NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume

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Updated at 2021-02-24


NCERT solutions for class 10 Maths Chapter 13 Surface Areas And Volume Exercise - 13.1

Q1 ) 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Let the length of each edge of the cube be a cm.

Then, Volume = 64cm3

? a3 = 64 
? a = 4 cm

When two cubes of equal volumes (i.e., equal edges) are joined end to end, we get a cuboid such that its,

l = Length = 4 + 4 = 8 cm

b = Breadth = 4 cm

and h = Height = 4 cm

? Surface area of the cuboid = 2(lb + bh + hl)

= 2(8 × 4 + 4 × 4 + 4 × 8)

=2(32 + 16 + 32)

=(2 × 80)

= 160 cm2

Q2 ) A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Here r, the radius of hemisphere = 7 cm ,

h , the height of cylinder = (13 - 7) = 6 cm



Clearly, radius of the base of cylindrical part is also r cm

Surface area of the vessel = Curved surface area of the cylindrical part + Curved surface area of hemispherical part

= 2?rh + 2?r2 = 2?r(h + r)

= 2 × 227 × 7 × 13 = 572 cm2

Q3 ) A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


We have VO = 15.5 cm, OA = OO' = 3.5 cm



Let r be the radius of the base of cone and h be the height of conical part of the toy.

Then r = OA = 3.5 cm

h = VO = VO' - OO = (15.5 - 3.5) cm = 12 cm

Also radius of the hemisphere = OA = r = 3.5 cm

l = Slant height, is given by :
?l=r2+h2

?l= OA2 + OV2 
= (3.5)2 + 122 
= 156.25 
= 12.5cm

Total surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere

= ?rl + 2?r2

= ?r(l + 2r) 
= 227 × 3.5 × 12.5 + 2 × 3.5

= 227 × 3.5 × 19.5 
= 214.5 cm2

Q4 ) A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter of the hemisphere can have? Find the surface area of the solid?



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


The greatest diameter that a hemisphere can have = 7 cm .

Surface area of the solid after surmounted hemisphere

= 6l2 ? ?R2 + 2?R2

= 6l2 + ?R2

= 6(7)2 + 227 × (72)2

= 6 × 49 + 11 × 72

= 294 + 38.5 = 332.5 cm2

Q5 ) A hemispherical depression is cut of one face of a cubical wooden block such the diameter l of the hemisphere is equal to the edge of the cube. Determine th surface area of the remaining solid.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Edge of the cube =l

Diameter of the hemisphere = l

? Radius of the hemisphere =l2

? Area of the remaining solid after cutting out the hemispherical depression.

= 6l2 ? ?(l2)2 + 2?(l2)2 
= 6l2 + ?(l2)2

= 6l2 + ? × l24 
= l24(24 + ?)

Q6 ) A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Let r cm be the radius and h cm be the height of the cylinder. Then.



r = 52 = 2.5mm

h = (14 ? 2 × 52) =14 ? 5 = 9mm

Also the radius of hemisphere = 52 cm

Now, surface area of the capsule = Curved surface of cylinder + Surface area of two hemispheres

= 2?rh + 2 × 2?r2 = 2?r(h + 2r)

= 2 × 227 × 52 × (9 + 2 × 52) = 220 mm2

Q7 ) A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2 . (Note that the base of the tent will not covered with canvas).



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




We have,

Height of the cylinderical part = 2.1 m

Diameter of the cylinderal part = 4 m

Radius of the cylinerical part = 2 m

Slant height of the conical part = 2.8 m

Thus, Total canvas used = Curved surface area of cylinder + Curved surface area of cone

= 2?rh + ?rl = ?r(2h + l)

= 227 × 2 × (2 × 2.1 + 2.8) 
= 44 m2

Now, cost of 1 m2 the canvas for the tent = Rs 500

So, cost of 44 m2 the canvas for the tent = Rs 44 × 500 = Rs 22000

Q8 ) From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Radius of the cylinder = 1.42 = 7cm

Height of the cylinder = 2.4 cm

Radius of the cone = 0.7 cm

Height of the cone = 2.4 cm

Slant height of the cone = (0.7)2 + (2.4)2 = 0.49 + 5.76

= 6.25 = 2.5 cm

Surface area of the remaining solid = Curved surface area of cylinder + Curved surface of the cone + Area of upper circular base of cylinder

= 2?rh + ?rl + ?r2 = ?r(2h + l + r)

= 227 × 0.7 × (2 × 2.4 + 2.5 + 0.7)

= 22 × 0.1 × (4.8 + 2.5 + 0.7)

= 2.2 × 8.0 = 17.6 = 18 cm2

Q9 ) A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm,and its base is of radius 3.5 m, find the total surface area of the article.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Surface area of the article when it is ready = Curved surface area of cylinder + 2 × Curved surface area of hemisphere.

= 2?rh + 2 × 2?r2 = 2?r(h + 2r)

where r = 3.5 m and h = 10 cm

= 2 × 227 × 3.5 × (10 + 2 × 3.5)

= 2 × 227 × 3.5 × 17 = 374 cm2

NCERT solutions for class 10 Maths Chapter 13 Surface Areas And Volume Exercise - 13.2

Q1 ) A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of ?.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Volume of the solid = Volume of the cone + Volume of the hemisphere

= 13?r2h + 23?R3

= 13?r2 × r + 23?r3 [? h = r and R = r]

= ?3r3(1 + 2) = ?r3

= ?(1)3 = ? cm3 [? r = 1 cm]

Q2 ) Rachel an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm,find the volume of air contained in the model the Rachel made. (Assume the outer and inner dimensions of the model be nearly the same).



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Volume of the air contained in the model = Volume of the cylindrical portion of the model + Volume of its two conical ends.

= ?r2h1 + 2 × 13?r2h2 
= ?r2(h1 + 23h2)

where r = 32 cm , h1 = 8 cm and h2 = 2 cm

= 227 × (32)2 × (8 + 23 × 2) = 227 × 94 × 24 + 43

= 227 × 94 × 283 = 66 cm3

Q3 ) A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemsipherical ends with length 5 cm and diameter is 2.8 cm (see figure).



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Volume of the gulab jamun = Volume of the cylindrical portion + Volume of the hemispherical ends

= ?r2h + 2(23?r3) 
= ?r2(h + 43r)

where r = 1.4 cm, h = 2.2 cm

= 227 × (1.4)2 × (2.2 + 43 × 1.4) 
= 227 × 1.96 × (6.6 + 5.63)

= 227 × 1.96 × 12.23 cm3

Volume of 45 gulab jamuns

= 45 × 227 × 1.96 × 12.23 cm3

Quantity of syrup in gulab jamuns = 30% of their volume

= 30100 × 45 × 227 × 1.96 × 12.23

= 9 × 11 × 1.96 × 12.27 
= 338.184

= 338 cm3

Q4 ) A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 m and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Volume of wood in the entire stand = Volume of the cuboid - 4 × Volume of a depression (i.e., cone)

= lbh ? 4 × 13?r2h

= 15 × 10 × 3.5 ? 4 × 13 × 227 × 0.5 × 0.5 × 1.4

= 525 ? 1.47 = 523.53 cm3

Q5 ) A vessel in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one- fourth of the water flows out. Find the number of lead shots dropped in the vessel.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Height of the conical vessel, h = 8 cm.

Its radius r = 5 cm

Volume of cone = Volume of water in cone = 13?r2h

= 13 × 227 × 5 × 5 × 8 = 440021 cm3

Volume of water flows out = Volume of lead shots

= 14 of the volume of water in the cone

= 14 × 440021 = 110021 cm3

Radius of the lead shot = 0.5 cm

Volume of one spherical lead shot = 43?r3 = 43 × 227 × (510)3

= 1121 cm3

? Number of lead shots dropped into the vessel

= Volume of water flows outVolume of one lead shot =1100211121

= 110021 × 2111 = 100

Q6 ) A solid iron pole consists of a cylindrical height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use ? = 3.14 )



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Volume of the solid iron pole = Volume of the cylindrical portion + Volume of the other cylindrical portion

= ?(r1)2h1 + ?(r2)2h2

= 3.14 × (12)2 × 220 + 3.14 × (8)2 × 60

= 3.14 × 144 × 220 + 3.14 × 64 × 60

= 99475.2 + 12057.6

= 111532.8 cm3

? The mass of the pole
= 111532.8 × 8 grams

= 111532.8 × 81000 = 892.26 kg

Q7 ) A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water circular such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Volume of the cylinder = ?r2h = 227 × (60)2 × 180

= 22 × 3600 × 1807 = 142560007 cm3

Volume of the solid = Volume of cone + Volume of hemisphere

= 13 × 227 × 602 × 120 + 23 × 227 × 603

= 31680007 + 31680007 = 63360007 cm3

Volume of water left in the cylinder = Volume of the cylinder - Volume of the solid

= 142560007 ? 63360007 =79200007

= 1131428.57142 = 1.131 m3

Q8 ) A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter ; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds , a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements and ? = 3.14



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Volume of spherical vessel = Volume of sphere + Volume of cylinder

= ?(1)2(8) + 43?(4.25)3

= 3.14(8 + 43 × 76.765625) 

= 3.14(8 + 102.353)

= 3.14 × 110.353 

= 346.51 cm3

? She is incorrect. The correct answer is 346.51 cm3

NCERT solutions for class 10 Maths Chapter 13 Surface Areas And Volume Exercise - 13.3

Q1 ) 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.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Volume of the sphere = 43?r3 = 43 × ? × (4.2)3 cm3

If h is the height of a cylinder of radius 6 cm. Then its volume = ?(6)2h = 36?h cm3

? The volume of metal in the form of sphere and cylinder remains the same , we have

?36?h = 43 × ? × (4.2)3

? h = 136 × 43 × (4.2)3

? h = 2.744 cm

Q2 ) 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.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Sum of the volumes of 3 gives spheres.

= 43?((r1)3 + (r2)3 + (r3)3)

= 43?(63 + 83 + 103)

= 43?(216 + 512 + 1000)

= 43?(1728) cm3

Let R be the radius of the new spheres whose volume is the sum of the volumes of 3 given spheres.

? 43?R3 = 43?(1728)

? R3 = 1728 => R = 12 cm

? The radius of the resulting sphere is 12 cm.

Q3 ) 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.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Let h be the required height of the platform.

The shape of the platform will be like the shape of a cuboid of dimensions 22 m × 14 m × h with a hole in the shape of cylinder of radius 3.5m and depth h.

The volume of the platform will be equal to the volume of the earth dug out from the well.

Now, the volume of the earth = Volume of the cylindrical well

= ?r2h = 227 × 12.25 × 20 = 770 m3

Also, the volume of the platform = 22 × 14 × h m3

But volume of the platform = Volume of the well

? 22 × 14 × h = 770

? h = 77022 × 14 = 2.5 m

? Height of the platform = 2.5 m

Q4 ) 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.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Let h be the required height of the embankment.



The shape of the embankment will be like the shape of a cylinder of internal radius 1.5 m and external radius 4 + 1.5 = 5.5 m

The volume of the embankment will be equal to the volume of earth dug out from the well.

Now, the volume of the earth = Volume of the cylindrical well

= ? × (1.5)2 × 14 = 31.5? m3

Also, the volume of the embankment

= ?(5.52 ? 1.52)h = ? × 7 × 4h

= 28?h m3

?28?h = 31.5?

? h = 31.528 = 1.125 m

Hence, the required height of the embankment = 1.125 m

Q5 ) 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 hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Volume of the cylinder
= ?r2h = ?(122)2 × 15 

= ? × 62 × 15 cm3

Volume of a cone having hemispherical shape on the top

= 13?r2h + 23?r3 

= 13?r2(h + 2r)

= 13?(62)2(12 + 2 × 62) 

= 13? × 32 × 18 cm3

Let the number of cone that can be filled with ice cream be n.

Then, 13? × 32 × 18 × n = ? × 62 × 15

? n = ? × 6 × 6 × 15 × 3 ? × 3 × 18 × 3 = 10

? 10 cones can be filled with ice cream.

Q6 ) 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?



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


The shape of the coin will be like the shape of a cylinder of radius
1.752 = 0.875 cm and of height 2 mm = 0.2 cm

Its volume = ?r2h 

= 227 × 0.875 × 0.875 × 0.2

= 0.48125 cm3

Volume of the cuboid = 5.5 × 10 × 3.5 = 192.5 cm3

Number of coins required to form the cuboid

= Volume of the cuboidVolume of the coin 

= 192.50.48125 = 400

? 400 coins must be melted to form a cuboid.

Q7 ) A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. The 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.



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :




Volume of the sand = Volume of the cylindrical bucket = ?r2h = ? × 18 × 18 × 32 cm3

Volume of the conical heap = 13?r2h where, h = 24 cm

= 13?r2 × 24 = 8?r2

The volume of the conical heap will be equal to that of sand.

therefore8?r2 = ? × 18 × 18 × 32

? r2 = 18 × 18 × 4 = 182 × 22

? r = 18 × 2 = 36

Here, slant height, l = r2 + h2

? l = 362 + 242 
= 129 + 4 
= 1213

? The radius of the conical heap is 36 cm and its slant height is 1213 cm

Q8 ) 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?



NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volume


Answer :


Width of the canal = 6 m

Depth of the canal = 1.5 m

Length of water column per hour = 10 km

Length of water column in 30 minutes or