A) 1.30
B) 1.47
C) 1.20
D) 1.25
Correct Answer: A
Solution :
[a] \[{{C}_{\max }}=60{}^\circ \] \[\therefore \]\[{}_{r}{{\mu }_{d}}=\frac{1}{\sin 60{}^\circ }\operatorname{or}\frac{{{\mu }_{g}}}{{{\mu }_{\ell }}}=\frac{2}{\sqrt{3}}\] \[\therefore \]\[{{\mu }_{\ell }}=\frac{\sqrt{3}}{2}{{\mu }_{g}}\frac{\sqrt{3}}{2}\times 1.5=1.3\]You need to login to perform this action.
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