# Problems on Volumes and Surface Area

Q11. The area of the base of the rectangular tank is 6500 cm^{2} and the volume of the water contained in it is 32.5 cubic meters. The depth of the water in the tank is

(a) 42m

(b) 50m

(c) 57m

(d) 62m

**Answer:** (b) 50m

Here l X b = 6500 cm^{2}

v = 32.5 m^{3} = 32.5 X 100 X 100 X 100 cm^{3}

We have v = l X b X h

⇒ 32.5 X 100 X 100 X 100 = 6500 X h

⇒ h = (32.5 X 100 X 100 X 100)/6500 cm = 5000 cm = 50m

Q12. A covered wooden box has the inner measures as 110cm, 70cm and 30cm and the thickness of wood is 2.5cm. The volume of the wood is

(a) 54982 cm^{3}

(b) 62547 cm^{3}

(c) 69832 cm^{3}

(d) 70875 cm^{3}

**Answer:** (d) 70875 cm^{3}

Inner volume of box = 110 X 70 X 30 cm^{3} = 231000 cm^{3}

Outer volume of box = (110 + 5) X (70 + 5) X (30 + 5) cm^{3}

= 115 X 75 X 35 cm^{3}

= 301875 cm^{3}

∴ Volume of the wood = (301875 - 231000) cm^{3} = 70875 cm^{3}

Q13. A hall is 15m long and 10m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of the 4 walls, the volume of the hall is

(a) 815 m^{3}

(b) 875 m^{3}

(c) 900 m^{3}

(d) 950 m^{3}

**Answer:** (c) 900 m^{3}

here l = 15m b = 10m

Area of floor + Area of ceiling = lXb + lXb = 15X10 + 15X10 = 300m^{2}

Sum of the areas of 4 walls = 2 lXh + 2 bXh = 2 X 15 X h + 2 X 10 X h = 30h + 20h = 50h

A/Q 50h = 300

⇒ h = 300/50 = 6m

∴ Volume of the ball = l X b X h = 15 X 10 X 6 m^{3} = 900 m^{3}

Q14. The sum of the length, breadth and depth of a cuboid is 20cm and its diagonal is 5√ 5 cm. Its surface area is

(a) 230 cm^{2}

(b) 275 cm^{2}

(c) 333 cm^{2}

(d) 350 cm^{2}

**Answer:** (b) 275 cm^{2}

Here l+b+h = 20

√ l^{2}+b^{2}+h^{2} = 5√ 5

Now (l+b+h)^{2} = 20^{2} = 400

(√l^{2}+b^{2}+h^{2})^{2} = (5√ 5)^{2}

⇒ l^{2}+b^{2}+h^{2} = 125

We have (l+b+h)^{2} = l^{2}+b^{2}+h^{2}+2 (lb+bh+hl)

⇒ 400 = 125 + 2 (lb+bh+hl)

⇒ 2 (lb+bh+hl) = 400 - 125

⇒ 2 (lb+bh+hl) = 275 cm^{2}

Q15. The size of a wooden block is (12cm X 10cm X 8cm). How many such blocks will be required to construct a solid wooden cube of minimum size ?

(a) 1300

(b) 1500

(c) 1600

(d) 1800

**Answer:** (d) 1800

L.C.M of 12, 10 and 8 is 120

Volume of a smallest cube = 120 X 120 X 120 cm^{3}

Volume of one block = 12 X 10 X 8 cm^{3} = 960 cm^{3}

∴ Required number of block = (120X120X120)/960 = 1800

Q16. The height of a cylinder is 12cm and its diameter is 10cm. The volume of the cylinder is

(a) 733.50 cm^{3}

(b) 856.21 cm^{3}

(c) 942.85 cm^{3}

(d) 993.88 cm^{3}

**Answer:** (c) 942.85 cm^{3}

Volume = π r^{2}h

= (22/7) X (10/2)^{2} X 12

= (22/7) X 5 X 5 X 12 = 942.85 cm^{3}

Q17. The height of a cylinder is 14cm and its curved surface area is 264cm^{2}. The volume of the cylinder is

(a) 396 cm^{3}

(b) 451 cm^{3}

(c) 482 cm^{3}

(d) 527 cm^{3}

**Answer:** (a) 396 cm^{3}

Here h = 14cm

2π rh = 264

⇒ 2 X (22/7) X r X 14 = 264

⇒ r = (264 X 7)/(2 X 22 X 14) = 3

∴ Volume = π r^{2}h

=(22/7) X 3^{2} X 14 = (22 X 9 X 14)/7 = 396 cm^{3}

Q18. A cylinder has a radius of 7 cm and the area of its curved surface is 176 cm^{2}. The volume of the cylinder is

(a) 577 cm^{3}

(b) 616 cm^{3}

(c) 659 cm^{3}

(d) 721 cm^{3}

**Answer:** (b) 616 cm^{3}

Here r = 7 cm

2πrh = 176

⇒ 2 X (22/7) X 7 X h = 176

⇒ h = 176/(2X22) = 4 cm

∴ Volume = πr^{2}h = (22/7) X 7^{2} X 4 = 616 cm^{3}

Q19. The ratio of the radius of two cylinders is 2:3 and the ratio of their heights is 3:5. The ratio of their volumes will be

(a) 3:7

(b) 3:17

(c) 4:15

(d) 4:17

**Answer:** (c) 4:15

Let, the radius be 2r and 3r

And the heights be 3h and 5h

∴ Ratio of their volume = | π X (2r)^{2} X 3h |

π X (3r)^{2} X 5h |

= | 4r^{2} X 3h |

9r^{2} X 5h |

= | 4 X 3 |

9 X 5 |

= 4/15 = 4:15

Q20. The number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is

(a) 375

(b) 390

(c) 420

(d) 450

**Answer:** (d) 450

Here for coins r = 1.5/2 cm

h = 0.2 cm

∴ Volume of one coin = πr^{2}h

= (π X (1.5/2) X (1.5/2) X 0.2)cm^{3} = 9π/80 cm^{3}

For cylinder R = 4.5/2 cm H = 10 cm

Volume = πR^{2}H

= (π X (4.5/2) X (4.5/2) X 10)cm^{3}

= (45 X 9π)/8 cm^{3}

∴ Required number of coins = | 45 X 9π | X | 80 |

8 | 9π |

= 450

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