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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Laws of Motion

BITSAT 2013 · Physics · Laws of Motion

A bob is hanging over a pulley inside a car through, a string. The...

A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontallyas shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically. The tension in the string is equal to - A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person

Options

  1. Amg2+a2m \sqrt{g^{2}+a^{2}}
  2. Bmg2+a2−mam \sqrt{g^{2}+a^{2}}-m a
  3. Cm\mathrmg2+\mathrma2+ma\mathrm{m} \sqrt{\mathrmg^{2}+\mathrma^{2}}+\mathrm{ma}
  4. Dm(g+a)\mathrm{m}(\mathrm{g}+\mathrm{a})

Correct answer

C. m\mathrmg2+\mathrma2+ma\mathrm{m} \sqrt{\mathrmg^{2}+\mathrma^{2}}+\mathrm{ma}

Step-by-step solution

Applying Newton's law along string \Rightarrow T−m\mathrmg2+\mathrma2=ma\mathrm{T}-\mathrm{m} \sqrt{\mathrmg^{2}+\mathrma^{2}}=\mathrm{ma} or T=m\mathrmg2+\mathrma2+ma\mathrm{T}=\mathrm{m} \sqrt{\mathrmg^{2}+\mathrma^{2}}+\mathrm{ma} A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person
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