Question

13. Find the number of coin each of which are
1.5 om in diameter and 0.2 om thick required
to form a notorolar yinder of height 10 cm
and diameter 4.5 om​

Answer 1

$\Large {\underline { \sf \orange{Appropriate \: Question :}}}$

Find the number of coins each of which are 1.5 cm in diameter and 0.2 cm thick required to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

$\Large {\underline { \sf \orange{Clarification :}}}$

Here, we are given that the thickness and diameter of each coin is 1.5 cm and 0.2 cm. Also, height and diameter of the right circular cylinder is 10 cm and 4.5 cm. We have to find the number of coins required to form a right circular cylinder.

In order to tackle this question, we'll have to find the volume of the cylinder and volume of each coin. Then, volume of total number of coins will be equal to the volume of the cylinder. We'll divide volume of each coin from volume of the cylinder in order to find the number of coins.

$\Large {\underline { \sf \orange{Explication \: of \: Steps :}}}$

Let the number of coins be x.

$\underline{ \sf {\maltese \; \; \; According \: to \: the \: question : \; \; \; }}$

$\longrightarrow \sf { Volume_{(Cylinder)}= Volume_{(1 \: Coin)} \times Number_{(Coins)} }$

So,

$\bigstar \: \boxed{\sf { Number_{(Coins)}= \dfrac{Volume_{(Cylinder)}}{Volume_{(1 \: Coin)}} }}$

$\underline{\small \sf {\maltese \; \; \; Finding \: volume \: of \: 1 \: coin : \; \; \; }}$

Given :

• Diameter = 1.5 cm
• Thickness (here, Height) = 0.2 cm

→ Diameter = 2 × Radius

→ $\sf{ \dfrac{Diameter}{2} }$ = Radius

→ $\sf{ \dfrac{1.5 }{2} }$ cm = Radius

→ 0.75 cm = Radius

Using formula to find volume :

Thickness of the coin can be considered as its height. Coins exist in the shape of cylinder.

★ Volume of cylinder = πr²h

→ Volume of 1 coin = π × (0.75)² × 0.2 cm³

→ Volume of 1 coin = ( π × 0.75 × 0.75 × 0.2 ) cm³

$\underline{\small \sf {\maltese \; \; \; Finding \: volume \: of \: cylinder : \; \; \; }}$

Given :

• Diameter = 4.5 cm
• Height = 10 cm

→ Diameter = 2 × Radius

→ $\sf{ \dfrac{Diameter}{2} }$ = Radius

→ $\sf{ \dfrac{4.5 }{2} }$ cm = Radius

→ 2.25 cm = Radius

Using formula to find volume :

★ Volume of cylinder = πr²h

→ Volume of cylinder = π × (2.25)² × 10 cm³

→ Volume of cylinder = ( π × 2.25 × 2.25 × 10 ) cm³

$\underline{\small \sf {\maltese \; \; \; Finding \: number \: of \: coins : \; \; \; }}$

$\bigstar \: \boxed{\sf { Number_{(Coins)}= \dfrac{Volume_{(Cylinder)}}{Volume_{(1 \: Coin)}} }}$

$\longrightarrow \sf {Number_{(Coins)} = \dfrac{ ( \cancel {\pi }\times 2.25 \times 2.25 \times 10) \; \cancel{cm^3} }{( \cancel{\pi} \times 0.75 \times 0.75 \times 0.2) \:\cancel{ cm^3} }}$

$\longrightarrow \sf {Number_{(Coins)} = \dfrac{ 2.25 \times 2.25 \times 10 }{ 0.75 \times 0.75 \times 0.2}}$

$\longrightarrow \sf {Number_{(Coins)} = \dfrac{ 2.25 \times 2.25 \times 10 \times 100 \times 100 \times 10 }{75 \times 75 \times 2}}$

$\longrightarrow \sf {Number_{(Coins)} = \dfrac{ 225 \times 225 \times 10 \times \cancel{100} \times \cancel{100} \times 10 }{ 75 \times 75 \times 2 \times \cancel{ 100} \times \cancel{100}}}$

$\longrightarrow \sf {Number_{(Coins)} = \dfrac{ \cancel{225 } \times \cancel{225} \times 10 \times 10 }{ \cancel{75} \times \cancel{75} \times 2 } }$

$\longrightarrow \sf {Number_{(Coins)} = \dfrac{ 3 \times 3 \times 10 \times 10 }{ 2 } }$

$\longrightarrow \sf {Number_{(Coins)} = \dfrac{ 9 \times 100 }{ 2 }}$

$\longrightarrow \sf {Number_{(Coins)} = \cancel{\dfrac{ 900 }{ 2 }}}$

$\longrightarrow \boxed {\pmb{ \rm \purple {Number_{(Coins)} = 450 }}}$

❝ Therefore, 450 coins are required to form a right circular cylinder of height 10 cm and diameter 4.5 cm. ❞

Answer 2

Answer:

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