Question : The height of a right circular cone is 35 cm and the area of its curved surface is four times the area of its base. What is the volume of the cone (in 10–3 m3 and correct up to three decimal places)?
Option 1: 3.316
Option 2: 2.994
Option 3: 2.625
Option 4: 3.384
Correct Answer: 2.994
Solution :
Let the radius of the base of the cone as $r$ (in cm) and the height of the cone as $h$ (in cm).
Given that, $h = 35$ cm, the slant height $l$ of the cone as $l = \sqrt{r^2 + h^2}$.
The curved surface area $A_{\text{curved}}=\pi r l$, and the base area $A_{\text{base}}=\pi r^2$
According to the problem,
$A_{\text{curved}} = 4A_{\text{base}}$
$⇒\pi r \sqrt{r^2 + h^2} = 4\pi r^2$
$⇒\sqrt{r^2 + h^2} = 4r$
$⇒r^2 + h^2 = 16r^2$
$⇒h^2 = 15r^2$
Substituting $h = 35$ cm,
$⇒35^2 = 15r^2$
$⇒r^2 = \frac{35^2}{15}$
The volume of the cone is given by $\frac{1}{3}\pi r^2 h$.
$V = \frac{1}{3}\pi \frac{35^2}{15} \times 35 \approx 2994 \text{ cm}^3$
To convert the volume from cubic centimetres to cubic meters, we divide by $10^6$:
$V = \frac{2994}{10^6} \approx 2994 \times 10^{-6} \text{ m}^3$
To express this in $10^{-3} \text{ m}^3$, we multiply by $10^3$:
$\therefore V = 2.994 \times 10^{-3} \text{ m}^3$
Hence, the correct answer is 2.994.
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