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        A simple pendulum is executing simple harmonic motion with a time period T; If the length of the pendulum. Is increased by 21%, the Increase in the time period of the pendulum of Increased length is -(1)1o% (2)21% (3) 30% (4) 50%
3 years ago

Rahul Verma
45 Points
							By formula $T_{1}=2\pi \sqrt{\frac{l}{g}}$\    now length is increased by 21% therefore new length = $l+l\frac{21}{100}= \frac{121l}{100}$          put in eq we get$T_{2}^{}= 2\pi \sqrt{\frac{\frac{121l}{100}}{g}}$$T_{2}= 2\pi \sqrt{\frac{\frac{121l}{100}}{g}}=2\pi\sqrt{} \frac{121l}{100g}=2\pi \sqrt{\frac{l}{g}}\frac{11}{10}=T_{1}\frac{11}{10}$      T2 - T1 = $\frac{T_{1}}{10}$so percentage change = $\frac{change in time period }{original time period}\times 100$ $= \frac{\frac{T_{1}}{10}}{T_{1}}\times 100$ $=\frac{T_{1}}{10T_{1}}\times 100$  $= 10%$%

3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions