Question

Find the volume of a bucket whose height is 84cm, diameter of top as 40cm and radius of bottom as 10cm.
${\text{A}}$. $30.4$ litre
${\text{B}}$. $61.6$ litre
${\text{C}}$. $30.8$ litre
${\text{D}}$. \[60\] litre

Answer 1

Hint: From the question, we have to find the volume of the bucket from the given data and choose the correct answer. We have to find the required result by using the formula of measurements.
The branch of mathematics which deals with the measurements of lengths, angles, areas, perimeters and volumes of plane and solid figures is called mensuration.

Formula used: The volume of the bucket or frustum cone is given by \[{\text{V = }}\dfrac{1}{3}{{\pi h}}\left[ {{\text{r}}_1^2 + {{\text{r}}_{\text{1}}}{{\text{r}}_{\text{2}}} + {\text{r}}_2^2} \right]\]${\text{c}}{{\text{m}}^{\text{3}}}$.
Where, ${\text{h}}$- height of bucket or frustum cone.
${{\text{r}}_{\text{1}}}$ - radius of the lower end of the bucket.
${{\text{r}}_{\text{2}}}$- radius of the upper end of the bucket.
${\text{Radius}} = \dfrac{{{\text{Diameter}}}}{2}$ .

Complete step-by-step solution:
From the given, we have the height of the bucket or frustum cone is ${\text{h}} = 84$ cm.
Now we have to find the radius of the upper end or top by using the diameter formula mentioned in the above formula.
${\text{Radius}} = \dfrac{{{\text{Diameter}}}}{2}$ .
From the data, we know the diameter of the top or upper bond is $40$ cm.
\[{\text{Radius of the top}}\left( {{{\text{r}}_{\text{1}}}} \right) = \dfrac{{{\text{Diameter of the top}}}}{2}\]
\[ \Rightarrow {{\text{r}}_{\text{1}}} = \dfrac{{40}}{2}\]
$ \Rightarrow {{\text{r}}_{\text{1}}} = 20$ cm
We know the radius of the lower end or bottom is ${{\text{r}}_{\text{2}}} = 10$ cm from the given data.
Now, we are going to substitute the above values in the formula of the volume of the bucket or frustum cone. Then, we get the required result.
\[{\text{V = }}\dfrac{1}{3}{{\pi h}}\left[ {{\text{r}}_1^2 + {{\text{r}}_{\text{1}}}{{\text{r}}_{\text{2}}} + {\text{r}}_2^2} \right]\]${\text{c}}{{\text{m}}^{\text{3}}}$
We know that, \[{{\pi }} = \dfrac{{22}}{7}\].
${\text{V}} = \dfrac{1}{3} \times \dfrac{{22}}{7} \times 84 \times \left[ {{{\left( {20} \right)}^2} + \left( {20} \right)\left( {10} \right) + {{\left( {10} \right)}^2}} \right]$
Now, squaring and cancelling the above terms. We get,
${\text{V}} = \dfrac{1}{3} \times \dfrac{{22}}{1} \times 12 \times \left[ {\left( {400} \right) + \left( {200} \right) + \left( {100} \right)} \right]$
\[{\text{V}} = \dfrac{1}{1} \times \dfrac{{22}}{1} \times 4 \times \left[ {700} \right]\]
\[{\text{V}} = 22 \times 4 \times 700\]
\[{\text{V}} = 61600\]\[{\text{c}}{{\text{m}}^3}\].
Therefore, we got the volume of the bucket, \[{\text{V}} = 61600\]\[{\text{c}}{{\text{m}}^3}\].
According to the question, we have to choose the correct answer but our choices of answer in the unit of litre. So, we have to convert the unit from (volume of the bucket) cubic centimetre to litre then we have to choose the correct option.
We know that, $1$${\text{c}}{{\text{m}}^{\text{3}}}$ = $0.001$ litre
$ \Rightarrow \left( {61600 \times 1} \right)$${\text{c}}{{\text{m}}^{\text{3}}}$= $\left( {61600 \times 0.001} \right)$ litre
Hence,
$ \Rightarrow 61600$${\text{c}}{{\text{m}}^{\text{3}}}$= $61.6$ litre.

$\therefore $ The correct option is ${\text{B}}$($61.6$ litre).

Note: We have to know that volume describes how much space a substance occupies and is given in litters (SI) or gallons (English). The volume of a substance is determined by how much material is present and how closely the particles of the material are packed together. A temperature and pressure can greatly affect the volume of a substance, especially gases. As with mass, increasing and decreasing the amount of material also increases and decreases the volume of the substance.