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# CBSE Class 9 Answered

Questions 23-25 are based on the information given below. Read the information and answer the questions that follow:A bob of mass ‘m’ is suspended by a string, g be the acceleration due to gravity the bob.When the bob is stationary, the tension ‘T’ in the string is T=mg.When the bob is pulled up with an acceleration ‘a’ the tension in the string is given by T= m (g + a)When the bob is allowed to move down with an acceleration ‘a’, the tension T in the string is given byT= m (g –a)

Asked by avishisanghai | 20 Feb, 2022, 05:19: PM
Let m is mass of the bob , g is acceleration due to gravity and T is tension force in the string .

When the bob is at rest , Net force acting on the bob is zero.

Hence , By applying Newton's second law , we get

T - mg = 0   or  T = mg

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When the bob is pulled up with uniform acceleration a , then by applying Newton's second law ,
we get

T - mg = ma    or   T =  m ( g + a)

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When the bob is moved down with uniform acceleration a , then by applying Newton's second law ,
we get

mg - T = ma    or   T =  m ( g - a)
Answered by Thiyagarajan K | 20 Feb, 2022, 08:02: PM

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