Question

The first measurement of the gravitational constant G was made by Cavendish in $1798$. In this experiment, two hard spheres, each of mass $m$ are attached to the ends of a rigid rod. This dumbbell shaped system is suspended with its axis horizontal by a fine invar fibre. The rod AB (of length $l$) connects the light masses $m_1=m_2=m$ (and of radius $r$). When two large spheres each of mass $M$ are brought nearer the small suspended sphere for producing torsion, the deflection in the suspended rod is $\theta$ radians. (Given that K is torsional constant). The expression for $G$ will be

• A. $\frac{k{\theta }^{2}r}{mMl}$
• B. $\frac{k\theta r^2}{\sqrt{mM}l}$
• C. $\frac{K\theta r^2}{mMl}$
• D. $\frac{K\theta r^3}{mMl}$

Answer 1

Answer: (C)

$\frac{K\theta r^2}{mMl}$

For the suspended masses
Moment of inertia $=I=m_1{r}_1^2+m_2{r}_2^2$
Period $T=2\pi \sqrt{\frac{I}{K}}$
Gravitation force $F=\frac{GmM}{r^2}$
Torque produced by moment of force $\tau =F\times l$
$\Rightarrow \tau =\frac{GmMl}{r^2}$
Also $\tau =K\theta$ , where K is constant
Equating $\Rightarrow G=\frac{K\theta r^2}{mMl}$