Question

The radius of a sphere is 7 cm. If an error of 0.08 sq.cm. is made in measuring its surface area, then the approximate error (in cubic cm) found in its volume is:

• A. 0.28
• B. 0.32
• C. 0.96
• D. 0.098

Hint:

Use differential calculus to estimate errors in dependent variables, especially in geometric contexts.

Answer 1

The Correct Option is A

1. Formulas: Surface Area of a sphere: $A = 4\pi r^2$
Volume of a sphere: $V = \frac{4}{3}\pi r^3$
2. Given information: Radius: $r = 7$ cm
Error in surface area: $\Delta A = 0.08$ sq.cm
3. Find the relationship between the errors: We want to find the error in volume ($\Delta V$). To relate $\Delta V$ and $\Delta A$, we can use differentials: Differentiate the surface area formula: $dA = 8\pi r \, dr$
Differentiate the volume formula: $dV = 4\pi r^2 \, dr$ Now, divide the equation for $dV$ by the equation for $dA$: $$\frac{dV}{dA} = \frac{4\pi r^2 \, dr}{8\pi r \, dr} = \frac{r}{2}$$ This gives us: $dV = \frac{r}{2} dA$ 4. Approximate the error in volume: We can approximate the differentials $dV$ and $dA$ with the small changes $\Delta V$ and $\Delta A$: $$\Delta V \approx \frac{r}{2} \Delta A$$ Substitute the given values: $$\Delta V \approx \frac{7}{2} \cdot 0.08 = 0.28 \text{ cubic cm}$$ Answer: The approximate error in the volume is (1) 0.28 cubic cm.