In Cavendish's experiment, let each small mass be 20g and each large mass be 5kg. The rod connecting the small massess is 50 cm long, while the small and the large speres are separated by 10.0 cm. The torsion constant is 4.8 xx 10^(-8) kg m^(2)s^(-2) and the resulting angular deflection is 0.4^(@). Calculating the value of universal gravitational constant G from this data.

In Cavendish's experiment, let each small mass be 20g and each large m

1.	In Cavendish's experiment, let each small mass be 20g and each large mass be 5kg. The rod connecting the small massess is 50 cm long, while the small and the large speres are separated by 10.0 cm. The torsion constant is 4.8 xx 10^(-8) kg m^(2)s^(-2) and the resulting angular deflection is 0.4^(@). Calculating the value of universal gravitational constant G from this data.
Answer» Solution :Here, m = 20g = 0.02 kg, M = 5 kg `R = 10 CM, 0.1m, 1 = 50 cm = 0.5 m` `THETA = 0.4^(@) = (0.4^(@))(2pi//360^(@)) = 0.007` rad, `K = 4.8 xx 10^(-8) kg m^(2)s^(-2)` Thus from `G = (k theta r^(2))/(Mml)` On substituting `G = 6.72 xx 10^(-11) Nm^(2)kg^(-2)`