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The volume of a sphere is \[1.76\,c{m^3}\]. The volume of 25 such spheres will be correctly represented by:
A) $0.44\, \times \,{10^2}\,c{m^3}$
B) $44.0\,c{m^3}$
C) $44\,c{m^3}$
D) $44.00\,c{m^3}$

Last updated date: 20th Jun 2024
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Hint: In this solution, we will be focusing on the concepts of significant digits. The answer should have as many significant digits as the terms used in the formula.
Formula used: In this question, we will use the following formula
-Volume of a sphere: $V = \dfrac{4}{3}\pi {R^3}$ where $R$ is the radius of the sphere

Complete step by step answer:
We’ve been given that the volume of a sphere is \[1.76\,c{m^3}\]. Then the volume of 25 such spheres will be the product of the number of spheres and the volume of one sphere.
So, we can write the new volume as
${V_{new}} = 25 \times 1.76$
$ \Rightarrow {V_{new}} = 44\,c{m^3}$
Now the answer to our calculation should have the same number of digits as the highest number of significant digits in the terms that we use in the equation. So, in the terms that we used, 25 has two significant digits, 2 and 5. However \[1.76\,\] has three significant digits, 1, 7, and 6. So our answer must also have three significant digits. Hence the volume of the spheres combined will be represented as
${V_{new}} = \,44.0\,c{m^3}$

Hence the volume of 25 spheres will be ${V_{new}} = \,44.0\,c{m^3}$ so, option (B) is the correct choice.

Note: The fact that all the options have the same magnitude can provide a hint that the question wants us to focus on the concepts of significant digits in the question. If after a decimal point we have a non-zero digit followed by a zero, it won’t be counted as a significant digit. However, any zeros directly after the decimal point are counted as significant digits.