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

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.