Simple Pendulum Length

Starting from the pendulum relation $T=2\pi \sqrt{L/g}$ presented in NCERT Class 11, Chapter 14 (Oscillations), we solve for the length by squaring and rearranging: $L=\frac{g{T}^2}{4{\pi }^2}$. Inserting $T=2$ s and $g=9.8$ m/s${}^2$ gives $L=\frac{9.8\times 4}{4{\pi }^2}=\frac{39.2}{39.48}\approx 0.99$ m. This is the familiar length of a seconds pendulum, which takes one second for each half-swing. The option 0.25 m would give a much shorter period near one second. The option 2.0 m confuses the numerical value of the period with the length. The option 3.9 m results from forgetting to divide by $4{\pi }^2$ and instead using $g{T}^2$ directly divided by four. To verify by substituting back, a length of $0.99$ m with $g=9.8$ m/s${}^2$ gives $T=2\pi \sqrt{0.99/9.8}=2\pi \sqrt{0.101}=2\pi (0.318)\approx 2.0$ s, which recovers the target period and confirms the algebra. Note that the result depends on the local value of $g$, so the same 2 s beat would need a slightly different length at a location with different gravity. A plausibility check: a length just under one metre for a 2 s pendulum is a standard laboratory result, and the units of $g{T}^2/(4{\pi }^2)$, namely $({\text{m/s}}^2)({\text{s}}^2)$, reduce to metres, confirming the answer is dimensionally and physically sound.