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:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mroot> <mrow> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </math> </span></p>
Option 4 - <p>3L<sub>1</sub> – 2L<sub>2</sub></p>
4 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
6 months ago
Correct Option - 3
Detailed Solution:

Spring constant of wire (k) =   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 ) ÷ ( 2 )      

L 1 L L 2 L = 1 2

L = 2L1 – L2

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