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Q.The kinetic energy and potential energy of a helicopter flying horizontally at a height 400m, are in the ratio 5:2. The velocity of the helicopter is (g = 9.8 m/s^2)

(a) 65 m/s
(b) 56 m/s
(c) 140 m/s
(d) 14 m/s
Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025MCQ· 1mImportance★★★★★
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Using PE=mghPE = mgh and KE=52PEKE = \tfrac{5}{2}PE, solving 12mv2=52mgh\tfrac{1}{2}mv^2 = \tfrac{5}{2}mgh gives v=140v = 140 m/s.

The helicopter flies horizontally at a constant height h=400h = 400 m, so its potential energy relative to the ground is:

PE=mgh=m×9.8×400=3920 m (J, per kg of mass)PE = mgh = m \times 9.8 \times 400 = 3920\,m \text{ (J, per kg of mass)}

Given KE:PE=5:2KE : PE = 5 : 2:

KE=52×PE=52×3920 m=9800 mKE = \frac{5}{2} \times PE = \frac{5}{2} \times 3920\,m = 9800\,m

Since KE=12mv2KE = \tfrac{1}{2}mv^2: …

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