Question

A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 1/4 times bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the new simple harmonic oscillations will be? #jeemainaug2021​

Answer 1

Answer:

Explanation:

T = 2π √ ℓ / g ℓ/g When bob is immersed in liquid mgeff = mg – Buoyant forceRead more on Sarthaks.com - https://www.sarthaks.com/1245482/bob-mass-suspended-thread-length-undergoes-simple-harmonic-oscillations-with-time-period