Young's modulus is determined by the equation given by where M is the mass and L is the extension of wire used in the experiment. Now error in Young's modules () is estimated by taking data from plot in graph paper. The smallest scale divisions are 5g an

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Solution

Q.
Young's modulus is determined by the equation given by $Y = 49000 \frac{m}{L} \frac{d y n e}{c m^{2}}$ where M is the mass and L is the extension of wire used in the experiment. Now error in Young's modules ( $Y$ ) is estimated by taking data from $M - L$ plot in graph paper. The smallest scale divisions are 5g and 0.02 cm along load axis and extension axis respectively. If the value of $M$ and are 500 g and 2 cm respectively then percentage error of $Y$ is [2024]

1 0.2%

2 0.02%

3 2%

4 0.5%

Ans.
(3)

   $\frac{∆ Y}{Y} = \frac{∆ m}{m} + \frac{∆ l}{l}$

   $= \frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01 \frac{∆ Y}{Y}$

   $% \frac{∆ Y}{Y} = 2 %$

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