ii) A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform vertical circular motion, under gravity. Minimum speed of a particle is 5 m/s. Consider following statements.
P) Maximum speed must be 5√5 m/s.
Q) Difference between maximum and minimum tensions along the string is 60 N.
Select correct option.
(A) Only the statement P is correct.
(B) Only the statement Q is correct.
(C) Both the statements are correct.
(D) Both the statements are incorrect.
Answers
Answered by
5
Answer:
Q IS CORRECT
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Answered by
5
Maximum speed will at A
maximum speed will at B
At point B,
mv^2/1.2 -mg=T1 i)
25/1.2-10=T1
20.8-10=T1
T1=10.8
using third equation of motion,v at top will be
v^2=u^2+2as
v^2=25+2×10×2.4
v=√73=8.5(approx)
This is the maximum speed
Maximum tension T will be at the bottom
T=mv^2/R +mg
T=73/1.2+10
T=60.8+10=70.8
Difference between maximum and minimum tension=>70.8-10.8=60
This shows that Q is correct
hence option B)is correct
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