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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 -
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Verified Answer
The correct answer is:
$\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}+\mathrm{ma}$
Applying Newton's law
along string
$$
\Rightarrow \mathrm{T}-\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}=\mathrm{ma}
$$
or $\mathrm{T}=\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}+\mathrm{ma}$
along string
$$
\Rightarrow \mathrm{T}-\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}=\mathrm{ma}
$$
or $\mathrm{T}=\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}+\mathrm{ma}$
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