Question

A spherical block of metal weighs 12 pounds. What is the weight in pounds of another block of the same metal if its radius is 3 times the radius of the 12-pound block?
(A) 324
(B) 15
(C) 36
(D) 108

Answer 1

Hint: In this question, we have two cases. In the \[{{1}^{st}}\] case, there is a spherical block of metal that weighs 12 pounds. Let us assume that the radius of the spherical block that weighs 12 pounds is x cm. Now, calculate the volume of the block that weighs 12 pounds using the formula, \[Volume=\dfrac{4}{3}\pi {{\left( radius \right)}^{3}}\] . Then, use the formula \[Density=\dfrac{Weight}{Volume}\] and calculate the density of the block that weighs 12 pounds. In the \[{{2}^{nd}}\] case, there is another spherical block whose radius is 3 times the radius of the 12-pound block. The radius of this block is 3x. Now, calculate the volume of this spherical block using the formula, \[Volume=\dfrac{4}{3}\pi {{\left( radius \right)}^{3}}\] . Then, use the formula \[Density=\dfrac{Weight}{Volume}\] and calculate the density. Since both of the spherical blocks are made of the same metal, so the density of both of the blocks must be the same. Therefore, the densities of the spherical block in the \[{{1}^{st}}\] case and \[{{2}^{nd}}\] case must be equal. Now, solve it further and get the value of the weight of the spherical block.

Complete step-by-step answer:
According to the question, we have two cases.
In the \[{{1}^{st}}\] case, there is a spherical block of metal that weighs 12 pounds.
Let us assume that the radius of the spherical block that weighs 12 pounds is x cm.
The radius of the sphere = x cm ………………………………..(1)
The weight of the sphere = 12 pounds ………………………………………(2)
We know the formula of volume of a sphere, \[Volume=\dfrac{4}{3}\pi {{\left( radius \right)}^{3}}\] …………………………(3)
From equation (1) we have the radius of the sphere.
Now, from equation (1) and equation (3), we get
\[Volume=\dfrac{4}{3}\pi {{\left( x \right)}^{3}}\,c{{m}^{3}}\] ……………………………………(4)
We know the formula, \[Density=\dfrac{Weight}{Volume}\] ……………………………….(5)
From equation (2), equation (3), and equation (5), we get
The density of the metal = \[\dfrac{12\,pounds}{\dfrac{4}{3}\pi {{\left( x \right)}^{3}}\,c{{m}^{3}}}\] …………………………………(6)
In the \[{{2}^{nd}}\] case, there is another spherical block whose radius is 3 times the radius of the 12-pound block.
From equation (1), we have the radius of the spherical block.
The radius of another sphere = 3x cm ………………………………..(7)
From equation (7), we have the radius of another sphere.
Now, from equation (3) and equation (7), we get
\[Volume=\dfrac{4}{3}\pi {{\left( 3x \right)}^{3}}\,c{{m}^{3}}\] ……………………………………(8)
From equation (5), and equation (8), we get
The density of the metal = \[\dfrac{Weight}{\dfrac{4}{3}\pi {{\left( 3x \right)}^{3}}\,c{{m}^{3}}}\] …………………………………(9)
From equation(6) and equation (9), we have the density of both spherical blocks.
Since both of the spherical blocks are made of the same metal, so the density of both of the blocks must be the same.
\[\begin{align}
  & \Rightarrow \dfrac{12\,pounds}{\dfrac{4}{3}\pi {{\left( x \right)}^{3}}\,c{{m}^{3}}}=\dfrac{Weight}{\dfrac{4}{3}\pi {{\left( 3x \right)}^{3}}\,c{{m}^{3}}} \\
 & \Rightarrow \dfrac{12\,pounds}{\dfrac{4}{3}\pi {{\left( x \right)}^{3}}\,c{{m}^{3}}}=\dfrac{Weight}{27\times \dfrac{4}{3}\pi {{\left( x \right)}^{3}}\,c{{m}^{3}}} \\
 & \Rightarrow 27\times 12\,pounds=Weight \\
\end{align}\]
\[\Rightarrow 324pounds=Weight\] ………………………………….(10)
Therefore, the weight of another spherical block is 324 pounds.
Hence, option (A) is the correct one.

Note:In this question, one might get confused about how to approach this question. The hidden clue to solve this question is to use the property that the density of the same metal is always equal. Therefore, we have to keep the formula of density in our minds, \[Density=\dfrac{Weight}{Volume}\] .