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`        a bullet loses 1/20 of its velocity in passing through a plank .the least number of planks required to stop the bullet is`
2 months ago

Arun
22281 Points
```							Let the thickness of one plank = dand the acceleration provided by the plank = av^2 = vo^2 + 2adIf n planks are required to stop the bullet, then0^2 = vo^2 + 2a*nd2and = -vo^2n = vo^2/(-2ad) -----------------(1)v = vo - vo/20 = 19 vo/20 in passing through one plank(19 vo/20)^2 = vo^2 + 2ad361/400 * vo^2 = vo^2 + 2ad-2ad = vo^2(1 - 361/400) -2ad = vo^2 * 39/400Substituting this value of -2ad into equation (1):n = vo^2/(vo^2 * 39/400) = 400/39 The minimum number of planks needed = smallest integer greater than 400/39 = 11.
```
2 months ago
Khimraj
3008 Points
```							No. of planks➾ n² / 2n - 1➧ Where n is loss in velocity.➧ There your answer will not consume time as time is marks in such exam,➾ 20² / 2 × 20 - 1➾ 10.25➾ 11 planks ...✔
```
one month ago
aswanth nayak
66 Points
```							Dear Student,Let v be the speed of the bullet incident on the first plank.Its speed after it passes the plank = 19/20 vIf x is the thickness of the plank, the deceleration a due to resistance of plank is given by,            (19/20)² v² - v² = 2ax => 2ax = -39/400 v²Let the bullet is stopped after passing through n such planks. Then the distance covered by bullet is x       => 0 - (19/20)² v² = 2anx      => -(19/20)v² = n × -39/400 v²      => n = 361/39 = 9.25 ≈ 9 ==> 9 Planks Hope it helps you regards
```
28 days ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions