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Simple Harmonic Motion question

2003 · Shift 0 · Q160
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Simple Harmonic Motion question

2003 · Shift 0 · Q160

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
The length of a simple pendulum executing simple harmonic motion is increased by 21%21\%21%. The percentage increase in the time period of the pendulum of increased length is
  1. A
    11%11\%11%
  2. B
    21%21\%21%
  3. C
    42%42\%42%
  4. D
    10%10\%10%
View written solutionFree

Correct answer: D

  1. For a simple pendulum, the time period is T=2πlgT = 2\pi \sqrt{\frac{l}{g}}T=2πgl​​

  2. Hence, T∝lT \propto \sqrt{l}T∝l​

  3. If the length is increased by 21%21\%21%, then the new length becomes l′=1.21ll' = 1.21ll′=1.21l

  4. Therefore, the new time period is T′=T1.21T' = T\sqrt{1.21}T′=T1.21​

  5. Now, 1.21=1.1\sqrt{1.21} = 1.11.21​=1.1 so T′=1.1TT' = 1.1TT′=1.1T

  6. Thus, the increase in time period is T′−TT×100=1.1T−TT×100=10%\frac{T' - T}{T} \times 100 = \frac{1.1T - T}{T}\times 100 = 10\%TT′−T​×100=T1.1T−T​×100=10%

  7. Therefore, the percentage increase in the time period is: 10%\boxed{10\%}10%​

  8. Checking options:

    • A: 11%11\%11% ❌
    • B: 21%21\%21% ❌
    • C: 42%42\%42% ❌
    • D: 10%10\%10% ✅

So the correct option is D.

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