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A mass M is hung with a light inextensible string as shown in Find the tension in the horizontal part of the string . |
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Answer» Correct Answer - `sqrt(3)Mg` `T_(2) cos 60^(@)=Mg` `T_(2)=2Mg` ...(i) and for horizontal equilibium of P `T_(1)=T_(2) sin 60^(@)=T_(2)(sqrt(3)//2)` ...(ii) Substituting the value of `T_(2)` from eqn (i) in (ii) `T_(1)=(2Mg)xx(sqrt(3)//2)` `=(sqrt(3))Mg` |
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