Question

A thick metallic cylinder hollow from inside has an outer diameter equal to 10cm and outer height equal to 14cm. If the thickness of the metal is 1.5 cm, find the volume of the metal contained in the cylinder. ​

Answer 1

Gɪᴠᴇɴ :-

• Outer diameter = 10cm.
• Outer Height = 14cm.
• Thickness of the metal = 1.5cm.

Tᴏ Fɪɴᴅ :-

• The volume of the metal contained in the cylinder. ?

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• Volume of Hollow cylinder = πh(r1² – r2²)
• Radius = (Diameter)/2
• Inner radius = outer radius - Thickness .

Sᴏʟᴜᴛɪᴏɴ :-

→ Outer Radius = 10/2 = 5cm. = r1

→ Inner Radius = 5 - 1.5 = 3.5cm. = r2

→ Height = 14cm = h .

Putting all values in formula we get :-

→ Volume = π * 14 * (5² - 3.5²)

→ Volume = 14π * (5 + 3.5)(5 - 3.5)

→ Volume = 14π * 8.5 * 1.5

→ Volume = 14 * 3.14 * 8.5 * 1.5

→ Volume = 560.49cm³ ≈ 561cm³ (Ans.)

Hence, volume of the metal contained in the Hollow cylinder is 561cm³.

Answer 2

${ \bold{ \underline{ \green{ \underline{ \huge{Solution:-}}}}}}$



${ \dagger{ {\blue{ \rm{ \underline{ \: \: \: GIVEN- }}}}}}$



${ \underline{ \pink{ \sf{dimension \: of \: hollow \: metallic \: cylinder}}}}$



${ \star{ \red{ \sf{ \: \: outer \: diameter = 10 \: cm}}}}$



then,



${ \rightarrow{ \sf{ \red{outer \: radius \: (r1) = \frac{10 \: cm}{2} = 5 \: cm}}}}$



${ \star{ \red{ \sf{ \: \: thickness \: of \: the \: metal = 1.5 \: cm}}}}$



${ \star{ \red{ \sf{ \: inner \: radius(r2) = r1 - thickness}}}}$



${ \rightarrow{ \red{ \sf{r2 = 5 \: cm - 1.5 \: cm \: = 3.5cm}}}}$



${ \star{ \red{ \sf{ \: \: \: height(h) = 14 \: cm}}}}$



${ \dagger{ \blue{ \underline{ \rm{ \: \: TO \: FIND- }}}}}$



${ \purple{ \sf{ volume \: of \: the \: hollow \: cylinder}}}$



${ \boxed{ \boxed{ \pink{ \sf{volume = \pi \: h( {r1}^{2} - {r2}^{2} )}}}}}$



${ \sf{ \blue{ volume = \pi \: 14 cm \: ( {(5 \: cm)}^{2} - {(3.5 \: cm)}^{2})}}}$



${ \implies{ \sf{ \blue{volume = \frac{22}{7} \times 14(25 - 12.25)}}}}$



${ \implies{ \sf{ \blue{volume = 44 \times 12.75 \: {cm}^{3} }}}}$



${ \implies{ \underline{ \boxed{\sf{ \red{volume = 561 {cm}^{3}}}}}}}$