The length, breadth and thickness of a strip are $(10.0 \pm 0.1)\; cm ,(1.00 \pm 0.01) \;cm$ and $(0.100 \pm 0.001)\; cm$ respectively. The most probable error in its volume will be?
$\pm \,0.03\, cm^{3}$
$\pm\, 0.111 \,cm^{3}$
$\pm\, 0.012\, cm^{3}$
એકપણ નહિ
We can reduce random errors by
A students measures the distance traversed in free fall of a body, the initially at rest, in a given time. He uses this data to estimate $g$ , the acceleration due to gravity . If the maximum percentage errors in measurement of the distance and the time are $e_1$ and $e_2$ respectively, the percentage error in the estimation of $g$ is
Durring Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale. The $20^{\text {th }}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \ kg$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale but now the $45^{\text {th }}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 m$. and its cross-sectional area is $8 \times 10^{-7} m ^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} m$. The maximum percentage error in the Young's modulus of the wire is
According to Joule's law of heating, heat produced $H = {I^2}\,Rt$, where I is current, $R$ is resistance and $t$ is time. If the errors in the measurement of $I, R$ and t are $3\%, 4\% $ and $6\% $ respectively then error in the measurement of $H$ is
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be $2.63 \;s , 2.56 \;s , 2.42\; s , 2.71 \;s$ and $2.80 \;s$. Calculate the absolute errors, relative error or percentage error.