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Shape Conversion of Solids

A hollow iron...

Question

A hollow iron pipe is $21 c m$ long and its external diameter is $8 c m$ . If the thickness of the pipe is $1 c m$ and iron weighs $8 g / c m^{3}$ , then the weight of pipe is :

3.6 k g

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3.696 k g

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36 k g

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36.9 k g

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Solution

The correct option is C $3.696 k g$
Volume of iron $= π (R^{2} - r^{2}) \times h$

$= π (4^{2} - 3^{2}) \times 21$

$= \frac{22}{7} \times (16 - 9) \times 21$

$= 462 c m^{2}$

Weight $= \frac{8 \times 462}{1000} k g = 3.696 k g$

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A hollow iron pipe is 21cm long and its external diameter is 8cm. If the thickness of the pipe is 1cm and iron weighs 8g/cm3, then the weight of pipe is :

The correct option is C 3.696kgVolume of iron =π(R2−r2)×h=π(42−32)×21=227×(16−9)×21=462cm2Weight =8×4621000kg=3.696kg

A hollow iron pipe is $21 c m$ long and its external diameter is $8 c m$ . If the thickness of the pipe is $1 c m$ and iron weighs $8 g / c m^{3}$ , then the weight of pipe is :

The correct option is C $3.696 k g$
Volume of iron $= π (R^{2} - r^{2}) \times h$

$= π (4^{2} - 3^{2}) \times 21$

$= \frac{22}{7} \times (16 - 9) \times 21$

$= 462 c m^{2}$

Weight $= \frac{8 \times 462}{1000} k g = 3.696 k g$