A particle is suspended from a string of length R. It is given a velocity `u = 3 sqrt(gR)` at the bottom Match the following `{:(,"Table-1",,"Table-2"),("(A)","Velocity at B","(P)","7 mg"),("(B)","Velocity at C","(Q)",sqrt(5gR)),("(C)","Tension in string at B","(R)",sqrt(7gR)),("(D)","Tension in string at C","(S)","5 mg"),(,,"(T)","None"):}`

A particle is suspended from a string of length R. It is given a veloc

1.	A particle is suspended from a string of length R. It is given a velocity `u = 3 sqrt(gR)` at the bottom Match the following `{:(,"Table-1",,"Table-2"),("(A)","Velocity at B","(P)","7 mg"),("(B)","Velocity at C","(Q)",sqrt(5gR)),("(C)","Tension in string at B","(R)",sqrt(7gR)),("(D)","Tension in string at C","(S)","5 mg"),(,,"(T)","None"):}`
Answer» Correct Answer - A::B::C::D `v_(B)^(2)=u_(A)^(2)-2gh_(AB)=(9gR)-(2gR)=7gR` `therefore" "v_(B)=sqrt(7gR)` Further, `T_(B)=(mv_(B)^(2))/(R)=7mg` Again, `v_(C)^(2)=v_(A)^(2)-2gh_(AC)=(9gR)-2(2R)=5gR` `therefore" "v_(C)=sqrt(5gR)` Further, `T_(C)+mg=(mv_(C)^(2))/(R)` `therefore" T_(C)=4mg`

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