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Question Number 23793 by math solver last updated on 06/Nov/17

Commented by math solver last updated on 06/Nov/17

q.2,3

$$\mathrm{q}.\mathrm{2},\mathrm{3} \\ $$

Answered by ajfour last updated on 06/Nov/17

(1/2)m(2gR)=(1/2)(3m)(((√(2gR))/3))^2 +                                         2mg(((nR)/6))  ⇒  1=(1/3)+(n/3)    ⇒  n=2 .

$$\frac{\mathrm{1}}{\mathrm{2}}{m}\left(\mathrm{2}{gR}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}{m}\right)\left(\frac{\sqrt{\mathrm{2}{gR}}}{\mathrm{3}}\right)^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{mg}\left(\frac{{nR}}{\mathrm{6}}\right) \\ $$$$\Rightarrow\:\:\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{{n}}{\mathrm{3}}\:\:\:\:\Rightarrow\:\:\boldsymbol{{n}}=\mathrm{2}\:. \\ $$

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