Question:
The curved surface area of a hemisphere is \(154\pi\text{ cm}^2\). Find its volume (in cubic cm).
एक अर्धगोले (hemisphere) का वक्र पृष्ठीय क्षेत्रफल \(154\pi\text{ cm}^2\) है। इसका आयतन (घन सेमी में) ज्ञात कीजिए।
Options:
- \(\dfrac{154}{3}\pi\sqrt{77}\)
- \(\dfrac{155}{3}\pi\sqrt{77}\)
- \(\dfrac{159}{3}\pi\sqrt{77}\pi\)
- \(\dfrac{154}{5}\pi\sqrt{77}\)
- \(\dfrac{154}{3}\pi\sqrt{77}\)
- \(\dfrac{155}{3}\pi\sqrt{77}\)
- \(\dfrac{159}{3}\pi\sqrt{77}\pi\)
- \(\dfrac{154}{5}\pi\sqrt{77}\)
Correct Answer: a)
Explanation:
Curved surface area of hemisphere \(= 2\pi r^2 = 154\pi \implies r^2 = 77 \implies r = \sqrt{77}\text{ cm}\).
Volume of hemisphere \(= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi (r^2 \cdot r) = \dfrac{2}{3}\pi (77\sqrt{77}) = \dfrac{154}{3}\pi\sqrt{77}\text{ cm}^3\).
अर्धगोले का वक्र पृष्ठीय क्षेत्रफल \(= 2\pi r^2 = 154\pi \implies r^2 = 77 \implies r = \sqrt{77}\text{ cm}\).
अर्धगोले का आयतन \(= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi (77\sqrt{77}) = \dfrac{154}{3}\pi\sqrt{77}\text{ cm}^3\)। अतः विकल्प (a) सही है।