A light string of length 10 m suspended from the ceiling of a hall can bear a maximum tension of 600N. A 50Kg man starts climbing up the string at lower end with zero initial velocity. The minimum time in which man can touch the ceiling of hall (g = 10m /sec] (a) 2 sec (b) 5.16 sec (c) 3.16 sec (d) 4.2 sec.
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Answers
OPTION C IS CORRECT THIS IS PAKKA
SEE THIS IS YOUR ANSWER WITH EXPLAINING
PLEASE MARK HAS BRAINLIST PLEASE SUPPORT...
Answer:
Option (c) 3.16 seconds
V90
Explanation:
We have been given that,
Length of the string attached (s) = 10m
• Maximum tension the rope can handle (Tmax) = 600N
Weight of the man climbing the
rope (m) = 50Kg
Initial velocity (u) = 0
• Time taken to climb the rope = t = To be found
We have 'u' and 't', and the relation s = ut + (1/2)at² can be used to figure out the time taken to climb up the rope. But first we'll have to find the acceleration of the man whilst he's climbing the rope so we can substitute the value in the formula.
Let's draw an FBD (free body diagram) for easier understanding. [Check the attachment]
We know that the force exerted (F = mg) by the man acts downward due to gravity, therefore a force of mg = 50 × 10 500N acts downward.
We have been given that the rope can withhold a maximum opposing force of 600N. 'mg' (500N) acts in the downward direction in the same line of force, therefore tension acts in the upward direction. [Tension acts in the direction opposite to that of the mass exerting force in the same line]
In order for the man to climb up the rope, he'll have to accelerate in the upward direction, let this be 'a'. Therefore the force exerted by the man to climb upward is F = ma.
To summarize,
• Tension acts upward. (T)
• Force exerted by the man's weight acts downward. (mg)
• Force exerted by the man's acceleration to climb acts upward. (ma)
With the help of Newton's Ilnd Law of Motion (F= ma) we can say that,
mg = T-ma
→ mg + ma = 600
m(g + a) = 600
50(10 + a) = 600
→ 500 + 50a = 600
→ 50a = 600 - 500
→ 50a = 100
Using the Second Equation of Motion we get,
s = ut + jat²
⇒ 10 = (0)t += (2)t²
⇒ 100+ t²
t² = 10
⇒ t =√10
⇒ t = 3.162 seconds (approx)
Therefore, the answer is Option (c) 3.16 seconds.
