Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A 2 \mathrm{~kg}$ steel rod of length $0.6 \mathrm{~m}  is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. lgnoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is___________ \mathrm{ms}^{-1}.
(Take \: \: \mathrm{g}=10 \mathrm{~ms}^{-2} )
 

Answers (1)

best_answer


PE_{lost}= KE_{gained}
mgh= \frac{1}{2}I\omega ^{2}
Mgh= \frac{1}{2}\frac{ML^{2}}{3}\times \omega ^{2}
\frac{6gh}{L^{2}}= \omega ^{2}
V\doteq \left ( \omega L \right )= \sqrt{6gh}= \sqrt{6\times 0\cdot 6\times 10}
    = 6\frac{m}{s}

 

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: Most important chapters for physics, chemistry, mathematics to score good marks
anam-khan

JEE Main 2025 session 2 dates clash with ISC, PSEB 12th exams; no impact on major science subjects
anam-khan

Delhi police EOW books FIITJEE for fraud; 192 complaints filed

Latest Question