Discussion Forum

In the determination of Young's  modulus $\left(Y=\frac{4MLg}{\pi ld^{2}}\right)$ Searle's method, a wire of length L = 2m and diameter d = 0.5 nm is used. For a load M = 2.5 kg, an extension l= 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometre, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. T1le contributions to the maximum probable error of the Y measurement is

Answer:

Explanation:

$\triangle d=\triangle l=\frac{0.5}{100}mm=0.005mm$

   $Y=\frac{4MLg}{\pi ld^{2}}$

 $\therefore$  $\left(\frac{\triangle Y}{Y}\right)_{max}=\left(\frac{\triangle l}{l}\right)+2\left(\frac{\triangle d}{d}\right)$

  $\left(\frac{\triangle l}{l}\right)=\frac{0.5/100}{0.25}=0.02$

and       $\frac{2 \triangle d}{d}=\frac{(2)(0.5/100)}{0.25}=0.02$

 or   $\frac{\triangle l}{l}=2 \frac{\triangle d}{d}$