Question 310: Units , Dimensions and Error Analysis

Question

In Searle's experiment, which is used to find Young's Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm ( measured by a scale of least count 0.001 cm ) and length is L = 110 cm ( measured by a scale of least count 0.1 cm ). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001 cm ). Find maximum possible error in the value of Young's modulus. Screw gauge and and meter scale are free from error.

Accepted Answer

The distance moved on the linear scale when circular scale makes one complete rotation is p = 1 mm (pitch). The number of divisions on the circular scale is N = 100. Thus, one division on the circular scale is LC = $${p \over N} = {1 \over {100}}$$ = 0.01 mm. The linear scale reading (LSR) is 1 mm and the circular scale reading (CSR) is 47. Thus, the diameter of the wire is
d = LSR + CSR $$	imes$$ LC
= 1 + 47 $$	imes$$ 0.01 = 1.47 mm = 0.147 cm.
The curved surface area is S = 2$$\pi$$rL. That is,
$$S = 2\pi \left( {{d \over 2}} ight)L$$
$$S = \pi dL = \pi \left( {{{1.47} \over {10}}} ight)5.6 = $$ 2.5848 cm
2
= 2.6 cm
2
which is corrected to two significant digits.

Units , Dimensions and Error Analysis

Integer / Numeric Answer Type