A ring has a uniformly distributed charge of 2π C and radius of 3 cm. A charge 10^–6 C is placed at the centre of the ring. Tension developed in the ring is 10^X N. Find x.

Let us analyze the problem step-by-step to find the tension developed in the charged ring.

We are given:
- Total charge on the ring, $Q=2\pi \text{C}$
- Radius of the ring, $R=3\text{cm}=3\times {10}^{-2}\text{m}$
- Charge placed at the center of the ring, $q={10}^{-6}\text{C}$
- The electrostatic repulsion between the central charge $q$ and the ring's charge $Q$ causes radial outward forces, which in turn develops a tension $T$ in the ring.

To find the relation between the electrostatic force and the tension, let us consider a small element of the ring subtending an angle $d\theta$ at the center. The length of this small element is:
$dl=Rd\theta$

Since the charge $Q$ is uniformly distributed along the circumference $2\pi R$ of the ring, the charge $dq$ on this small element is:
$dq=\frac{Q}{2\pi R}dl=\frac{Q}{2\pi R}\left(Rd\theta \right)=\frac{Q}{2\pi }d\theta$

Substituting $Q=2\pi \text{C}$ into the equation for $dq$:
$dq=\frac{2\pi }{2\pi }d\theta =d\theta \text{C}$

The electrostatic repulsive force $dF$ between the central charge $q$ and this element charge $dq$ acts radially outwards and is given by Coulomb's law:
$dF=\frac{1}{4\pi {\epsilon }_0}\frac{q·dq}{R^2}$

This outward radial force $dF$ is balanced by the components of the tension $T$ acting at the two ends of the element. The tension forces pull tangentially at each end of the element. The components of the tension perpendicular to the radial bisector cancel each other, while the radial components point inwards, toward the center, balancing the electrostatic force:
$dF=2T\sin \left(\frac{d\theta }{2}\right)$

For an infinitesimally small angle $d\theta$, we can use the small-angle approximation $\sin \left(\frac{d\theta }{2}\right)\approx \frac{d\theta }{2}$. Therefore:
$dF=2T\left(\frac{d\theta }{2}\right)=Td\theta$

Now, let us equate the two expressions for $dF$:
$Td\theta =\frac{1}{4\pi {\epsilon }_0}\frac{q·dq}{R^2}$

Substitute $dq=d\theta$ into the equation:
$Td\theta =\frac{1}{4\pi {\epsilon }_0}\frac{q·d\theta }{R^2}$

Dividing both sides by $d\theta$ gives the tension $T$:
$T=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^2}$

Now, substitute the given numerical values:
- $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9{\text{N·m}}^2{\text{/C}}^2$
- $q={10}^{-6}\text{C}$
- $R=3\times {10}^{-2}\text{m}$

Calculating the tension:
$T=(9\times {10}^9)\times \frac{{10}^{-6}}{{(3\times {10}^{-2})}^2}$

$T=\frac{9\times {10}^3}{9\times {10}^{-4}}$

$T={10}^7\text{N}$

Comparing this with the given format $T={10}^x\text{N}$, we find:
$x=7$

Therefore, the correct answer is 7.