Category: Class 9

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:

1. The area of the sheet required for making the box.
2. The cost of sheet for it, if a sheet measuring 1m² costs ₹ 20.

Solution:
Given
l = 1.5 m = 150 cm
b = 1.25 m = 125 cm
h = 65 cm

1. Total area of plastic sheet required = LSA + Area of base
= 2h(l + b) + l x b
= 2 x 65 (150 + 125) + 150 x 125
= 130 (275) + 18750
= 35750 + 18750
= 54500 cm²
= 5.45 m²

2. Cost of sheet = area of plastic sheet x rate = 5.45 x 20
= 109.00
= ₹ 109.

Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m².
Solution:
Given
l = 5m
b = 4m
h = 3 m
Area of the room to be white washed = LSA + Area of ceiling
= 2h(l + b) + l x b
= 2 x 3 (5 + 4) + 5 x 4
= 6(9) + 20
= 54 + 20 = 74 m²
Cost of painting = 7.50 x 74 m²
= ₹ 555

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15000, find the height of the hall.
Solution:
Given
P = 250 m
Rate of painting = ₹ 10/ m²
Cost of painting = ₹ 15000
Area of walls to be painted = \(\frac{15000}{10}\) = 1500 m²
Perimeter = 2 (l + b) = 250
∴ l + b = 250/2 = 125m
Area of walls = LSA = 2h (l + b) = 1500
= h (l + b) = \(\frac{1500}{10}\) = 750 m² …..(i)
Putting the value of (l + b) in (i), we get
h(125) = 750
h = \(\frac{750}{125}\) = 6 m.

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9315 m². How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?
Solution:
Given
l = 22.5 cm
h = 7.5 cm = 0.075 m
b = 10 cm = 0.1 m
Area to be painted = 9.375 m² = 93750 cm²
Area of one brick = 2(lb + bh + hl)
= 2(22.5 x 10 + 10 x 7.5 + 7.5 x 22.5)
= 2(225 + 75 + 168.75)
= 2 x 468.75
= 937.5 cm²
No. of bricks which can be painted = \(\frac{93750}{937.5}\) = 100

Question 5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

1. Which box has the greater lateral surface area and by how much?
2. Which box has the smaller total surface area and by how much? Sol.

Solution:
1. LSA of cubical box = 4a²
= 4 x 10 x 10 = 400 cm²
LSA of cuboidal box = 2h (l + b)
= 2 x 8 (12.5 x 10)
= 16 x 22.5
= 360 cm²
LSA of cubical box is more than cuboidal box by 40 cm².

2. TSA of cubical box = 6a² = 6 x 10 x 10
= 600 cm²
TSA of cuboidal box = 2 (lb + bh + hl)
= 2(12.5 x 10 + 10 x 8 + 12.5 x 8)
= 2(125 + 80 + 100)
= 2 x 305 = 610 cm².
∴ TSA of cuboidal box is more than cubical box by 10 cm².

Question 6.
A small indoor green house (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

1. What is the area of the glass?
2. How much of tape is needed for all the 12 edges?

Solution:

Given
l = 30 cm
b = 25 cm and
h = 25 cm
Area of glass required = 2(30 x 25 + 25 x 25 x 30)
= 2(750 + 625 + 750)
= 2 x 2125 = 4250 cm²
Length of tape required = 4(l + b + h)
= 4(30 + 25 + 25)
= 4 x 80 = 320 cm

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹ 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
Big Box :
l = 25 cm
b = 20 cm
h = 5 cm
Small Box:
l = 15 cm
b = 12 cm
h = 5 cm
Number of boxes required = 250
Rate = ₹ 4 per 1000 cm²
TSA of bigger box = 2 (25 x 20 + 20 x 5 + 5 x 25)
= 2(500 + 100 + 125)
= 2 x 725 = 1450 cm²
TSA of smaller box = 2 (15 x 12 +12 x 5 + 5 x 15)
= 2 (180 + 60 + 75)
= 2 x 315 = 630 cm²
Area of cardboard required one bigger = TSA + 5% of TSA
= 1450 + \(\frac{5}{100}\) x 1450
= 1522.5 cm²
Area of cardboard required for 250 boxes of bigger size = 250 x 1522.5
= 380625 cm²
Area of cardboard required for one small box = 630 + \(\frac{5}{100}\) x 630
= 661.5 cm²
Total area of cardboard required for 250 boxes of smaller size = 250 x 661.5
= 165375 cm²
Total area = (165375 + 380625) cm²
= 546000 cm²
Total cost of each kind of cardboard = \(\frac{4}{1000}\) x 546000
= ₹ 2184/-

Question 8.
Parveen wanted to make a temporary shelter for her car, by making a box – like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rollpd up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?
Solution:
Given
l = 4m
b = 3 m and
h = 2.5 m
Area of the tarpaulin required = 2h(l + b) + l x b
= 2 x 2.5(4 + 3) + 4 x 3
= 5 x 7 + 12
= 35 + 12= 47m

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 12 Heron’s Formula Ex 12.2

Question 1.
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD, = 5 m and AD = 8 m. How much area does it. occupy?
Solution:
In ∆DCB
DB² = DC² + BC²

In ∆DBA P = (8 + 9 + 13)m = 30m
s = \(\frac{P}{2}\) = 15 m
s – a = 15 – 13 = 2
s – b = 15 – 9 = 6
s – c = 15 – 8 = 7
Area of ∆DBA = \(\sqrt{15x 2x6x7}\)
\(\sqrt{3x5x2x2x3x7}\)
= 2 x 3\(\sqrt{5×7}\) x 7 = 6\(\sqrt{35}\)m²
Area of quadrilateral ABCD = (30 + 6\(\sqrt{35}\)) m²
= 30 + 6 x 5.91
= 30 + 35.46
= 65.46 m2

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
In ∆ABC

P = 3 + 4 + 5 = 12cm
s = \(\frac{12}{2}\) = 6 cm
s – a = 6 – 5 = 1 cm
s – b = 6 – 4 = 2cm
s – c = 6 – 3 = 3 cm
Area of ∆ABC = \(\sqrt{6x1x2x3}\)
= \(\sqrt{2x3x2x3}\)
= 2 x 3 = 6 cm²
In ∆ADC P = 5 + 5 + 4 = 14 cm
s = \(\frac{P}{2}\) = 7 cm
s – a = 7 – 5 = 2cm
s – b = 7 – 5 = 2cm
s – c = 7 – 4 = 3cm
Area of ∆ADC = \(\sqrt{7x2x2x3}\) = 2\(\sqrt{21}\) cm²
Area of ∆BCD = Area of ∆ABC + Area of ∆ADC
= (6 + 2\(\sqrt{21}\)) cm²
= (6 + 2 x 4.58)
= 15.16 cm²

Question 3.
Radha made a picture of an aeroplane with coloured paper is shown in Fig. Find the total area of the paper used.
Solution:
The figure is divided into five parts as shown in Fig.

• Part I – Triangle having sides 5 cm, 5 cm and 1 cm
• Part II – Rectangle having sides 1 cm and 6.5 cm
• Part III – Trapezium having sides 2,1,1,1.
• Part IV and V. Right angled triangles having sides 6 cm and 1.5 cm.

For Part I:
a = 5 cm
b = 5 cm
c = 1 cm
s = \(\frac{P}{2}\) = \(\frac{5+5+1}{2}\) = \(\frac{11}{2}\) = 5.5 cm
s – a = 5.5 – 5 = 0.5
s – b = 5.5 – 5 = 0.5
s – c = 5.5 – 1 = 4.5

Area of triangle IV and V = \(\frac{1}{2}\) x 1.5 x 6 = 4.5 cm²
∴ Area of paper required = Area of part I + Area of part II + Area of part III + Area of part IV + Area of part V

= 2.49 + 1.27 + 15.5
= 19.26 cm²

Question 4.
A triangle and a parallelogram have the same base and the &me area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:

In ∆ABE
a = 30 cm
b = 28 cm
c = 26 cm
s = \(\frac{P}{2}\) = \(\frac{30+28+26}{2}\) = 42 cm
s – a = 42 – 30 = 12
s – b = 42 – 28 = 14
s – c = 42 – 26 = 16
Area of ∆ABE = \(\sqrt{42x12x14x16}\)
= \(\sqrt{2x3x7x2x2x3x2x7x4x4}\)
= 2 x 3 x 7 x 2 x 4
= 336 cm²

Area of parallelogram ABCD = area of ∆ABE = 336 cm² (given)
Area of parallelogram = b x h
336 = 28 x h
⇒ \(\frac{336}{28}\)
h= 12 cm.

Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Solution:
We know that the diagonal of a rhombus divide it into two triangles of equal area.

a = 48m
b = 30m
c = 30m
s = \(\frac{P}{2}\) = \(\frac{48+30+30}{2}\) = \(\frac{108}{2}\) = 54
s – a = 54 – 48 = 6 cm
s – b = 54 – 30 = 24 cm
s – c = 54 – 30 = 24 cm
Area of ∆ABD
= \(\sqrt{54x6x24x24}\)
= \(\sqrt{2x3x3x3x2x3x2x2x2x3x2x2x2x3}\)
= 2 x 3 x 3 x 2 x 2 x 3 x 2 = 432 cm²
Area of rhombus = 2 x 432 = 864 cm2
Area of grass field for each cow = \(\frac{864}{18}\) = 48 cm²

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.), each piece measurig 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umberella?
Solution:

a = 50 cm
b = 50cm
c = 20 cm
s = \(\frac{P}{2}\) = \(\frac{50+50+20}{18}\) = 60 cm
s – a = 60 – 50 = 10
s – b = 60 – 50 = 10
s – c = 60 – 20 = 40
Area of a ∆ = \(\sqrt{60x10x10x40}\)
= \(\sqrt{2x3x10x10x10x2x2x10}\)
= 10 x 10 x 52\(\sqrt{6}\)
= 200\(\sqrt{6}\) cm²
Area of cloth of each type = 200\(\sqrt{6}\) x 5 = 1000\(\sqrt{6}\) cm²

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. How much paper of each shade has been used in it?
Solution:

∠AOB = 90°
BD = AC = 32 cm
OA = OC = 16 cm
Area (ABD) = area (DBC) = \(\frac{1}{2}\) x 32 x 16 = 256 cm²
In ∆CEF
a = 6 cm
b = 6 cm
c = 8 cm
s = \(\frac{P}{2}\) = \(\frac{6+6+8}{2}\) = 10
s – a = 10 – 6 = 4
s – b = 10 – 6 = 4
s – c = 10 – 8 = 2

Area of ∆CEF = \(\sqrt{10x4x4x2}\)
= \(\sqrt{5x2x4x4x2}\)
= 8\(\sqrt{5}\) cm²
= 8 x 2.24 = 17.92 cm²

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig.). Find the cost of polishing the tiles at the rate of the field.
Solution:

a = 35 cm
b = 28 cm
c = 9 cm
s = \(\frac{P}{2}\) = \(\frac{35+28+9}{2}\)
s = a = 36 – 35 = 1
5 = b = 36 – 28 = 8
c = c = 36 – 9 = 27
Area of a tile = \(\sqrt{36x1x8x27}\)
= \(\sqrt{2x2x3x3x2x2x2x3x3x3}\)
= 2 x 3 x 2 x 3\(\sqrt{6}\) = 36\(\sqrt{6}\) cm²
= 36 x 2.45 = 88.2 cm²
Total area of tiles = 16 x 88.2 = 1411.2 cm²
Cost of polishing the tiles per cm² = 50 P
Total cost of polishing the tiles = 1411.2 x 50 P
= ₹ \(\frac{1411.2×50}{100}\) = ₹ 705.60

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
ABCD is a trapezium. Draw BE parallel to AD. Draw BF ⊥ DC. ABED is a parallelogram
AB = DE = 10 m and AD = BE = 14 m
EC = DC – DE = 25 – 10 = 15 m
In ∆BEC
a = 15m
b = 14m
c = 13m
s = \(\frac{15+14+13}{2}\) = 21

s – a = 21 – 15 = 6
s – b = 21 – 14 = 7
s – c = 21 x 13 = 8

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 12 Heron’s Formula Ex 12.1

Question 1.
A traffic signal borard, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’ Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution:

Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig.) the advertisements yield an earning of ₹ 5000/m² per year. Acompany hired one of its walls for 3 months. How much rents did it pay?

Solution:

s = \(\frac{122+120+22}{2}\) = \(\frac{264}{2}\) = 132 m
s – a = 132 – 122 = 10 m
s – b = 132 – 22 = 12m
s – c = 132 – 22 = 110 m
area of triangular portion of wall = \(\sqrt{32x10x12x110}\)
= \(\sqrt{2x2x3x11x10x2x2x3x11x10}\)
= 10 x 2 x 2 x 3 x 11 = 1320 m²
Rate = ₹ 5000/m² per year
Rent for 3 months = 1320 x \(\frac{5000×3}{12}\)
= 330 x 5000
= ₹ 16,50,000

Question 3.
There is a slide in a park. One of its side walls has been painted in some colour with a message “Keep The Park Green And Clean” (see Fig.). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Solution:
a = 15 m
b = 11 m
c = 6m
p = a + b + c
= 15 + 11 + 6
= 32 m P

s = \(\frac{P}{2}\) = \(\frac{32}{2}\) = 16m
s – a = 16 – 15 = 1 m
s – b = 16 – 11 = 5 m
s – c = 16 – 6 = 10m
Area of triangular park = \(\sqrt{16x1x5x10}\)
= \(\sqrt{2x2x2x2x1x5x2x5}\)
= 2 x 2 x 5√2 = 2o\(\sqrt{2m^{2}}\)

Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:

a = 18 cm
b = 10 cm
Let the third side be c
p = 42 cm
18 + 10 + C = 42
C = 14
S = \(\frac{P}{2}\) = \(\frac{42}{2}\) = 21 cm
s – a = 21 – 18 = 3
s – b = 21 – 10 = 11
s – c = 21 – 14 = 7
Area of ∆ = \(\sqrt{21x3x11x7}\) = \(\sqrt{3x7x3x11x7}\)
= 3 x 7\(\sqrt{11}\) = 21\(\sqrt{11}\) cm²

Question 5.
Sides of triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area
Solution:
a = 12x
b = 17x
c = 25x
p = 12x + 11x + 25x = 540
54x = 540
x = \(\frac{540}{54}\) = 10
a = 12 x 10 = 120 cm
b = 17 x 10 = 170 cm
c = 25 x 10 = 250 cm
s = \(\frac{P}{2}\) = \(\frac{540}{2}\) = 270 cm
s – a = 270 – 120 = 150 cm
s – b = 270 – 170 = 100 cm
s – c = 270 – 250 = 20 cm
Area of ∆ = \(\sqrt{270x150x100x20}\)
= \(\sqrt{3x3x3x10x3x5x10x10x10x2x10}\)
= 3 x 3 x 10 x 10 x 10
= 9000 cm²

Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:

a = 12 cm
b = 12cm
Let the third side be c.
p = a + b + c
30 = 12 + 12 + c
c = 30 – 24 = 6 cm
s = \(\frac{P}{2}\) = \(\frac{30}{2}\) = 15
s – a = 15 – 12 = 3
s – b = 15 – 12 = 3
s – c = 15 – 6 = 9
Area of ∆ = \(\sqrt{5x3x3x9}\)
= \(\sqrt{5x3x3x3x3x3}\)
= 3 x 3\(\sqrt{15}\) = 9\(\sqrt{15}\) cm²

Area of Quadrilaterals:
To find the area of quadrilaterals divide the quadrilateral into two triangles using a diagonal and then use heron’s formula.

Example 1:
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, And the height of the parallelogram.
Solution:

Here, a = 13 cm,
b = 14 cm,
c = 15 cm
Base of parallelogram =14 cm.

= 7 x 3 x 2 x 2
= 84 cm²
Let h be the height of the parallelogram ADEC.
Area of parallelogram ADEC = Area of ∆ABC (given)
Base x height = 84
14 x height = 84
h = \(\frac{84}{14}\)
= 6 cm.

Example 2:
The sides of a quadrilateral, taken in order are 5, 12, 14 and 15 meters respectively, and the angle contained by the first two sides is a right angle. Find its area.
Solution:

Here, AB = 5m
BC = 12m
CD = 14m
DA = 15 m
Join AC. The ABCD is divided into two triangles ABC and ACD. The area of the quadrilateral is equal to sum of areas of ∆ABC and ∆ADC.
In ∆ABC

area of ABCD
= ar (∆ABC) + ar (∆ACD) –
= (84 + 30)m² = 114m²

Example 3:
In a parallelogram measure of adjacent sides are 34 cm and 20 cm. One of the diagonals is 42 cm. Find the area of the parallelogram.
Solution: In Fig.

AB = DC = 34 cm
AD = BC = 20 cm
AC =42 cm.
We know that the diagonal of a parallelogram divides it into two tri¬angles of equal area.
∴ Area of parallelogram ABCD = 2 x Area of (∆ABC)
Consider ∆ABC
a = 34cm
b = 20cm
c = 42cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{34+20+42}{2}\)
= \(\frac{96}{2}\) = 48 cm
s – a = 48 – 34 = 14 cm
s – b = 48 – 20 = 28 cm
s – c = 48 – 42 = 6 cm
By Heron’s formula,

= 2 x 2 x 6 x 14
= 336 cm²
∴ Area of (∥^gm ABCD) = 2 x 336
= 672 cm²

Example 4:
A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of? 5 per m2. Find the cost of painting.
Solution:

Given P = 32m
BD = 10m
Rate of painting = ₹ 5/m²
Let the sides of rhombus be x m.
P = Ax
⇒ 32 = 4x
∴ x = 8m
So, AB = BC = CD = DA = 8 m
We know that the diagonal of a rhombus divides it into two triangles of equal area. .
∴ Area of rhombus ABCD = 2 x area of ∆ABD
Consider ∆ABD
a = 8m
b = 8m
c = 10m
S = \(\frac{a+b+c}{2}\) = \(\frac{8+8+10}{2}\)
= 13 m.
s – a = 13 – 8 = 5 m
s – b = 13 – 8 = 5m
s – c = 13 -10 = 3 m
By Heron’s formula,
Area of ∆ABD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{3x5x5x3}\)
= 5 x \(\sqrt{39}\)
= 5 x 6.24 = 31.2 m²
Area of rhombus ABCD = 2 x 31.2 = 62.4 m²
Area of rhombus to be painted = 2 x area of rhombus (∴ Painting is to be done on both sides)
= 2 x 62.4 = 124.80
Cost of painting = Rate x Area
= 5 x 124.80
= ₹ 624.

Example 5:
Two parallel sides of a trapezium are 60 cm and 77 cm other sides are 25 cm and 26 cm. Find the area of trapezium.
Solution:

Given AB = 60 cm
DC = 77 cm
AD = 25 cm
BC =26 cm
Draw a line BE ∥ AD from point B.
In ABED
AB ∥ DE
AD ∥ BE
ABED is a parallelogram.
AB = DE = 60
EC = 77 – 60 = 17 cm.
AD = BE = 25 cm
In ∆BEC
a = EC = 17 cm
b = BE = 25 cm
c = BC = 26cm
s = \(\frac{17+25+36}{2}\)
s – a = 34 – 17 = 17 cm
s – b = 34 – 25 = 9 cm
s – c = 34 – 26 = 8cm
By Heron’s formula,

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:

1. Draw a ray OA and with O as centre and any radius, draw an arc, cutting OA at B.
2. With B as centre and the same radius draw an arc cutting the arc drawn in step (i) at C.
3. With C as centre draw another arc with same radius cutting the arc drawn in step (i) at D.
4. with C as centre and the same radius draw an arc.
5. With D as centre and the same radius draw an arc, cutting the arc drawn in step (iv) at E.
6. Draw OE ∴ ∠AOF = 90°

Question 2.
Construct an angle of 45° at the initial point of a given ray justify the construction.
Solution:

1. Draw ∠AOF = 90° by following the same steps for constructing a 90° angle.
2. Draw OG, the bisector of ∠AOF.
3. ∠AOF= 45°

Question 3.
Construct the angle of the following measurements:
Solution:

1. 30°
2. 22 \(\frac{1}{2}\)°
3. 15°

1. 30°

1. Draw a ray OA.
2. With O as centre and any radius draw an arc which intersect OA at B.
3. With B as centre and same radius draw an arc cutting the arc drawn is step (ii) at C.
4. Join OC and produce upward
5. ∠BOC = 60°
6. Draw the bisector OD of ABOC.
7. ∠BOD = ∠COD = 30°

2. 22 \(\frac{1}{2}\)°

1. Draw a ray OA.
2. With O as centre and any radius draw an arc which intersect intersect CM at B.
3. With B as centre and same radius draw an arc intersecting the arc drawn in step (ii) at C.
4. With C as centre and same radius draw an arc cutting the arc drawn in step (ii) at D.
   
5. With C and D as centres and same radius draw arcs to intersect at E.
6. Join OE.
7. ∠AOG – 90°
8. Draw the bisector OE of ∠AOE, to get ∠AOF = 45°
9. Draw the bisector OG of ∠AOF.
10. ∠AOG = 22\(\frac{1}{2}\)°

3. 15°

1. Draw a ray CM.
2. With O as centre and any radius draw an arc which intersect CM at B.
3. With B as centre and same radius draw an arc intersecting the arc drawn in step (ii), at C.
4. Draw OD as the bisector of ZAOC.
5. ∠BOD = 30°
6. Draw the bisector OE of ∠AOD.
7. ∠AOE = 15°

Question 4.
Construct the following angles and verily by measuring them by a protractor.

(i) 75°
(ii) 105°
(iii) 135°

Solution:

Question 5.
Construct an equilateral triangle given its side and justify the construction.
Solution:

1. Draw a ray AY with intial point A.
2. With centred and radius equal to length of a side of the A draw an arc BY, cutting the ray AX at B.
3. With centre B and the same radius draw an arc cutting are BY at C.
4. Join AC and BC to obtain the required A.

Construction of Triangles:

Example 1:
Construction of a Triangle when its base, a base angle and sum of other two sides are given
Solution:
Step of Construction:

1. Draw the base PQ.
2. At point P draw an angle, MPQ equal to the given angle.
3. Cut a line segment PM equal to sum of sides i.e., (PR + RQ) from point P.
4. Join MQ.
5. Draw the perpendicular bisector of MQ which intersect PM at R.
6. Join QR. PQR is the required triangle.

Justification:
Mark points S as shown in Fig.
QS = MS (∴ RS is the perpendicular bisector of MQ)
RS = RS (Common)
∠QSR = ∠MSR (Each \({ 90 }^{ \underline { 0 } }\))
∆RSQ ≅ ∆RSQ (By SAS)
and so PQ = RM (By CPCT)
Now PR = PM – RM = PM – RQ
PM = PR + RQ.

Example 2:
Construct a AABC in which BC = 3.6 cm, AB + AC = 4.8 and A RSM QS – MS RS = RS cm and B = ∠60°.
Solution:
Steps of Construction:

1. Draw BC = 3.6 cm.
2. Draw ∠CBX= \({ 60 }^{ \underline { 0 } }\) at B.
3. From BX, cut off line segment BD = 4.8 cm.
4. Join DC.
5. Draw the perpcndicub: bisector ofDC meeting BD at A.
6. joinAC.
7. ABC is the required Mangle.

Justification:
‘A‘ lies on the perpendicular bisector of DC
∴ AD = AC
Now BD = 4.8 cm
⇒ BA + AD = 4.8
BA + AC = 4.8 (∴ AD = AC)

Example 3:
Construction of a Triangle when its Base Angle and the Difference of the other two sides are given.
Solution:
Case I:
Given the base BC, a base angle, say ∠B and AB – AC. Steps of Construction:

1. Draw the base BC.
2. At point B draw an angle ∠CBX equal to the given angle.
3. Cut a line segment BD = AB – AC from point B.
4. Join DC.
5. Draw the perpendicular bisector of . DC which intersect BX at A.
6. Join AC.
7. ΔABC is the required triangle.

Justification:
As point A lies on the perpendicular bisector of DC
∴ AD = AC
BD = AB – AD = AB – AC (∴ AD = AC)

Case II:
Given the base BC, a base angle say ∠B and (AC – AB).
Steps of Construction:

1. Draw the base SC.
2. At point S, draw an angle, ∠CBX equal to the given angle and extend the arm XB backward.
3. Cut a line segment BD equal to (AC – AB) from the extended arm.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersect BX at A.
6. Join AC.
7. ΔABC is the required triangle.

Justification:
As the perpendicular bisector of DC passes through A.
∴ AD – AC
BD = AD – AB
∴ BD = AC – AB

Example 4:
Construct a ∆ABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw base BC = 3.4 cm.
2. Draw ∠CBX = \({ 45 }^{ \underline { 0 } }\).
3. From BX, cut line segment BD =1.5 cm.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersect BX at A.
6. Join AC.
7. ABC is the required triangle.

Justification:
As A lies on perpendicular bisector of DC.
AD = AC
BD = 1.5 cm
⇒ AB – AD =1.5 cm
AB – AC =1.5 cm

Example 5:
Construct a ∆PQR in which QR = 5.8 cm, PR – PQ = 1.8 cm and ∠Q = \({ 45 }^{ \underline { 0 } }\). Justify your construction.
Solution:
Steps of Construction:

1. Draw base QR = 5.8 cm.
2. Draw ∠RQP = \({ 45 }^{ \underline { 0 } }\).
3. Produce arm XQ backward and cut a line segment QS = 1.8 cm.
4. Join SR.
5. bisector of SR which intersect QX at P.
6. Join PR.
7. PQR is the required triangle.

Justification:
As P lies on perpendicular bisector of SR
PS = PR
QS = 1.8 cm
⇒ PS – PQ = 1.8 cm
∴ PR – PQ = 1.8 cm

Example 6:
Construct a ∆ABC in which BC = 5.8 cm, ∠C = \({ 60 }^{ \underline { 0 } }\) and AB – AC = 2.5 cm.
Solution:
Steps of Construction:

1. Draw base BC = 5.8 cm
2. Draw ∠ACB = \({ 60 }^{ \underline { 0 } }\)
3. Produce arm CX backward and cut a line segment CD = 2.5 cm.
4. Join BD.
5. Draw perpendicular bisector of BD which intersect CX at A.
6. Join AB.
7. ABC is the required triangle.

Justification:
A lies on the perpendicular bisector of BD.
∴ AB =AD
CD = 2.5 cm
⇒ AD – AC = 2.5 cm
⇒ AB – AC = 2.5 cm

Example 7:
Construct a ∆ABC in which AC – AB = 3.5 cm, BC = 6.2 cm and ∠C = \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw BC = 6.2 cm
2. Draw ∠ACB = \({ 45 }^{ \underline { 0 } }\)
3. Cut CD = 3.5 cm from CX.
4. Join BD and draw the perpendicular bisector of BD which intersect CX at A.
5. Join AB.
6. ABC is the required triangle.

Justification:
As A lies on the perpendicular bisector of BD.
∴ AB = AD
CD = 3.5 cm
⇒ AC – AD = 3.5 cm
∴ AC – AB = 3.5 cm

Example 8:
Construction of a Triangle when its Perimeter and Two Base Angles are given.
Solution:
Given the base angles, ∠B and ∠C and perimeter, i.e., AB + BC + AC.
Steps of Construction:

1. Draw a line segment, DE equal to AB + BC + CA.
2. Draw ∠EDM and ∠DEN equal to the base angles ∠B and ∠C respectively.
3. Draw bisectors of ∠MDE and ∠NED which intersect at A.
4. Draw the perpendicular bisectors of AD and AE which intersect DE at B and C respectively.
5. Join AE and AC.
6. ABC is the required triangle.

Justification:
As B lies on the perpendicular bisector of AD
∴ AB = DB
In ∆ADB
AB = DB
∠ADB = ∠DAB
(∴ Angles opposite to equal sides of a A are equal)
Similarly, AC = CE
and ∠CAE = ∠CEA
Now DE = DB + BC + CE
= AB + BC + AC
In ∆ABD, ∠ABC = ∠DAB + ∠ADB = ∠ADB + ∠ADB = 2∠ADB
In ∆ACE, ∠ACB = ∠CAE + ∠CEA = ∠CEA + ∠CEA = 2∠CEA

Example 9:
Construct a triangle whose perimeter is 6.4 cm and angles at the base are \({ 60 }^{ \underline { 0 } }\) and \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw line segment DE equal to (AB + BC + CA) = 6.4 cm
2. Draw ∠EDM = \({ 60 }^{ \underline { 0 } }\) and ∠DEN = \({ 45 }^{ \underline { 0 } }\).
3. Draw AD and AE as bisectors of ∠MDE and ∠NED which inter sect at A.
4. Draw the perpendicular bisector of AD and AE which intersect DE at B and C respectively.
5. Join AB and AC.
6. ABC is the required triangle.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 11 Work and Energy

Work and Energy Intext Questions

Work and Energy Intext Questions Page No. 148

Question 1.
A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. below). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer:
The work done W on the body by the force is given by:
Work done = Force × Displacement
W = F × s
Given:
F = 7 N
s = 8 m
Hence, work done, W = 7 × 8
= 56 Nm
= 56 J.

Work and Energy Intext Questions Page No. 149

Question 1.
When do we say that work is done?
Answer:
We can say a work is done whenever the conditions given below are satisfied:

1. A force is applied over the body.
2. A displacement of the body is caused by the applied force, along the direction of the applied force.

Question 2.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
When a force ‘F’ displaces a body by a distance ‘d’ in the direction of the applied force, then the work done ‘W’ on the body is given by:
Work done = Force × Displacement
W = F × d.

Question 3.
Define 1 J of work.
Answer:
1 J is the amount of work done when an object is provided with a force of 1 N that displaces it through a distance of 1 m in the direction of the applied force.

Question 4.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Answer:
Here,
Applied force, F = 140 N
Displacement, d = 15 m
We know,
Work done is given by the expression:
Work done = Force × Displacement
W = F × d
So,
W= 140 × 15 = 2100 J
Hence, 2100 J of work is done in ploughing the length of the field.

Work and Energy Intext Questions Page No. 152

Question 1.
What is the kinetic energy of an object?
Answer:
The energy attained by or generated in a body due to its action or motion is called kinetic energy. Every object which possesses motion contain a kinetic energy. A body uses kinetic energy to do work. Kinetic energy can be used for any work to be performed. Kinetic energy is useful to generate other forms of energy too. It is expressed by KE and can be calculated by the following formula:
E_k = \(\frac { 1 }{ 2 }\) mv²
Here, m represents mass of object.
And ‘V’ gives the velocity by which object is shifting or working.

Question 2.
Write an expression for the kinetic energy of an object.
Answer:
Energy E_k is proportional to:

• Object’s mass and
• Square of its velocity.

Energy E_k due to a moving object with a body mass ‘m’ which is moving with a velocity v, can be given by the expression,
E_k = \(\frac { 1 }{ 2 }\) mv²
Its S.I. unit is joule (J).

Question 3.
The kinetic energy of an object of mass, m moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
Given:
K.E. of the object = 25 J
Velocity of the object, v = 5 m/s
Putting value to formula:
∵ K.E = \(\frac { 1 }{ 2 }\) mv²

• m = 2 × K.E / v²
• m = 2 × \(\frac { 25 }{ 25 }\) = 2 kg

Condition 1:
If velocity is double, v = 2 × 5 = 10 m/s
∴ K.E. (for v = 10 m/s) = \(\frac { 1 }{ 2 }\) mv² = \(\frac { 1 }{ 2 }\) × 2 × 100 = 100 J

Condition 2:
If velocity is tripled, v = 3 × 5 = 15 m/s
∴ K.E. (for v = 10 m/s) = \(\frac { 1 }{ 2 }\) mv² = \(\frac { 1 }{ 2 }\) × 2 × 225 = 225 J.

Work and Energy Intext Questions Page No. 156

Question 1.
What is Power?
Answer:
Work done is calculated by the amount of power consumption. Power can be understood by the term efficiency of an object to consume or generate energy. So, power is the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by the expression,
Power = \(\frac { Work }{ Time }\) = \(\frac { Energy }{ Time }\)
or
P = \(\frac { W }{ T }\)
It is calculated in watt (W).

Question 2.
Define 1 watt of power.
Answer:
As we know that:
Power = \(\frac { Work }{ Time }\)
Hence,
A body is said to have power of 1 watt if its work is equal to 1 joule in 1 s, i.e.,
1 W = \(\frac { 1J }{ 1s }\)

Question 3.
A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer:
Power = \(\frac { Work }{ Time }\)
Given,
Work done = Energy consumed by the lamp = 1000 J
Time = 10 s
Putting values,
Power = \(\frac { 1000 }{ 10 }\) = 100 Js^-1
= 100 W.

Question 4.
Define average power.
Answer:
When efficiency of an operator is changed with time, average power is calculated.
The average power of an object is defined as the total work done by it in the total time taken.
Total time taken

Work and Energy NCERT Textbook Exercises

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

• Suma is swimming in a pond.
• A donkey is carrying a load on its back.
• A wind – mill is lifting water from a well.
• A green plant is carrying out photosynthesis.
• An engine is pulling a train.
• Food grains are getting dried in the sun.
• A sailboat is moving due to wind energy.

Answer:
Work is done whenever the two given conditions are satisfied:

• A force is applied over the body.
• A displacement of the body is caused by the applied force, along the direction of the applied force.

Hence, work is done in case:

• Suma is swimming in a pond.
• A wind – mill is lifting water from a well.
• An engine is pulling a train.
• A sailboat is moving due to wind energy.

Work is not being done in case:

• A donkey is carrying a load on its back.
• A green plant is carrying out photosynthesis.
• Food grains are getting dried in the sun.

Explanation:

1. Suma applies a force to push the water backwards which causes a displacement. Hence, work is done by Suma while swimming.
2. While carrying a load, the donkey has to apply a force in the upward direction. But, displacement is exchanged with shifting, so the work done is zero.
3. A wind – mill works against the gravitational force to lift water. Hence, work is done.
4. In this case, chemical change occurs not a physical. Therefore, the work done is zero.
5. An engine applies force to pull the train. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train.
6. This is an example of evaporation. Hence, the work done is zero.
7. Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

Question 2.
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Gravitational forces are proportional to ‘h’ which is vertical displacement and work done by the force of gravity is considered only if vertical displacement occurs. Vertical displacement is given by the difference in the initial and final positions / heights of the object which is zero. In this case work done by gravity is given by thy expression,
W = mgh
Where,
h = Vertical displacement = 0
W = mg × 0 = 0 J
Therefore, the work done by gravity on the given object is zero joule.

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
When a battery lights a bulb, then the chemical energy of the battery is converted into electrical energy. When the bulb receives this electrical energy, then it converts it into light and heat energy. Hence, the transformation of energy is as follows:
Chemical Energy → Electrical Energy → Light Energy + Heat Energy

Question 4.
Certain force acting on a 20 kg mass changes its velocity from 5 ms^-1 to 2 ms^-1. Calculate the work done by the force.
Answer:
Kinetic energy is given by the expression, (E_k) = \(\frac { 1 }{ 2 }\) mv².
Where,
E_k = Kinetic energy of the object moving with a velocity,
v Kinetic energy when the object was moving with a velocity 5 ms^-1.
(E_k)₅ = \(\frac { 1 }{ 2 }\) × 20 × (5)² = 250 J
Kinetic energy when the object was moving with a velocity 2 ms^-1.
(E_k)₂ = \(\frac { 1 }{ 2 }\) × 20 × (2)² = 40 J

Question 5.
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body.
Therefore, by the expression,
W = mgh
Here,
Vertical displacement, h = 0
W = mg × 0 = 0
Hence, the work done by gravity on the body is zero.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
No. In freely falling object only potential energy decreases progressively, but at same time kinetic energy increases and total of both remains equal to initial energy.
Total energy = Potential energy + Kinetic energy
So, this process does not violate the law of conservation of energy. During the process, total mechanical energy of the body remains equal.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:
When we ride a bicycle, the chemical energy of muscles of rider’s body gets transferred into heat energy and kinetic energy of the bicycle. Heat energy is changed to physical energy. Kinetic energy provides a velocity to the bicycle. The transformation can be shown as:
Muscular energy → Kinetic energy + Heat energy
During the transformation, the total energy remains conserved.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
When we push a huge rock, there is no transfer of muscular energy to the stationary rock. Here, the energy is completely spent doing work (pushing) against friction between the ground and the rock.

Question 9.
A certain, household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer:
1 unit of energy is equal to 1 kilowatt hour (kWh).
1 unit = 1 kWh
1 k Wh = 3.6 × 10⁶ J
Therefore, 250 units of energy
= 250 × 3.6 × 10⁶ = 9 × 10⁸ J.

Question 10.
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fell, find its kinetic energy when it is half – way down.
Answer:
Gravitational potential energy is given by the expression,
W = mgh
Where,
h = Vertical displacement = 5 m
m = Mass of the object = 40 kg
g = Acceleration due to gravity = 9.8 ms^-2
∴ W = 40 × 5 × 9.8 = 1960 J.
At half – way down, the potential energy of the object will be \(\frac { 1960 }{ 2 }\) = 980 J.
At this point, the object has an equal amount of potential and kinetic energy.
This is due to the law of conservation of energy.
Hence, half – way down, the kinetic energy of the object will be 980 J.

Question 11.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:
Work is done whenever the two given conditions are satisfied:

1. A force acts on the body.
2. There is a displacement of the body by the application of force in or opposite to the direction of force.

If the direction of force is perpendicular to displacement, then the work done is zero. When a satellite moves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Hence, the work done on the satellite by the Earth is zero.

Question 12.
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer:
Yes. For a uniformly moving object, suppose an object is moving with constant velocity. The net force acting on it is zero. But, there is a displacement along the motion of the object. Hence, there can be a displacement without a force.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
When a person holds a bundle of hay over his head, then there is no displacement in the bundle of hay. And since
displacement of the body by the application of force is required to prove that work is done, no work is done here.
Here, force of gravity is acting on the bundle, but the person 1 is not applying any force on it. Hence, in the absence of force, work done by the person on the bundle is zero.

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer:
Energy consumed by an electric heater can be obtained with the help of the expression,
P = \(\frac { W }{ t }\)
Where,
Power rating of the heater, P = 1500, W = 1.5 kW
Time for which the heater has operated, t = 10 h
Work done = Energy consumed by the heater
Therefore,
Energy consumed = Power × Time = 1.5 × 10 = 15 kWh
Hence, the energy consumed by the heater in 10 h is 15 kWh.

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer:
According to law of conservation of energy, energy can be neither created nor destroyed. It can only be converted from one form to another. Considering the case of an oscillating pendulum.

When a pendulum moves from its actual position P to either of its extreme point A or B, it rises through a height h above the mean level P. Here at this point, the kinetic energy of the bob changes into potential energy and the kinetic energy becomes zero, and the bob possesses only potential energy.

When it moves towards point P, its potential energy decreases progressively. And the kinetic energy increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates.

The pendulum loses its kinetic energy to overcome atmospheric friction and stops after some time. Hence law of conservation of energy is not violated and the total energy of the pendulum and the surrounding system remain conserved.

Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer:
We know that Kinetic energy of an object of mass m, moving with a velocity, v is given by:
E_k = \(\frac { 1 }{ 2 }\) mv²
To bring the object to rest, total energy must be consumed.
So, \(\frac { 1 }{ 2 }\) mv² amount of work is required to be done on the object.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Answer:
Kinetic energy, E_k = \(\frac { 1 }{ 2 }\) mv²
Where,
Mass of car, m = 1500 kg
Velocity of car, v = 60 km/h
= 60 × \(\frac { 5 }{ 18 }\) ms^-1
E_k = \(\frac { 1 }{ 2 }\) × 1500 × [60 × \(\frac { 5 }{ 18 }\)]²
= 20.8 × 10⁴ J.
Hence, 20.8 × 10⁴ J of work is required to stop the car.

Question 18.
In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Answer:
Case I: Here, the direction of force acting on the block is perpendicular to the displacement. Therefore, work done will be zero.

Case II: Here, the direction of force acting on the block is in the direction of displacement. Therefore, work done will be positive.

Case III: Here, the direction of force acting on the block is opposite to the direction of displacement. Therefore, work done will be negative.

Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
When all the forces cancel out each other, acceleration in an object will be zero even when several forces are acting on it. And as for a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

Question 20.
Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.
Answer:
We know,
P = \(\frac { W }{ t }\)
Given,
Power of the device (P) = 500 W = 0.50 kW
Total Time (t) = 10 h
Since,
Work done = Energy consumed by the device
Therefore, energy consumed
= Power × Time = 0.50 × 10 = 5 kWh
Hence, the energy consumed by four equal rating devices in 10 h will be:
4 × 5 kWh = 20 kWh
= 20 Units.

Question 21.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer:
As the object hits the ground, its kinetic energy gets converted into heat energy and sound energy. Sometimes, it also deform the ground and itself depending upon the nature of the ground and the amount of kinetic energy of the object. Freely falling object towards the ground feels following changes:

• Its potential energy decreases and kinetic energy increases.
• When the object touches the ground, all its potential energy gets converted into kinetic energy.

Work and Energy Additional Questions

Work and Energy Multiple Choice Questions

Question 1.
S.I. unit of work is ____________ .
(a) Newton
(b) Joule
(c) Watt
(d) All.
Answer:
(b) Joule

Question 2.
Work done is applied and displacement occurred is acute ____________ .
(a) Negative
(b) Zero
(c) Positive
(d) None.
Answer:
(c) Positive

Question 3.
Work done is negative, if angle formed between force applied and displacement occurred is ____________ .
(a) Obtuse
(b) Right angle
(c) Acute
(d) Zero.
Answer:
(a) Obtuse

Question 4.
Work is done if displacement of object occurs in ____________ .
(a) Opposite the direction of force
(b) The direction of force
(c) Zero
(d) Infinite.
Answer:
(b) The direction of force

Question 5.
Force is to work done ____________ .
(a) Inverse
(b) Zero
(c) Proportional
(d) Infinite.
Answer:
(c) Proportional

Question 6.
If direction of applied force and displacement are perpendicular then resultant work will be ____________ .
(a) Zero
(b) Positive
(c) Negative
(d) Infinite.
Answer:
(a) Zero

Question 7.
Writing for two hours is equal to work ____________ .
(a) 1 joule
(b) 2 joule
(c) Zero
(d) None.
Answer:
(c) Zero

Question 8.
Falling ball will have ____________ .
(a) Kinetic energy
(b) Potential energy
(c) Both
(d) None.
Answer:
(c) Both

Question 9.
If mass of an object is doubled over a pully, force required to displacement upto previous destination will be ____________ .
(a) Double
(b) Half
(c) Four times
(d) Equal.
Answer:
(a) Double

Question 10.
Change in kinetic energy, if velocity of an object is doubled will be ____________ .
(a) Double
(b) Half
(c) Equal
(d) 4 times.
Answer:
(d) 4 times.

Question 11.
Rate at which work is done is called ____________ .
(a) Work
(b) Power
(c) K.E.
(d) P.E.
Answer:
(b) Power

Question 12.
Running wings of a fan shows an example of ____________ .
(a) K.E.
(b) P.E.
(c) Work
(d) Power.
Answer:
(a) K.E.

Work and Energy Very Short Answer Type Questions

Question 1.
Give a formula / expression to give total energy.
Answer:
Total energy = Potential energy + Kinetic energy.

Question 2.
How displacement is related to work?
Answer:
Displacement is proportionally related to work.

Question 3.
What is a positive work?
Answer:
When displacement occurs in direction of force applied and object shifted forms an acute angle with force, work is termed as positive.

Question 4.
What kind of force is applied in a lift?
Answer:
Upward and gravitational force is applied in a lift.

Question 5.
What kind of quantity is power?
Answer:
It is scalar.

Question 6.
What will be the potential energy of an object at ground?
Answer:
PE = mgh
Here, h = 0
So, PE = mg.

Question 7.
What will be the work done if no force is being applied?
Answer:
Since, W = F × S
If F = 0;
W = 0 × S
Hence, W = 0
∴ No work will be done.

Question 8.
Name three forms of energy.
Answer:
Potential energy, Kinetic energy and Gravitational energy.

Question 9.
Write expression to calculate kinetic energy if mass of an object is m and its velocity is v.
Answer:
K.E. = \(\frac { 1 }{ 2 }\) mv².

Question 10.
Write expression for potential energy for an object being shifted to ‘d’ height and ‘a’ mass.
Answer:
PE = agd.

Question 11.
If a machine consumes 1000 J of energy in a second and runs for 5 hours, then calculate total energy consumed.
Answer:
P = \(\frac { W }{ t }\)
= 1000 × (5 × 3600)
= 18000000 joule
or
= 1.8 × 10⁷ joule.

Question 12.
What happen to potential and kinetic energy of a falling object?
Answer:
P.E. = decrease to zero
K.E. = increase from zero
Total P.E. converts to K.E.

Work and Energy Short Answer Type Questions

Question 1.
If a pulley is pulling a box towards itself with a force equal to 10 N and have drawn the box upto 2 m. Calculate the work done by it.
Answer:
Given, F = 10 N and s = 2 m
We know W = F × s
Putting values = 10 × 2
W = 20 Nm
or
20 J.

Question 2.
If a hammer operates over a pully machine with 21 J and pulls the balls to 7 m, Calculate the force applied by the hammer.
Answer:
Given:
W = 21 J and s = 7 m
Using formula W = \(\frac { F }{ s }\)
or
F = \(\frac { W }{ s }\)
Putting values F = \(\frac { 21 }{ 7 }\) = 3 N
Hence, F = 3 N was applied by the hammer.

Question 3.
Calculate the work done by a pair of bullocks if an object is pulled by them for 10 m which has mass equal to 20 kg and an acceleration of 20 ms^-1.
Answer:
Given, m = 20 kg, a = 20 ms^-1, s = 10 m
Using formula W = F × s
or
W = m × a × s
Putting values = 20 × 20 × 10
= 4000 kg ms^-2 m = 4000 J.

Question 4.
Give two examples of energy produced due to work or action.
Answer:
Examples showing production of kinetic energy:

1. Heat energy in body after exercise.
2. Generation of kinetic energy in a ball after falling through a point.

Question 5.
Read the following examples and tell in which case work is being done?

1. Manish is riding bicycle in a circular path.
2. John throws a pebble which comes back to initial height of ground.
3. Reading books for 2 hours.
4. Cooking pizza.

Answer:
Work is being done in none case because:

1. Case 1: In a circular path displacement is zero, hence work done is zero.
2. Case 2: Again no displacement occurs.
3. Case 3: Any kind of force is not being used, hence zero work.
4. Case 4: No force is applied here, so zero work.

Question 6.
(a) Under what conditions work is said to be done?
(b) A porter lifts a luggage of 1.5 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.
Answer:
(a) Conditions for work to be done:

1. Force should be applied.
2. Body should move in the line of action of force.
3. Angle between force and displacement should not be 90°.

(b) Given,
Mass of luggage, m = 15 kg
Displacement, s = 1.5 m
Using formula:
Work done i.e., W = F × s = mg × s
= 15 × 10 × 1.5 = 225 J.

Question 7.
Four persons jointly lift a 250 kg box to a height of 1 m and hold it.

1. Calculate the work done by the persons in lifting the box.
2. How much work is done for just holding the box?
3. Why do they get tired while holding it? (g = 10 ms²)

Answer:

1. Given, F = 250 × 10 = 2500 N
   s = 1 m
   W = F × s = 2500 × 1 = 2500 J.
2. Zero work, as there is no displacement.
3. Men are applying a force which is opposite and equal to the gravitational force acting on the box. Muscular effort is involved and therefore persons feel tried.

Question 8.
(i) Justify that “a body at a greater height has larger energy”.
(ii) A body of mass 2 kg is thrown up at a velocity of 10 m/s. Find the potential energy at the highest point.
Answer:
(i) When an object is placed at a greater height, the height increases from the reference level (Velocity remains constant i.e., zero). Hence by comparing the potential energy at two points, we can see that the P.E. at greater height will be larger.

(ii) Given,
Using formula m = 2 kg, v = 10 m/s
Initial KE = \(\frac { 1 }{ 2 }\) mv²
= \(\frac { 1 }{ 2 }\) × 2 × (10)² = 100 J
Height will be
h = \(\frac { { v }^{ 2 }-{ u }^{ 2 } }{ 2g } \) = \(\frac { { 0 }^{ 2 }-{ 10 }^{ 2 } }{ 2×10 } \)
= \(\frac { -100 }{ 20 }\) = -5 m
So, magnitude of height = 5 m and
P.E. at highest point = mgh
= 2 × 10 × 5
= 100 J.

Question 9.
A light and a heavy object have the same momentum. What is the ratio of their kinetic energies? Which one has a larger kinetic energy?
Answer:
p₁ = m₁v₁, p₂ = m₂v₂
But p₁= p₂  or  m₁v₁ = m₂v₂
and If m₁ < m₂ then v₁ > v₂
(K.E.)₁ = \(\frac { 1 }{ 2 }\) m₁v₁²
(K.E.)₂ = \(\frac { 1 }{ 2 }\) m₂v₂²
(K.E.)₁ = \(\frac { 1 }{ 2 }\) (m₁v₁)v₁ = \(\frac { 1 }{ 2 }\) p₁v₁
(K.E.)₂ = \(\frac { 1 }{ 2 }\) (m₂v₂)v₂ = \(\frac { 1 }{ 2 }\) p₂v₂.

Question 10.
(i) Define 1 kWh. Relate it to joules.
(ii) Find the energy in kWh in the month of September by four devices of power 100 W each, if each one of them is used for 10 hours daily.
Answer:
(i) 1 kWh is the energy used in 1 hour at the rate of 1000 J/s (or 1 kW)
1 kWh = 3.6 × 10⁶ J

(ii) Energy consumed by four devices in the month of September
= 4 × \(\frac { 100 }{ 1000 }\) × 10 × 30
= 120 kWh.

Work and Energy Long Answer Type Questions

Question 1.
(i) Give SI unit and commercial unit of electrical energy.
(ii) Calculate the power of an electric motor that can lift 800 kg of water to store in a tank at a height of 1500 cm in 20 s. (g = 10 m/s²)
(iii) A lamp consumes 500 J of electrical energy in 20 seconds. What is the power of the lamp?
Answer:
(i) SI unit of electrical energy is Joules (J).
Commercial unit of electrical energy is kilowatt hour (kWh)

(ii) Given,
m = 800 kg,
h = 1500 cm = 15 m,
t = 20 sec, g = 10 m/s²
Using formula,
P = \(\frac { W }{ t }\) = \(\frac { mgh }{ t }\)
= \(\frac { 800×10×15 }{ 20 }\) = 600 W.

(iii) Given,
E = 500 J, t = 20 sec
Using formula,
power P = \(\frac { W }{ E }\) = \(\frac { 500 }{ 20 }\) = 25W.
& r E 20

Question 2.
(i) Name the commercial unit of electrical energy.
(ii) Establish the relationship between the SI unit and the commercial unit of electric energy.
(iii) If 4 bulbs of 50 W for 6 hours, 3 tube lights of 40 W for 8 hours, a T.V of 100 W for 6 hours, a refrigerator of 300 W for 24 hours are used. Calculate the electricity bill amount for a month of 30 days. The cost per unit is ? ₹2.50.
Answer:
(i) The commercial unit of energy is kilowatt hour (kWh)
(ii) The SI unit of energy is joule.
Now, 1 kWh = 1 kW × 1 h
= 1000 W × 1 h
= 1000 W × 3600 s
= 3600000 J
= 3.6 × 10⁶ J
1 kWh = 3.6 × 10⁶ J

(iii)

Total energy consumed = 9960 Wh = 9.960 kWh
Electricity bill amount = 9.960 units × ₹2.50 = ₹24.90
For 30 days = 30 × 24.90 = ₹747.

Question 3.
(i) Name the physical quantity defined by rate of doing work. Define its SI unit.
(ii) Why is concept of average power useful? How is it determined?
(iii) A boy of mass 45 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power.
(g = 10 m/s²)
Answer:
(i) Power: Watt or J/s. 1 watt is the power of an object which does work at the rate of 1 J per second.

(ii) Power of an object may vary. Hence, average power is important in the case when the average power of the entire process within a given time is calculated.

(iii) P = \(\frac { mgh }{ t }\) = \(\frac { (50×10×45×0.15) }{ 9 }\) = 375 W.

Question 4.
(i) A body of mass 15 kg possesses kinetic energy of 18.75 kJ. Find the velocity.
(ii) An electric bulb of 100 W is used for 4 hours a day. Calculate the energy consumed by it in a day in joules and kilowatt hour unit.
Answer:
(i) K.E. = \(\frac { 1 }{ 2 }\) mv²
= 18.75 kJ = 18750
v² = \(\frac { 18750×2 }{ 15 }\)
v = √2500 m/s = 50 m/s

(ii) E = \(\frac { P }{ t }\) = 100 W × 4 h = 0.4 kWh
Energy consumed by it in a day = 0.4 × 3.6 × 10⁶ J
= 1.44 × 10⁶ J

Question 5.
(a) A stone is thrown upwards from a point A, as shown in the figure. After reaching the highest point B, it comes down. Explain the transformation of energy from A to B and B to A and also mention the type of energy possessed by the stone at point A, B and C of its journey.
(b) A body of mass 20 kg is dropped from a height of 100 m. Find its K.E. and P.E. after,
(i) First second
(ii) Second second
(iii) Third second
Answer:
(a) While moving upward (A to B) K.E → P.E. and while moving downward (B to A)
P.E.→ K.E.
At,
A → K.E.
B → P.E.
C → K.E. + P.E.

(b) Total Energy = mgh
= 20 × 10 × 100 = 2 × 10⁴ J

(i) After first, second: v = u + gt ms^-1
= 10 × 1 = 10 m/s
K.E. = \(\frac { 1 }{ 2 }\) mv².
= \(\frac { 1 }{ 2 }\) × 20 × 10 × 10
= 1000 J
P.E.= T.E. – K.E.
= 20,000 – 1000 = 19,000 J

(ii) After second, second: v = 20 ms^-1
K.E. = mgh + \(\frac { 1 }{ 2 }\) mv².
= 4,000 J
P.E. = T.E – K.E.
= 20,000 – 4,000 = 16,000 J

(iii) After third, second: v = 30 ms^-1
K.E. = mgh + \(\frac { 1 }{ 2 }\) mv².
= 9,000 J
P.E. = T.E.- K.E.
= 20,000 – 9,000 = 11,000 J

Work and Energy Higher Order Thinking Skills (HOTS)

Question 1.
Earth and other planets moves continuously around the sun. Do they work?
Answer:
No, Earth and other planets do not work while moving around the earth because they move in a circular path and reach the initial point after sometimes, so shows no displacement, hence no work is done by earth and other planets.

Question 2.
Generally heavy objects exert more power over other objects. Give reason.
Answer:
As expression, P = wit
On expanding, P = \(\frac { m.a.s }{ t } \)
Here, P is proportional to mass in p × m
Hence, heavy object exert more pressure or power.

VI. Value-Based Question

Question 1.
Ravi saw a lady labour who carried stones on her head from one point of the construction site to the other end which was some 500 m far. He prepares a trolley for the labour to carry the stones, to make her work easier:

1. Is any work done by the labour while carrying the stones from point A to point B on head by lady labour in the construction site?
2. Is any work done by pulling the trolley of stones from point A to point B?
3. What value of Ravi is seen in the above act?

Answer:

1. No work is said to be done in carrying the stones from point A to B on head by the lady.
2. Work is said to be done by pulling the trolley of stones.
3. Ravi showed kindness, general awareness and sympathy.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.5

Question 1.
In the figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution:
We have a circle with centre O, such that ∠AOB = 60° and ∠BOC = 30°

∴ ∠AOB + ∠BOC = ∠AOC
∠AOB = 60° + 30° = 90°
Now, tne arc ABC subtends ∠AOC = 90° at the centre and ∠ADC at a point D on the circle other than the arc ABC.
∴ ∠ADC = \(\frac{1}{2}\) [∠AOC]
∠ADC = \(\frac{1}{2}\) (90°) = 45°

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
We have a circle having a chord AB equal to radius of the circle.
∴ AO = BO = AB
∆AOB is an equilateral triangle.
Since, each angle of an equilateral triangle = 60
∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ABC at a point on the minor arc of the circle.

∠ACB = \(\frac{1}{2}\) [reflex ∠AOB]
= \(\frac{1}{2}\) [300°] = 150°
Similarly, ∠ADB = \(\frac{1}{2}\) [∠AOB]
= \(\frac{1}{2}\) x [60°] = 30°
Thus, the angle subtended by the chord on the minor arc = 150° and on the major arc = 30°.

Question 3.
In the figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Solution:
The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

∴ reflex ∠POR = 2∠PQR,
But ∠PQR = 100°
reflex ∠POR = 2 x 100° = 200°
Since, ∠POR + reflex ∠POR = 360°
∠POR + 200° = 360°
∠POR = 360° – 200°
∠POR = 160°
Since, OP = OR [Radii of the same circle]
∴ In ∆POR, ∠OPR = ∠ORP [Angles opposite to equal sides of a triangle are equal]
Also, ∠OPR + ∠ORP + ∠POR = 180° [Sum of the angles of a triangle = 180°]
∠OPR + ∠OPR + 160° = 180° [∴ ∠OPR = ∠ORP]
2∠OPR = 180° – 160° = 20°
∠OPR = \(\frac{20^{\circ}}{2}\) =10°

Question 4.
In the Figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
Solution:
We have, in ∆ABC,
∠ABC = 69° and
∠ACB = 31°

But ∠ABC + ∠ACB + ∠BAC = 180°
∴ 69° +31° + ∠BAC = 180°
∠BAC = 180° – 69° – 31°
= 80°
Since, angles in the same segment are equal.
∴ ∠BDC = ∠BAC
∠BDC =80°

Question 5.
In the figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
Solution:
In ∆CDE,
Exterior ∠BEC – ∠AED = 130°

130° = ∠EDC + ∠ECD
130° = ∠EDC + 20°
∠EDC = 130° – 20° = 110°
∠BDC = 110°
Since, angles in the same segment are equal.
∠BAC = ∠BDC
∠BAC = 110°

Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution:
Given:
∠BAC = 30°, ∠DBC = 70°
To find:
∠BCD and ∠ECD.
∠BDC = ∠BAC = 30° (∠s on the same segment of a circle are equal)

In ∆BCD, 70° + 30° + ∠C = 180° (ASP)
100° + ∠C = 180°
∠C = 80°
AB = BC (Given)
∠BAC = ∠BCA = 30°
∠BCD = ∠BCA + ∠ECD
80° = 30° + ∠ECD
∠ECD – 50°

Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Given
ABCD is a cyclic quadrilateral in which AC and BD are diagonals.
To prove:
ABCD is a rectangle.

Proof:
BD is the diameter of the circle.
∠BAD = 90°(angle in a semicircle)
Similarly ∠BCD = 90°
AC is the diameter of the circle
∠ABC = 90° (angle in a semicircle)
Similarly ∠ADC = 90°
In quadrilateral ABCD, ∠A = ∠B = ∠C = ∠D = 90°
ABCD is a rectangle.

Question 8.
If the non-parallel sides of h trapezium are equal, prove that it is cyclic.
Solution:
Given:
AD = BC, AB ∥ DC.
To prove:
ABCD is cyclic quadrilateral.
Construction:
Draw AE and BF As on DC.

Proof:
In ∆ADE and ∆BCF
∴ AE = BF
(∴ Distance between two parallel lines are equal)
∠E = ∠F (Each 90°)
AD = BC (Given)
∆ADE = ∆BCF (By RHS)
and so ∠D = ∠C (By CPCT)
AB ∥ DC and AD is the transversal.
∴ ∠BAD + ∠ADC =180° (CIA’s)
⇒ ∠BAD + ∠BCD = 180° (∠ADC = ∠BCD)
∴ ABCD is cyclic quadrilateral.

Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are J drawn to intersect the circles at A, D and Q respectively (see Fig.). Prove that ∠ACP = ∠QCD.

Solution:
Given
C (O, r) and C (O₁, r₁) are two circles. Two lines ABD and PBQ are drawn which intersect at B.
To prove:
∠1 = ∠3

Proof:
∠1 = ∠2 …(1) (Z s on the same segment of a circle are equal.)
∠3 = ∠4 …..(2) (Z s on the same segment of a circle are equal)
∠2 = ∠4 …(3) (OA’s)
From (1), (2) and (3), we get
∠1 = ∠3 …(2)

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Given
C (O, r) and C (O₁, r₁) are two Circles in whichAB andAC are diameter. These circles intersect at point A and D.
To prove:
BDC is a line.
Construction: JoinAD.
Proof:
∠ADB = 90° (∠s in a semicircle)
∠ADC) = 90° (∠s in a semicircle)

Adding (1) and (2), we get
∠ADB + ∠ADC = 90° + 90°
∠BDC = 180°
∴ BDC is a line and hence D lies on the third side.

Question 11.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD – ∠CBD.
Solution:
Given:
ABC and ADC are two right ∆’s on common base AC. ∠B – 90° and ∠D = 90°.
To prove: ∠CAD = ∠CBD
Proof:
∠ABC + ∠ADC = 90° + 90°
= 180°

ABCD is a cyclic quadrilateral.
∠CAD and ∠CBD are angles on the same segment of a circle.
∴ ∠CAD = ∠CBD.

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
Given: ABCD is a ∥^gm
To prove: ABCD is a rectangle.

Proof:
∠A = ∠C
∠A + ∠C =180°
∠A + ∠A = 180°
∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ ABCD is a rectangle.
∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ ABCD is a rectangle.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 10 Gravitation

Gravitation Intext Questions

Gravitation Intext Questions Page No. 134

Question 1.
State the universal law of gravitation.
Answer:
Suppose there are two objects having mass M and m respectively.
The distance between their centres is equal to d.
The force of attraction is F.
Thus, F ∝ M . m … (i)
and, F ∝ 1/d² … (ii)
Joining equation (i) and (ii)
we get F ∝ M . m/d²
⇒ F = G . M . m/d² … (iii)
where, G is the proportionality constant and called Universal Gravitation Constant.
The expression (iii) is called expression for Universal Law of Gravitation.
The universal law of gravitation is represented as:
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
Where, G is the universal gravitation constant given by:
G = 6.67 × 10^-11 Nm² kg^-2.

Question 2.
Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the earth.
Answer:
Let M_e be the mass of the Earth and m be the mass of an object on its surface.
And say R is the radius of the Earth,
then according to the universal law of gravitation,
the gravitational force (F) acting between the Earth and the object is given by the relation:
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
F = GM_em/R

Gravitation Intext Questions Page No. 136

Question 1.
What do you mean by free fall?
Answer:
Its a phenomenon of gravity. When an object falls from any height under the influence of gravitational force only, it is said to have a free fall. In the case of free fall, no change in direction takes place but the magnitude of velocity changes because of acceleration.

Question 2.
What do you mean by acceleration due to gravity?
Answer:
Change in velocity due to variation in height produces acceleration which is due to gravity in the object and is known as acceleration due to gravity denoted by letter g. The value of acceleration due to gravity is g = 9.8 m/s².

Gravitation Intext Questions Page No. 138

Question 1.
What are the differences between the mass of an object and its weight?
Answer:

Question 2.
Why is the weight of an object on the moon \(\frac { 1 }{ 6 }\)th its weight on the earth?
Answer:
The mass of moon is \(\frac { 1 }{ 100 }\) times and its radius \(\frac { 1 }{ 4 }\) times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is \(\frac { 1 }{ 6 }\)th of its weight on the earth.

Gravitation Intext Questions Page No. 141

Question 1.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

Question 2.
What do you mean by buoyancy?
Answer:
The upward force exerted by a liquid on an object that is immersed in it is known as buoyancy.

Question 3.
Why does an object float or sink when placed on the surface of water?
Answer:

1. An object sinks in water if its density is greater than that of water.
2. An object floats in water if its density is less than that of water.

Gravitation Intext Questions Page No. 142

Question 1.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
When we weigh our body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

Question 2.
You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?
Answer:
The cotton bag is heavier than the iron bar. The cotton bag experiences larger up – thrust of air than the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass.

Gravitation NCERT Textbook Exercises

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
According to Universal Law of gravitation, the gravitational force of attraction between any two objects of mass M and m is proportional to the product of their masses and inversely proportional to the square of distance r between them. So, force F is given by
F = G\(\frac { M\times m }{ { r }^{ 2 } } \)
Now, when the distance ‘r’ is reduced to half then force between two masses becomes
F’ = G\(\frac { M\times m }{ { (\frac { r }{ 2 } ) }^{ 2 } } \)
Or
F’ = 4F
Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

Question 2.
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Answer:
All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

Question 3.
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m).
Answer:
Given that,
Mass of the body, m = 1 kg
Mass of the Earth, M = 6 × 10²⁴ kg
Radius of the Earth, R = 6.4 × 10⁶ m
Now, magnitude of the gravitational force (F) between the Earth and the body can be given as,
F = G\(\frac { M\times m }{ { r }^{ 2 } } \) = \(\frac { 6.67 × 10 × 6 × 10 × 1 }{ (6.4 × 6.4) }\)
= \(\frac { 6.67 × 6 × 10 }{ (6.4 × 6.4) }\) = 9.8N (approx.)

Question 4.
The Earth and the Moon are attracted to each other by gravitational force. Does the Earth attract the Moon with a force that is greater or smaller or the same as the force with which the Moon attracts the Earth? Why?
Answer:
According to the Universal Law of Gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the Moon with an equal force with which the Moon attracts the Earth.

Question 5.
If the Moon attracts the Earth, why does the Earth not move towards the Moon?
Answer:
The Earth and the Moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the Moon. Hence, it accelerates at a rate lesser than the acceleration rate of the Moon towards the Earth. For this reason, the Earth does not move towards the Moon.

Question 6.
What happens to the force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?
Answer:
(i) From Universal Law of Gravitation, force exerted on an object of mass at by Earth is given by
F = G\(\frac { M\times m }{ { R }^{ 2 } } \) ….1
When nws of the object say ne is doubled than
F’ = G\(\frac { M\times 2m }{ { R }^{ 2 } } \)  = 2F
So as the mass of any one of the object is doubled the force is also doubled,

(ii) The force F is inversely proportional to the distance between the objects. So if the distance between two objects es doubled, then the gravitational force of attraction between them is reduced to one fourth of its original value, Similarly, if the distance between two objects is tripled. then the gravitational force of attraction becomes one ninth of its original value.

(iii) Again from Universal Law of Attraction, from equation 1, force ‘F’ is directly proportional to the product of both the masses, So, if both the masses are doubled then, the gravitational force of attraction become four times the original value.

Question 7.
What in the importance of Universal Law of Gravitation?
Answer:
Universal Law of Gravitation is important because it tells us about:

1. the force that is responsible for binding us to Earth.
2. the motion of Moon around the Earth.
3. the motion of planets around the Sun.
4. the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both Sun and Moon on the Earth.

Question 8.
What in the acceleration of free fall?
Answer:
Acceleration of free tall is the acceleration produced when a body falls under the influence of the force of gravitation of the Earth alone. It is denoted by ‘g’ and its value on the surface of the Earth is 9.8 ms^-2.

Question 9.
What do we call the gravitational force between the Earth and an object?
Answer:
Gravitational force between the Earth and an object is known as the weight of the object.

Question 10.
Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator}.
Answer:
Weight of a body on the Earth is given by:
W = mg
Where,
m = Mass of the body
g = Acceleration due to gravity
The value of g is greater at poles than at the equator.
Therefore, gold at the equator weighs less than at the poles.
Hence, Amit’s friend will not agree with the weight of the gold bought.

Question 11.
Why does a sheet of paper fell slower than one that is crumpled into a ball?
Answer:
When a sheet of paper is crumpled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.

Question 12.
Gravitational force on the surface of the Moon is only \(\frac { 1 }{ 6 }\) as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?
Answer:
Weight of an object on the Moon = \(\frac { 1 }{ 6 }\) × Weight of an object on the Earth.
Also,
Weight = Mass × Acceleration
Acceleration due to gravity, g = 9.8 m/s²
Therefore, weight of a 10 kg object on the earth = 10 × 9.8 = 98 N
And, weight of the same object on the Moon = 1.6 × 9.8 = 16.3 N.

Question 13.
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.
Answer:
According to the equation of motion under gravity:
-u² = 2 gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0, u = 49 m/s.
During upward motion, g = – 9.8 ms^-2.
(i) Let ‘h’ be the maximum height attained by the ball.
Hence,
0 – 49² = 2 × 9.8 × h
h = \(\frac { 49 × 49 }{ 2 × 9.8}\) = 122.5 m

(ii) Let ‘t’ be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:
v = u + gt
We get,
= 49 + t × (- 9.8)
9.8 t = 49
t = \(\frac { 49 }{ 9.8}\) = 5 s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

Question 14.
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer:
According to the equation of motion under gravity: v² – u² = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 ms^-2
∴ v² – 0² = 2 × 9.8 × 19.6
v² = 2 × 9.8 × 19.6 = (19.6)²
v = 19.6 ms^-1
Hence, the velocity of the stone just before touching the ground is 19.6 m s^-1.

Question 15.
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:
According to the equation of motion under gravity:
v² – u² = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0
s = Height of the stone
g = Acceleration due to gravity = -10 ms^-2
Let h be the maximum height attained by the stone.
Therefore,
0 – (40)² = 2 × h × (-10)
h = \(\frac { 40 × 40 }{ 20 }\) = 80 m
Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey
= 80 + (-80) = 0.

Question 16.
Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.
Answer:
According to question,
M_Sun = Mass of the Sun = 2 × 10³⁰ kg
M_Earth = Mass of the Earth = 6 × 10²⁴ kg
R = Average distance between the Earth and the Sun = 1.5 × 10¹¹ m.
From Universal Law of Gravitation,
F = G\(\frac { M\times m }{ { R }^{ 2 } } \)
Therefore, putting all the values given in question in above equation we get
F = 6.67 × 10^-11 \(\frac { (6\times { 10 }^{ 24 })\times (2\times { 10 }^{ 30 }) }{ { (1.5\times { 10 }^{ 11 }) }^{ 2 } } \) = 3.56 × 10²² N.

Question 17.
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:
Let be the point at which two stones meet and let ‘h’ be their height from the ground. It is given in the question that height of the tower is H = 100 m
Now, first consider the stone which falls from the top of the tower.
So, distance covered by this stone at time ‘t’ can be calculated using equation of motion:
x – x₀ = u₀t + \(\frac { 1 }{ 2 }\)gt²
Since, initial velocity u = 0,
so we get
100 =  x \(\frac { 1 }{ 2 }\)gt² ………… (1)
The distance covered by the same stone that is thrown in upward direction from ground is
x = 25t –\(\frac { 1 }{ 2 }\)gt²
In this case intitial velocity is 25 m/s.
So, x = 25t – \(\frac { 1 }{ 2 }\)gt²  ………… (2)
Adding equations (1) and (2) we get,
100=25t
or,
t = 4s
Putting value in equation (2).
x = 25 × 4 – \(\frac { 1 }{ 2 }\) × 9.8 × (4)²
= 100 – 78.4
= 21.6 m.

Question 18.
A ball thrown up vertically returns to the thrower after 6 s. Find:
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
Answer:
(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height. Final velocity of the ball at the maximum height, v = 0 Acceleration due to gravity, g = -9.8 m s^-2
Equation of motion, v = u + gt
will give,
0 = u + (-9.8 × 3)
u = 9.8 × 3
= 29.4 ms^-1
Hence, the ball was thrown upwards with a velocity of 29.4 ms^-1.

(b) Let the maximum height attained by the ball be ‘h’
Initial velocity during the upward journey, u = 29.4 ms^-1
Final velocity, v = 0
Acceleration due to gravity, g = -9.8 ms^-2
From the equation of motion, s = ut + \(\frac { 1 }{ 2 }\) at²
h = 29.4 × 3 + \(\frac { 1 }{ 2 }\) × – 9.8 × (3)² = 44.1 m.

(c) Ball attains the maximum height after 3 s.
After attaining this height, it will start falling downwards.
In this case, Initial velocity, u = 0
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in
4 s – 3 s = 1 s.
Equation of motion, s = ut + \(\frac { 1 }{ 2 }\) g²
will give,
s = 0 x t + \(\frac { 1 }{ 2 }\) x 9.8 x 12 = 4.9 m
Total height  = 44.1 m.
This means that the ball is 39.2 m (44.1 m – 4.9 m) above the ground after 4 seconds.

Question 19.
In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
An object immersed in a liquid experiences buoyant force in the upward direction.

Question 20.
Why does a block of plastic released under water come up to the surface of water?
Answer:
For an object immersed in water two forces act on it:

1. Gravitational force which tends to pull object in downward direction
2. Buoyant force that pushes the object in upward direction.

Here, in this case buoyant force is greater than the gravitational pull on the plastic block. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water.

Question 21.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm^-3, will the substance float or sink?
Answer:
If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.

= \(\frac { 50 }{ 20 }\)
= 2.5 g cm^-3
The density of the substance is more than the density of water (1 g cm^-3).
Hence, the substance will sink in water.

Question 22.
The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm^-3? What will be the mass of the water displaced by this packet?
Answer:
Density of the 500 g sealed packet

= \(\frac { 500 }{ 350 }\)
= 1.428 g cm^-3
The density of the substance is more than the density of water (1 g cm^-3). Hence, it will sink in water.
The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.

Gravitation Additional Questions

Gravitation Multiple Choice Questions

Question 1.
Two objects of different masses falling freely near the surface of moon would __________ .
(a) have same velocities at any instant.
(b) have different accelerations.
(c) experience forces of same magnitude.
(d) undergo a change in their inertia.
Answer:
(c) experience forces of same magnitude.

Question 2.
The value of acceleration due to gravity __________ .
(a) is same on equator and poles.
(b) is least on poles.
(c) is least on equator.
(d) increases from pole to equator.
Answer:
(c) is least on equator.

Question 3.
The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become __________ .
(a) \(\frac { F }{ 4 }\)
(b) \(\frac { F }{ 2 }\)
(c) F
(d) 2 F.
Answer:
(a) \(\frac { F }{ 4 }\)

Question 4.
A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone __________ .
(a) will continue to move in the circular path.
(b) will move along a straight line towards the centre of the circular path.
(c) will move along a straight line tangential to the circular path.
(d) will move along a straight line perpendicular to the circular path away from the boy.
Answer:
(c) will move along a straight line tangential to the circular path.

Question 5.
An object is put one by one in three liquids having different densities. The object floats with 1 2 3, and 9 11 7 parts of their volumes outside the liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following statement is correct?
(a) d₁ > d₂ > d₃
(b) d₂₁ > d₂ < d₃
(c) d₁ < d₂ > d₃
(d) d₁ < d₂ < d₃.
Answer:
(d) d₁ < d₂ < d₃.

Question 6.
In the relation F = GM m/d₂, the quantity G __________ .
(a) depends on the value of ‘g’ at the place of observation.
(b) is used only when the Earth is one of the two masses.
(c) is greatest at the surface of the Earth.
(d) is universal constant of nature.
Answer:
(d) is universal constant of nature.

Question 7.
Law of gravitation gives the gravitational force between __________ .
(a) the Earth and a point mass only.
(b) the Earth and Sun only.
(c) any two bodies having some mass.
(d) two charged bodies only.
Answer:
(c) any two bodies having some mass.

Question 8.
The value of quantity G in the law of gravitation __________ .
(a) depends on mass of Earth only.
(b) depends on radius of Earth only.
(c) depends on both mass and radius of Earth.
(d) is independent of mass and radius of the Earth.
Answer:
(d) is independent of mass and radius of the Earth.

Question 9.
Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be __________ .
(a) 14 times
(b) 4 times
(c) 12 times
(d) unchanged.
Answer:
(b) 4 times

Question 10.
The atmosphere is held to the earth by __________ .
(a) gravity
(b) wind
(c) clouds
(d) Earth’s magnetic field.
Answer:
(a) gravity

Question 11.
The force of attraction between two unit point masses separated by a unit distance is called __________ .
(a) gravitational potential.
(b) acceleration due to gravity.
(c) gravitational field.
(d) universal gravitational constant.
Answer:
(d) universal gravitational constant.

Question 12.
The weight of an object at the centre of the Earth of radius R is __________ .
(a) zero.
(b) infinite.
(c) R times the weight at the surface of the Earth.
(d) \(\frac { 1 }{ { R }^{ 2 } } \) times the weight at surface of the Earth.
Answer
(a) zero.

Gravitation Very Short Answer Type Questions

Question 1.
Why Moon revolves around the Earth?
Answer:
Gravitational force of Earth.

Question 2.
What is the SI unit of gravitational force?
Answer:
Newton (N).

Question 3.
Which law of physics is represented by the statement every object attract other object in universe towards itself’?
Answer:
Universal Law of Gravitation.

Question 4.
State the relation between gravitational force and distance among objects.
Answer:
Inversely proportional.

Question 5.
In which conditions free fall of an object occur?
Answer:
When an object falls from a height under the influence of gravity and no other force, it is said to have a free fall.

Question 6.
What is the SI unit of gravitational constant?
Answer:
Nm²kg^-2.

Question 7.
Express the relation between thrust and pressure.
Answer:
Pressure = thrust / area.

Question 8.
What kind of force is exerted by a liquid?
Answer:
Equal and unidirectional.

Question 9.
In what condition an object sinks?
Answer:
If the weight of the object is more than 9.8 N, then the object will sink.

Question 10.
Which material is taken as standard to calculate any object’s relative density?
Answer:
Water.

Gravitation Short Answer Type Questions

Question 1.
What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?
Answer:
Gravitational force. This force depends on the product of the masses of the planet and Sun, and the distance between them.

Question 2.
On the Earth, a stone is thrown from a height in a direction parallel to the Earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?
Answer:
Both the stones will take the same time to reach the ground because the two stones fall from the same height.

Question 3.
Suppose gravity of Earth suddenly becomes zero, then in which direction will the Moon begin to move if no other celestial body affects it?
Answer:
The Moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of Moon is due to centripetal force provided by the gravitational force of Earth.

Question 4.
Two identical packets are dropped from two Aeroplanes, one above the equator and the other above the north pole, both at height h. Assuming all conditions are identical, will those packets take same time to reach the surface of Earth. Justify your answer.
Answer:
The value of ‘g’ at the equator of the Earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.

Question 5.
The weight of any person on the Moon is about \(\frac { 1 }{ 6 }\) times that on the Earth. He can lift a mass of 15 kg on the Earth. What will be the maximum mass, which can be lifted by the same force applied by the person on the moon?
Answer:
The value of ‘g’ at the equator of the Earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.

Gravitation Long Answer Type Questions

Question 1.
State ‘Archimedes’ Principle and write its two applications.
Answer:
‘Archimedes’ Principle:
Force exerted by liquid on wholly or partly immersed object is equal to the weight of the fluid displaced by the object. Applications based on ‘Archimedes’ principle are:

1. designing of water transport vehicles.
2. hydrometers used for determining the density of liquids

Question 2.
What is relative density? What is the density of water?
Answer:
Relative Density (RD) or Specific Gravity (SG) is the ratio of either densities or weights. Hence, when we compare or divide value of an objects’ density with water’s density, it is called Relative Density of a substance to water.

1. In SI units, the density of water is (approximately) 1000 kg/m³ Or 1 g/cm³.

Question 3.
Mass of a rectangular copper solid piece is 300 g. With dimensions 5 × 2 × 5 cm³, what should be its specific gravity, calculate? Will the bar float or sink in water?
Answer:

Given:
Mass of copper = 300 g
5 × 2 × 5 = 50 cm³
Density of copper = mass / volume
\(\frac { 300 }{ 50 }\) = 6 g / cm³
Density of water, = 1 g/cm³
Specific gravity of iron = \(\frac { 6 }{ 1 }\) = 6.
Hence the bar will sink.

Question 4.
How does the weight of an object vary with respect to mass and radius of the earth. In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected?
Answer:
We know, weight of an object is directly proportional to the mass of the earth and inversely proportional to the square of the radius of the earth, i.e.,.
Weight of a body ∝ \(\frac { M }{ { R }^{ 2 } } \)
Original weight, W₀  = mg = mG\(\frac { M }{ { R }^{ 2 } } \)
When hypothetically M becomes 4 M and R becomes \(\frac { R }{ 2 }\) then weight becomes
W₀ = mG \(\frac { 4M }{ { (\frac { R }{ 2 } ) }^{ 2 } } \) = (16 m G) M
R² = 16 × W₀
The weight will be 16 times heavier.

Question 5.
(a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force. If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water? Give reason for each case.
(b) A ball weighing 4 kg of density 4000 kg m^-3 is completely immersed in water of density 10³ kg m^-3. Find the force of buoyancy on it. (Given: g = 10 ms^-2.)
Answer:
(a)

1. The cube will experience a greater buoyant force in the saturated salt solution because the density of the salt solution is greater than that of water.
2. The smaller cube will experience lesser buoyant force as its volume is lesser than the initial cube.

(b) Buoyant force = weight of the liquid displaced = density of water x volume of water displaced xg 4
= 1000 × \(\frac { 4 }{ 4000 }\) × 10 = 10N.
4000

Gravitation Higher Order Thinking Skills (HOTS)

Question 1.
How will the weight of a body of mass 250 g of changes, if it is taken from equator to the poles? Give reasons.
Answer:
As we move from equator to poles, acceleration due to gravity increases. It is because radius of earth is less at poles than at equator. Therefore, its weight will increase.

Question 2.
Aman tried to immerse an empty plastic bottle in a bucket of water. But each time he fails. Why does this happen?
Answer:
When Aman tried to immerse an empty plastic bottle in a bucket of water, it comes above the surface of water. It is due to the upward force (upthrust or buoyant force). The upthrust exerted by water on the bottle is greater than its own weight.

Gravitation Value Based Question

Question 1.
Rashmi was wearing a high heel shoes for a beach party. Her friend told her to wear flat shoes as she will be tired soon with high heel and will not feel comfortable.

1. What is the reason of one’s feeling tired with high heel shoes on a beach?
2. Name the unit of pressure.
3. What value of Rashmi’s friend is reflected in the above act?

Answer:

1. Because the high heel shoes would exert lot of pressure on the loose sand of beach and will sink more in the soil as compared to flat shoes. Therefore, large amount of force will be required to walk with high heels.
2. Pascal.
3. Rashmi’s friend showed the value of being intelligent, concerned and helpful.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.2

Question 1.
Construct a triangle ABC in which BC = 7 cm, B = ∠75° and AB + AC = 13 cm.
Solution:
BC = 7 cm
∠B = 15°
AB + BC = 13 cm.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° AB – AC = 3.5 cm.
Solution:
BC = 8 cm
∠B = 45°
AB – AC = 3.5 cm.

Question 3.
Construct a triangle PQR in which QR = 6 cm. ∠Q = 60° and PR – PQ = 2 cm.
Solution:
QR = 6 cm
∠Q =60°
PR – PQ = 2 cm

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution:
XY + YZ + ZX = 11 cm
∠Y = 30°
∠Z =90°

1. Draw a line segment BC =11 cm.
2. At B construct an angle of 30° and at C, draw angle of 90°.
3. Bisect these angles. Let the bisectors of these angles intersect atX.
4. Draw perpendicular bisectors AC of BX to intersect BC at Y and DZ of XC to intersect BC at Z.
5. Join XY and XZ. XYZ is the required D.

Question 5.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Solution:
Steps of construction:

1. Draw \(\overline { BC } \) = 12 cm.
2. Construct ∠CBY = 90°.
3. From \(\overline { BY } \), cut off BX = 18 cm.
4. Join CX.
5. Draw PQ, the perpendicular bisector of CX, such that PQ meets BX at A.
6. JoinAC.

Thus, ABC is the required triangle.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.4

Question 1.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:
Given
Let O and O₁ be the centre of bigger and smaller circle respectively
OA = OB = 5 cm
O₁A – O₁B = 3 cm
OO₁ = 4 cm

To find: AB.
Construction:
Join OA, OB, O₁A and O₁B join AB also.
In ∆OAO₁ and ∆OBO₁
OA = OB (Radii of a circle)
O₁A = O₁B (Radii of a circle)
OO₁ = OO₁ (Common)
so ∆OAO₁ = ∆OBO₁ (By SSS)
and so ∠1 = ∠2 (By CPCT)
In ∆OCA and ∆OCB,
OA = OB (Radii of a circle)
∠1 = ∠2 (Proved)
OC = OC (Common)
∆OCA = ∆OCB (By SAS)
so AC = BC (By CPCT)
and ∠ACO = ∠BCO (By CPCT)
∠ACO + ∠BCO = 180° (LPA’s)
⇒ ∠ACO + ∠BCO = 180°
2∠ACO = 180°
∠ACO = 90°
ar (OAO₁) = \(\frac{1}{2}\) x OO₁ x AC
= \(\frac{1}{2}\) x 4 x AC = 2ACcm² …..(i)
In ∆QAO₁, a = 5 cm, bc = 4 cm, c = 3 cm
12
s = \(\frac{5+4+3}{2}\) = \(\frac{12}{2}\) = 6 cm
s – a = 6 – 5 = 1 cm
s – b = 6 – 4 = 2 cm
s – c = b – 3 = 3 cm
ar(OAO₁) = \(\sqrt{s(s-a)(s -b)(s- c)}\).
= \(\sqrt{6x1x2x3}\)
= 6 cm² …..(ii)
From (i) and (ii), we get
2AC =6
AC = 3 cm
Now, AB = 2AC (∴ AC = BC)
= 2 x 3 = 6 cm.

Method II. By Construction:
Geometrically, AB is the diameter of the circle of radius 3 cm as it passes through centre O₁
AB = 2 x 3 = 6 cm.

Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segment of the other chord.
Solution:
Given
C (O, r) is a circle in which AB and CD are two equal chords which intersect at P.
To prove:
CP = BP and AP = DP.
Construction:
Draw OE and OF perpendiculars on AB and CD respectively. Join OP.
Proof:
In ∆OPF and ∆OPE,
OP = OP (Common)
OE = OF (∴ AB = CD)
∠F = ∠E (Each 90°)
∆OPF = ∆OPE (By RHS)
and so PE = PF …..(1) (By CPCT)
AB = CD (Given)

\(\frac{1}{2}\) AB = \(\frac{1}{2}\) CD
BE = CF
and AE = DF …..(2)
Adding (1) and (2), we get,
PE + AE = PF + DF
∴ AP = DP
Subtracting (1) and (2) we get,
BE – PE = CF – PF
∴ BP = CP

Question 3.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
Given
AB and CD are two equal chords of a circle which intersect at E.
To prove:
∠1 = ∠2
Construction:
Draw OL ⊥ AB and OM ⊥ CD. Join OE.

Proof:
In ∆OLE and ∆OME,
OE = OE
OL = OM (∴ AB = CD)
∠L = ∠M (Each 90°)
∆OLE = ∆OME (By RHS)
and so∠1 = ∠2 (By CPCT)

Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD. (see Fig. adjacent)

Solution:
Given:
C (O, r) and C (O, r) are two concentric circles.
To prove: AB = CD

Construction: Draw OP ⊥ AD.
Proof:
In circle I, AD is the chord and OP ⊥ AD.
AP = DP …(1)
In circle II, BC is the Chord and OP L BC.
∴ BP = CP …(2)
Subtracting (1) and (2), we get
AP – BP = DP – CP
AB = CD

Question 5.
Three girls Reshma. Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Solution:
Given:
OR = OM = 5 m and SR = SM = 6 m.
To find: MR.

Constrution:
Join OR, OM and OS. Draw ON ⊥ SR. In AORS and AOMS,
OS = OS (Common)
RS = MS (Given)
OR = OM (Given)
∆ORS = ∆OMS (By SSS)
and ∠1 = ∠2 (By CPCT)
SP = SP (Common)
SR = SM (Given)
∠1 = ∠2 (Proved)
∆SPR = ∆SPM (By SAS)
and so PR = PM (By CPCT)
and ∠3 = ∠4 (By CPCT)
∠3 + ∠4 = 180° (LPA’s)
2∠3 = 180°
∠3 = \(\frac{180^{\circ}}{2}\) = 90°
ar (∆OSR) = \(\frac{1}{2}\) x OS x PR …(i)
= \(\frac{1}{2}\) x 5 x PR
ON ⊥ SR
RN = \(\frac{1}{2}\) SR
(Perpendicular drawn from the centre of a circle to a chord bisects the chord)
= \(\frac{6}{2}\) = 3m
In ∆ONR ON² = \(\sqrt{O R^{2}-N R^{2}}\) (Using Pythagoras Theorem)
= \(\sqrt{5^{2}-3^{2}}\) = \(\sqrt{4^{2}}\) = 4m
ar (∆OSR) = \(\frac{1}{2}\) x SR x ON
= \(\frac{1}{2}\) x 6 x \(\frac{1}{2}\) x 4 = 12m² …..(ii)
From (i) and (ii), we get
PR = \(\frac{2×12}{2}\) = 4.8m
MR = 2 PR
= 2 x 4.8
= 9.6 m

Question 6.
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Solution:
Given: OS = OA = 20 m and AS = SD = AD
To find: AS, SD and AD.
Construction:
Draw AE ⊥ SD. Join OS.
Let AS = SD = AD = 2x (say)

In equilateral ∠ASD, AE ⊥ SD
⇒ E is the mid-point of SD
SE = \(\frac{2x}{2}\) = x
In ∆AES, AS² = AE² + SE²
(2x)² = AE² + x²
4x² – x² = AE²
AE = \(\sqrt{3x^{2}}\) = \(\sqrt{3}\)
OE = AE – AO
= (\(\sqrt{3}\) – 20)m
In ∆OES, OS² = OE² + SE²
(20)² = [(\(\sqrt{3}\)x) 20]² + x²
400 = (\(\sqrt{3}\)x)² – 2 x \(\sqrt{3}\)x × 20 + (20)² + x²
= 3x² + 400 – 40\(\sqrt{3}\)x + x²
400 – 400 = 4x² – 40\(\sqrt{3}\)x
0 = 4x² – 40\(\sqrt{3}\)x
40\(\sqrt{3}\)x = 4x²
40\(\frac { 40\sqrt { 3 } }{ 4 } \) = x
x = 10\(\sqrt{3}\)m .
2x = 2 x 10\(\sqrt{3}\) = 20\(\sqrt{3}\)m
AS = SD = AD = 20\(\sqrt{3}\)m.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

Question 1.
In Fig. given below E is any point on median AD of a ∆ABC. Show that ar (ABE) = ar (ACE).

Solution:
Given.
E is any point on median AD of ∆ABC.
To prove
ar (ABE) = ar (ACE)

Proof:
In ∆ABC, AD is the median
ar (ABD) = ar(ACD) ….(1)
In ∆EBC, ED is the median
∴ ar (BDE) = ar (CDE) ….(2)
Subtracting (2) from (1), we get
ar (ABD) – ar (BDE) = ar (ACD) – ar (CDE)
ar (ABE) = ar (ACE)

Question 2.
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = \(\frac{1}{4}\) ar (ABC).

Solution:
Given
E is the mid point of – median AD of ∆ABC.
To prove:
ar (BED) = \(\frac{1}{4}\) ar (ABC)
Proof:
In ∆ABC, AD is the media
ar (ABD) = \(\frac{1}{2}\) ar (ABC) …..(1)
In ∆ABD, BE is the median

ar (BED) = \(\frac{1}{2}\) ar (ABD)
= \(\frac{1}{2}\) [\(\frac{1}{2}\) ar (ABC)]
= \(\frac{1}{4}\) ar (ABC)

Question 3.
Show that the diagonals of parallelogram divide it into four triangles of equal area.
Solution:
Given.
ABCD is a parallelogram.
To prove:
ar (AOB) = ar (BOC) = ar (COD) = ar (AOD)

Proof:
In ∆ABC, OA is the median
∴ ar (AOB) = ar (AOD) …..(1)
In AABC, BO is the median
∴ ar (AOB) = ar (BOC) …..(2)
In ABCD, CO is the median
ar (BOC) = ar (COD) …(3)
From (1), (2) and (3), we get
ar (AOB) ar (BOC) = ar (COD) = ar (AOD)

Question 4.
In Fig. given below, ABC and ABC are two triangles on the same base AB. If line – segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

Solution:
Given.
∆ABC and ∆ABD have a common base AB.
OC = OD
To prove
ar (ABC) = ar (ABD)

Proof:
As OC = OD,
O is the mid-point of CD
In ∆ACD, O is the median
ar (AOC) = ar(AOD)
In ABCD, BO is the median
ar (BOC) = ar (BOD)
Adding (1) and (2), we get
ar (AOC) + (BOC) = ar (AOD) + ar (BOD)
ar (ABC) = ar (ABC)

Question 5.
D, E and F are respectively the mid-points of the sides BC, CA and AB of a AABC. Show that

1. BDEF is a parallelogram.
2. ar (DEF) = \(\frac{1}{2}\) ar (ABC)
3. ar (BDEF) = \(\frac{1}{4}\) ar (ABC).

Solution:
D, E, F are the mid points of sides BC, CA and AB of ∆ABC.
To prove:

1. BDEF is a parallelogram.
2. ar (DEF) = \(\frac{1}{4}\) ar (ABC)
3. ar (BDEF) = \(\frac{1}{2}\) ar (ABC)

Proof:
In ∆ABC, F is the mid-point of AB, E is the mid point of AC.
∴ FE ∥ BC and FE = – BC (By MPT)
⇒ FE ∥ BD and FE =BD

1. BDEF is a ∥^gm
Similarly CDFE and AEDF are ∥^gm
In parallelogram BDEF, DF is the diagonal
∴ ax (BDF) = ar (DEF)
[In a ∥^gm diagonal divides it into 2∆s of equal areas]
Similarly, In ∥^gm CDFE, DE is a diagoilal …(1)
ar (CDE) = (DEF) …(2)
In ∥^gm AEDF, FE is a diagonal
ar (AEF) = ar (DEF) …(3)
From (1), (2) and (3), we get
ar (BDF) = ar (CDF) = ar (AEF) = (DEF) …(4)

2. ar (ABC) = ar (AEF) + ar (ADF) + ar (CDE) + ar (DEF)
ar (ABC) = 4 ar (DEF) [Using (4)]
ar (DEF) = \(\frac{1}{4}\) ar (ABC)

3. ar (BDEF) = ar (BDF) + ar (DEF)
ar (BDEF) = 2ar (DEF) [∴ ar (BDF) = ar (DEF)]
= 2 x \(\frac{1}{4}\) ar (ABC) = \(\frac{1}{2}\) ar (ABC)

Question 6.
In the Fig. diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB – CD, then show that:

1. ar (DOC) = ar (AOB)
2. ar DCB = ar (ACB)
3. DA ∥ CB or ABCD is a parallelogram.

[Hint: From D and B, draw perpendiculars to AC.]

Solution:
Given
ABCD is a quadrilateral in which OB = OD and AB = CD.
To prove:

1. ar (DOC) = ar (AOB)
2. ar (DCB) = ar (ACB)
3. DA ∥ CB or ABCD is a ∥^gm

Construction:
Draw DE and BF perpendicular from point D and B on AC.
Proof:
1. In ∆OED and ∆OFB
∠1 = ∠2 (V.O.A.’s)
OD = OB (given)
∠E = ∠F (each 90°)
∆ OED = ∆ OFB (by AAS)
and so DE = BF (by CPCT)
In ∆DEC and ∆BFA, DE = BF (proved)
∠E = ∠F (each 90°)
DC = BA (given)
∆DEC = ∆BFA (by RHS)
and so ∠3 = ∠4 (by CPCT)
ar (OED) = ar (OFB) [∴ AOED = AOFB] …(1)
ar (DEC) = ar (BFB) [∴ ADEC = ABEA] …..(2)
Adding (1) and (2), we get
ar (OED) + ar (DEC) = ar (OFB) = ar (BFA)
ar (OCD) = ar (OAB) …..(3)

2. Adding ar (OBC) on both sides of equation (3)
ar (OCD) + (OBC) = ar (OAB) + ar (OBC)
ar (DCB) = ar (ACB)

3. ∆DCB and ∆ACB have the same base BC and have equal area
∴ they will lie between the same parallels BC and AD
and so BC ∥ AD
∠3 and ∠4 are A.I.A’s and are equal
∴ AB ∥ DC
In quadrilateral ABCD, AB ∥ DC and AB = DC (given)
ABCD is a parallelogram.

Question 7.
D and E are points on sides AB and AC respectively of ∆ABC such that ar (DBC) = ar (EBQ. Prove that DE ∥ BC.
Solution:
Given
ar (DBC) = ar (EBC)
To prove:
DE ∥ BC

Proof:
∆ DBC and ∆EBC have the same base BC and
ar (DBC) = ar (EBC)
∴ They will lie between the same parallel lines DE and BC.
and so DE ∥ BC

Question 8.
XY is a line parallel to side BC of a triangle ABC. If BE ∥ AC and CF ∥ AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).
Solution:
Given
XY ∥ BC, BE ∥ AC and CF ∥ AB.
To prove
ar (ABE) = ar (ACF)
Proof:
In quadrilateral ACBE, AE ∥ CB (∴ BC ∥ XY)
and AC ∥ EB
ACBE is a ∥^gm

Similarly ABCF is a parallelogram.
∥^gm ACBE and ABCF have the same base BC and are between the same parallels BC and AY.
∴ ar (ACBE) = ar (ABCF)
ar (ABE) + ar (ABC) = ar (ABC) + ar (ACF)
∴ ar (ABE) = ar (ACF)

Question 9.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. below). Show that ar (ABCD) = ar (PBQR).
[Hint: Joint AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Solution:
Given
ABCD and BPRQ are parallelograms. CP ∥ AQ
To prove:
ar (ABCD) = ar (PBQR)
Construcion:
Join AC and PQ
Proof:
∆ACQ and ∆APQ lie on the same base AQ and are between the same parallels AQ and CP.
∴ ar (ACQ) = ar (APQ)
⇒ ar (ABQ) + ar (ABC) = ar (ABQ) + ar (BQP)
∴ ar (ABC) = ar (BQP) …..(i)

Multiplying (1) by 2 on both sides
2 ar (ABC) = 2 ar (BQP)
∴ ar (ABCD) = ar (PBQR)
[∴ ABCD and PBQR are parallelogram]

Question 10.
Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at C. Prove that ar (AOD) = ar (BOC).
Solution:
Given
ABCD is a trapezium in which AB ∥ DC.
To prove:
ar (AOD) = ar (BOC).

Proof:
∆ADC and ∆BCD lie on the same base DC and between the same parallels AB and CD.
∴ ar (ADC) = ar (BCD)
Subtracting ar (DOC) from both sides
ar (ADC) – ar (DOC) = ar (BCD) – ar (DOC)
ar (AOD) = ar (BOC)

Question 11.
In Fig. ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that:

1. ar (ACB) = ar (ACF)
2. ar (AEDF) = ar (ABCDE).

Solution:
Given
ABCDE is a pentagon.
To prove:
1. ar (ACB) = ar (ACF)

2. ar (AEDF) = ar (ABCDE)
Proof:
∆ACB and ∆ACF lie on the same base AC and between the same parallels AC and BE.
ar (ACB) = ar (ACF)

3. Adding ar (AEDC) on both sides
ar (ACB) + ar (AEDC) = ar (ACF) + ar (AEDC)
ar (ABCDE) = ar (AEDF)

Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Solution:
Given.
ABCD is a quadrilateral.
Construction:
Join AC. Draw DE ∥ CA which intersect BA produced at E.

Proof:
∆ADC and ∆ACE both lie on the same base AC and between the same parallels AC and DE.
ar (ADC) = ar (ACE)
Adding ar (ABC) on both sides
ar (ADC) + ar (ABC) = ar (ACE) + ar (ABC)
ar (ABCD) = ar (EBC)

Question 13.
ABCD is a trapezium with AB ∥ DC. Aline parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
Solution:
Given
ABCD is trapezium with AB ∥ DC and AC ∥ XY.
To prove:
ar (ADX) = ar (ACY).
Construction:
Join CX.
Proof:
∆ADX and ∆ACX both lie on the same base AX and between the same parallel AN and DC.

∴ ar (ADX) = ar (ACX) …(1)
∆ACX and ∆ACT both lie on the same base AC and between the same parallels AC and AT.
ar (ADY) = ar (ACX) …(2)
From (1) and (2), we get
ar (ADX) = ar(ACT)

Question 14.
In Fig. below AP ∥ BQ ∥ CR. Prove that ar (AQC) = ar (PBR).

Solution:
Given:
AP∥BQ∥CR.
To prove:
ar (AQC) = ar (PBR)
Proof:
∆ABQ and ∆PBQ both lie on the same base BQ and between the same parallels AP and BQ.
∴ ar (ABQ) = ar (PBQ) …(1)
∆BCQ and ∆BRQ both lie on the same base BQ and between the same parallels BQ and CR.

∴ ar (BCQ) = ar (BRQ) …(2)
Adding (1) and (2), we get
ar (ABQ) tar (BCQ) = ar (PBQ) + ar (BRQ)
∴ ar (AQC) = ar (PBR)

Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
Solution:
Given:
ar (AOD) = ar (BOC)
To prove:
ABCD is a trapezium.

Proof:
ar ∠AOD = ar (BOC) (given)
Adding ar (AOB) on both sides
ar (AOD) + ar (AOB) = ar (BOC) + ar (AOB)
ar (ABD) = ar (ABC)
∆ABD and ∆ABC both lie on same base AB and have equal area.
∴ they will lie between the same parallels.
⇒ AB ∥ DC
∴ ABCD is a trapezium.

Question 16.
In Fig. below, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:
Given
ar (DRC) = ar (DPC)
ar (BDP) = ar (ARC)
To prove:
ABCD and DCPR are trapezium.

Proof:
∆DRC and ∆DPC both lie on the same base DC and have equal area .•. They will lie between the same parallels.
⇒ DC ∥ BP
and so DCPR is a trapezium.
ar (BDP) = ar (ARC) [given] …(1)
ar (DPC) = ar (DRC) [given] …(2)
Subtracting (1) and (2), we get
ar (BDP) – ar (DPC) = ar (ARC) – (DRC)
ar (BDC) = ar (ADC)
∆ADC and ∆BDC both lie on same base DC and have equal area.
∴ they will lie between the same parallels
⇒ AB ∥ DC
and so ABCD is a trapezium.

MP Board Class 9th Maths Solutions