The approximate value of $\log _{10} 99$ is (Given $\log _{10} e = 0.4343$)

There is an error of $\pm 0.04 \text{ cm}$ in the measurement of the diameter of a sphere. When the radius is $10 \text{ cm}$,the percentage error in the volume of the sphere is

Consider the following statements:
$A$ is the relative error in the area of a square when the relative error in its side is $0.4$.
$B$ is the relative error in the volume of a sphere when the relative error in its radius is $0.3$.
$C$ is the relative error in the surface area of a closed cylinder whose height is equal to its radius,when the relative error in its height is $0.2$.
$D$ is the approximate error in $y = x^2 + x - 3$ when $x = 2$ and $\delta x = 0.1$.
The ascending order of the values of errors in these statements is:

If the percentage error in the radius of a circle is $3\%$,then the percentage error in its area is (in $\%$)

Given $f(x) = x^3 - 4x$,if $x$ changes from $2$ to $1.99$,then the approximate change in the value of $f(x)$ is

The approximate value of $x^3-2x^2+3x+2$ at $x=2.01$ is