Question

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that $1c{m^3}$of iron has approximately 8 g mass. (Use $\pi = 3.14$)

Answer 1

Hint: In this question remember to use the given information and take the help of figure also remember that volume of cylinder is given as; $V = \pi {r^2}h$, use this information to approach the solution of the question.

Complete step-by-step answer:
According to the given information we have a solid iron pole which consist of cylinder of 220 cm height and 24 cm diameter of base there is another cylinder surmounting it which have height 60 cm and radius 8 cm
So, the given dimensions are height of large cylinder (H) = 220 cm and diameter = 24 cm
$ \Rightarrow $ Radius (R) = $\dfrac{{diameter}}{2} = \dfrac{{24}}{2}$=12 cm
Height of small cylinder (h) = 60 cm and radius of small cylinder (r) = 8 cm
We have to find the mass of iron pole and we have given that mass of $1c{m^3}$ is equal to 8 g
So, to find the mass we have to find the volume of pole which is equal to volume of large cylinder + volume of small cylinder
We know that volume of cylinder is given as; $V = \pi {r^2}h$ here r is the radius of cylinder and h is the height of cylinder
Now substituting the values of the dimensions of large cylinder in the above formula we get
Volume of large cylinder = $\pi {\left( {12} \right)^2}\left( {220} \right)$
$ \Rightarrow $Volume of large cylinder = $3.14 \times 144 \times \left( {220} \right)$
$ \Rightarrow $Volume of large cylinder = $99475.2c{m^3}$
Now substituting the values of dimensions of small cylinder in the formula of volume of cylinder we get
Volume of small cylinder = $\pi {\left( 8 \right)^2}\left( {60} \right)$
$ \Rightarrow $Volume of small cylinder = $3.14 \times 64 \times 60$
$ \Rightarrow $Volume of small cylinder = $12057.6c{m^3}$
So, Volume of iron pole = Volume of small cylinder + volume of large cylinder
Substituting the values in the above equation we get
Volume of iron pole = $12057.6c{m^3}$+ $99475.2c{m^3}$
Volume of iron pole = $111532.8c{m^3}$
Now we know that $1c{m^3}$ of iron has 8 g mass
So, mass of $111532.8c{m^3}$ iron has mass = $8 \times 111532.8$
$ \Rightarrow $Mass of $111532.8c{m^3}$ iron has mass = 892262.4 g
$ \Rightarrow $Mass of $111532.8c{m^3}$ iron has mass = \[\dfrac{{892262.4}}{{1000}}kg\]
$ \Rightarrow $Mass of $111532.8c{m^3}$ iron has mass = 892.26 kg
Therefore, the mass of the iron pole will be 892.26 kg

Note: Whenever we face such a type of question first find total volume. Then for finding mass apply the formula, and always keep in mind that the units of density in the given question and the volume we have to find, should be the same.