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JEE Advanced
Exam
JEE Advanced
Subject
Physics
Chapter
Experimental Physics

JEE Advanced · Physics · Experimental Physics

Time period of a simple pendulum is given by T=2 π ( )/(g) where is...

Time period of a simple pendulum is given by T=2πℓg\mathrm{T}=2 \pi \sqrt{\frac{\ell}{g}} where ℓ\ell is the length of pendulum and g is acceleration due to gravity. If percentage increase in the length and acceleration due to gravity are 69%69 \% and 21%21 \% respectively then find out percentage increase in time period.

Correct answer

18.18

Step-by-step solution

T′−TT×100=(ℓ′gℓg′−1)×100=(1.691.21−1)×100=(1.31.1−1)×100=0.21.1×100=20011=18.18%\frac{\mathrm{T}^{\prime}-\mathrm{T}}{\mathrm{T}} \times 100=\left(\sqrt{\frac{\ell^{\prime} \mathrm{g}}{\ell \mathrm{g}^{\prime}}}-1\right) \times 100=\left(\sqrt{\frac{1.69}{1.21}}-1\right) \times 100=\left(\frac{1.3}{1.1}-1\right) \times 100=\frac{0.2}{1.1} \times 100=\frac{200}{11}=18.18 \%
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