Find Fractional Error Using Error Analysis Formula

 

❓ Question

Given,

$$Y=\frac{MgL^3}{4bd^3\delta}$$

The measured quantities and their errors are:

$$g=9.8\,m/s^2,\qquad \Delta g=0$$
$$M=2\,kg,\qquad \Delta M=1\,g$$
$$L=1\,m,\qquad \Delta L=1\,mm$$
$$b=4\,cm,\qquad \Delta b=0.1\,mm$$
$$d=0.4\,cm,\qquad \Delta d=0.01\,mm$$
$$\delta=5\,mm,\qquad \Delta\delta=0.01\,mm$$

Find the maximum fractional error in $Y$.

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✍️ Solution

Given,

$$Y=\frac{MgL^3}{4bd^3\delta}$$

For maximum fractional error,

$$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta g}{g} + 3\frac{\Delta L}{L} + \frac{\Delta b}{b} + 3\frac{\Delta d}{d} + \frac{\Delta\delta}{\delta}$$

Now calculate each fractional error:

$$\frac{\Delta M}{M} = \frac{10^{-3}}{2} = 0.0005$$
$$\frac{\Delta g}{g}=0$$
$$3\frac{\Delta L}{L} = 3\left(\frac{10^{-3}}{1}\right) = 0.003$$

For $b$,

$$\Delta b=0.1\,mm=0.01\,cm$$
$$\frac{\Delta b}{b} = \frac{0.01}{4} = 0.0025$$

For $d$,

$$\Delta d=0.01\,mm=0.001\,cm$$
$$3\frac{\Delta d}{d} = 3\left(\frac{0.001}{0.4}\right) = 0.0075$$

For $\delta$,

$$\frac{\Delta\delta}{\delta} = \frac{0.01}{5} = 0.002$$

Therefore,

$$\frac{\Delta Y}{Y} = 0.0005+0+0.003+0.0025+0.0075+0.002$$
$$\boxed{\frac{\Delta Y}{Y}=0.0155}$$

✅ Final Answer

$$\boxed{\frac{\Delta Y}{Y}=0.0155}$$

or, in percentage form,

$$\boxed{1.55\%}$$