Mathematics in Physics

JEE Main level, single correct, medium

111 of 121

Question 111 of 121

A thin 1m1m long rod has a radius of 5mm5mm. A force of 50π×103N50{\text{π×10}}^{3}N is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01mm0.01mm, which of the following statements is false?

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Answer: (A) The maximum value of YY that can be determined is 2×1014Nm−22\times {10}^{14}N{m}^{-2}

Given, l=1ml=1m, r=5×10−3mr=5\times 1{0}^{-3}m and F=50π×103NF=50\text{π}\times 1{0}^{3}\text{N}.
Least count in length is =0.01mm=0.01×10−3m=0.01mm=0.01\times {10}^{-3}m
Young's modulus is given by Y=FA∆llY=\dfrac{\dfrac{F}{A}}{\dfrac{∆l}{l}}
For, ∆l→Min∆l\rightarrow Min, Y→MaxY\rightarrow Max
⇒YΔll=FA\Rightarrow Y\dfrac{\Delta l}{l}=\dfrac{F}{A}
Y=50π×103π×(5×10−3)2×lΔlY=\dfrac{50\text{π}\times 1{0}^{3}}{\text{π}\times {\left(5\times 1{0}^{-3}\right)}^{2}}\times \dfrac{l}{\Delta l}
Y=50×10325×10−6×1Δl⇒Y=2×109ΔlY=\dfrac{50\times 1{0}^{3}}{25\times 1{0}^{-6}}\times \dfrac{1}{\Delta l}\Rightarrow Y=\dfrac{2\times 1{0}^{9}}{\Delta l}
Here Δl\Delta l is unknown so we can not find the YY
The change in young’s modulus ratio is ∆YY=2∆rr+∆ll\dfrac{∆Y}{Y}=2\dfrac{∆r}{r}+\dfrac{∆l}{l}.
As seen from above, the minimum contribution to change in young’s modulus is given by the uncertainty in the length and the maximum contributing to the uncertainty in young’s modulus is given by uncertainty in radius.
Hence, correct option is AA.

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