Optional Exercise – 13.5 - 10th Mathematics –Surface area and Volume

Question 1:  A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.

Solution:  It can be observed that 1 round of wire will cover 3 mm height of cylinder.

Length of wire required in 1 round = Circumference of base of cylinder= 2pr = 2p  × 5 = 10p

Number of round = Height of cylinder/Demeter of wire = 12/0.3= 40 round

Length of wire in 40 rounds = 40 × 10p = 40x10 x3.14=1256cm=12.56m

Radius of wire  = 0.3/2 = 0.15cm

Volume of wire = Area of cross-section of wire × Length of wire = p (0.15)² × 1256 = 3.14 x 0.0225 x 1256   = 88.736 cm³

Mass = Volume × Density = 88.736 × 8.88 = 787.979 gm

Question 2:  A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of p as found appropriate.)

 Solution: The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

AC = √ (3² + 4²) = 5 cm      

Area of DABC = ½ x AC x r =1/2 x BC x AB

Þ 5 x r = 3 x 4 Þ 12/5=2.4cm

Volume of double cone = Volume of cone 1 + Volume of cone 2

= 1/3 p r²H  + /3 p r²h = 1/3 p r² (H  + h)

= 1/3 x 3.14 x  2.4  x 2.4 x  5 =  30.14 cm³

 Surface area of double cone = Surface area of cone 1 + Surface area of cone 2

          = p r L  + p r l  

          = p r ( L + l )

           = 3.14 x 2.4 (4+3) 

            = 3.14 x 2.4 x7 

             = 52.75 cm2

 Question 3:  A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Solution: Volume of cistern = 150 × 120 × 110 = 1980000 cm3

Volume to be filled in cistern = 1980000 – 129600 = 1850400 cm3

Volume of n bricks = 22.5 × 7.5 × 6.5 = 1096.875

As each brick absorbs one-seventeenth of its volume,

Þ  Volume absorbed by 1 bricks =  1/17 x 1096.875 =64.52

Actual volume of bricks without water = 1096.875 - 64.52 =1032.375

Number of bricks = 1850400/1032.375 =1792.37

Therefore, 1792 bricks were placed in the cistern.

Q 4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Solution: Area of the valley, A = 97280 km2

Level in the rise of water in the valley, h = 10 cm = (10/100000) km = (1/10000) km

Thus, amount of rain fall in 14 day = Ah = 97280 km2 × (1/10000) km = 9.828 km3

Amount of rain fall in 1 day = 9.828 /14 = 0.702km3

Volume of water in 3 rivers = length × breadth × height 

= 1072 km × 75 m × 3 m 

= 1072 km × (75/1000) km × (3/1000) km = 0.2412 x3 

= 0.7236 km3

This shows that the amount of rain fall is approximately equal to the amount of water in three rivers.

Question 5:  An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the given figure).

Solution:  Radius (R) of upper circular end of frustum      

=18/2 = 9 cm                                           

Radius (r) of lower circular end of frustum 

= Radius of circular end of cylindrical part    

= 8/2 = 4 cm           

    

Height (H) of frustum 

= 22 − 10 = 12 cm

Height (h) of cylindrical = 10 cm

Slant height (l) of frustum

 = √(R-r)² + H² =√(9-4)² + 12² =13 cm                                 

Area of tin sheet required = CSA of frustum t + CSA of cylindrical

= p (R+r)l + 2prh = p [{(9+4)x13} + {2x4x10}]=22/7{169+80}=782.57 cm2

Question 6: Derive the formula for the volume of the frustum of a cone

Solution:  Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let R and r be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

In D  ABG and  D ADF, DF||BG

<A = <A and <AGB = <AFD
∴ D ABG ∼ D ADF (AA similarity) 

 DF/BG = AF/AG =AD/AB

Þ r/R = H-h/H=L-l/L

Þif r/R = H-h/HÞ H =(Rh/R-r)

Volume of frustum of cone = Volume of cone ABC − Volume of cone ADE

= 1/3xpR²H - 1/3xpr² (H-h)

 = 1/3p [R²H - r²(H-h)]

= 1/3p [R²(Rh/R-r) - r²{(Rh/R-r)-h}]

= 1/3p [(R³h/R-r) - r²{(Rh –Rh+rh) /(R-r)}]

= 1/3p [(R³h  - r³ h) /(R-r)]

= 1/3ph [(R³  - r³ ) /(R-r)]

= 1/3ph [{(R  - r )( R²  + r²  -Rr)} /(R-r)]

= 1/3ph ( R²  + r²  -R r) 

Further Study links

X Maths Surface Areas and Volumes Test paper-1

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X Maths Surface Areas and Volumes Test paper-2

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Mensuration Surface area and volume solved test paper-3

X Maths Surface Areas and Volumes Concepts for class 10th

10th surface area and volume Cuboid and a Cube Cylinder Cone 

CBSE MATHEMATICS (Class-9,10) ch-Surface Area and Volumes

10th Surface area and volume practice paper for CBSE Exam

1.Lead spheres of diameter 6cm are dropped into a cylindrical beaker containing some water and are completely submerged. If the diameter is 18cm and the water rises by 40cm, find the number of lead spheres dropped in the  water.                (Ans = 90)                                                                                                                   

2. A circus tent is cylindrical to a height of 3m and conical above it. If its diameter is 105m and the slant height of the conical portion is 53m, calculate the length of the canvas cloth 5m wide required to make the tent.(Ans = 1947m)

3.  A cone, a hemi-sphere and a cylinder stand on equal bases and have the same height. Find the ratio of their volumes as well the ratio of their total surface areas  (Ans = 1:2:3, (√2 + 1):3:4)

4.  A cone of radius 10cm is divided into two parts by drawing a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of the two parts of the cone (Ans = 1:7)

5. A building is in the shape of a cylinder surmounted by a hemi-spherical vaulted dome. The internal diameter of the building is equal to the total height of the building. If the volume of air space inside the building is 880/21 m3, find the height of the crown of the vault above the floor.             (Ans = 4m)

6. An inverted cone of vertical height 12cm and radius of the base 9cm has water to a depth of 4cm. Find the area of the internal surface of the cone not in contact with water.    (Ans = 376.8cm2)

7. The mass of a spherical iron shot-put 12cm in diameter is 5kg. Find the mass of a hollowcylindrical pipe 12cm long (made of the same metal), if it’s internal and external diameters are 20cm and 22cm, respectively.                                    (Ans = 4.375kg)

For full paper: 

MATH ADDA By Guru:JSUNIL": 10th Surface area and volume practice paper for CB...: 1. A solid iron rectangular block of dimensions 4.4 m, 2.6m, and 1m is cast into a hollow cylindrical pipe of internal radius 30cm and t...

X Surface area and Volume Sample paper

Q1.If the radii of the circular ends of a conical bucket, which is 16 cm high, are 20 cm and 8 cm, Find the capacity and the total surface area of the bucket. (10459.43cu.cm, 1961.14 sq. cm) 

Q2. Find the volume of right circular cylinder which has a height of 21cm and base radius 5cm.Find also the curved surface area. (1650 cu cm, 660 sq cm.) 

Q3. The base radii of two right circular cones of same height are in the ratio 3:5.Find the ratio of their volumes. (9:25) 

Q4. The circumference of the base of a 16m high solid cone is 3m.Find the volume of cone.(3.818 m3)

Q5 A circus tent is cylindrical upto a height of 3m and conical above it. If the diameter of the base is 105 m and the slant height of the conical part is 53 m. Find the total canvas used in making the tent. (9735 sq m.)


 Q6. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10cm and its base is of radius 3.5 cm. Find the total surface area of the article. (374 sq. cm)


Q7. A solid consists of a cylinder with a cone on one end and a hemisphere on the other end. If the length of the entire solid is 12.8cm and the diameter and height of the cylinder are 7cm and 6.5cm respectively. Find the total surface area of the solid. (269.28sq cm) 

Q8. A cylindrical container is filled with ice cream, whose radius is 6cm and height 15cm. The whole ice cream is distributed among 10 children in equal cones having hemispherical top .If the height of the conical portion is four times the radius of its base .Find the radius of the base of the cone. (3cm) 

Q9. A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 2cm and the diameter of the base is 4cm.If a right circular cylinder circumscribes the solid, find how much more space it will cover. (π = π) (8 π cm3) 
Q 10. Water in a canal, 6m wide and 1.5m deep is flowing with a speed of 10km/hr.How much area will it irrigate in 30 minutes, if 8cm of standing water is needed. (56.25 hectares).



10TH MATHS BY JSUNIL": X Surface area and Volume Excel exercise: 1. Find the edge of a cube of volume equal to the volume of a cuboid of dimensions 63 cm × 56 cm × 21 cm. 2. Find the number of 5 cm cubes ...
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Height and Distance	Surface Areas and Volumes	Sample papers