
In the determination of Young's modulus $\dfrac{4MLg}{\pi l d^2}$ by using Searle's method, a wire of length L=2 m and diameter d=0.5mm is used. For a load M=2.5kg, an extension l=0.25mm in the length of the wire is observed. Quantities D and l are measured using a screw gauge and a micrometer, respectively. they have the same pitch of 0.5mm. The number of divisions on their circular scale is 100. the contribution to the maximum probable error of the Y measurement?
A. Due to the errors in the measurements of d and l are the same.
B. Due to the error in the measurement of dd is twice that due to the error in the measurement of ll.
C. Due to the error in the measurement of l,l is twice that due to the error in the measurement of dd.
D. Due to the error in the measurement of dd is four times that due to the error in the measurement of ll.
Answer
597k+ views
Hint: Use the error formula and calculate the contribution of terms l and d to the total error. Check if the contribution of both the terms l and d is the same for the value of the total error. Infer about the cases which lead to maximum probable error.
Step by step solution:
Given pitch p = 0.5 mm
number of divisions on the circular scale = 100
length of wire L = 2 m
diameter d = 0.5 mm
therefore we can get
least count = ratio of the pitch and the number of the divisions on circular scale =$\dfrac{0.5}{100}$mm= 0.005 mm
We know that the maximum possible error in Y due to l and d are given as
$\dfrac{\Delta Y}{Y} = \dfrac{\Delta l}{l} + \dfrac{2\Delta d}{d}$
Which is obtained by differentiating the given function.
We can calculate the error contribution of $l = \dfrac{\Delta l}{l} = \dfrac{0.005}{0.25} = \dfrac{1}{50}$
Similarly the error contribution of $d =\dfrac{2\Delta d}{d} =\dfrac{ 2(0.005)}{0.5 } = \dfrac{1}{50}$
As we can see the error contribution of l and d is equal.
Hence maximum probable error in Y measurement is due to the errors in the measurements of d and l being the same.
Note: The possible mistake that one can make is that one may consider the negative sign for the error due to d term because it is in the denominator, which is not true. For the maximum total error possible we need to consider the value of error due to d term with a positive sign.
Step by step solution:
Given pitch p = 0.5 mm
number of divisions on the circular scale = 100
length of wire L = 2 m
diameter d = 0.5 mm
therefore we can get
least count = ratio of the pitch and the number of the divisions on circular scale =$\dfrac{0.5}{100}$mm= 0.005 mm
We know that the maximum possible error in Y due to l and d are given as
$\dfrac{\Delta Y}{Y} = \dfrac{\Delta l}{l} + \dfrac{2\Delta d}{d}$
Which is obtained by differentiating the given function.
We can calculate the error contribution of $l = \dfrac{\Delta l}{l} = \dfrac{0.005}{0.25} = \dfrac{1}{50}$
Similarly the error contribution of $d =\dfrac{2\Delta d}{d} =\dfrac{ 2(0.005)}{0.5 } = \dfrac{1}{50}$
As we can see the error contribution of l and d is equal.
Hence maximum probable error in Y measurement is due to the errors in the measurements of d and l being the same.
Note: The possible mistake that one can make is that one may consider the negative sign for the error due to d term because it is in the denominator, which is not true. For the maximum total error possible we need to consider the value of error due to d term with a positive sign.
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