How many silver coins, 2cm in diameter and of thickness 2 mm, must be melted to form a showpiece in the form of a frustum of a cone having height 9 cm and radii of the outer circular ends as 5cm and 3cm?
if the cone were complete, it would have height 45/2, so the missing part has height 27/2.
So, if there are x coins,
(π*0.2*1^2)x = π/3 (5^2*45/2 - 3^2*27/2)
x = 735
l asked for 3 marks ,please reput in step bystep
Are you stupid
I think Right answer
To find the number of silver coins required to form a frustum of a cone, we need to calculate the volume of the frustum and the volume of a single silver coin.
First, let's find the volume of the frustum of a cone using the formula:
V = (1/3) * π * h * (R^2 + r^2 + R*r)
where:
V = volume
π = pi (approximately 3.14)
h = height of the frustum (9 cm)
R = radius of the larger circular end (5 cm)
r = radius of the smaller circular end (3 cm)
V = (1/3) * 3.14 * 9 * (25 + 9 + 15)
= (1/3) * 3.14 * 9 * 49
= 9.62 * 49
= 471.38 cubic cm (approximately)
Next, let's find the volume of a single silver coin. Since the coin is a cylinder, we can calculate its volume using the formula:
V_coin = π * r^2 * h_coin
where:
V_coin = volume of a single silver coin
π = pi (approximately 3.14)
r = radius of the coin (2 cm = 2/10 cm = 0.2 cm)
h_coin = thickness of the coin (2 mm = 2/10 mm = 0.2 cm)
V_coin = 3.14 * (0.2)^2 * 0.2
= 3.14 * 0.04 * 0.2
= 0.02512 cubic cm (approximately)
Finally, we can calculate the number of silver coins required by dividing the volume of the frustum by the volume of a single silver coin:
Number of coins = V / V_coin
= 471.38 / 0.02512
= 18,763 coins (approximately)
Therefore, approximately 18,763 silver coins should be melted to form a showpiece in the form of a frustum of a cone with the given dimensions.