Where should an object be placed on the axis of a convex lens of focal length $8 \ cm$ so as to achieve magnification of $-4$ (in $cm$)? (Distances are measured from the optic centre of the lens)

An illuminated object and a screen are placed $90 \, cm$ apart. What is the focal length of the lens required to produce an image on the screen,twice the size of the object?.......$cm$

Given below are two statements:
Statement $(I)$: When an object is placed at the centre of curvature of a concave lens,the image is formed at the centre of curvature of the lens on the other side.
Statement $(II)$: $A$ concave lens always forms a virtual and erect image.
In the light of the above statements,choose the correct answer from the options given below:

The radii of curvature of both the surfaces of a convex lens of focal length $f$ and focal power $P$ are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens are:

The optical power of a thin glass lens (refractive index $1.5$) in air is $-5 \ D$. What will be its optical power in a liquid with a refractive index of $1.6$ (in $D$)?

$A$ convex lens of focal length $0.15 \,m$ is made of a material of refractive index $\frac{3}{2}$. When it is placed in a liquid, its focal length is increased by $0.225 \,m$. The refractive index of the liquid is