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A bob weighing 50 gram hangs vertically ...

A bob weighing `50` gram hangs vertically at the end of a string `50 cm long`. If `20` gram force is applied horizontally, by how much distance the bob is pulled aside from its initial position when it reaches in equilibrium position?

Text Solution

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Here, `m=50 gram , `F=20 g f=20 xx 980 dyne Let the bob be in equilibrium when it is at location `B`, Fig. 2(c ) . 17. Then
`F/(CB0 =(mg)/(OC) =T/(BO) or (CB)/(OC) =F/(mg) =(20 xx 980)/(50 xx 980) =0.4`
tan `theta =(CB)/(OC) =0.4 =tan 21^(0)48' so theta =21^(0)48'`
In `Delta OCB, sin 21^_(0) 48' =(CB)/(OB)
or `CB=OB sin 21 ^)(00 48' =50 xx 0. 3714 =17. 57 cm.
.
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