InterviewSolution
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InterviewSolution
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A massless rod of length `l` is hinged at one end and is held horizontal by two identical vertical wires, which are tied at distances `a` and `b` form the hinged end. `A` load `P` si applied at the free end of the rod. The tension in the secound wire isA. `(Pl)/(a^(2) + b^(2))a`B. `(Pl)/(a^(2) + b^(2))b`C. `(Pl)/(a + b)a`D. `(Pl)/(a + b)b` |
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Answer» Correct Answer - B The rod is in rotational equilibrium taking torque about the hinge. sum tau `= T_(1) a + T_(2)b - Pl = 0......(1)` As the wires are elastic, `T_(1) = k( Delta l_(1))` and `T_(2) - k(Delta l_(2))` Hence, `(T_(1))/(T_(2)) = (Delta l_(1))/(Delta l_(2)) = (a)/(b)` On solving equs (1) and (2) for `T_(1)` and `T_(2)`, we get `T_(1) = (Pl)/(a^(2) + b^(2))a, T_(2) = (pl)/(a^(2) + b^(2))b` |
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