# An iron pillar consists of a cylindrical portion $2.8m$ high and $20cm$ in diameter and a cone $42cm$ high is surmounted on it. Find the weight of the pillar, given that $1c{m^3}$ of iron weights $7.5gm$.

Answer

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Hint: Find the volume of the pillar by finding the volumes of cylindrical and conical portions. And then calculate its weight as per the given information.

Given, the height of the cylindrical portion of the pillar, ${h_{cylinder}} = 2.8m = 280cm$.

And the height of the conical portion ${h_{cone}} = 42cm$.

Since, the cone is surmounted on the cylinder, the radius of the base will be the same for cylinder and cone. And diameter is given as $20cm$ in the question. So, radius will be:

$

\Rightarrow r = \dfrac{d}{2} = \dfrac{{20}}{2}, \\

\Rightarrow r = 10cm \\

$

And we know formulae of volume of cylinder and cone, then:

$

\Rightarrow {V_{cylinder}} = \pi {r^2}{h_{cylinder}}, \\

\Rightarrow {V_{cone}} = \dfrac{1}{3}\pi {r^2}{h_{cone}} \\

$

Thus, the total volume of the pillar will be:

\[

\Rightarrow V = {V_{cylinder}} + {V_{cone}}, \\

\Rightarrow V = \pi {r^2}{h_{cylinder}} + \dfrac{1}{3}\pi {r^2}{h_{cone}}, \\

\Rightarrow V = \pi {r^2}\left[ {{h_{cylinder}} + \dfrac{{{h_{cone}}}}{3}} \right] \\

\]

Putting values of respective heights and radius, we’ll get:

$

\Rightarrow V = \dfrac{{22}}{7} \times {\left( {10} \right)^2}\left[ {280 + \dfrac{{42}}{3}} \right], \\

\Rightarrow V = \dfrac{{22}}{7} \times 100 \times 294, \\

\Rightarrow V = 92400 \\

$

Thus the volume of the pillar is $92400c{m^3}$. Now, we have to determine the weight of the pillar. So, according to question:

Weight of $1c{m^3}$of iron $ = 7.5gm$,

$\therefore $Therefore, the weight of $92400c{m^3}$of iron will be:

$

\Rightarrow w = 92400 \times 7.5gm, \\

\Rightarrow w = 693000gm, \\

\Rightarrow w = 693kg. \\

$

Thus, the total weight of the pillar is $693kg$.

Note: If we have to calculate the weight of a solid when its density is given, we always have to calculate the volume of the solid first because weight is directly related to volume and density as:

$

\Rightarrow Density = \dfrac{{weight}}{{Volume}}, \\

\Rightarrow weight = Density \times Volume. \\

$

Given, the height of the cylindrical portion of the pillar, ${h_{cylinder}} = 2.8m = 280cm$.

And the height of the conical portion ${h_{cone}} = 42cm$.

Since, the cone is surmounted on the cylinder, the radius of the base will be the same for cylinder and cone. And diameter is given as $20cm$ in the question. So, radius will be:

$

\Rightarrow r = \dfrac{d}{2} = \dfrac{{20}}{2}, \\

\Rightarrow r = 10cm \\

$

And we know formulae of volume of cylinder and cone, then:

$

\Rightarrow {V_{cylinder}} = \pi {r^2}{h_{cylinder}}, \\

\Rightarrow {V_{cone}} = \dfrac{1}{3}\pi {r^2}{h_{cone}} \\

$

Thus, the total volume of the pillar will be:

\[

\Rightarrow V = {V_{cylinder}} + {V_{cone}}, \\

\Rightarrow V = \pi {r^2}{h_{cylinder}} + \dfrac{1}{3}\pi {r^2}{h_{cone}}, \\

\Rightarrow V = \pi {r^2}\left[ {{h_{cylinder}} + \dfrac{{{h_{cone}}}}{3}} \right] \\

\]

Putting values of respective heights and radius, we’ll get:

$

\Rightarrow V = \dfrac{{22}}{7} \times {\left( {10} \right)^2}\left[ {280 + \dfrac{{42}}{3}} \right], \\

\Rightarrow V = \dfrac{{22}}{7} \times 100 \times 294, \\

\Rightarrow V = 92400 \\

$

Thus the volume of the pillar is $92400c{m^3}$. Now, we have to determine the weight of the pillar. So, according to question:

Weight of $1c{m^3}$of iron $ = 7.5gm$,

$\therefore $Therefore, the weight of $92400c{m^3}$of iron will be:

$

\Rightarrow w = 92400 \times 7.5gm, \\

\Rightarrow w = 693000gm, \\

\Rightarrow w = 693kg. \\

$

Thus, the total weight of the pillar is $693kg$.

Note: If we have to calculate the weight of a solid when its density is given, we always have to calculate the volume of the solid first because weight is directly related to volume and density as:

$

\Rightarrow Density = \dfrac{{weight}}{{Volume}}, \\

\Rightarrow weight = Density \times Volume. \\

$

Last updated date: 28th Sep 2023

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Total views: 363.6k

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