During Searle’s experiment, zero of the Vernier scale lies between and of the main scale. The division of the Vernier scale exactly coincides with one of the main scale divisions. [2014]

Question

During Searle’s experiment, zero of the Vernier scale lies between $3.20 \times 10^{- 2} m$ and $3.25 \times 10^{- 2} m$ of the main scale. The $20^{th}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{- 2} m$ and $3.25 \times 10^{- 2} m$ of the main scale but now the $45^{th}$ division of the 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 $8 \times 10^{- 7} m^{2}$ . The least count of the Vernier scale is $1.0 \times 10^{- 5} m$ . The maximum percentage error in Young’s modulus of the wire is _____. [2014]

Smart Achievers · Accepted Answer

(4)

$Young's modulus Y = \frac{F L}{A \times l}$

$Here F, A and L are accurately known.$

$∴$ $3.20 \times 10^{- 2} m$

$l = (45 - 20) \times L.C. = 25 \times 10^{- 5} m$

$\frac{Δ Y}{Y} \times 100 = \frac{Δ l}{l} \times 100 = \frac{1.0 \times 10^{- 5}}{25 \times 10^{- 5}} \times 100 = 4 %$

Solution

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Solution

Ans. (4) Young's modulus Y=FLA×l Here F,A and L are accurately known. ∴ Percentage error in Young's modulus l=(45-20)×L.C.=25×10-5 m ΔYY×100=Δll×100=1.0×10-525×10-5×100=4%

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