Problems on Volumes and Surface Area
Q21. Find the area of the iron sheet required to prepare a cone 20 cm height with base radius 15 cm.
(a) 1126.60 cm2
(b) 1365.25 cm2
(c) 1750.35 cm2
(d) 1885.71 cm2
Answer: (d) 1885.71 cm2
Here r = 15 cm
h = 20 cm
∴ Slant height l = | = | = | = | = 25 cm | ||||
√r2 + h2 | √152 + 202 | √225 + 400 | √625 |
Area of the sheet = πrl + πr2
= πr (l + r)
= 22/7 X 15 (25 + 15)
= 22/7 X 15 x 40
= 1885.71 cm2
Q22. A rectangular water tank is open at the top. Its capacity is 24m3. Its length and breadth are 4 m and 3 m respectively. Ignoring the thickness of the material used for building the tank, the total cost of painting the ineer and outer surfaces of the tank at Rs. 10 per m2 is
(a) Rs. 650
(b) Rs. 750
(c) Rs. 800
(d) Rs. 900
Answer: (c) Rs. 800
Let the height of the tank be x meter
∴ Volume of the tank = 4 X 3 X x m3
⇒ 24 m3 = 12x m3
⇒ x = 2 m
Area of the surface to be painted = 2 [2(l+b) X h + (lXb)]
= 2 [2 X (4+3) X 2 + 4 X 3 ]
= 2 [28 + 12] = 80 m2
∴ Cost of painting = Rs. 80 X 10 = Rs. 800
Q23. What will be the area of an iron sheet which can be used to form a cone having base radius 7 cm and height 24 cm ?
(a) 704 cm2
(b) 712 cm2
(c) 726 cm2
(d) 740 cm2
Answer: (a) 704 cm2
r = 7 cm h = 24 cm
∴ l = √(r2+h2) = √(72+242) = √(49+576) = √(625) = 25 cm
Required area = Total surface area of the cone
= πr(l+r) = 22/7 X 7(25+7) = 22 X 32 = 704 cm2
Q24. The ratio of the radius and height of a cone is 5:12. Its volume is 314 2/7 cc. Its slant height is.
(a) 13 cm
(b) 15 cm
(c) 16 cm
(d) 21 cm
Answer: (a) 13 cm
Let the radius and height of the cone be 5x, 12x
∴ Volume = 1/3πr2h
⇒ 314 | 2 | = | 1 | X | 22 | X (5x)2 X 12x |
7 | 3 | 7 |
⇒ | 2200 | = | 22 | X 25x2 X 12x |
7 | 21 |
⇒ x3 = | 2200 X 21 |
7 X 22 X 25 X 12 |
⇒ x3 = 1
⇒ x = 1
∴ r = 5 h = 12
∴ l = √(r2+h2) = √(52+122) = √(25+144) = 13cm
Q25. If the ratio of volumes of two cones is 2:3 and the ratio of the raddi of their bases is 1:2, then the ratio of their heights will be
(a) 5:4
(b) 5:3
(c) 7:3
(d) 8:3
Answer: (d) 8:3
Here | V1 | = | 2 |
V2 | 3 |
R1 | = | 1 |
R2 | 2 |
Now | V1 | = | 1/3 π R12H1 |
V2 | 1/3 π R22H2 |
Now | 2 | = | R12H1 |
3 | R22H2 |
⇒ | 2 | = ( | R1 | )2 | H1 |
3 | R2 | H2 |
⇒ | 2 | = ( | R1 | )2 | H1 |
3 | R2 | H2 |
⇒ | 2 | = ( ½ )2 | H1 |
3 | H2 |
⇒ | H1 | = | 2 | X 4 = | 8 |
H2 | 3 | 3 |
∴ H1 : H2 = 8 : 3
Q26. The volume of a hemisphere of radius 7 cm is
(a) 712.5 cm3
(b) 718.6 cm3
(c) 723.6 cm3
(d) 728.3 cm3
Answer: (b) 718.6 cm3
Volume = 2/3 πr3
= | 2 | X | 22 | X 73 |
3 | 7 |
= | 44 X 49 |
3 |
= 718.6 cm3
Q27. How many balls of diameter 0.4 cm each are formed a cuboid of measuring (4cm X 5cm X 6cm) ?
(a) 3512
(b) 3580
(c) 3620
(d) 3675
Answer: (b) 3580
Volume of the cuboid = 4cm X 5cm X 6cm = 120 cm3
Volume of each ball = 4/3πr3
= | 4 | X | 22 | X | 0.4 | X | 0.4 | X | 0.4 |
3 | 7 | 2 | 2 | 2 |
= | 0.704 |
21 |
= 0.03352
Number of balls = | 120 |
0.03352 |
= 3580 (approx)
Q28. The volume of a sphere is 2145 11/21 cm3. Its diameter is
(a) 16 cm
(b) 18 cm
(c) 21 cm
(d) 23 cm
Answer: (a) 16 cm
Volume = 4/3πr3
⇒ | 45056 | = | 4 | X | 22 | X | r3 |
21 | 3 | 7 |
⇒ r3 = | 45056 | X | 3 X 7 |
21 | 4 X 22 |
⇒ r3 = 512
⇒ r3 = 83
⇒ r = 8
∴ Diameter = 2 X 8 = 16 cm
Q29. A cone of height 15 cm and base diameter 30 cm is carved out of a wooden sphere of radius 15 cm. The percentage of wasted wood is
(a) 61%
(b) 65%
(c) 72%
(d) 75%
Answer: (d) 75%
Volume = 4/3πr3
= | 4 | X | 22 | X 153 |
3 | 7 |
Volume of cone =1/3πr2h = | 1 | X | 22 | X 152 X 15 = | 1 | X | 22 | X 153 |
3 | 7 | 3 | 7 |
Wasted wood = | 4 | X | 22 | X 153 - | 1 | X | 22 | X 153 |
3 | 7 | 3 | 7 |
= | 22 | X 153 = π153 |
7 |
Percentage of wasted wood = | π X 15 3 | X 100 % |
4/3 X 153 |
= ¾ X 100% = 75%
Q30. 12 sphere of the same size are made by melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is
(a) 4 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Answer: (a) 4 cm
Here H = 2 cm R = 8 cm
Volume of the cylinder = πR2h = π82 X 2 = 128π cm3
Let, the radius of each sphere be r
then
12 X | 4 | πr3 = 128π |
3 |
⇒ r3 = | 128π X 3 |
48π |
⇒ r3 = 8
⇒ r = 2
∴ Diameter = 2 X 2 = 4 cm
Practice Test Exam