Question

In Searle’s experiment to find Young’s modulus the diameter of the wire is measure as $ d = 0.05cm $ ,length of the wire is $ l = 125cm $ and when a weight, $ m = 20.0kg $ is put, extension in wire was found to be $ 0.100cm $ . Find the permissible error in Young’s modulus $ (Y) $ . Use: $ Y = \dfrac{{mgl}}{{\dfrac{\pi }{4}{d^2}x}} $ .
(A) 6.3%
(B) 5.3%
(C) 2.3%
(D) 1%

Answer 1

Hint: The maximum permissible error of a quantity is nothing but the maximum error that it can possess after calibration. It can be 2 units more or less than the maximum permissible limit your manufacturer has specified. It can be theoretically calculated for a function $ u(x,y,z) = {x^\alpha }{y^\beta }{z^\gamma } $ as
 $ \dfrac{{\Delta u}}{u} = \alpha \dfrac{{\Delta x}}{x} + \beta \dfrac{{\Delta y}}{y} + \gamma \dfrac{{\Delta z}}{z} $ .

Formulas used: The formula used will be $ Y = \dfrac{{mgl}}{{\dfrac{\pi }{4}{d^2}x}} $ where $ Y $ is the Young’s Modulus, $ m $ is the mass of the body, $ g $ is acceleration due to gravity, $ l $ is the length of the wire, $ d $ is the diameter, and $ x $ is the extension the wire experiences.

Complete Step by Step answer
The maximum permissible error for a quantity can be calculated theoretically and determined easily. However, it is not necessary that the theoretical and practical values match. To determine them theoretically consider using the formula $ \dfrac{{\Delta u}}{u} = \alpha \dfrac{{\Delta x}}{x} + \beta \dfrac{{\Delta y}}{y} + \gamma \dfrac{{\Delta z}}{z} $ for the function $ u(x,y,z) = {x^\alpha }{y^\beta }{z^\gamma } $ .
Also, we know that the formula to find Young’s modulus is $ Y = \dfrac{{mgl}}{{\dfrac{\pi }{4}{d^2}x}} $ . Applying the theoretical formula for maximum permissible error here we get, $ \dfrac{{\Delta Y}}{Y} = \dfrac{{\Delta m}}{m} + \dfrac{{\Delta l}}{l} + 2\dfrac{{\Delta d}}{d} + \dfrac{{\Delta x}}{x} $ .
Using this formula, we will be able to find the maximum permissible error for the Young’s modulus of the given problem. However we only know $ m = 20.0kg,l = 125cm,d = 0.05cm,x = 0.100cm $ but we do not have values of $ \Delta m,\Delta l,\Delta d,\Delta x $ .
Since we do not have values of the error in measurement of mass $ m $ , length $ l $ ,diameter $ d $ or extension in wire $ x $ .
Thus, we get, $ \Delta m = 0.1,\Delta l = 1,\Delta d = 0.01,\Delta x = 0.001 $ . Substituting the values in the formula to find maximum permissible error we get,
 $ \dfrac{{\Delta Y}}{Y} = \dfrac{{0.1}}{{20.0}} + \dfrac{1}{{125}} + 2\dfrac{{0.01}}{{0.05}} + \dfrac{{0.001}}{{0.100}} $
 $ \Rightarrow \dfrac{{\Delta Y}}{Y} = 0.005 + 0.008 + 2(0.2) + 0.01 $
Solving R.H.S we get,
 $ \Rightarrow \dfrac{{\Delta Y}}{Y} = 0.043 $
 $ \Rightarrow \dfrac{{\Delta Y}}{Y}\% = 4.3\% $ .

Note:
Usually the percentage error is calculated from dimensional formula but here we can see that there is more than one entity that represents a single entity in the dimension formula which is why we don't do that here.