Answer :

Let r be the radius of sphere and V be the volume

Let r_{1} = 10 cm and r_{2} = 9.8 cm

The change dr is given by dr = r_{2} – r_{1}

⇒ dr = 9.8 – 10 = -0.2 cm …(i)

Now volume of sphere is given by

As we have to find the change in V (that is dV) with respect to r, we have to differentiate V with respect to r

⇒ dV = 4πr^{2}dr

The radius is shrinked from 10 cm hence r will be 10 and put dr value from (i)

⇒ dV = 4π(10)^{2}(-0.2)

⇒ dV = -4π(20)

⇒ dV = -80π

The negative sign indicates that the volume is decreased

Hence the change in volume or decrease in volume is 80π cm^{3}

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