JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Elasticity & SHM

During Searle's experiment, zero of the Vernier scale lies between

3.20 \times 1 0^{- 2} m

and

3.25 \times 1 0^{- 2} m

of the main scale. The

2 0^{th}

division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of

2 k g

is applied to the wire, the zero of the Vernier scale still lies between

3.20 \times 1 0^{- 2} m

and

3.25 \times 1 0^{- 2} m

of the main scale but now the

4 5^{th}

division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is

2 m

and its cross-sectional area is

8 \times 1 0^{- 7} m^{2}

. The least count of the Vernier scale is

1.0 \times 1 0^{- 5} m

. The maximum percentage error in the Young's modulus of the wire is

Answer: 4

Solution

The vernier least count is

1 0^{- 5}

m, so the elongation produced by the added load is

Δ L = (45 - 20) \times 1 0^{- 5} = 2.5 \times 1 0^{- 4}

m. Young's modulus is

Y = \frac{F ℓ}{A Δ L}

, and since the load, the wire length and the cross-section are given without stated uncertainties, the whole measurement error is carried by

Δ L

. The smallest change that can be resolved in

Δ L

is one least count,

1 0^{- 5}

m, so the maximum percentage error is

\frac{1 0 ^{- 5}}{2.5 \times 1 0 ^{- 4}} \times 100 = 4%

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer