$A$ and $B$ are the two pages separated by $13 c m$ . A body of $169 k g w t$ is suspended by thread of 17 $c m$ connecting to A $& B$ , such that the two segments of strings are perpendicular . Then tensions in shorter and longer parts of strings

100 k g w t, 69 k g w t

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65 k g w t, 156 k g w t

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156 k g w t, 65 k g w t

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69 k g w t, 100 k g w t

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Solution

The correct option is C $156 k g w t, 65 k g w t$
$\begin{matrix} \frac{169 g}{sin 90^{0}} = \frac{T_{1}}{sin (90 + θ)} = \frac{T_{2}}{sin (180 - θ)} T_{1} = 169 g \times cos θ T_{2} = 169 g \times \frac{5}{13} = 169 g \times \frac{12}{13} = 65 k g w t = 156 k g w t ∴ O p t i o n (C) i s c o r r e c t . \end{matrix}$

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A and B are the two pages separated by 13cm . A body of 169kgwt is suspended by thread of 17cm connecting to A &B , such that the two segments of strings are perpendicular . Then tensions in shorter and longer parts of strings

The correct option is C 156kgwt,65kgwt169gsin900=T1sin(90+θ)=T2sin(180−θ)T1=169g×cosθT2=169g×513=169g×1213=65kgwt=156kgwt∴Option(C)iscorrect.

$A$ and $B$ are the two pages separated by $13 c m$ . A body of $169 k g w t$ is suspended by thread of 17 $c m$ connecting to A $& B$ , such that the two segments of strings are perpendicular . Then tensions in shorter and longer parts of strings