
The height of a solid metal cylinder is 20cm. Its radius is 1.5cm. The cylinder is melted and cast into spheres of each radius 1.5cm. How many such spheres can be cast from the cylinder?
Answer
585k+ views
Hint: First find the volume of the cylinder using formula $\pi {{r}^{2}}h$ using data given in the question, then find the volume of one sphere using the dimension given. Then divide the volume of the cylinder by the volume of the sphere you will get your answer.
Complete step-by-step answer:
At first we will find the volume of the cylinder by using $\pi {{r}^{2}}h$ formula.
In the formula $\pi =3.14$, ‘r’ depicts radius and ‘h’ describes height.
From the given information, the radius of the cylinder is 1.5 cm and height is 20 cm.
Therefore, the volume of cylinder after substituting the values, we get
\[{{\pi }^{2}}h=3.14\times 1.5\times 1.5\times 20=141.3c{{m}^{2}}\]
It is given that the cylinder was melted down to form spheres. We can say that ‘n’ numbers of spheres were made so the total volume of the cylinder is equal to the total sum of volume of all spheres.
We know that the volume of one sphere can be written as $\dfrac{4}{3}\pi {{r}^{3}}$.
Here ‘r’ is the radius of the sphere.
As per the given information, the radius of the sphere is 1.5 cm.
So the volume of one sphere will be,
$=\dfrac{4}{3}\times 3.14\times {{\left( 1.5 \right)}^{3}}$
So we can write as,
$\text{volume of cylinder = n }\!\!\times\!\!\text{ volume of 1 sphere}$
Here we know the volume of 1 cylinder is $\text{141}\text{.3c}{{\text{m}}^{3}}$.
So we can write it as,
$\text{141}\text{.3=n}\times \dfrac{4}{3}\times 3.14\times {{\left( 1.5 \right)}^{3}}$
$\therefore n=\dfrac{141.3}{\dfrac{4}{3}\times 3.14\times {{\left( 1.5 \right)}^{3}}}=10$
So, there are a total 10 spheres that can be cast out of the given cylinder.
Note: Students should know that the formulas of volume of sphere and cylinder by heart. They should be careful about the calculation mistake as it includes lots of calculations. Students often make mistakes in calculating the formula $\text{volume of cylinder = n }\!\!\times\!\!\text{ volume of 1sphere}$. In this way they get the wrong answer.
Complete step-by-step answer:
At first we will find the volume of the cylinder by using $\pi {{r}^{2}}h$ formula.
In the formula $\pi =3.14$, ‘r’ depicts radius and ‘h’ describes height.
From the given information, the radius of the cylinder is 1.5 cm and height is 20 cm.
Therefore, the volume of cylinder after substituting the values, we get
\[{{\pi }^{2}}h=3.14\times 1.5\times 1.5\times 20=141.3c{{m}^{2}}\]
It is given that the cylinder was melted down to form spheres. We can say that ‘n’ numbers of spheres were made so the total volume of the cylinder is equal to the total sum of volume of all spheres.
We know that the volume of one sphere can be written as $\dfrac{4}{3}\pi {{r}^{3}}$.
Here ‘r’ is the radius of the sphere.
As per the given information, the radius of the sphere is 1.5 cm.
So the volume of one sphere will be,
$=\dfrac{4}{3}\times 3.14\times {{\left( 1.5 \right)}^{3}}$
So we can write as,
$\text{volume of cylinder = n }\!\!\times\!\!\text{ volume of 1 sphere}$
Here we know the volume of 1 cylinder is $\text{141}\text{.3c}{{\text{m}}^{3}}$.
So we can write it as,
$\text{141}\text{.3=n}\times \dfrac{4}{3}\times 3.14\times {{\left( 1.5 \right)}^{3}}$
$\therefore n=\dfrac{141.3}{\dfrac{4}{3}\times 3.14\times {{\left( 1.5 \right)}^{3}}}=10$
So, there are a total 10 spheres that can be cast out of the given cylinder.
Note: Students should know that the formulas of volume of sphere and cylinder by heart. They should be careful about the calculation mistake as it includes lots of calculations. Students often make mistakes in calculating the formula $\text{volume of cylinder = n }\!\!\times\!\!\text{ volume of 1sphere}$. In this way they get the wrong answer.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
Write an application to the principal requesting five class 10 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the median of the first 10 natural numbers class 10 maths CBSE

Write examples of herbivores carnivores and omnivo class 10 biology CBSE
