The approximate value of $\log _{10} 1002$ is (Given $\log _{10} e = 0.4343$)

The time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$,where $g$ denotes the acceleration due to gravity. If the length of the pendulum is increased by $1 \%$,then the approximate change in its time period is (in $\%$)

Electric current $(I)$ is measured by a galvanometer,where the current is proportional to the tangent of the angle $( \theta)$ of deflection. If the deflection is read as $45^{\circ}$ and an error of $1\%$ is made in reading the angle,the percentage error in the current is:

The time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$. If the length is increased by $2 \%$,then the approximate change in the time period is:

Using differentials,find the approximate value of $(33)^{-\frac{1}{5}}$.

The approximate value of $\log _{10} 998$ is (given that $\log _{10} e=0.4343$ )