A compound microscope is designed with two symmetric biconvex...

Focal length of the objective after cutting:
For a symmetric biconvex lens with both radii of magnitude $R$:
$\dfrac{1}{f_o} = (\mu - 1)\left(\dfrac{1}{R} - \dfrac{1}{-R}\right) = \dfrac{2(\mu - 1)}{R}$

When the lens is cut vertically (perpendicular to the principal axis), each half becomes a plano-convex lens with one flat surface:
$\dfrac{1}{f_o'} = (\mu - 1)\left(\dfrac{1}{R} - 0\right) = \dfrac{\mu - 1}{R}$

Comparing the two: $f_o' = 2f_o$, so the focal length of the objective doubles.

Magnification of a compound microscope:
$M \approx \dfrac{L}{f_o} \cdot \dfrac{D}{f_e}$

where $L$ is the tube length, $f_o$ is the objective focal length, $f_e$ is the eyepiece focal length and $D$ is the least distance of distinct vision.

Since the eyepiece and $D$ are unchanged, equating the initial and final magnifications:
$\dfrac{L_{\text{initial}}}{f_o} \cdot \dfrac{D}{f_e} = \dfrac{L_{\text{final}}}{2f_o} \cdot \dfrac{D}{f_e}$

$\dfrac{L_{\text{initial}}}{f_o} = \dfrac{L_{\text{final}}}{2f_o}$

$L_{\text{final}} = 2\,L_{\text{initial}}$

Hence, to maintain the same magnification, the tube length must be doubled.

The correct option is $(1)$ increased two times.