In the determination of the Young-s modulus of a given wire- the force- length- radius and extension in the wire are measured as 1000- 0-01 N- 1-25 0- 0-002 m- 0-001 0- 0-00002 m- and 0-01 0- 0-00002 m respectively- The percentage error in the measurement of Youngs modulus is -

The correct option is A 4.36 Y #160;= #160; stressstrain #160; = #160; FA×Ll ΔYY #160; = #160; ( Δ FF #160; + #160; Δ AA #160 ...

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Standard VII

Physics

Effects of Heat

In the determ...

Question

In the determination of the Young's modulus of a given wire, the force, length, radius and extension in the wire are measured as $(1000 \pm 0.01)$ N, $(1.25 0 \pm 0.002)$ m, $(0.001 0 \pm 0.00002)$ m, and $(0.01 0 \pm 0.00002)$ m respectively. The percentage error in the measurement of Youngs modulus is :

4.36

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Solution

The correct option is A $4.36$
$Y = \frac{stress}{strain} = \frac{F}{A} \times \frac{L}{l}$

$\frac{Δ Y}{Y} = (\frac{Δ F}{F} + \frac{Δ A}{A} + \frac{Δ L}{L} + \frac{Δ l}{l}) \times 100$

$= (\frac{0.01}{1000} + \frac{2 \times 0.00002}{0.0010} + \frac{0.002}{1.250} + \frac{0.00002}{0.01}) \times 100$

$= 0.001 + 4 + 0.16 + 0.2 = 4.361$

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In the determination of the Young's modulus of a given wire, the force, length, radius and extension in the wire are measured as (1000±0.01) N, (1.250±0.002) m, (0.0010±0.00002) m, and (0.010±0.00002) m respectively. The percentage error in the measurement of Youngs modulus is :

The correct option is A 4.36Y=stressstrain=FA×LlΔYY=(ΔFF+ΔAA+ΔLL+Δll )×100 =(0.011000+2×0.000020.0010+0.0021.250+0.000020.01)×100 =0.001+4+0.16+0.2=4.361