The elongation of a wire on the surface of the earth is 10-4m. The same wire of same dimensions is elongated by 6 × 10-5 m on another planet. The acceleration due to gravity on the planet will be__________ms-2. (Take acceleration due to gravity on the surface of earth = 10 ms-2)

Δ L (change in length) = F l A E = m g l A E Δ l ∝ g Δ l 1 g 1 = Δ l 2 g 2 ⇒ 1 0 − 4 1 0 = 6 × 1 0 − 5 g 1 g 1 = 6 × 1 0 − 4 1 0 − 4 = 6 m/sec2

Ask & Answer Question

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics Class 11th Physics Mechanical Properties of Solids

The elongation of a wire on the surface of the earth is 10^-⁴m. The same wire of same dimensions is elongated by 6 × 10^-⁵ m on another planet. The acceleration due to gravity on the planet will be__________ms^-². (Take acceleration due to gravity on the surface of earth = 10 ms^-²)

2 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

2 weeks ago

$Δ L$ (change in length) $= \frac{F l}{A E} = \frac{m g l}{A E}$

$Δ l \propto g$

$\frac{Δ l_{1}}{g_{1}} = \frac{Δ l_{2}}{g_{2}} \Rightarrow \frac{1 0^{- 4}}{10} = \frac{6 \times 1 0^{- 5}}{g_{1}}$

$g_{1} = \frac{6 \times 1 0^{- 4}}{1 0^{- 4}} =$ 6 m/sec²

<math> <mrow> <mi>Δ</mi> <mi>L</mi> </mrow> </math> (change in length) <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mi>F</mi> <mi mathvariant="script">l</mi> </mrow> <mrow> <mi>A</mi> <mi>E</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>m</mi> <mi>g</mi> <mi mathvariant="script">l</mi> </mrow> <mrow> <mi>A</mi> <mi>E</mi> </mrow> </mfrac> </mrow> </math>                <math> <mrow> <mi>Δ</mi> <mi mathvariant="script">l</mi> <mo>∝</mo> <mi>g</mi> </mrow> </math>                <math> <mrow> <mfrac> <mrow> <mi>Δ</mi> <msub> <mrow> <mi mathvariant="script">l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>Δ</mi> <msub> <mrow> <mi mathvariant="script">l</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> <mtext> </mtext> <mo>⇒</mo> <mfrac> <mrow> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>6</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>5</mn> </mrow> </msup> </mrow> <mrow> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>                <math> <mrow> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>6</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> </mrow> </math> 6 m/sec2