 0
⇒ x² – 40x +400 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0, we get
a = 1, b = -40, c = 400

Since, Length = Breadth
⇒ This rectangle is a square.
Thus, it is not possible to design a rectangular park of given perimeter and area.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.
Solution:
Since, diameter of the cylinder = 10 cm 10
∴ Radius of the cylinder (r) = \(\frac{10}{2}\) cm = 5cm
⇒ Length of wire in one round = 2πr
= 2 × 3.14 × 5 cm = 31.4 cm
∵ Diameter of wire = 3 mm = \(\frac{3}{10}\) cm
∴ The thickness of cylinder covered in one round = \(\frac{3}{10}\) cm
⇒ Number of rounds (turns) of the wire to cover 12 cm = \(\frac{12}{3 / 10}=12 \times \frac{10}{3}\) = 40
∴ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds
= l = 40 × 31.4 cm = 1256 cm
Now, radius of the wire = \(\frac{3}{2}\) mm = \(\frac{3}{20}\) cm
∴ Volume of wire = πr²l
= 3.14 × \(\frac{3}{20} \times \frac{3}{20}\) × 1256 cm³
∵ Density of wire = 8.88 g/cm³
∴ Mass of the wire = [Volume of the wire] × density

Question 2.
A right triangle, whose sides are 3cm and 4cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).
Solution:
Let us consider the right ABAC, right angled at A such that AB = 3 cm, AC = 4 cm
∴ Hypotenuse BC = \(\sqrt{3^{2}+4^{2}}\) = 5cm
Obviously, we have obtained two cones on the same base AA’ such that radius = DA or DA’.
Now,

Question 3.
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Solution:
Dimensions of the cistern are 150 cm, 120 cm and 100 cm.
∴ Volume of the cistern = 150 × 120 × 110 cm³ = 1980000 cm³
Volume of water contained in the cistern = 129600 cm³
∴ Free space (volume) which is not filled with water = (1980000 – 129600) cm³
= 1850400 cm³
Now, volume of one brick
= (22.5 × 7.5 × 6.5) cm³ = 1096.875 cm³
∴ Volume of water absorbed by one brick
= \(\frac{1}{17}\) × 1096.875 cm³
Let n bricks can be put in the cistern.
∴ Volume of water absorbed by n bricks
= \(\frac{n}{17}\) × 1096.875 cm³
∴ Volume occupied by n bricks = [free space in the cistern + volume of water absorbed by n bricks]
⇒ [n × 1096.875] = [1850400 + \(\frac{n}{17}\)(1096.875)]
⇒ 1096.875 n – \(\frac{n}{17}\)(1096.875) = 1850400
⇒ (n – \(\frac{n}{17}\)) × 1096.875 = 1850400
⇒ \(\frac{16}{17} n=\frac{1850400}{1096.875} \Rightarrow n=\frac{1850400}{1096.875} \times \frac{17}{16}\)
= 1792.4102 ≈ 1792
Thus, 1792 bricks can be put in the cistern.

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the norma water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Volume of three rivers = 3 {(Surface area of a river) × Depth}

Since 0.7236 km³ ≠ 9.728 km³
∴ The additional water in the three rivers is not equivalent to the rainfall.

Question 5.
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see figure).

Solution:
We have,
For the cylindrical part:
Diameter = 8 cm ⇒ Radius (r) = 4 cm
Height = 10 cm

Question 6.
Derive the formula for the curved surface area and total surface area of the frustum of a cone.
Solution:

Since, ∆OC₁Q ~ ∆OC₂S [By AA similarity]

Now, the total surface area of the frustum = (curved surface area) + (base surface area) + (top surface area)
= πl(r₁ + r₂) + πr₂² + πr₁² = π [(r₁ + r₂)l + r₁² + r₂²]

Question 7.
Derive the formula for the volume of the frustum of a cone.
Solution:
We have,
[Volume of the frustum RPQS] = [Volume of right circular cone OPQ] – [Volume of right circular cone ORS]

Since, ∆OC₁Q ~ ∆OC₂S [By AA similarity]

From (1) and (2), we have
Volume of the frustum RPQS