A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is

To find the magnification of the compound microscope, let us first identify the given parameters from the problem:
Focal length of the objective lens, $f_o=\; 2$ cm
Focal length of the eyepiece, $f_e=\; 4$ cm
Tube length of the microscope, $L=\; 40$ cm
Least distance of distinct vision, $D=\; 25$ cm

For a compound microscope, the total magnification (when the final image is formed at the near point or distance of distinct vision, $D$) is given by the formula:
$m=\; -\frac{L}{f_o}(1+\frac{D}{f_e})$

Considering the magnitude of the magnification, we write:
$\left|m\right|=\frac{L}{f_o}(1+\frac{D}{f_e})$

Now, let us substitute the given values into the formula:
$\left|m\right|=\frac{40}{2}(1+\frac{25}{4})$

Simplify the terms step-by-step:
$\left|m\right|=\; 20$ × (1+6.25)
$\left|m\right|=\; 20$ × 7.25
$\left|m\right|=\; 145$

Note: In some standard physics textbook approximations, the magnification is approximated as:
$m=\frac{L}{f_o}·\frac{D}{f_e}$
Using this approximation, we get:
$m=\frac{40}{2}·\frac{25}{4}=\; 20$ × 6.25 = 125
Thus, following the standard approximation corresponding to the options provided, the magnifying power is 125.

Therefore, the correct answer is 125.