radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m Volume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³. Hence, the volume of heap of wheat is 86.625 m³. Let, the slant height of heap of wheat = l m We know that, l² = h² + r² ⇒ l² = 3² + (5.25)²Read more
radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m
Volume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³.
Hence, the volume of heap of wheat is 86.625 m³.
Let, the slant height of heap of wheat = l m
We know that, l² = h² + r² ⇒ l² = 3² + (5.25)² ⇒ l² = 9 + 27.5625 ⇒ l² = 36.5625
⇒ l = √36.5625 = 6.05 m (approx.)
Required area of canvas = πrl
= 22/7 × 5.25 × 6.05 = 22 × 0.75 × 6.05 = 99.825 m²
Hence, the required area of canvas to protect wheat is 99.825 m².
Volume of cone V = 1570 cm³ and height h = 15 cm Let, the radius of base of cone = r cm Volume of cone = 1/3πr²h ⇒ 1570 = 1/3 × 3.14 × r² × 15 ⇒ 1570 = 3.14 × r² 5 ⇒ r² = 1570/(3.14×5) = 100 ⇒ r = √100 = 10 cm Hence, the radius of base of cone is 10 cm.
Volume of cone V = 1570 cm³ and height h = 15 cm
Let, the radius of base of cone = r cm
Volume of cone = 1/3πr²h
⇒ 1570 = 1/3 × 3.14 × r² × 15 ⇒ 1570 = 3.14 × r² 5 ⇒ r² = 1570/(3.14×5) = 100
⇒ r = √100 = 10 cm
Hence, the radius of base of cone is 10 cm.
Volume of cone V = 48π cm³ and height h = 9 cm Let, the radius of base of cone = r cm Volume of cone = 1/3πr²h ⇒ 48π = 1/3 × π × r² × 9 ⇒ 48π = π × r² × 3 ⇒ r² = 48π/π×3 = 16 ⇒ r = √16 = 4 cm Therefore, the diameter of base = 2 × 4 = 8 cm Hence, the diameter of base of cone is 8 cm.
Volume of cone V = 48π cm³ and height h = 9 cm
Let, the radius of base of cone = r cm
Volume of cone = 1/3πr²h
⇒ 48π = 1/3 × π × r² × 9 ⇒ 48π = π × r² × 3 ⇒ r² = 48π/π×3 = 16 ⇒ r = √16 = 4 cm
Therefore, the diameter of base = 2 × 4 = 8 cm
Hence, the diameter of base of cone is 8 cm.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Radius of hemispherical bowl r = 10.5/2 = 5.25 cm Therefore, the volume of hemispherical bowl = 2/3πr³ =2/3 × 22/7 × 5.25 × 5.25 × = 2 × 22 × 0.25 × 5.25 × 5.25 = 303 cm³ (Approx.) = 303/1000Litre [∵ 1 cm³ = 1/1000 litre] = 0.303 litre Hence, the hemispherical bowl holds 0.303 litres of milk.
Radius of hemispherical bowl r = 10.5/2 = 5.25 cm
See lessTherefore, the volume of hemispherical bowl = 2/3πr³
=2/3 × 22/7 × 5.25 × 5.25 × = 2 × 22 × 0.25 × 5.25 × 5.25 = 303 cm³ (Approx.)
= 303/1000Litre [∵ 1 cm³ = 1/1000 litre]
= 0.303 litre
Hence, the hemispherical bowl holds 0.303 litres of milk.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m Volume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³. Hence, the volume of heap of wheat is 86.625 m³. Let, the slant height of heap of wheat = l m We know that, l² = h² + r² ⇒ l² = 3² + (5.25)²Read more
radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m
See lessVolume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³.
Hence, the volume of heap of wheat is 86.625 m³.
Let, the slant height of heap of wheat = l m
We know that, l² = h² + r² ⇒ l² = 3² + (5.25)² ⇒ l² = 9 + 27.5625 ⇒ l² = 36.5625
⇒ l = √36.5625 = 6.05 m (approx.)
Required area of canvas = πrl
= 22/7 × 5.25 × 6.05 = 22 × 0.75 × 6.05 = 99.825 m²
Hence, the required area of canvas to protect wheat is 99.825 m².
Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m
Radius of sphere r = 7 cm Therefore, volume of sphere = 4/3πr³ = 4/3 × 22/7 × 7 × 7 × 7 = 4/3 × 22 × 7 × 7 = 1437(1/3) cm³ Hence, the volume of sphere = 4/3πr³ = 4/3 × 22/7 × 0.63 × 0.63 × 0.63 = 4 × 22 × 0.03 × 0.63 × 0.63 = 1.05 m³(approx.) Hence, the volume of sphere is 1.05 m³.
Radius of sphere r = 7 cm
See lessTherefore, volume of sphere = 4/3πr³
= 4/3 × 22/7 × 7 × 7 × 7 = 4/3 × 22 × 7 × 7 = 1437(1/3) cm³
Hence, the volume of sphere = 4/3πr³
= 4/3 × 22/7 × 0.63 × 0.63 × 0.63 = 4 × 22 × 0.03 × 0.63 × 0.63 = 1.05 m³(approx.)
Hence, the volume of sphere is 1.05 m³.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base.
Volume of cone V = 1570 cm³ and height h = 15 cm Let, the radius of base of cone = r cm Volume of cone = 1/3πr²h ⇒ 1570 = 1/3 × 3.14 × r² × 15 ⇒ 1570 = 3.14 × r² 5 ⇒ r² = 1570/(3.14×5) = 100 ⇒ r = √100 = 10 cm Hence, the radius of base of cone is 10 cm.
Volume of cone V = 1570 cm³ and height h = 15 cm
See lessLet, the radius of base of cone = r cm
Volume of cone = 1/3πr²h
⇒ 1570 = 1/3 × 3.14 × r² × 15 ⇒ 1570 = 3.14 × r² 5 ⇒ r² = 1570/(3.14×5) = 100
⇒ r = √100 = 10 cm
Hence, the radius of base of cone is 10 cm.
If the volume of a right circular cone of height 9 cm is 48 π cm³, find the diameter of its base.
Volume of cone V = 48π cm³ and height h = 9 cm Let, the radius of base of cone = r cm Volume of cone = 1/3πr²h ⇒ 48π = 1/3 × π × r² × 9 ⇒ 48π = π × r² × 3 ⇒ r² = 48π/π×3 = 16 ⇒ r = √16 = 4 cm Therefore, the diameter of base = 2 × 4 = 8 cm Hence, the diameter of base of cone is 8 cm.
Volume of cone V = 48π cm³ and height h = 9 cm
See lessLet, the radius of base of cone = r cm
Volume of cone = 1/3πr²h
⇒ 48π = 1/3 × π × r² × 9 ⇒ 48π = π × r² × 3 ⇒ r² = 48π/π×3 = 16 ⇒ r = √16 = 4 cm
Therefore, the diameter of base = 2 × 4 = 8 cm
Hence, the diameter of base of cone is 8 cm.