Question

# In Searle's exp to find Young's modulus, the diameter of wire is measured as $$D = 0.05 cm,$$ length of wire is $$L = 125 cm$$, and when a weight, $$m = 20.0 kg$$ is put, extension in the wire was found to be $$0.100 cm.$$ Find a maximum permissible error in young's modulus (Y).

A
43 %
B
6.3%
C
0.63%
D
630%

Solution

## The correct option is B $$6.3$$%Given: $$D = 0.05 cm$$ ;  $$ΔD = 0.01 cm$$$$L = 125.0 cm$$;   $$ΔL = 0.1 cm$$Weight , $$F= 20\times 10= 200 N$$$$δL = 0.100 cm$$;  $$Δ(δL) = 0.001 cm$$So, Young Modulus , $$Y=\dfrac{FL}{\pi (\cfrac D2)^2 \delta L}= \dfrac{200\times 1.25}{3.14\times( \cfrac{0.05\times 10^{-2}}2)^2\times 0.1\times 10^{-2} }= 1.273\times 10^{12}\ Nm^{-2}$$By error method $$\dfrac{\Delta Y}{Y}= \dfrac{\Delta L}L+ 2\dfrac{\Delta D}D + \dfrac{\Delta (\delta L)}{\delta L}$$By Substituting the values $$\therefore \dfrac{\Delta Y}Y=0.063$$ $$\therefore$$ Maximum permissible error in young's modulus $$=\dfrac{\Delta Y}Y\times 100 \text %=6.3\text %$$Physics

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