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The length of a metal wire is `l_1` when the tensionin it is `T_1 and is l_2` when the tension is `T_2`. The natural length of the wire isA. `(T_2)/(T_1)(l_1+l_2)`B. `T_2l_1+T_2l_2`C. `(l_1T_2-l_2T_1)/(T_2-T_1)`D. `(l_1T_2+l_2T_1)/(T_2+T_1)` |
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Answer» Correct Answer - C `Y=(Fl)/(atrianglel)` `Y`, `l` and `a` are constants `(Fl)/(trianglel)=`constant or `trianglelpropF` Now, `l_1lpropT_1` and `l_2-lT_2` Dividing, `(l_1-1)/(l_1-1)=(T_1)/(T_2)` or `l_1T_2-lT_2=l_2T_1-lT_1` or `l(T_1-T_2)=l_2T_1-l_1T_2` or `l=(l_1T_2-l_1T_2)/(T_1-T_2)` or `l=(l_1T_2-l_2T_1)/(T_3-T_1)` |
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