If the luminous intensity of a $100 \, W$ unidirectional bulb is $100 \, \text{candela}$, then the total luminous flux emitted from the bulb is ....... $lumen$.

$A$ point source is placed at a distance of $15 \, cm$ from a converging (convex) lens of focal length $10 \, cm$. At what distance (in $cm$) should a convex mirror of focal length $12 \, cm$ be placed so that the image is formed on the object itself?

The mirrors are perpendicular to each other as shown in the figure. $A$ light ray $AB$ is incident on the mirror $M_1$. The reflected ray then undergoes a reflection from the mirror $M_2$. The final ray after reflection from $M_2$ will be parallel to the incident ray, if

Two plane mirrors of length $L$ are separated by a distance $L$,and a man $M_2$ is standing at a distance $L$ from the connecting line of the mirrors,as shown in the figure. $A$ man $M_1$ is walking in a straight line at a distance $2L$ parallel to the mirrors at a speed $u$. Then,the man $M_2$ at $O$ will be able to see the image of $M_1$ for a total time of:

$A$ convex lens of focal length $30\,cm$,a concave lens of focal length $120\,cm$,and a plane mirror are arranged as shown. For an object kept at a distance of $60\,cm$ from the convex lens,the final image,formed by the combination,is a real image at a distance of:

$A$ beaker of radius $r$ is filled with water (refractive index $\frac{4}{3}$) up to a height $H$ as shown in the figure. The beaker is kept on a horizontal table rotating with angular speed $\omega$. This makes the water surface curved so that the difference in the height of the water level at the center and at the circumference of the beaker is $h$ $(h \ll H, h \ll r)$,as shown in the figure. Take this surface to be approximately spherical with a radius of curvature $R$. Which of the following is/are correct? ($g$ is the acceleration due to gravity)
$(A)$ $R=\frac{h^2+r^2}{2 h}$
$(B)$ $R=\frac{r^2}{2 h}$
$(C)$ Apparent depth of the bottom of the beaker is close to $\frac{3 H}{4}\left(1+\frac{\omega^2 H}{4 g}\right)^{-1}$
$(D)$ Apparent depth of the bottom of the beaker is close to $\frac{3 H}{2}\left(1+\frac{\omega^2 H}{2 g}\right)^{-1}$