While measuring the acceleration due to gravity by a simple pendulum,a student makes a positive error of $1\%$ in the length of the pendulum and a negative error of $3\%$ in the value of time period. His percentage error in the measurement of $g$ by the relation $g = 4{\pi ^2}(l/T^2)$ will be ........ $\%$

Students $A, B$ and $C$ measure the length of a room using a $25 \,m$ long measuring tape of least count $0.5 \,cm$,a meter-scale of least count $0.1 \,cm$,and a foot-scale of least count $0.05 \,cm$,respectively. If the actual length of the room is $9.5 \,m$,which of the following students will report the lowest relative error in the measured length?

In the expression for time period $T$ of a simple pendulum $T = 2 \pi \sqrt{\frac{l}{g}}$,if the percentage error in time period $T$ and length $l$ are $2 \%$ and $2 \%$ respectively,then the percentage error in acceleration due to gravity $g$ is equal to ......... $\%$

If $P = \frac{A^3}{B^{5/2}}$ and $\Delta A$ is the absolute error in $A$ and $\Delta B$ is the absolute error in $B$,then the absolute error $\Delta P$ in $P$ is:

If $\theta_1 = 25.5 \pm 0.1 \, ^\circ C$ and $\theta_2 = 35.3 \pm 0.1 \, ^\circ C$,find $\theta_1 - \theta_2$.

If the length of rod $A$ is $3.25 \pm 0.01 \,cm$ and that of rod $B$ is $4.19 \pm 0.01 \,cm$,then the rod $B$ is longer than rod $A$ by: