NCERT Solutions For Class 10 Maths Chapter 13 - Surface Areas and Volumes

 NCERT Solutions For  Class 10 Maths Chapter 13 - Surface Areas and Volumes

Exercise 13.1 

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

Solution

`r = 7/2` cm 

Total height of the toy `= 15.5` cm 

Height of the cone `= (15.5 - 3.5)` cm `= 12`cm  

`l = sqrt{r^2 + h^2}`

`= sqrt{(7/2)^2 + 12^2}`

`= sqrt {49/4 + 144}`

`=  sqrt {frac{49+ 576}{4}}`

`= sqrt {625/4}`

`= 25/2`cm

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

`= pirl + 2pir^2`

`= pir(l + 2r)`

`= 22/7 × 7/2 (25/2 + 2× 7/2 )`

`= 11 (frac{25+ 14}{2})`

` = 11 × 39/2`

`= 214.5 cm^2`

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.

Solution 

The greatest diameter the hemisphere  `= 7` cm

Side = 7 cm 

`r = 7/2`

SA of the solid `=` CSA of the hemisphere `-` The base area of the hemisphere `+` TSA of the cubical block 

`= 2pir^2 - pir^2 + 6a^2`

`= pir^2 + 6a^2`

`= 22/7 ×7/2 ×7/2 + 6 × 7×7`

` = 77/2 + 294`

`= 38.5 + 294`

`= 332.5 cm^2`

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.

Solution

Edge `= l`

`r = l/2`

SA of remaining solid `=` CSA of the hemisphere `-` base area of the hemisphere `+` TSA of Cubical wooden block    

`= 2pir^2 - pir^2 + 6a^2`

`= pir^2 + 6a^2`

`= pi(l/2)^2 + 6 × l^2`

`= pifrac{l^2}{4} + 6l^2`

`= frac{pil^2 + 24l^2}{4}`

`= frac{l^2}{4} (pi + 24)`

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.

Solution

height `= (14 - 5)`mm =  9mm

`r = 2.5`mm

 Surface area of the capsule `= 2×` CSA of hemisphere `+` CSA of cylinder

`= 2 × 2pir^2 + 2pirh`

`= 2pir (2r + h)`

`= 2× 22/7 ×5/2 (2× 5/2 + 9)` 

`= 110/7 × 14`

`= 110 × 2`

`= 220 mm^2`

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 `₹ 500` per `m^2`. (Note that the base of the tent will not be covered with canvas.)

Solution

Height of the cylinder = 2.1m

`r = 2`m

`l = 2.8`m  

Area of the canvas `=` CSA of the cone `+` CSA of the cylinder

`= pirl + 2pirh`

`= pir (l + 2h)` 

`= 22/7 × 2 (2.8 + 2× 2.1)`

`= 44/7 (2.8 + 4.2)`

`= 44/7 × 7`

`= 44 m^2` 

`therefore`the cost of the canvas of the tent at the rate of `₹ 500` per `m^2`

= 44 × 500 = ₹22000

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 `cm^2`.

Solution

Height of the cylinder = height of the conical cavity = 2.4 cm 

`r = 0.7` cm

`l = sqrt{r^2 + h^2}`

`= sqrt {0.7^2 + 2.4^2}`

`= sqrt{0.49 + 5.76}`

`= sqrt{6.25}`

`= 2.5` cm 

TSA of remaining solid  `=` CSA of conical cavity `+` CSA of cylinder `+` Base area of the cylinder 

`= pirl + 2pirh + pir^2`

`= pir(l + 2h + r )`

= `22/7` ×` 0.7 (2.5 + 2× 2.4 + 0.7)`

`= 2.2 (2.5 + 4.8 + 0.7)`

`= 2.2 ×8`

`= 17.6 cm^2`

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.

Solution

Height of the cylinder = 10 cm

`r = 3.5` cm 

Total surface area of the wooden article `=` CSA of the cylinder + 2 × CSA of the hemisphere 

`= 2pirh + 2× 2pir^2`

`= 2pir(h + 2r)`

`= 2 × 22/7× 7/2(10 + 2× 7/2)`

`= 22× 17`

`= 374 cm^2`

Exercise 13.2

1. 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 `pi`

Solution 

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

`= 1/3pir^2h + 2/3 pir^3`

`= 1/3pir^2 (h + 2r)`

`= 1/3 pi×1^2 (1 + 2× 1)`

`= 1/3pi × 3`

`= pi`

7. 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 such that it touches the bottom. Find the volume

Solutions 

Radius of the cone, cylinder and hemisphere = 60 cm 

Height of the cone = 120 cm 

Height of the cylinder = 180 cm 

Total volume of the solid = volume of the cone + volume of the hemisphere 

`= 1/3pir^2h + 2/3pir^3`

`= 1/3pir^2 (h + 2r)`

`= 1/3 × pi × 60^2 (120 + 2× 60)`

`= 1200 × pi (240)`

`= 288 × 10^3 pi`

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

`= pir^2h -  288 × 10^3 pi`

`= pi(r^2h -  288 × 10^3)`

`= 22/7 (60^2 × 180 - 288 × 10^3)`

`= 22/7 × 36000`

`= 1131428.5 cm^3`

`= 1.131 m^3`

8. 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 `cm^3`. Check whether she is correct, taking the above as the inside measurements, and `pi = 3.14`.

Solutions 

Height of the cylinder = 8cm 

radius of the cylinder = 1cm 

radius of the sphere = 4.25 cm 

Volume of the glass vessel = volume of the cylinder + volume of the sphere

`= pir^2h + 4/3pir^3`

`=  pi (r^2h + r^3)`

`= 3.14 (1^2 × 8 + 4/3 × 4.25^3)`

`= 3.14 (8 + 102)`

`= 3.14(110)`

`= 345.4 cm^3`

Work in progress