Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Class 10 Mathematics by VP Classes

Class 10 : Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

The document Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
All you need of Class 10 at this link: Class 10

NCERT TEXTBOOK QUESTIONS SOLVED
Page No. 251
EXERCISE 13.3

[Take π = 22/7, (Unless stated otherwise)]
Q 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.
Sol. Radius of the sphere (r1) = 4.2 cm
∴ Volume of the sphere Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Radius of the cylinder (r2) = 6 cm
Let ‘h’ be the height of the cylinder
∴ Volume of the cylinder =Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Since, Volume of the metallic sphere = Volume of the cylinder.
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Q 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.
Sol. Radii of the given spheres are:
r1 = 6 cm
r2 = 8 cm
r3 = 10 cm
⇒ Volume of the given spheres are:
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
∴ Total volume of the given spheres
= V1 + V2 + V3
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRevEx 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Let the radius of the new big sphere be R.
∴ Volume of the new sphere
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Since, the two volume must be equal.
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
⇒ R3 = 1728
⇒ R3 =23 × 23 × 33
⇒ R3 = (2 × 2 × 3)3
⇒ R = 2 × 2 × 3
⇒ R = 12 cm
Thus, the required radius of the resulting sphere = 12 cm.

Q 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.
Sol. Given, diameter of the well = 7m
⇒ Radius = 7/2m
and depth of the well = 20 m
Volume of the earth taken out from the well = πr2h
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Earth is evenly spread on a platform 22 m by 14 m
Let height of platform be hm
∴ Volume of platform = 22 x 14 x h
According to the question,
Volume of the platform = Volume of the earth taken out from the well
⇒ 22 x 14 x h
⇒ π x 49 x 5
⇒22/7 x 49 x 5
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Q 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 shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Sol. Diameter of cylindrical well (d) = 3 m
⇒ Radius of the cylindrical well (r) = 3/2 m = 1.5m
Depth of the well (h) = 14 m
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Let the height of the embankment = ‘H’ meter
Internal radius of the embankment (r) = 1.5 m.
External radius of the embankment R = (4 + 1.5) m = 5.5 m.
∴ Volume of the embankment
= πR2H − πr2H = πH [R2 − r2] = πH (R + r) (R − r)
= 22/7 × H (5.5 + 1.5) (5.5 − 1.5) = 22/7 × H × 7 × 4 m3
Since, Volume of the embankment = Volume of the cylindrical well
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Thus, the required height of the embankment = 1.125 m.

Q 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.
Sol. For the circular cylinder:
Diameter = 12 cm ⇒ Radius = 12/2 =  6 cm
Height (h) = 15 cm
∴ Volume = πr2h
⇒ Volume of total ice cream = 22/7 × 6 × 6 × 15 cm3
For conical + hemispherical ice-cream cone:
Diameter = 6 cm ⇒ radius (R) = 3 cm
Height of ice cream cone (H) = 12 cm
Volume = (Volume of the conical part) + (Volume of the hemispherical part)
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Let number of ice-cream cone required to fill the total ice cream = n.
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Thus, the required number of cones is 10.

Q 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 × 10 cm × 3.5 cm?
Sol. For a circular coin:
Diameter = 1.75 cm
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev   |∵ A coin is like a cylinder
For a cuboid:
Length (l) = 10 cm, Breadth (b) = 5.5 cm
and Height (h) = 3.5 cm
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Number of coins
Let the number of coins need to melt be ‘n’
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Thus, the required number of coins = 400.

Page No. 252
Q 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.
Sol. For the cylindrical bucket:
Radius (r) = 18 cm
Height (h) = 32 cm
Volume = πr2h
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

⇒ Volume of the sand = Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

For the conical heap:
Height (H) = 24 cm
Let radius of the base be (R).
∴ Volume of conical heap =  Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Radius of the conical heap of sand:
∵ Volume of the conical heap of sand = Volume of the sand
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRevEx 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Slant Height
Let ‘l’ be the slant height of the conical heap of the sand.
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
∴ l2 = R2 + H2
⇒ l2 =242 + 362
⇒ l2 = (12 × 2)2 + (12 × 3)2
⇒ l2 =122 [22 + 32]
⇒ l2 =122 × 13
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Thus, the required height = 36 cm and slant height   Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Q 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?
Sol. Width of the canal = 6 m
Depth of the canal = 1.5 m
Length of the water column in 1 hr = 10 km
∴ Length of the water column in 30 minutes  Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Since the above amount (volume) of water is spread in the form of a cuboid of height as
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Let the area of the cuboid = a
∴ Volume of the cuboid=Area × Height Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Thus,
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Thus, the required area = 56.25 hectares.

Q 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, 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 ?
Sol.  Given, diameter of pipe = 20 cm
⇒ radius = 10 cm
Water flowing from the pipe at rate = 3 km/h
Let it filled the tank in t hours.
Volume of water flowing in t hours
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Volume of the water in cylindrical tank = πr2h = π(5)x 2 m3
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev  
According to the question,
Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of Class 10

Dynamic Test

Content Category

Related Searches

Summary

,

shortcuts and tricks

,

Previous Year Questions with Solutions

,

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

,

Extra Questions

,

Free

,

Semester Notes

,

Viva Questions

,

Objective type Questions

,

Important questions

,

mock tests for examination

,

Sample Paper

,

ppt

,

video lectures

,

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

,

Exam

,

study material

,

practice quizzes

,

past year papers

,

Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

,

pdf

,

MCQs

;