JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Units & Measurement

In the determination of Young's modulus

(Y = \frac{4 M L g}{π l d ^{2}})

by using Searle's method, a wire of length

L = 2 m

and diameter

d = 0.5 mm

is used. For a load

M = 2.5 kg

, an extension

l = 0.25 mm

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 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the

Y

measurement

(A)due to the errors in the measurements of $d$ and $l$ are the same.✓ Correct
(B)due to the error in the measurement of $d$ is twice that due to the error in the measurement of $l$ .
(C)due to the error in the measurement of $l$ is twice that due to the error in the measurement of $d$ .
(D)due to the error in the measurement of $d$ is four times that due to the error in the measurement of $l$ .

Answer: (A)

Solution

The least count is

L.C. = \frac{0.5}{100} = 0.005

mm. The fractional error in

Y

\frac{Δ Y}{Y} = \frac{Δ l}{l} + \frac{2Δ d}{d}

. Computing:

\frac{Δ l}{l} = \frac{0.005 \times 1 0 ^{- 3}}{0.25 \times 1 0 ^{- 3}} = \frac{1}{50}

and

\frac{2Δ d}{d} = \frac{2 \times 0.005 \times 1 0 ^{- 3}}{0.5 \times 1 0 ^{- 3}} = \frac{1}{50}

. Both contributions are equal.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option

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JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Units & Measurement

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