Get Answers to all your Questions

header-bg qa

A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The
maximum tension in the string is found to be  \frac{x}{40}N. The value of x is __________.

 

Option: 1

99


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

T_{max}=mg+\frac{mv^{2}}{L}                ......(1)

Given   m=\frac{1}{4}kg,L=1m,g=9.8m/s^{2}

and amplitude   A=\frac{1}{10}m

For SHM,  \mathrm{KE}_{\max }=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2 

\begin{aligned} & \text { using } \omega=\sqrt{\frac{g}{L}} \\ & m^2=m\left(\sqrt{\frac{g}{L}}\right)^2 A^2=\frac{m_{g A}^2}{L} \end{aligned}

using (2) in (1)

\begin{aligned} & \mathrm{T}_{\max }=2 \mathrm{mg}+\frac{\mathrm{mgA^{2}}}{\mathrm{L}^2} \\ & =\mathrm{mg}\left[1+\frac{1}{10^2}\right]=\frac{1}{4} \times 9.8 \times \frac{101}{100} \\ & \text { or } T_{\max }=\frac{98.98}{40} \end{aligned}

Therefore x=99

 

Posted by

jitender.kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE