A hollow cylindrical tube $20$ cm long is make of iron and its external diameter is $8$ cm. The volume of iron used in making the tubes is $440$ $c m^{3}$ , what is the thickness of the tube?

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Solution

The correct option is A $1$ cm
External radius $= \frac{8}{2} = 4 c m$

Let internal radius $= r c m$

Volume of tube $= π (4^{2} - r^{2}) \times 20$

$\Rightarrow 440 c m^{3} = \frac{22}{7} \times 20 (16 - r^{2})$

$\Rightarrow (16 - r^{2}) = \frac{440 \times 7}{22 \times 20} = 7$

$\Rightarrow r^{2} = 16 - 7 \Rightarrow r = \sqrt{9} = 3 c m$

$∴$ Thickness $= 4 - 3 = 1 c m$

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A hollow cylindrical tube 20 cm long is make of iron and its external diameter is 8 cm. The volume of iron used in making the tubes is 440 cm3, what is the thickness of the tube?

The correct option is A 1 cmExternal radius=82=4cmLet internal radius=rcmVolume of tube=π(42−r2)×20⇒440cm3=227×20(16−r2)⇒(16−r2)=440×722×20=7⇒r2=16−7⇒r=√9=3cm∴Thickness=4−3=1cm

A hollow cylindrical tube $20$ cm long is make of iron and its external diameter is $8$ cm. The volume of iron used in making the tubes is $440$ $c m^{3}$ , what is the thickness of the tube?