Current topics for essays in hindi

Let BS be perpendicular to HG, and let HB be drawn; also put the radius AO $=a$, OH $=b$, AH $(b-a)=c$, H$n = y$ and HB $=c+x$; then A$n = y-c$, G$n=2a-y+c$, and consequently $\overline{y-c}\times \overline{2a-y+c}~ (=~$A$n~\times$ G$n =~$B$n^2 =~$BH$^2-$H$n^2) = \overline{c+x}^2 -y^2$. From which Equation we get $\displaystyle{ y=\frac{2ac+2c^2+2cx+x^2}{2a+2c}}$ (because $a+c=b$.) Whence also $\displaystyle{ 2p\times \overline{1-\frac{Hn}{HB}}= 2p\times \overline{1-\frac{2bc+2cx+x^2}{2b\times\overline{c+x}}}}$: Which multiply'd by $\displaystyle{\frac{c\dot{x}+x\dot{x}}{b}=\dot{y}}$ gives $\displaystyle{\frac{p\times\overline{2ax\dot{x}-x^2\dot{x}}}{b^2}}$ for the Fluxion of the required Force; whereof the Fluent $\displaystyle{\frac{p\times\overline{ax^2-\frac{1}{3}x^3}}{b^2}}$ will be the Attraction of the Segment ABS: Which therefore, when B coincides with G and $x$ is $=2a$, becomes $\displaystyle{\frac{4pa^3}{3b^2}}$, for the Measure of the Attraction of the whole Sphere. .

Current topics for essays in hindi

current topics for essays in hindi

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