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$ R \cos \theta- f\sin \theta=mg$

$ R \sin \theta+ f\cos \theta=\large\frac{mv^2}{r}$

$f=\mu R$

$=>\large\frac{v^2}{rg}=\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}$

$=>\large\frac{\tan \theta+\mu}{1-\mu \sin \theta}$

$v=\sqrt {\large\frac{750 \times 10(0.087 +0.5)}{1-(0.5 \times 0.087)}}$

$\quad= 67 \;m/s$

Hence b is the correct answer.

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