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  • A mass of is suspended vertically by a rope of length <img alt="5 \mathrm{~m}" src="htt

A mass of 10 \mathrm{~kg} is suspended vertically by a rope of length 5 \mathrm{~m} from the roof. A force of 30 \mathrm{~N} is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is \mathrm{\theta=\tan ^{-1}\left(x \times 10^{-1}\right)}. The value of \mathrm{x} is__________.
(Given, \mathrm{g=10 \mathrm{~m} / \mathrm{s}^{2}} )

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer



\mathrm{T\cos \theta = 100\: ---(1)}
\mathrm{T\sin \theta = 30\: ---(2)}

eqn (2) / eqn (1)

\mathrm{\tan \theta = \frac{30}{100}= \frac{3}{10}}
\mathrm{ \theta = \tan^{-1}\left [ 3\times 10^{-1} \right ]}
\mathrm{x = 3}

Posted by

Kuldeep Maurya

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