In five successive measurements, the mass of a ball is measured to be $2.61 \,g, 2.58 \,g, 2.40 \,g, 2.73 \,g$ and $2.80 \,g$. The mean absolute error in the measurement is (in $\,g$)

$A$ physical parameter $a$ can be determined by measuring the parameters $b, c, d$ and $e$ using the relation $a = \frac{b^\alpha c^\beta}{d^\gamma e^\delta}$. If the maximum percentage errors in the measurement of $b, c, d$ and $e$ are $b_1\%, c_1\%, d_1\%$ and $e_1\%$ respectively,then the maximum percentage error in the value of $a$ determined by the experiment is:

$A$ physical quantity $P$ is related to four observables $a, b, c$ and $d$ as $P = \frac{\sqrt{a b} \cdot d^\alpha}{\sqrt{c}}$ (where $\alpha$ is a constant). The percentage errors in $a, b, c$ and $d$ are $0.5 \%$ each. If the percentage error in $P$ is $2 \%$,then the value of $\alpha$ is:

An experiment is performed to obtain the value of acceleration due to gravity $g$ by using a simple pendulum of length $L$. In this experiment,the time for $100$ oscillations is measured by using a watch of $1$ second least count,and the value is $90.0$ seconds. The length $L$ is measured by using a meter scale of least count $1$ mm,and the value is $20.0$ cm. The error in the determination of $g$ would be ........... $\%$.

$A$ physical quantity $p$ is described by the relation $p = a^{1/2} b^2 c^3 d^{-4}$. If the relative errors in the measurement of $a, b, c$ and $d$ are $2\%$,$1\%$,$3\%$ and $5\%$ respectively,then the relative error in $p$ will be ........... $\%$.

The mean time period of a second's pendulum is $2.00 \, s$ and the mean absolute error in the time period is $0.05 \, s$. To express the maximum estimate of error,the time period should be written as: