33just by observation it seems to be 2pi(m/(k1+k3))^.5
14 Answers
1actually options r
√(3/2)
1/√3
√23
∞
so ur ans isnt matching[2]
btw how u got that[1]
49let the mass when moved by dist x downwards, the pulley ''of many strings'' rotate through angle θ...
thus..
x=rθ........r=radius of pulley..
k1=k2=k3=k(suppose)
restoring torque=3kxθ=Iα..
or 3krθ2=xmr2α..
so θ/α=0....since the pulley is massless....
so T=2Ï€√1/(θ/α)
giving T=∞............whch is in options luckily...[1][1]
49ye ye ye....i did it....i did it...i do not ask for it generally...but i want a pink 4 dis....[1]
neways someone please verify...[1]
1actually der is no one to verify here
site is becoming slack day by day [2]
49dont even dare to say so about target iit ever again...never again i warn you...arre mujhe sab pata hota toh har ek sawaal ka hi soln deta...but wat to do...i dont kno everything...and even our beloved admins are having great deal of problem dont they.....???otherwise wouldnt they come if they had even a little bit of time???
and please che dont tell those sort of thing abt t iit ...please donot tell...at least in front of me..[2][2][2]
33put values of k1 k3 in my answer and m=1 you will get 1/√3 ... thats in option
you should have checked that..
for a small movement of block or say rotation theta of P1 draw fbd of p1 and then find tension in ropw... now T=ma you are done...