Question

There is a possible error of 0.02 cm in measuring the base diameter of a right circular cone as 14 cm. If the semi-vertical angle of the cone is 45°, then the approximate error in its volume is (in cu. cm)

• A. 1.078
• B. 3.08
• C. 1.54
• D. 6.16

Hint:

For problems involving approximate errors, use differentials. If a quantity \(Q\) depends on a variable \(x\), the error \(dQ\) is approximately \(Q'(x) \cdot dx\), where \(dx\) is the error in measuring \(x\).

Answer 1

The Correct Option is C

Let \(r\) be the base radius and \(h\) be the height of the cone.
The semi-vertical angle is \( \alpha = 45^\circ \). In a right circular cone, \( \tan\alpha = \frac{r}{h} \).
Since \( \tan(45^\circ) = 1 \), we have \( r = h \).
The volume of the cone is \( V = \frac{1}{3}\pi r^2 h \). Since \(h=r\), this becomes \( V = \frac{1}{3}\pi r^3 \).
We need to find the approximate error in volume, \(dV\), which can be found using differentials.
\( dV = \frac{dV}{dr} dr \).
\( \frac{dV}{dr} = \frac{d}{dr}(\frac{1}{3}\pi r^3) = \pi r^2 \).
So, \( dV = \pi r^2 dr \).
We are given the diameter \(D=14\) cm, so the radius is \( r = 7 \) cm.
The error in the diameter is \(dD = 0.02\) cm. The error in the radius is half of this.
\( dr = \frac{1}{2} dD = \frac{1}{2}(0.02) = 0.01 \) cm.
Now, substitute the values into the differential equation for volume error.
\( dV = \pi (7^2) (0.01) = 49 \pi (0.01) = 0.49\pi \).
Using the approximation \( \pi \approx \frac{22}{7} \).
\( dV \approx 0.49 \times \frac{22}{7} = 0.07 \times 22 = 1.54 \).
The approximate error in the volume is 1.54 cu. cm.