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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-13-ex-13-3/

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

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

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Given: radius of metallic sphere = 4.2 cm
∴ Volume = \(\frac { 4 }{ 3 }\)π(4.2)³ …. (i)
∵ Sphere is melted and recast into a cylinder of radius 6 cm and height h.
∴ Volume of the cylinder =πr²h = π(6)² x h … (ii)
According to question,
Volume of the cylinder = Volume of the sphere

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radius of 1st metallic sphere = 6 cm
∴ Volume of 1^st metallic sphere = \(\frac { 4 }{ 3 }\)π(6)³ cm³
Radius of 2^nd metallic sphere = 8 cm
∴ Volume of 2^nd metallic sphere = \(\frac { 4 }{ 3 }\)π(8)³ cm³
Radius of 3^rd metallic sphere = 10 cm
∴ Volume of 3^rd metallic sphere = \(\frac { 4 }{ 3 }\)π(10)³ cm³
Volume of all three metallic spheres = \(\frac { 4 }{ 3 }\)π(6³+8³+10³) cm³
∵ 3 spheres are melted and recast into a new metallic sphere of radius r.
∴ Volume of new metallic sphere = \(\frac { 4 }{ 3 }\)πr³

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Given: diameter of the well = 7 m Radius = \(\frac { 7 }{ 2 }\)m
and depth of the well = 20 m
Volume of the earth taken out from the well = πr²

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Given: diameter of the well = 3 m
⇒ Radius = \(\frac { 3 }{ 2 }\)m
Depth of the well = 14 m
Volume of the earth taken out from the well = πr²h
= π(\(\frac { 3 }{ 2 }\))² x 14 = \(\frac { π×9×14 }{ 4 }\) = \(\frac { 63 }{ 2 }\)πm³
∵ Earth taken out from the well evenly spread to form an embankment having height h and width of embankment around the well is 4 m.
∴ External radius (R) = radius of well + width of the embankment
= \(\frac { 3 }{ 2 }\)m + 4m = \(\frac { 11 }{ 2 }\)m
Internal radius = \(\frac { 3 }{ 2 }\)m = radius of well

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:

Let total number of ice cream cones are n.
∴ All ice cream cones are filled from ice cream in the cylinder.
Total volume of n number of ice cream cones = Volume of ice cream in the cylinder
n x π x 54 = π(36)15
⇒ n x 54 = 36 x 15
⇒ n = \(\frac { 36 × 15 }{ 54 }\) = 10

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
Solution:
Given: diameter of each coin = 1.75 cm ⇒ radius = \(\frac { 1.75 }{ 2 }\)cm
and thickness of each coin = 2 mm
Let n number of coins are melted to form a cuboid.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Given: radius of the cylindrical bucket = 18 cm
and height = 32 cm
∴ Volume of the cylindrical bucket = π²h = π(18)² x 32 cm³
Let radius of the conical heap = r cm
Given: height of the conical heap = 24 cm
∴ Volume of the conical heap = \(\frac { 1 }{ 3 }\)π(r²) 24cm³
According to question,
Volume of the cylinderical bucket = Volume of the conical heap

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Given: width of canal = 6m, depth = 1.5 m
Rate of flowing water – 10 km/h
Volume of the water flowing in 30 minutes = \(\frac { 6×1.5×30×10 }{ 60 }\)km³
= \(\frac { 6×1.5×10×1000×30 }{ 10×60 }\)km³ = 45000 m³
We require water for standing up to height = 8 cm = \(\frac { 8 }{ 100 }\) m
Let the required area he A
∴ Volume of water required = A(\(\frac { 8 }{ 100 }\))m³
According to question. 45000 = \(\frac { A×8 }{ 100 }\)
⇒ \(\frac { 45000×100 }{ 8 }\) = A ⇒ A = 562500 m²

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his Held, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Given: diameter of the pipe = 20 cm ⇒ radius of the pipe = 10 cm
Water flowing from the pipe at rate = 3 km
Let it filled the tank in ‘t’ hours.
Volume of the water flowing in ‘t’ hours.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-14-ex-14-2/

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2

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

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
For Mode:

For Mean:

We conclude that the maximum number of patients in the hospital are of the age 36.8 years. While on an average the age of patient admitted to the hospital is 35.37 years.

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

Determine the modal lifetimes of the components.
Solution:

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

Solution:

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

Solution:

Hence, we conclude that most states/U.T. have a student teacher ratio of 30.6 and on an average this ratio is 29.2.

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

Find the mode of the data.
Solution:

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

Solution:

We hope the NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-12-ex-12-3/

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Question 1.
Find the area of the shaded region in the given figure, if PQ = 24cm, PR = 7cm and O is the centre of the circle.
Solution:

Question 2.
Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 400.
Solution:
∠AOC = 40° (given)
Radius of the sector AOC = 14 cm

Question 3.
Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Solution:
ABCD is a square

Given: side of the square = 14 cm
∴ Area of the square = (side)² = (14)² = 196 cm²
Radius of the semicircle APD = \(\frac { 1 }{ 2 }\)(side of square) = \(\frac { 1 }{ 2 }\) x 14 = 7 cm
Area of the semicircle APD = \(\frac { 1 }{ 2 }\) πr² = \(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 = 11 × 7 = 77cm²
Similarly, area of the semicircle BPC = 77 cm²
Total area of both the semicircles = 77 + 77 = 154 cm²
Area of the shaded region = Area of square – area of both semicircles
= 196 – 154 = 42 cm²

Question 4.
Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
Solution:
Area of the equilateral triangle OAB

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.

Solution:
Given: side of the square ABCD = 4 cm
Area of the square ABCD = 4 x 4 = 16 cm²
Radius of the quadrant at corner = 1 cm

∴ Area of the remaining portion = Area of the square – Area to be cut from square
= 16 – (\(\frac { 44 }{ 7 }\)) = 16 – \(\frac { 44 }{ 7 }\)
= \(\frac { 112-44 }{ 7 }\) = \(\frac { 68 }{ 7 }\)cm²

Question 6.
In a circular table cover of the radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).

Solution:
Radius of the circle(r) = 32cm
Area of the circle = πr²
= \(\frac { 22 }{ 7 }\) × 32 × 32 = \(\frac { 22528 }{ 7 }\)cm²
∴ An equilateral triangle is formed in the circle as shown
Angle subtended by et ch side at centre

Question 7.
In the figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:
Side of the square ABCD = 14 cm
Area of the squat e = (side)² = 14 x 14 = 196²

Question 8.
The given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

(i) the distance around the track along its inner edge.
(ii) the area of the track.
Solution:

Area of the tracks at both semicircular ends = 2 x 1100 = 2200 cm²
Area of the 2 rectangular portions = 2 x l x h = 2 x 106 x 10 = 2120 cm²
Total area of the track = area of the track at semicircular ends + area of the rectangular portions
= 2200 + 2120 = 4320 cm²

Question 9.
In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
Given: OA = 7 cm
Radius of the semicircle ABC = OA = 7 cm
Area of the semicircle ABC = \(\frac { 1 }{ 2 }\)πr² = \(\frac { 1 }{ 2 }\) x \(\frac { 22 }{ 7 }\) x 7 x 7 = 11 x 7 = 77 cm²
Diameter AB = 2(OA) = 2 x 7 = 14 and OA = OC = 7 cm (radius)

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.
(Use π = 3.14 and \(\sqrt{3}\) = 1.73205).

Solution:
Given: area of an equilateral triangle ABC = 17320.5 cm²
Let side of the triangle AB’C be ‘a’
∴ Area of the ∆ABC = \(\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\)
\(\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\) = 17320.5

Now,
area of the all 3 sectors = \(\frac { 3×31400 }{ 6 }\) = 15700 cm²
∴ Area of the shaded portion = Area of the equilateral triangle
– Area of the three sectors formed at each vertex)
= 17320.5 – 15700 = 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of the radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Solution:
Radius of the one circular design = 7 cm
Area of the one circular design = πr² = \(\frac { 22 }{ 7 }\) x 7 x 7 = 154 cm²
Now, area of the 9 circular designs = 9 x 154 = 1386 cm²
Diameter of the circular design = 7 x 2 = 14 cm
Side of the square = 3(diameter of one circle) = 3 x 14 = 42 cm
Area of the square = 42 x 42 = 1764 cm²
Area of the remaining portion of handkerchief
= Area of the square – (Area of the 9 circular designs)
= 1764 – 1386
= 378 cm²

Question 12.
In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region.

Solution:
(i) Radius of the quadrant OACB = 3.5 cm

(ii) OD = 2cm and OB = 3.5 cm

Question 13.
In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use re = 3.14)
Solution:
Given: side of the square OABC = OA = 20 cm
Area of the square = 20 x 20 = 400 cm²
(Diagonal of the square)² = (side of the square)² + (side of the square)² (By pythagoras theorem)
Diagonal of the square = \(\sqrt{2}\) x (side of the square)
= \(\sqrt{2}\) x (20) = 20\(\sqrt{2}\)cm
Radius of the quadrant of circle = Diagonal of square = 20\(\sqrt{2}\)

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB=30°, find the area of the shaded region.
Solution:
Given: ∠AOB = 30°
Radius of the sector AOB = 21 cm

Question 15.
In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Radius of the quadrant of the circle = 14 cm

Area of the shaded region = Area of the semicircular region – Area of the region I
= 154 cm² – 56 cm² = 98 cm²

Question 16.
Calculate the area of the designed region in the figure common between the two quadrants of the circles of the radius 8 cm each.
Solution:
Side of the square = 8 cm
Area of the square = 8 x 8 = 64 cm²
Radius of the quadrant (formed at vertex) = 8 cm

We hope the NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3, drop a comment below and we will get back to you at the earliest.