Question
If the radius of a sphere is increased by 10%, then the volume will be increased by :
Answer: Option A
If R is the radius of sphere, volume of the sphere = $$\frac{4}{3}\pi {R^3}$$
When radius of sphere is increased by 10%
New volume :
$$\eqalign{
& = \frac{4}{3}\pi {\left( {1.1R} \right)^3} \cr
& = \frac{4}{3}\pi {R^3}\left( {1.331} \right) \cr} $$
Difference :
$$\eqalign{
& = \frac{4}{3}\pi {R^3}\left( {1.331} \right) - \frac{4}{3}\pi {R^3} \cr
& = \frac{4}{3}\pi {R^3}\left( {1.331 - 1} \right) \cr
& = \frac{4}{3}\pi {R^3}\left( {0.331} \right) \cr} $$
Increase % :
$$\eqalign{
& = \frac{{\frac{4}{3}\pi {R^3}\left( {0.331} \right)}}{{\frac{4}{3}\pi {R^3}}} \times 100 \cr
& = 33.1\% \cr} $$
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If R is the radius of sphere, volume of the sphere = $$\frac{4}{3}\pi {R^3}$$
When radius of sphere is increased by 10%
New volume :
$$\eqalign{
& = \frac{4}{3}\pi {\left( {1.1R} \right)^3} \cr
& = \frac{4}{3}\pi {R^3}\left( {1.331} \right) \cr} $$
Difference :
$$\eqalign{
& = \frac{4}{3}\pi {R^3}\left( {1.331} \right) - \frac{4}{3}\pi {R^3} \cr
& = \frac{4}{3}\pi {R^3}\left( {1.331 - 1} \right) \cr
& = \frac{4}{3}\pi {R^3}\left( {0.331} \right) \cr} $$
Increase % :
$$\eqalign{
& = \frac{{\frac{4}{3}\pi {R^3}\left( {0.331} \right)}}{{\frac{4}{3}\pi {R^3}}} \times 100 \cr
& = 33.1\% \cr} $$
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