$\frac{1}{l_1 l_2}$
Step 1: The spring constant of a spring is inversely proportional to its length when the spring is cut into smaller sections. Mathematically, \[ k' = \frac{k}{l} \] where $k'$ is the spring constant of a smaller piece and $l$ is its length.
Step 2: When the original spring of constant $k$ is divided into two sections of lengths $l_1$ and $l_2$, their respective spring constants are: \[ k_1 = \frac{k}{l_1}, \quad k_2 = \frac{k}{l_2} \] Step 3: The ratio of $k_1$ to $k_2$ is: \[ \frac{k_1}{k_2} = \frac{\frac{k}{l_1}}{\frac{k}{l_2}} = \frac{l_2}{l_1} \] Step 4: Therefore, the correct answer is (A).
Values of dissociation constant \( K_a \) are given as follows:
Correct order of increasing base strength of the conjugate bases \( {CN}^-, {F}^- \) and \( {NO}_2^- \) is:
The major product of the following reaction is: