A particle of mass 1 kg tied to 1.2 m long string is whirled to perform vcm under gravity. minimum speed of a particle is 5mpersec
Answers
Answer:
Explanation:
it's interesting question....
minimum velocity will be at the maximum height...
Let the tension be t1 at the highest postion...
then,
clealry we can observe that
mg force will be in the downward direction...and at highest point, centripetal force will be in opposite direction....
so their, resultant,will be the tension...of the string
t1=Mv^2/R -Mg
t1=1×5×5×10/12 -1×10
t1=250/12-10
t1=130/12=1.8(approx)
since the acceleration,will be constant g,everywhere,so,we can take out the maximum velocity,which will be at downward position...
V^2=u^2+2×10×2×1.2
V^2=25+48
V=√73
let the tension be t2 at tension at the bottom most point,
t2=mg+mv^2/R
t2=1×10+1×73/1.2
t2=10+6.8
t2=7.8
t2-t1=7.8-1.8
t2-t1=60
from this we get,that,option Q is exactly right....
also,we take out maximum velocity as √73,=8.5
so option p is incorrect...
(hence option A is right)