Question 1

A conical flask is full of water. The flask has base radius r and height h, This water is poured into a cylindrical flask of base radius mr. The height of water in the cylindrical flask is?

Accepted Answer

Volume of water = Volume of conical flask = (1/3)πr2h Now, the water is poured into cylindrical flask. ∴ Volume of cylinder = Volumes of water ⇒ π (mr)2 x Height = (1/3)πr2h ∴ Height = h/3m2

Question 2

A hemispherical basin of 150 cm diameter holds water 120 times as much as a cylindrical tube. If the height of the tube is 15 cm, then the diameter of the tube is?

Accepted Answer

Given, diameter = 150 cm ∴ r = 150/2 cm According to the question, 2/3π (150/2)3 = 120 πr2 x 15 ⇒ (2/3) x 150 x 150 x 150 / 8 = 120 x 15 x r2 ⇒ r2 = (150 x 150 x 150) / (12 x 120 x 5) ⇒ r2 = 625/4 ⇒ r = √625/4 = 25/2 ∴ Diameter = 2r = 2 x 25/2 = 25 cm

Question 3

A copper sphere of diameter 36 cm is drawn into a wire of diameter 4 mm. Find the length of the wire.?

Accepted Answer

Let length of the wire be h According to the question, Volume of sphere = Volume of wire (4/3) x π x 18 x 18 x 18 = π x (2/10) x (2/10) x h ∴ h = (100 x 1944) cm = (100 x 1944)/100 m = 1944 m [∵ 1 cm = 1/100 m]

Question 4

Find the area of the iron sheet required to prepare a cone double the height of 12 cm with diameter of the base 14 cm.?

Accepted Answer

r = Radius of the cone = 14/2 = 7 cm h = Height of the cone = 12 x 2 = 24 cm ∴ Slant height (l) = √r2 + h2 = √(7)2 + (24)2 = √49 + 576 = √625 = 25 cm Area of the sheet = Total surface area = (πrl + πr2) = πr(l + r) = (22/7) x 7 x (25 + 7) = 704 sq cm

Question 5

What part of a ditch 48 m long, 16.5 m broad and 4 m deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4 m and length 56 m?

Accepted Answer

Volume of the earth dugout as a tunnel = πr2h = (22/7) x 2 x 2 x 56 = 704 m3 Volume of the ditch = (48 x 33)/(2 x 4) = 24 x 33 x 4 = 3168 m3 ∴ Part required = 704/3168 = 2/9

Question 6

Water flows into a tank 100 m x 150 m x 2 m through a rectangular pipe 1.5 m x 1.25 m at a speed of 20 km/h. In what time, will the water rise by 2 m?

Accepted Answer

Volume of water in the tank = 100 x 500 x 2 = 30000 m3 Speed of water = 20 x 1000/60 = 1000/3 m/min water flow per minute = (15/10) x (125/100) x (1000/3) = 625 m3 Time taken = 30000/625 = 48 min

Question 7

A hospital room is to accommodate 56 patients. It should be done in such a way that every patient gets 2.2 m 2 of floor and 8.8 3 of space. If the length of the room is 14 m, then breadth and height of the room are respectively?

Accepted Answer

Let the breadth and height of room be m and h m, respectively. Then, according to the question, l x b = n x Area occupied by one patient ⇒ 14 x b = 56 x 22 ∴ b = (56 x 22)/14 = 8.8 m3 ∴ h x Area of occupied by one patient = 8.8 h = 8.8/2.2 h = 4 m

Question 8

A right circular metal cone (solid) is 8 cm high and the radius is 2 cm. It is metaled and recast into a sphere. What is the radius of the sphere?

Accepted Answer

Given that, the height and radius of a right circular ,etal cone (solid) are 8 cm and 2 cm, respectively. i.e., h = 8 cm and r = 2 cm Let the radius of the sphere is R Then, by condition, 1/3 π r2 h = 4/3 π R3 h ⇒ 4 x 8 = 4 R3 ⇒ R3 = (2)3 = ⇒ R = 2 ∴ Radius odf the sphere = 2 cm

Question 9

A cone has height which is half of 16.8 cm, while diameter of its base is 4.2 cm. It is melted and recast into a sphere. Find the surface area of the sphere.?

Accepted Answer

Given that, Height = 1 m Diameter = 140 cm or 140/100 m = 1.40 m ∴ r = 1.40/2 = 0.7 m Now, Required sheet = Total surface area = 2πr( h + r) = (22/7) x 1.4 x (1 + 0.7) = 7.48 sq m ∴ Sheet required = 7.48 sq m

Question 10

A cylindrical box of radius 5 cm contains 10 solid spherical balls, each of radius 5 cm. If the top most ball touches the upper cover of the box, then volume of the empty space in the box is?

Accepted Answer

Height of the cylinder = 10 x Diameter of each ball = 10 x 10 = 100 cm ∴ Required empty space = π (5)2 (100) - 4/3 π (5)3 x 10 = π (5)2 x 10 [10 - 20/3] = π (5)2 x 10 [30 - 20/3] = 2500/3 π cm3

Question 11

If the height, curved surface area and the volume of a cone are h, c and V respectively, then 3πVh 3 - c 2 + 9V 2 will be equal to?

Accepted Answer

Let h be the height, Volume of the cone = V and curved surface area of the cone = c ∴ 1/3 πr2h = V, πrl = c i.e., πr√r2 + h2 = c π2r2(r2 + h2) = c2 Consider 3πVh3 - c2h2 + 9V2 = 3π x 1/3πr2h x h3 - π2r2(r2 + h2)h + 9 x 1/9 x π2r4 h2 = π2r2h4 - π2r4h2 - π2r2h4 + π2r4 = 0

Question 12

Three cubes with sides in the ration 3 : 4 : 5 are melted to form a single cube whose diagonal is 12√ 3 cm. The sides of the cubes are?

Accepted Answer

Let the side of the cube be 3k, 4k and 5k, respectively. So, volumes of these cubes are 27 k3, 64k3, 125 k3 respectively. ⇒ Volume of the new bigger cube = 27 k3 + 64k3 + 125 k3 = 216k3 So, side of the new cube = 6k Since, diagonal of the cube = √(6k)2 + (6k)2 + (6k)2 = 12√3 ⇒ 108k2 = 432 ⇒k = 2 So, sides of the three cubes were 6 cm, 8 cm and 10 cm respectively.

Question 13

If 1 cubic cm of cast iron weights 21 gm, then the weight of a cast iron pipe of length 1 m with a bore of 3 cm and in which the thickness of the metal is 1 cm, is?

Accepted Answer

External radius = 2.5 cm, length = 100 cm ∴ External volume = [π x (2.25)2 x 100] cm2 internal radius = 1.5 cm ∴ internal volume = [π x (1.5)2 x 100]cm3 Volume of metal = [π x (2.5)2 x 100 - π x (1.5)2 x 100] cm3 = π x 100 x [(2.5)2 - (1.5)2] cm3 = (22/7 x 100 x 4 x 1 x 21/1000) kg = 26.4 kg.

Question 14

The curved surface area and the total surface area of a cylinder are in the ratio 1 : 2. If the total surface area of the right cylinder is 616 cm 2, then its volume is?

Accepted Answer

Curved surface area of cylinder = 2πrh cm2 Total surface area of cylinder = 2πr(h + r) cm2 Now, According to the question 2πrh/2πr(h + r) = 1/2 ⇒ h/(h + r) = 1/2 ⇒ 2h = h + r ∴ Now, Total surface area = 616 cm2 ⇒ 2π (h + r) = 616 ⇒ 2πr (r + r) = 616 (∵ h = r) ⇒ 2πr(2r) = 616 ⇒ 4πr2 = 616 ⇒ r = √style="text-decoration:overline">616/4π = √616 x 7/4 x 22 = √7 x 7 = ∴ r = h = 7 cm ∴ Volume = π r2 h = (22/7) x (7 x 7) x 7 = 1078 cm3

Question 15

10 circular plates each of thickness 3 cm, each are placed one above the other and a hemisphere of radius 6 cm is placed on the top just to cover the cylindrical solid. What is the volume of the solid so for formed?

Accepted Answer

If 10 circular plates each of thickness 3 cm, each are placed one above the other, then it forms a cylinder with height (3 x 10 = 30 cm) and a hemisphere of radius 6 cm is placed on the top just to cover the cylinder that means its radius is 6 cm also ∴ Radius of hemisphere (R) = 6 cm Radius of cylinder (r) = 6 cm and height of cylinder (h) = 30 cm ∴ Volume of the solid = Volume of cylinder + Volume of hemisphere = πr2h + (2/3)πR3 = π(6)2 x 30 + (2/3)π(6)3 = π x 36 x 30 + (2/3)π x 216 = 1080π + 2π x 72 = 1080π + 144π = 1224π cm3 Which is the required volume of solid.

Question 16

Find the area of a square, one of whose diagonals is 3.8 m long.

Accepted Answer

Area of the square = 1 2 × d i a g o n a l 2 = 1/2 * 3.8 * 3.8 = 7.22 sq.m

Question 17

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

Accepted Answer

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. Area of each tile = (41 x 41) sq.cm Required number of tiles = 1517 x (902/41) x 41 = 814

Question 18

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

Accepted Answer

Let breadth = x metres. Then, length = (x + 20) metres. Perimeter = 5300 m = 200 m. 26.50 2[(x + 20) + x] = 200 2x + 20 = 100 2x = 80 x = 40. Hence, length = x + 20 = 60 m.

Question 19

The lengths of diagonals of a rhombus are 24cm and 10cm the perimeter of the rhombus (in cm ) is :

Accepted Answer

52

Question 20

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

Accepted Answer

Let breadth = x metres. Then, length = (x + 20) metres. Perimeter = 5300 m = 200 m. 26.50 2[(x + 20) + x] = 200 2x + 20 = 100 2x = 80 x = 40. Hence, length = x + 20 = 60 m.

Volume and Surface Area Questions