Question

find the number of coins 1.5 CM in diameter and 2 millimetre to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 CM​

Answer 1

Answer:

• The number of coins = 450 coins.

Given :

• Diameter of coins = 1.5 cm

• Height of coins = 2 mm

• Diameter of right circular cylinder = 4.5 cm

• Height of right circular cylinder = 10 cm

To find :

• The number of coins =?

Step-by-step explanation:

Diameter of coin = 1.5 cm [Given]

• So, its radius = 1.5/2 = 0.75 cm

• Height of coin = 2 mm = 0.2 cm

Now,

Volume of coin = volume of cylinder = πr²h

= π x (0.75)² x 0.2

= π x 0.75 x 0.75 x 0.2

= 0.1125π cm³

Diameter of right circular cylinder = 4.5 cm [Given]

• So, it's radius = 4.5/ 2 = 2.25 cm

Volume of melted cylinder = volume of cylinder = πr²h

= π x (2.25)² x 10

= π x 2.25 x 2.25 x 10

= 50.625π cm³

Number of coin = Volume of melted cylinder / Volume of coin

= 50.625π/0.1125π

= 450

So, the number of coins = 450 coins.

Answer 2

$\huge\underline\mathbb {SOLUTION:-}$

Answer:

• The required number of coins = 450

Given:

• Diameter of coins = 1.5 Cm
• Height of coins = 2 mm
• Diameter of right circular cylinder = 4.5 Cm
• Height of right circular cylinder = 10 Cm

To Find:

• The required number of coins = ?

Explanation:

Let N be the number of 1.5 Cm diameter coins required to form a right circular cylinder of height 10 Cm and diameter 4.5 Cm.

Now,

Volume of right circular cylinder = N × Volume of 1.5 Cm diameter coin.

Therefore:

$N \: = \frac{\pi( \frac{4.5}{2}) {}^{2} \times 10 }{\pi( \frac{1.5}{2}) {}^{2} \times \frac{2}{10} }$

$\implies \frac{45 \times45 }{20 \times 20} \times \frac{10 \times 20 \times 20 \times 10}{15 \times 15 \times 12}$

$\implies \mathsf {9 \times 10\times 5}$

$\implies \mathsf {450}$

Diameter of a coin = 1.5 Cm

Radius = $\mathsf {\frac{15}{2}\:Cm}$

Thickness of coin = 0.2 Cm

Volume of one coin = $\mathsf {\pi r^2h = \pi \bigg(\frac{1.5}{2}\bigg)^2\times 0.2\:Cm^3}$

Volume of N coins = Volume of the right circular cylinder to be formed:

$\implies \mathsf {\pi\bigg(\frac{1.5}{2}\bigg)^2 \times 0.2 \times N = \pi\bigg(\frac{4.5}{2}\bigg)^2 \times 10}$

$\implies \mathsf {N = \frac{4.5 \times 4.5 \times 10 \times 2 \times 2}{2 \times 2 \times 1.5 \times 1.5 \times 0.2} = 450}$

• Hence, the required number of coins = 450