Question

The volume of metal in a metallic cylindrical pipe is 748 cm
3
. Its length is
14 cm and its external radius is 9 cm. Find its thickness.

Answer 1

Hope u like my process
=====================
=> Let the inner radius be r cm

=> External Radius = R = 9 cm

=> Height of pipe = h= 14 cm

=> Volume of metallic cylinder

$= \pi {r}^{2} h = 748 \: cm {}^{3} \\ \\ where \: \: \pi = \frac{22}{7} \\ \\ \underline{ \: \: \: \: \: so \: \: \: \: \: \: }\\ \\ = > \it \green{\frac{22}{7} \times {r}^{2} \times 14 }= \green{748} \\ \\ = > \it \green{ {r}^{2} } = \green{ \frac{748}{44} } = \green{ \frac{68}{4} } = \green{17} \\ \\ = > \boxed{ \bf \: r = \green{ \sqrt{17} } = \underline{ \orange{4.12 \: \: cm}}}$

Now the thickness can be determined by

=> R - r = 9 - 4.12 = 4.88 cm

_____________________________
Hope this is ur required answer

Proud to help you

Answer 2

Answer:

1 cm

Step-by-step explanation:

Exernal Radious = 9 cm

Heiht = 14 cm

Volume = 748 cm^3

$\star$ Let internal radious be R

$\implies \sf \: \pi \: R ^{2} h - \pi {r}^{2} h = 748 {cm}^{3}$

$\implies( \dfrac{22}{7} \times 9 \times 9 \times 14 )- (\dfrac{22}{7} \times r \times r \times 14) = 748 {cm}^{3}$

$\implies \sf \: 44 \times 81 - 44 {r}^{2} = 748 {cm}^{3}$

$\implies \sf44(81 - {r}^{2} ) = 748 {cm}^{3}$

$\implies \sf81 - {r}^{2} = \dfrac{748}{44}$

$\implies \sf81 - {r}^{2} = 17$

$\implies \sf{r}^{2} = - 17 + 81$

$\implies \sf {r}^{2} = 64$

$\implies \sf {r} = \sqrt{64}$

$\implies \sf r = 8cm$

Thickness = External Radious - Internal Radious

Thickness = 9 cm - 8 cm

Thickness = 1 cm