In an experiment to determine the Young’s modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm. The error in the measurement of Young’s modulus $\left(\Delta Y\right)$  is found to be x × 10¹⁰Nm^-2. The value of x is _________.

(take g = 10 ms^-2)

<p>L = 1 m</p><p><math><mrow><mi>Δ</mi><mi>L</mi><mo>=</mo><mn>0</mn><mo>.</mo><mn>4</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>3</mn></mrow></msup><mi>m</mi></mrow></math></p><p><math><mrow><mi>m</mi><mo>=</mo><mn>1</mn><mi>k</mi><mi>g</mi></mrow></math></p><p>d = 0.4 × 103 m</p><p><math><mrow><mfrac><mrow><mi>F</mi></mrow><mrow><mi>A</mi></mrow></mfrac><mo>=</mo><mi>y</mi><mfrac><mrow><mi>Δ</mi><mi>L</mi></mrow><mrow><mi>L</mi></mrow></mfrac></mrow></math></p><p><math><mrow><mi>y</mi><mo>=</mo><mfrac><mrow><mi>F</mi><mi>L</mi></mrow><mrow><mi>A</mi><mi>Δ</mi><mi>L</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mrow><mo> (</mo><mrow><mi>m</mi><mi>g</mi></mrow><mo>)</mo></mrow><mo>×</mo><mn>1</mn></mrow><mrow><mrow><mo> (</mo><mrow><mfrac><mrow><mi>π</mi><msup><mrow><mi>d</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mn>4</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mo>×</mo><mn>0</mn><mo>.</mo><mn>4</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>3</mn></mrow></msup></mrow></mfrac></mrow></math></p><p><math><mrow><mi>Δ</mi><mi>y</mi><mo>=</mo><mn>0</mn><mo>.</mo><mn>1</mn><mo>×</mo><mn>0</mn><mo>.</mo><mn>1</mn><mn>9</mn><mn>9</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>1</mn><mn>2</mn></mrow></msup><mo>=</mo><mn>1</mn><mo>.</mo><mn>9</mn><mn>9</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>1</mn><mn>0</mn></mrow></msup></mrow></math></p><p>= 1.99</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment