Measurement of two quantities along with the error is: $A = 2.5 \ ms^{-1} \pm 0.5 \ ms^{-1}$,$B = 0.10 \ s \pm 0.01 \ s$. The value of $AB$ will be:

The length of a cylinder is measured with a metre rod having a least count of $0.1 \; cm$. Its diameter is measured with vernier calipers having a least count of $0.01 \; cm$. If the length and diameter of the cylinder are $5.0 \; cm$ and $2.00 \; cm$ respectively,then the percentage error in the calculated value of volume will be (in $\%$)

The length of a rod is measured by a meter scale having a least count of $0.1 \text{ cm}$ and the diameter is measured by vernier callipers having a least count of $0.01 \text{ cm}$. The length of the rod is $5.0 \text{ cm}$ and the radius is $2.0 \text{ cm}$. The percentage error in the calculated value of the volume is: (in $\%$)

Resistance of a rod is calculated by measuring its length with the help of a meter scale of least count $1 \ mm$. Its radius is measured with the help of a screw gauge having $50$ divisions on the circular scale and a pitch of $1 \ mm$. Resistivity of the material is exact. The length of the wire is found to be $20 \ cm$ and the diameter of the wire is $4 \ mm$. Find the percentage error in the calculation of resistance.

If two resistors have resistances $R_1 = (350 \pm 3) \text{ } \Omega$ and $R_2 = (140 \pm 4) \text{ } \Omega$, then the percentage error for the sum and difference of $R_1$ and $R_2$ are respectively:

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: