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Physics · MOTION IN A STRAIGHT LINE
हिन्दीA drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is \( 1\text{ m} \) long and requires \( 1\text{ s} \). There is a pit located \( 13\text{ m} \) away from the starting point. How long does the drunkard take to fall into the pit?
Correct answer · Option 3
\( 37\text{ s} \)
Explanation
Net displacement per cycle of 8 s is \( 5-3 = 2\text{ m} \). After 4 full cycles (32 s), net displacement is \( 8\text{ m} \). In the next cycle, taking 5 steps forward reaches \( 8 + 5 = 13\text{ m} \) at \( 32 + 5 = 37\text{ s} \).
