Question

If the radius of sphere is increased by 100 percent then the volume of corresponding sphere is increased by

Answer 1

If the radius of sphere is increased by 100 per cent then the volume of corresponding sphere is increased by 700%.

Step-by-step explanation:

Let the initial radius of the sphere be “r₁” and the initial volume be “V₁”.

So, the Initial Volume of the sphere, V₁ = $\frac{4}{3}$ * πr₁³ …… (i)

It is given that, the radius of sphere is increased by 100% i.e.,  

The new radius, r₂ = r₁ + (100% of r₁) = 2r₁

Now,  

The New Volume of the sphere,  

V₂ = $\frac{4}{3}$ * πr₂³

⇒ V₂ = $\frac{4}{3}$ * π * (2r₁)³

⇒ V₂ = $\frac{4}{3}$ * π * 8 * r₁³

⇒ V₂ = 8 * V₁ ……. [substituting from (i)]

Thus,

The volume of the corresponding sphere is increased by,

= [(V₂ – V₁)/V₁] * 100

= [(8V1 – V₁)/V₁] * 100

= [7V₁/V₁] * 100

= 7 * 100

= 700%

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Answer 2

Answer:

700%

Step-by-step explanation:

Case 1

Let radius = r

so,

volume =

$\frac{4}{3} \times \pi \times {r}^{3}$

Case 2

Now

radius increased by 100%

so,

radius =

$r +r \times \frac{100}{100} = 2r$

Now,

Volume =

$\frac{4}{3} \times \pi \times {(2r)}^{3} = \frac{4}{3} \times \pi \times 8 {r}^{3}$

now, increase in volume =

$\frac{4}{3} \pi \times8 {r}^{3} - \frac{4}{3} \pi \times {r}^{3}$

$= \frac{4}{3} \pi {r}^{3} \times 7$

Now,

volume increase in % =

$\frac{ \frac{4}{3} \pi {r}^{3} \times 7}{ \frac{4}{3} \pi {r}^{3} } \times 100$

$= 700\%$