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Standard XII

Physics

Poisson's Effect

In the Searle...

Question

In the Searle's method to determine the Young's modulus of a wire, a steel wire of length $156 c m$ and diameter $0.054 c m$ is taken as experimental wire. The average increase in length for $1.5 k g w t$ is found to be $0.050 c m$ . Then the Young's modulus of the wire is

3.002 \times 10^{11} N / m^{2}

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1.002 \times 10^{11} N / m^{2}

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2.002 \times 10^{11} N / m^{2}

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2.5 \times 10^{11} N / m^{2}

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Solution

The correct option is B $1.002 \times 10^{11} N / m^{2}$
Young’s modulus is given by

$Y = \frac{F L}{A Δ l}$

$Y = \frac{1.5 \times 10 \times 156}{3.14 \times 2.7 \times 10^{- 4} \times 5 \times 10^{- 4}}$

$Y = 1.002 \times 10^{- 11} N / m^{2}$

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In the Searle's method to determine the Young's modulus of a wire, a steel wire of length 156 cm and diameter 0.054 cm is taken as experimental wire. The average increase in length for 1.5 kg wt is found to be 0.050 cm. Then the Young's modulus of the wire is

The correct option is B 1.002×1011N/m2Young’s modulus is given by Y=FLAΔl Y=1.5×10×1563.14×2.7×10−4×5×10−4 Y=1.002×10−11N/m2

In the Searle's method to determine the Young's modulus of a wire, a steel wire of length $156 c m$ and diameter $0.054 c m$ is taken as experimental wire. The average increase in length for $1.5 k g w t$ is found to be $0.050 c m$ . Then the Young's modulus of the wire is

The correct option is B $1.002 \times 10^{11} N / m^{2}$
Young’s modulus is given by

$Y = \frac{F L}{A Δ l}$

$Y = \frac{1.5 \times 10 \times 156}{3.14 \times 2.7 \times 10^{- 4} \times 5 \times 10^{- 4}}$

$Y = 1.002 \times 10^{- 11} N / m^{2}$