A) same
B) half
C) double
D) four times
Correct Answer: B
Solution :
Here, \[{{I}_{1}}={{I}_{2}},{{n}_{1}}=n,{{n}_{2}}=\frac{n}{2}\] Using the relation \[nIBA=C\theta \] \[\Rightarrow \] \[I=\frac{C\theta }{nBA}\] \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{\theta }_{1}}}{{{n}_{1}}}\times \frac{{{n}_{2}}}{{{\theta }_{2}}}\] \[\therefore \] \[\frac{{{\theta }_{1}}}{{{\theta }_{2}}}=\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{n}{n/2}=2\] \[[\because {{I}_{1}}={{I}_{2}}]\] \[\Rightarrow \] \[{{\theta }_{2}}=\frac{{{\theta }_{1}}}{2}\] Hence, the deflection will become half.
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