Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

The document Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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NCERT TEXTBOOK QUESTIONS SOLVED

Page No. 244
EXERCISE 13.1

[Unless stated otherwise, take π = 22/7]

Q 1. Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid
Sol.  Volume of each cube = 64 cm3
∴ Total volume of the two cubes = 2 × 64 cm3 = 128 cm3
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Let the edge of each cube = x
∴ x3 = 64 = 43
⇒ x =4 cm
Now, Length of the resulting cuboid l =2x cm
Breadth of the resulting cuboid b = x cm
Height of the resulting cuboid h = x cm
∴ Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(2x · x) + (x · x) + (x · 2x)]
= 2 [(2 × 4 × 4) + (4 × 4) + (4 × 2 × 4)] cm2
= 2 [32 + 16 + 32] cm2 = 2 [80] cm2 = 160 cm2.

Q 2. 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.
Sol. For cylindrical part:
Radius (r) = 7 cm
Height (h) = 6 cm
∴ Curved surface area
= 2πrh
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
For hemispherical part: 
Radius (r) = 7 cm
∴ Surface area = 2πr2
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
∴ Total surface area
= (264 + 308) cm2 = 572 cm2.

Q 3. 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.
Sol. Here, r = 3.5 cm
∴ h = (15.5 − 3.5) cm = 12.0 cm
Surface area of the conical part = πrl
Surface area of the hemispherical part =2πr2
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Q 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Sol. Side of the block = 7 cm
⇒ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid
= [Total S.A. of the cubical block] + [S.A. of the hemisphere] − [Base area of the hemisphere]
= (6 × l2) + 2πr2 − πr2
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

  Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRevEx 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Q 5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Sol. Let ‘l’ be the side of the cube.
∴ The greatest diameter of the curved hemisphere = l
⇒ Radius of the curved hemisphere  Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
∴ Surface area of hemisphere = 2πr2
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Base area of the hemisphere   Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Surface area of the cube = 6 × l2 = 6l2
∴ Surface area of the remaining solid
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Q 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig.). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Sol. Radius of the hemispherical part
= 5/2 mm = 2.5 mm
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
∴ Surface area of one hemispherical part = 2πr2
⇒ Surface area of both hemispherical parts
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Area of cylindrical part = Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
∴ Total surface area
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Page No. 245
Q 7. 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 be covered with canvas.)
Sol. For cylindrical part:
Radius (r) = 4/2 m = 2 m
Height (h) = 2.1 m
∴ Curved surface area = 2πrh =  Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
For conical part:
Slant height (l) = 2.8 m
Base radius (r)= 2m
∴ Curved surface area  Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Total surface area 
= [Surface area of the cylindrical part] + [Surface area of conical part]
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Cost of the canvas used: 
Cost of 1 m2 of canvas = Rs 500
∴ Cost of 44 m2 of canvas = Rs 500 × 44 = Rs 22000.

Q 8. 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.
Sol. For cylindrical part:
Height = 2.4 cm
Diameter = 1.4 cm
⇒ Radius (r) = 0.7 cm
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
For conical part: 
Base area (r) = 0.7 cm
Height (h) = 2.4 cm
∴ Slant height Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRevEx 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
∴ Curved surface area of the conical part
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRevEx 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Base area of the conical part
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Total surface area of the remaining solid
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRevEx 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev

Q 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Sol. Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
∴ Total surface area = 2πrh + 2πr2 = 2πr (h + r)
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Curved surface area of a hemisphere = 2πr2
∴ Curved surface area of both hemispheres
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
Base area of a hemisphere = πr2
∴ Base area of both hemispheres = 2πr2
Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 10 Notes | EduRev
∴ Total surface area of the remaining solid
= 297 cm2 + 154 cm2 − 77 cm2 
= (451 − 77) cm2 = 374 cm2.

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