One end of a long metallic wire of length L, area of cross-section A and Young’s modulus Y is tied to the ceiling. The other end is tied to a massless spring of force constant k. A mass m hangs freely from the free end of the spring. It is slightly pulled down and released. Its time period is given by

The system consists of a metallic wire and a spring connected in series, supporting a mass $m$. Let us find the effective force constant of this combination to determine the time period of oscillation.

For a metallic wire of length $L$, area of cross-section $A$, and Young's modulus $Y$, the equivalent force constant (stiffness) $k_w$ is given by the formula:
$k_w=\frac{YA}{L}$

The wire is tied to a massless spring of force constant $k$ in a series arrangement. For two springs (or elastic bodies) connected in series, the effective force constant $k_{\mathrm{eff}}$ satisfies:
$\frac{1}{k_{\mathrm{eff}}}=\frac{1}{k_w}+\frac{1}{k}$

Substituting the value of $k_w$ into the equation:
$\frac{1}{k_{\mathrm{eff}}}=\frac{L}{YA}+\frac{1}{k}$

Finding a common denominator to simplify:
$\frac{1}{k_{\mathrm{eff}}}=\frac{kL+YA}{kYA}$

Thus, the effective force constant $k_{\mathrm{eff}}$ is:
$k_{\mathrm{eff}}=\frac{kYA}{kL+YA}$

The time period $T$ of a simple harmonic oscillator of mass $m$ suspended from an effective spring constant $k_{\mathrm{eff}}$ is given by:
$T=2\pi \sqrt{\frac{m}{k_{\mathrm{eff}}}}$

Substituting $k_{\mathrm{eff}}$ into the time period formula:
$T=2\pi \sqrt{\frac{m\left(kL+YA\right)}{kYA}}$

Therefore, the correct option is 2π √{m(KL+YA)/KYA}.