A bob of mass 'm' suspended by a thread of length I undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 1 4  times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be:-

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mi>T</mi> </mrow> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>T</mi> </mrow> </math> </span></p>
Option 3 - <p>T</p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>T</mi> </mrow> </math> </span></p>
2 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
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6 months ago
Correct Option - 4
Detailed Solution:

Time period T = 2π l g e f f

T i = T = 2 π l g    

T f = T ' = 2 π 4 l / 3 g ( 1 ρ σ ) = 2 π 1 6 l 9 g

T ' = 4 3 T

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Physics Oscillations 2025

Physics Oscillations 2025

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