Question

Can any one solve it ???

Answer 1

0.1mg. wt =10× 0.1 ×10^-6=10^-5kg

So we are given

10^-5=g*40*15/20*10^-4

10^-5*10^-4×2/3=g

g=6.55*10^-11

Answer 2

$f =g \frac{ {m}^{1} \times {m}^{2} }{ {r}^{2} }$





$= >$



GIVEN………



F=0.1mg wt





$= 0.1 \times {10}^{ - 3} \times 980 \: dyne$






$= 98 \times {10}^{ - 3} dyne$








$98 \times {10}^{ - 8} \: newton$






$= > 9.8 \times {10}^{ - 7} newton$








${m}^{1} = 40kg$









${m}^{2} = 15kg$








$r = 20cm = 20 \times \frac{1}{100} = 0.2m$









so
${r}^{2} = {0.2}^{2} = 0.04m$

PUT IN FORMULA


WE GET













=>>>>
$\frac{f \times {r}^{2} }{ {m}^{1} \times {m}^{2} } = g$











=>>>>
$\frac{9.8 \times {10}^{ - 7} \times 0.2 \times 0.2 }{40 \times 15}$





$= > > \frac{98 \times 2 \times 2 \times {10}^{ - 7} }{4 \times 15 \times 10 \times 10 \times 10 \times10 }$





$= = = > \frac{392 \times {10}^{ - 7 - 4} }{60}$






$= = = = > \frac{392 \times {10}^{ - 11} }{60}$





$= = = > 6.5333 \times {10}^{ - 11} {m}^{2} {kg}^{ - 1} \times {s}^{ - 2}$