First slide

Accuracy; precision of instruments and errors in measurements

Question

In an experiment, the following observation's were recorded: L=2.820 m, M=3.00kg, l=0.087cm, Diameter D=0.041 cm. Taking g=9.81 m/s²,using the formula $Y = \frac{4 M g L}{π D^{2} l},$ the maximum permissible error in Y is

Difficult

A

7.96%
B

4.56%
C

6.50%
D

8.42%

Solution

$\begin{array}{l} Y = \frac{4 M g L}{π D^{2} l} so maximum permissible error in \\ \begin{matrix} Y & = \frac{Δ Y}{Y} \times 100 = (\frac{Δ M}{M} + \frac{Δ g}{g} + \frac{Δ L}{L} + \frac{2 Δ D}{D} + \frac{Δ l}{l}) \times 100 \\ = (\frac{0.01}{3.00} + \frac{0.01}{9.81} + \frac{0.001}{2.820} + 2 \times \frac{0.001}{0.041} + \frac{0.001}{0.087}) \times 100 \\ = 0.0649 \times 100 = 6.50 % \end{matrix} \end{array}$

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