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

## 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.
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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 Radius of the cylinder (r2) = 6 cm
Let ‘h’ be the height of the cylinder
∴ Volume of the cylinder = Since, Volume of the metallic sphere = Volume of the cylinder. 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: ∴ Total volume of the given spheres
= V1 + V2 + V3   Let the radius of the new big sphere be R.
∴ Volume of the new sphere Since, the two volume must be equal.  ⇒ 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
and depth of the well = 20 m
Volume of the earth taken out from the well = πr2h 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 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  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 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) Let number of ice-cream cone required to fill the total ice cream = n. 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  |∵ A coin is like a cylinder
For a cuboid:
Length (l) = 10 cm, Breadth (b) = 5.5 cm
and Height (h) = 3.5 cm Number of coins
Let the number of coins need to melt be ‘n’ 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:
Height (h) = 32 cm
Volume = πr2h ⇒ Volume of the sand = For the conical heap:
Height (H) = 24 cm
Let radius of the base be (R).
∴ Volume of conical heap = Radius of the conical heap of sand:
∵ Volume of the conical heap of sand = Volume of the sand    Slant Height
Let ‘l’ be the slant height of the conical heap of the sand. ∴ l2 = R2 + H2
⇒ l2 =242 + 362
⇒ l2 = (12 × 2)2 + (12 × 3)2
⇒ l2 =122 [22 + 32]
⇒ l2 =122 × 13 Thus, the required height = 36 cm and slant height 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    Since the above amount (volume) of water is spread in the form of a cuboid of height as Let the area of the cuboid = a
∴ Volume of the cuboid=Area × Height Thus,  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
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 Volume of the water in cylindrical tank = πr2h = π(5)x 2 m3 According to the question, Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

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