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Motion in A Straight Line question

2022 · 28 Jul · Shift 1 · Q48
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  5. /2022 · 28 Jul · Shift 1 · Q48

Motion in A Straight Line question

2022 · 28 Jul · Shift 1 · Q48

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
A NCC parade is going at a uniform speed of 9 km/h9 \mathrm{~km} / \mathrm{h}9 km/h under a mango tree on which a monkey is sitting at a height of 19.6 m19.6 \mathrm{~m}19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given g=9.8 m/s2g=9.8 \mathrm{~m} / \mathrm{s}^{2}g=9.8 m/s2 )
  1. A
    5 m
  2. B
    10 m
  3. C
    19.8 m
  4. D
    24.5 m
View written solutionFree

Correct answer: A

  1. Convert the speed of the parade into SI units

Given speed of cadet/parade: 9 km/h=9×10003600 m/s=2.5 m/s9\,\text{km/h} = 9\times \frac{1000}{3600}\,\text{m/s} = 2.5\,\text{m/s}9km/h=9×36001000​m/s=2.5m/s

  1. Find the time taken by the mango to fall

The monkey drops the mango from rest from height: h=19.6 mh=19.6\,\text{m}h=19.6m

Using h=12gt2h = \frac{1}{2}gt^2h=21​gt2

So, 19.6=12(9.8)t219.6 = \frac{1}{2}(9.8)t^219.6=21​(9.8)t2 19.6=4.9t219.6 = 4.9t^219.6=4.9t2 t2=4t^2 = 4t2=4 t=2 st = 2\,\text{s}t=2s

  1. Find which cadet receives the mango

In time t=2 t=2\,t=2s, the cadet moves horizontally by: x=vt=2.5×2=5 mx = vt = 2.5 \times 2 = 5\,\text{m}x=vt=2.5×2=5m

So the cadet who is 5 m away from the tree at the instant the mango is dropped will receive it.

  1. Check options
  • A: 5 5\,5m ✅
  • B: 10 10\,10m ❌
  • C: 19.8 19.8\,19.8m ❌
  • D: 24.5 24.5\,24.5m ❌

Therefore, the correct option is A.

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