91影视

• Radius of the sphere is $R=2000\mathrm{m}$
• Density of sphere is$p=2.6脳{10}^3\mathrm{kg}/{\mathrm{m}}^3$
• Hanging distance of plumb is 3R
• Length of the plumb line, I =0.5m

We use the concept of density to find the mass of the sphere and Newton鈥檚 second law to write the equation of net force for equilibrium. We can write the net force equation for x and y directions and by solving them, we got in terms of mass and radius. We can write the length component to find the value of x.

Formulae:

Density of a material,$p=\frac{M}{V}$P (i)

Force according to Newton鈥檚 law of motion, F=ma (ii)

Gravitational Force, $F=\frac{GMm}{r^2}$ (iii)

We have density and radius of the sphere. Using this information, we can calculate mass,

$$\begin{gathered} M=pV\left(\mathrm{using}\mathrm{eduation}\left(\mathrm{i}\right)\right) \\ =p\left(\frac{4蟺}{3}\right)R^3 \\ =2.6\mathrm{kg}/{\mathrm{m}}^3{10}^3\left(4脳\frac{3.14}{3}\right){\left(2脳{10}^3\mathrm{m}\right)}^3 \\ =8.7脳{10}^{13}\mathrm{kg} \end{gathered}$$

Using equation (iii), the force between spherical mountain and plumb is,

$F=\frac{GMm}{r^2}$

For equilibrium net force along x should be zero, hence, we can write the equation as:

$$\begin{gathered} \underset{}{鈭�F_{xnet}=0} \\ T\sin 胃-F=0\left(\mathrm{from}\mathrm{body}\mathrm{diagram}\right) \end{gathered}$$

Using the value of equation (iii) we get,

$T\sin 胃-F=0....................\left(1\right)$

Similarly, we can write the net force along y direction,

$$\begin{gathered} \underset{}{鈭�}{F}_{ynet}=0 \\ T\cos -mg=0\left(\mathrm{from}\mathrm{body}\mathrm{diagram}\right) \\ T=\frac{mg}{\cos 胃}......................\left(2\right) \end{gathered}$$

Putting equation (2) in equation (1) we get,

$$\begin{gathered} \frac{mg}{\cos 胃}\sin 胃-\frac{GMm}{r^2}=0 \\ mg\tan 胃=\frac{GMm}{r^2} \\ \tan 胃=\frac{GM}{g{r}^2}.....................\left(3\right) \end{gathered}$$

Lower end of the plumb moves towards the sphere,

$$\begin{gathered} x=I\tan 胃 \\ =\frac{IGM}{g{r}^2}\left(\mathrm{from}\mathrm{equation}\left(3\right)\right) \\ =\frac{\left(0.50\right)\left(6.67脳脳{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(8.7脳{10}^{13}\mathrm{kg}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2{\left(3脳2.00脳{10}^3\mathrm{m}\right)}^2} \\ =8.2脳{10}^{-6}\mathrm{m} \end{gathered}$$

Hence,

the lower end moves at a distance of $8.2脳{10}^{-6}\mathrm{m}$.