Question

a hollow cylindrical pipe is 21cm long. it's inner and outer diameters are 6cm and 10cm respectively. find the volume of the metal used in making the pipe.​

Answer 1

Answer

1056 cm³

Explanation:

Given that length of pipe is 21cm.

Let the outer radius of pipe be R cm and inner radius of pipe be r cm.

A hollow cylindrical pipe is 21cm long. It's inner and outer diameters are 6cm and 10cm respectively.

Here, d = 6cm and D = 10cm

r = d/2 = 6/2 = 3 cm and R = D/2 = 10/2= 5 cm

Volume of metal used in making the pipe = π(R² - r²)l

Substitute the known values

→ 22/7 [(5)² - (3)²] × 21

→ 22/7 × 21 (25 - 9)

→ 22 × 3 (16)

→ 66(16)

→ 1056

Therefore, 1056 cm³ is the volume of metal used in making the pipe.

Answer 2

Answer:

${\boxed{\bold{Volume\:of\:metal=1056\:{cm}^{3} }}}$

Explanation:

Given:-

• Length of pipe = 21 cm
• Outer diameter = 10 cm
• Inner diameter = 6 cm
• Volume of metal= ?

Let the inner radius be r1 and outer radius be r2.

$\tt r_1 = \frac{Diameter_1}{2} \\\rightarrow\tt r_1 = \frac{6}{2}\:cm \\\rightarrow{\boxed{\tt{r_1 = 3\:cm}}}$

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$\tt r_2 = \frac{Diameter_2}{2} \\\rightarrow\tt r_2 =\frac{10}{2}\:cm \\\rightarrow{\boxed{\tt{r_2 = 5\:cm}}}$

Now, formula to be used in this case:-

${\boxed{\bold{Volume=\pi ({r_2}^{2} - {r_1}^{2} )\times l }}}$

$\rightarrow\tt Volume = \frac{22}{7} ({5}^{2} - {3}^{2} ) \times 21 \\\\\rightarrow\tt Volume = \frac{22}{7} (25-9)\times 21 \\\\\rightarrow\tt Volume = \dfrac{22}{7} \times 16 \times 21 \\\\\rightarrow\tt Volume = \dfrac{22}{\cancel{7}} \times 16 \times \cancel{21} \\\\\rightarrow{\boxed{\tt {Volume = 1056\:{cm}^{3} }}}$