Question

The surface area of a sphere is \(49\pi\) sq.cm. If it is increased by \(0.016\) sq.cm., then the approximate increase in its volume (in c.c.) is:

• A. \(0.07\)
• B. \(0.04\)
• C. \(0.032\)
• D. \(0.028\)

Hint:

For small changes in geometrical quantities, differential approximation is extremely useful: \[ dV \approx \frac{dV}{dr}dr \] and \[ dS \approx \frac{dS}{dr}dr \] This avoids lengthy exact calculations and gives very accurate approximations.

Answer 1

The Correct Option is D

Step 1: Find the radius from the surface area.
$S=4\pi r^2=49\pi$, so $4r^2=49$, $r^2=\dfrac{49}{4}$ and $r=\dfrac72$.
Step 2: Set up the differential for area.
$S=4\pi r^2$ gives $dS=8\pi r\,dr$. Small changes are well handled by differentials.
Step 3: Find $dr$ from the small area change.
With $dS=0.016$ and $r=\dfrac72$: \[ 0.016=8\pi\cdot\frac72\,dr=28\pi\,dr\Rightarrow dr=\frac{0.016}{28\pi}. \]
Step 4: Set up the differential for volume.
$V=\dfrac43\pi r^3$ gives $dV=4\pi r^2\,dr$.
Step 5: Substitute the values.
\[ dV=4\pi\cdot\frac{49}{4}\cdot\frac{0.016}{28\pi}=\frac{49\times0.016}{28}. \]
Step 6: Compute.
$\dfrac{49\times0.016}{28}=\dfrac{0.784}{28}=0.028$. \[ \boxed{0.028} \]