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In Cavendish's experiment, let each smal...

In Cavendish's experiment, let each small mass be `20g` and each large mass be `5kg`. The rod connecting the small masses is `50cm` long, while the small and the large spheres are separated by `10.0cm.` The torsion constant is `4.8xx10^(-8) kg m^2 s^(-2)` and the resulting angular deflection is `0.4^@`. Calculate the value of universal gravitational constant `G` from this data.

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Here, `m=20 g=0.02 kg,M=5 kg`
`r=10 cm=0.1 m, l=50 cm=0.5 m`
`theta=0.4^(@)=(0.4^(@))(2pi//360^(@))=0.007 rad`, `K=4.8xx10^(-8)kgm^(2)s^(-2)`
Thus, from `G=(k thetar^(2))/(Mml)`
`G=((4.8xx10^(-8))(0.007)(0.1)^(2))/(5xx0.02xx0.5)`
`=6.72xx10^(-11) Nm^(2)kg^(-2)`
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NARAYNA-GRAVITATION-EXERCISE -IV
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