Question 1

The diameter of the base of a cone is 6 cm and altitude is 4 cm, What is the approximate curved surface area of the cone?

Accepted Answer

Radius of cone (r) = 6/2 = 3 cm and height of cone (h) = 4cm Slant height (l) = √h2 + r2 = √(4)2 + (3)2 = √16 + 9 = 5 cm Now, curved surface area = πrl = (22/7) x 3 x 5 = 330/7 = 47 cm2

Question 2

The volume of a right circular cone is 100π cm 3 and its height is 12 cm. Find its slant height.?

Accepted Answer

Volume = (1/3)πr2h According to the question, (1/3)πr2h = 100π ⇒ (1/3)πr2 x 12 = 100π ⇒ r2 = 25 ∴ r = √25 = 5 cm ∴ Slant height (l) = √h2 + r2 = √122 + 52 = √169 = 13 cm

Question 3

If the volume of a right circular cone of height 9 cm is 48π cm 3, then find the diameter of its base.?

Accepted Answer

volume = 48 π cm3 and h = 9 cm ∴ (1/3)πr2h = 48π ⇒ (1/3) x π x r2 x 9 = 48π ∴ r2 = (48π x 3) / (π x 9) = 16 ∴ r = √16 = 4 cm ∴ Diameter = 2r = 2 x 4 = 8 cm

Question 4

Shantanu's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 5 such caps.?

Accepted Answer

Slant height (l) = √r2 + h2 =√72 + 242 = √49 + 576 = √625 = 25 cm Curved surface area = πrl = (22/7) x 7 x 25 = 550 sq cm ∴ Area of 5 caps = 550 x 5 = 2750 sq cm

Question 5

The radius of the base of a right circular cone is doubled. To keep the volume fixed, the height of the come will be?

Accepted Answer

In first situation. Radius = r1, height = h1 and volume = v1 In second situation, Radius =2r1, height = h2 and volume = v2 In the volume is fixed, then v1 = v2 ⇒ (1/3)πr21h1 = (1/3)π(2r1)2h2 ⇒ h1 = 4h2 ∴ h2 = h1/4 Therefore, height of the the cone will be one-fourth of the previous height.

Question 6

The diameter of a right circular cone is 14 m, while its slant height is 9 m, Find the volume of the cone.?

Accepted Answer

Given, l = 9 m and diameter = 14 m ⇒ r = 14/2 = 7 m Now, volume = (1/3)πr2h = (1/3)π x 49 x √l2 - r2 = (1/3)π x 49 x √81 - 49 = (1/3) x 49π x √32 = 49π√32 /3 m3

Question 7

If the ratio of volumes of two cones is 2 : 3 and the ratio of the radii of their bases is 1 : 2, then the ration of their heights will be?

Accepted Answer

[(1/3)πr12h1]/[(1/3)πr22h2] = 2/3 ⇒ (r1/r2)2 x (h1/h2) = 2/3 ⇒ (1/2)2 x (h1/h2) = 2/3 [ &bacaus; r1/r2 = 1/2] ⇒ h1/h2 = 8/3 ∴ h1 : h2 = 8 : 3

Question 8

The ratio of the radius and height of a cone is 5 : 12. Its volume is 314 2/ 7 cm 3. Its slant height is?

Accepted Answer

Let radius = 5k, height = 12k According to the question = 1/3 x 22/7 x (5k)2 x 12k = 2200/7 ⇒ k = 1 ∴ r = 5, h = 12 ∴ Slant height (l) = √r2 + h2 = √25 + 144 = √169 =13 cm

Question 9

If the height of the right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone?

Accepted Answer

Here, x = -50%, y = 200% According to the formula Net effect = [2x + y + x2 + 2xy/100 + x2y/1002] % = [-100 + 200 + 2500 - 20000/100 + 500000/10000]% = [100 - 175 + 50] % = -25%

Question 10

If the volume of two right circular cones are in the ratio 1 : 3 and their diameters are in the ratio 3 : 5, then the ratio of their heights is?

Accepted Answer

Let diameter, radius and height of first cone are d1, r1 and h1, respectively and that of second cone are d2, r2 and h2, respectively. r1/r2 = d1/d2 = 3/5, h1/h2 = ? Given, [(1/3)πr21h1] / [(1/3)πr22h2] = 1/3 ⇒ (3/5)2 x h1/h2 = 1/3 ⇒ h1/h2 = (1/3) x (25/9) = 25/27

Question 11

The diameters of two cones are equal. If their slant heights be in the ratio of 5 : 7, then find the ratio of their curved surface areas.?

Accepted Answer

Given, l1/l2 = 5/7 Now, curved surface area of the first cone = π rl1 and curved surface area of second cone = πrl2 ∴ Ratio = πrl1/πrl2 = l1/l2 = 5 : 7

Question 12

The radius of the base of a right circular cone is increased by 15% keeping the height fixed. The volume of the cone will be increased by.?

Accepted Answer

Let the fixed height of a right circular cone is h and initial radius is r Then, initial volume of cone, V1 = (1/3)πr2h After increasing 15% radius of a cone = (r + 3r/20) = 23r/20 New volume become, V2 = (1/3)π(23/20)2r2h ∴ Increasing percentage = [(V2 - V1) / V1] x 100 = {[(1/3)πr2h] / [(1/3)πr2h]} {(23/20)2 - 1} x 100 = (23/20 + 1)(23/20 - 1) x 100 = 43/20 x 3/20 x 100 = 32.25%

Question 13

What is the diameter of the largest circle lying on the surface of a sphere of surface area 616 sq cm?

Accepted Answer

Surface area of sphere = 6616 cm2 4πr2 = 6166 ⇒r2 = (6166 x 7)/(4 x 22) ⇒ r2 = 7 x 7 ⇒ r = 7 cm ∴ Diameter of largest circle lying on sphere = 2 x r = 2 x 7 = 14 cm

Question 14

The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What is the ratio (approximate) of their volumes?

Accepted Answer

Given that, the diameter of moon is approximately one-fourth of the diameter of Earth. Let radius on moon = r Then, radius of Earth = 4r ∴ Volume of Moon/Volume of Earth = [(4/3)πr3] / [(4/3)π(4r)3] = [r3] / [64r3] = 1/64 = 1 : 64

Question 15

A sphere and a hemisphere have the same surface area. The ratio of their volumes is?

Accepted Answer

According to the question, Surface area of sphere = Surface area of hemisphere 4πr12 =3πr22 ⇒ r1/r2 = √3/2 ∴ Ratio in volume = [(4/3)πr13] / [(4/3)πr23] = 3√3/8 : 1

Question 16

Find the number of lead balls of diameter 2 cm each, that can be made from a sphere of diameter 16 cm.?

Accepted Answer

Radius of the sphere = 16/2 = 8 cm Volume of the sphere = (4/3) x π x 8 x 8 x 8 cm3 Radius of each lead ball = 2/2 = 1 cm Volume of each lead ball = Volume of sphere / Volume of lead ball = (4/3) π x 1 x 1 x 1 = 4π/3 cm3 ∴ Number of lead balls = [(4/3) x π x 8 x 8 x 8 x 3] / [4 π] = 8 x 8 x 8 = 512

Question 17

A hemispherical bowl has 3.5 cm radius. It is to be painted inside as well as outside. Find the cost of painting it at the rate of ₹ 5 per 10 sq cm.?

Accepted Answer

Curved surface area of the hemisphere = 2πr2 = 2 x (22/7) x (7/2) x (7/2) = 77 sq As bowl is to painted inside and outside. ∴ Total surface to be painted = 77 x 2 = 154 sq cm ∴ Cost of painting 154 sq cm = (5/10) x 154 = 1/2 x 154 = ₹ 77

Question 18

If the surface area of a sphere is 616 sq cm, what is its volume?

Accepted Answer

Curved surface area of the sphere = 4πr2 or 616 = 4πr2 ⇒ πr2 = 616/4 = 154 ⇒ r2 = (154 x 7) / 22 = 49 ∴ r = √49 = 7 cm ∴ Volume of the sphere = (4/3)πr3 = (4/3) x (22/7) x 7 x 7 x 7 = 4312 / 3 cm3

Question 19

If the ratio of the diameters of two spheres is 3: 5, then what is the ratio of their surface areas?

Accepted Answer

Let the diameter's of two sphere are d1 and d2, respectively. ∴ Ratio of their surface areas = 4πr12/4πr22 = (2r1)2/(2r2)2 = d12/d22 = (d1/d2)2 = (3/5)2 = 9/25 = 9 : 25

Question 20

If 64 identical small spheres are made out of a big sphere of diameter 8 cm, what is surface area of each small sphere?

Accepted Answer

Volume of small spheres = Volume of bigger sphere / Number of small spheres = [(4/3)π(4)3] / 64 = [(4/3) x π x 4 x 4 x 4] / 64 = 4/3 π cm3 Let radius of small sphere be r ∴ 4/3πr3 = 4π/3 ⇒ r2 = 1 cm Now, surface area of small sphere = 4πr2 = 4π cm2

Volume and Surface Area Questions