A student performs an experiment to determine young's modulus of a wire, exactly 2 m long, by searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0 . 8 mm with an uncertainty of ± 0.05 m m at a load of exactly 1 . 0 kg . The student also measures the diameter of the wire to be 0 . 4 mm with an uncertainty of ± 0.01 m m . The young's modulus obtained from the reading is [ Take g = 9 .8 ms - 2 ]

Young's modulus Y = F L A l = 4 F L π d 2 l = 4 1.0 × 9.8 2 π 0.4 × 10 - 3 2 0.8 × 10 - 3 = 2.0 × 10 11 N m - 2 Further, ∆ Y Y = 2 ∆ d d + ∆ l l ∴ ∆ Y = 2 ∆ d d + ∆ l l Y = 2 × 0.01 0.4 + 0.05 0.8 × 2.0 × 10 11 = 0.2 × 10 11 N m - 2 Or Y + ∆ Y = 2 + 0.2 × 10 11 N m - 2

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A student performs an experiment to determine young's modulus of a wire, exactly $2 m$ long, by searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8 mm$ with an uncertainty of $\pm 0.05 m m$ at a load of exactly $1.0 kg$ . The student also measures the diameter of the wire to be $0.4 mm$ with an uncertainty of $\pm 0.01 m m .$ The young's modulus obtained from the reading is $[$ Take $g = 9 .8 {ms}^{- 2}]$

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Answer:

B

Solution:

Young's modulus $Y = \frac{F L}{A l} = \frac{4 F L}{π d^{2} l}$

$= \frac{(4) (1.0 \times 9.8) (2)}{π {(0.4 \times 10^{- 3})}^{2} (0.8 \times 10^{- 3})}$

$= 2.0 \times 10^{11} {N m}^{- 2}$

Further, $\frac{∆ Y}{Y} = 2 (\frac{∆ d}{d}) + (\frac{∆ l}{l})$

$∴ ∆ Y = \{2 (\frac{∆ d}{d}) + (\frac{∆ l}{l})\} Y$

$= \{2 \times \frac{0.01}{0.4} + \frac{0.05}{0.8}\} \times 2.0 \times 10^{11}$

$= 0.2 \times 10^{11} {N m}^{- 2}$

Or $(Y + ∆ Y) = (2 + 0.2) \times 10^{11} {N m}^{- 2}$

Stream:NTA_ABHYASSubject:PhysicsTopic:Practical physicsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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