Answer:5.47 s
Explanation:
Given
mass of solid sphere [tex]m=71 kg[/tex]
radius of sphere [tex]r=12 cm[/tex]
Torque Applied [tex]T=0.41 N-m[/tex]
Inclination Produced [tex]\theta =0.76 rad[/tex]
Now we know that time Period of Torsion Pendulum
[tex]T=2\pi \sqrt{\frac{I}{C}}[/tex]
where [tex]C=torque\ constant[/tex]
[tex]I=moment\ of\ inertia [/tex]
C is torque required to produce unit deflection
[tex]C=\frac{T}{\theta }=\frac{0.41}{0.76}=0.539 N.m/rad[/tex]
Moment of Inertia of solid sphere [tex]I=\frac{2}{5}\times mr^2[/tex]
[tex]I=\frac{2}{5}\times 71\times (0.12)^2=0.4089\ kg-m^2[/tex]
[tex]T=2\pi \sqrt{\frac{0.4089}{0.539}}[/tex]
[tex]T=2\times 3.142\times 0.7589[/tex]
[tex]T=5.47 s[/tex]
