Question:
A wet open umbrella is held vertical and it whirls about the handle at a uniform rate of 21 revolutions in \(44\ \text{s}\). If the rim of the umbrella is a circle of \(1\ \text{m}\) in diameter and the height of the rim above the floor is \(4.9\ \text{m}\), the locus of the drop is a circle of radius
A wet open umbrella is held vertical and it whirls about the handle at a uniform rate of 21 revolutions in \(44\ \text{s}\). If the rim of the umbrella is a circle of \(1\ \text{m}\) in diameter and the height of the rim above the floor is \(4.9\ \text{m}\), the locus of the drop is a circle of radius
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The locus is a circle larger than the umbrella rim due to tangential projection.
Updated On: Apr 23, 2026
- \(\sqrt{2.5}\ \text{m}\)
- \(1\ \text{m}\)
- \(3\ \text{m}\)
- \(1.5\ \text{m}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Water drops fly off tangentially with tangential speed \(v = r\omega\). Time to fall \(t = \sqrt{2h/g}\). Horizontal range = \(v t\). Locus radius = \(\sqrt{(\text{range})^2 + r^2}\).
Step 2: Detailed Explanation:
Radius of umbrella rim \(r = 0.5\ \text{m}\).
Angular velocity \(\omega = \frac{21 \times 2\pi}{44} = \frac{42\pi}{44} = \frac{21\pi}{22}\ \text{rad/s}\).
Tangential speed \(v = r\omega = 0.5 \times \frac{21\pi}{22} = \frac{21\pi}{44}\ \text{m/s}\).
Time to fall: \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 4.9}{9.8}} = \sqrt{1} = 1\ \text{s}\).
Horizontal range = \(v \times t = \frac{21\pi}{44} \times 1 \approx 1.5\ \text{m}\).
Locus radius = \(\sqrt{(\text{range})^2 + r^2} = \sqrt{(1.5)^2 + (0.5)^2} = \sqrt{2.25 + 0.25} = \sqrt{2.5}\ \text{m}\).
Step 3: Final Answer:
Thus, radius = \(\sqrt{2.5}\ \text{m}\).
Water drops fly off tangentially with tangential speed \(v = r\omega\). Time to fall \(t = \sqrt{2h/g}\). Horizontal range = \(v t\). Locus radius = \(\sqrt{(\text{range})^2 + r^2}\).
Step 2: Detailed Explanation:
Radius of umbrella rim \(r = 0.5\ \text{m}\).
Angular velocity \(\omega = \frac{21 \times 2\pi}{44} = \frac{42\pi}{44} = \frac{21\pi}{22}\ \text{rad/s}\).
Tangential speed \(v = r\omega = 0.5 \times \frac{21\pi}{22} = \frac{21\pi}{44}\ \text{m/s}\).
Time to fall: \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 4.9}{9.8}} = \sqrt{1} = 1\ \text{s}\).
Horizontal range = \(v \times t = \frac{21\pi}{44} \times 1 \approx 1.5\ \text{m}\).
Locus radius = \(\sqrt{(\text{range})^2 + r^2} = \sqrt{(1.5)^2 + (0.5)^2} = \sqrt{2.25 + 0.25} = \sqrt{2.5}\ \text{m}\).
Step 3: Final Answer:
Thus, radius = \(\sqrt{2.5}\ \text{m}\).
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