Young's modulus is determined by the equation given by Y = 49000 l m cm 2 dyn where M is the mass and l is the extension of wire used in the experiment. Now error in Young modules ( Y ) is es

Question

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

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Answer:

B

Solution:

$\frac{Δ Y}{Y} \frac{Δ Y}{Y} = \frac{Δ m}{m} + \frac{Δ ℓ}{ℓ} = \frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01 = 0.02 \Rightarrow % \frac{Δ Y}{Y} = 2%$

TestHub · Accepted Answer

Y Δ Y ​ Y Δ Y ​ ​ = m Δ m ​ + ℓ Δ ℓ ​ = 500 5 ​ + 2 0.02 ​ = 0.01 + 0.01 = 0.02 ⇒ % Y Δ Y ​ = 2% ​

Physics - Error Question with Solution | TestHub

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