A wire of length L is hanging from a fixed support. The length changes to L1 andL2 when masses 1kg and 2kg are suspended respectively from its free end. Then the value of L is equal to:

Spring constant of wire (k) = <!- [if gte mso 9]><xml> </xml><! [endif]-> A E L Case 1 k [ Δ ] = 1 g <!- [if gte mso 9]><xml> </xml><! [endif]-> k [ L 1 − L ] = 1 g - (1) Case 2 <!- [if gte mso 9]><xml> </xml><! [endif]-> k [ L 2 − L ] = 2 g - (2) ( 1 ) ÷ ( 2 ) <!- [if gte mso 9]><xml> </xml><! [endif]-> L 1 − L L 2 − L = 1 2 L = 2L1 – L2

A wire of length L is hanging from a fixed support. The length changes to L1 andL2 when masses 1kg and 2kg are suspended respectively from its free end. Then the value of L is equal to:

3L1 – 2L2

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Physics NCERT Exemplar Solutions Class 11th Chapter Five Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics Mechanical Properties of Solids

A wire of length L is hanging from a fixed support. The length changes to L₁ andL₂when masses 1kg and 2kg are suspended respectively from its free end. Then the value of L is equal to:

Option 1 -

$\sqrt[]{L_{1} L_{2}}$

Option 2 -

$\frac{L_{1} + L_{2}}{2}$

Option 3 -

$2 L_{1} - L_{2}$

Option 4 -

3L₁ – 2L₂

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Spring constant of wire (k) = $\frac{A E}{L}$

Case 1

$k [Δ] = 1 g$

$k [L_{1} - L] = 1 g$ - (1)

Case 2 $k [L_{2} - L] = 2 g$ - (2)

$(1) \div (2)$

$\frac{L_{1} - L}{L_{2} - L} = \frac{1}{2}$

L = 2L₁ – L₂

Spring constant of wire (k) = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814258907"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mfrac> <mrow> <mi>A</mi> <mi>E</mi> </mrow> <mrow> <mi>L</mi> </mrow> </mfrac> </mrow> </math> Case 1      <math> <mrow> <mi>k</mi> <mrow> <mo> [</mo> <mrow> <mi>Δ</mi> </mrow> <mo>]</mo> </mrow> <mo>=</mo> <mn>1</mn> <mi>g</mi> </mrow> </math>         <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814258908"> </o:OLEObject></xml><! [endif]->                       <math> <mrow> <mi>k</mi> <mrow> <mo> [</mo> <mrow> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <mi>L</mi> </mrow> <mo>]</mo> </mrow> <mo>=</mo> <mn>1</mn> <mi>g</mi> </mrow> </math>  - (1)              Case 2   <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1814258910"> </o:OLEObject></xml><! [endif]->      <math> <mrow> <mi>k</mi> <mrow> <mo> [</mo> <mrow> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>−</mo> <mi>L</mi> </mrow> <mo>]</mo> </mrow> <mo>=</mo> <mn>2</mn> <mi>g</mi> </mrow> </math>             - (2)          <math> <mrow> <mrow> <mo> (</mo> <mrow> <mn>1</mn> </mrow> <mo>)</mo> </mrow> <mo>÷</mo> <mrow> <mo> (</mo> <mrow> <mn>2</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math>    <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1029" DrawAspect="Content" ObjectID="_1814258911"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <mi>L</mi> </mrow> <mrow> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>−</mo> <mi>L</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> L = 2L1 – L2