1. Given data

• Two wires $W_1$ and $W_2$ have the same radius $r$.
• Their material densities are $\rho_1$ and $\rho_2$ with $\rho_2 = 4\rho_1.$
• They are joined at point $O$.
• The point $O$ is midway between the two bridges, so the vibrating lengths of the two parts are equal: $L_1 = L_2.$
• The whole composite wire is under the same tension $T$.
• In the stationary wave formed, the joint $O$ is a node.

We need the ratio of the number of antinodes in $W_1$ and $W_2$.

2. Linear mass densities of the two wires

Since both wires have the same radius, their cross-sectional area is the same: $A = \pi r^2.$

Hence linear mass density is $\mu = \rho A.$

So, $\mu_1 = \rho_1 A, \qquad \mu_2 = \rho_2 A = 4\rho_1 A = 4\mu_1.$

Thus, $\mu_2 = 4\mu_1.$

3. Wave speeds in the two wires

Wave speed on a stretched string is $v = \sqrt{\frac{T}{\mu}}.$

Therefore, $v_1 = \sqrt{\frac{T}{\mu_1}}, \,\, v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{4\mu_1}} = \frac{v_1}{2}.$

So, $v_1 : v_2 = 2 : 1.$

4. Condition for stationary waves in each segment

Since the bridges are fixed ends and the joint $O$ is also a node, each part behaves like a string fixed at both ends.

If $n_1$ and $n_2$ are the numbers of antinodes in $W_1$ and $W_2$ respectively, then for a string fixed at both ends: $L = n\frac{\lambda}{2}.$

So, $L_1 = n_1 \frac{\lambda_1}{2}, \qquad L_2 = n_2 \frac{\lambda_2}{2}.$

Since $L_1 = L_2$, $n_1 \lambda_1 = n_2 \lambda_2.$

5. Use same frequency in both segments

Because the two parts are joined and vibrate together, the frequency is the same: $f = \frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2}.$

Hence, $\frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2} = 2.$

So, $\lambda_1 = 2\lambda_2.$

Substitute into $n_1 \lambda_1 = n_2 \lambda_2:$

$n_1(2\lambda_2) = n_2\lambda_2$ $2n_1 = n_2$ $n_1 : n_2 = 1 : 2.$

6. Answer

The ratio of the number of antinodes formed in $W_1$ to $W_2$ is $\boxed{1:2}.$

So the correct option is B.