Two blocks of mass `m_1 =10 kg and m_2=5 kg`, connected to each other by a massless inextensible string of length 0.3 m are placed along a diameter of a turn table. The co-efficent of friction between the table and `m_1` is 0.5 while there is no friction between `m_2` and the table. The table is rotating with an angular velocity opf 10 rewds about a veritical axis passing through its center O. The masses are palecd along the dimeter of the table on either side of the center O such that the mass `m_1` is at a distance of 0.124 m from O. The masses are observe to be at rest with respect to an observer on the turn table.
The minimum angular speed of the turn table the masses will slip from this pasition

Options

A. 10.53 red/sec
B. 11.67 red/sec
C. 7.87 red/sec
D. 12.79 red/sec

Answer

B

Explanation

Max friction force on `m_1, f_"max"=mum_1g=(5.0)(10)(9.8)=49N`
`:.T+f="centerpetal force on" m_1=m_1r_1omega^2" "......(1)`
when `f=f_"max",omega=omega.="minimum angular speed, let T be T"`
for mass `m_2,T.=m_2r_2(omega.)^2" ".......(2)`
`:. T.+f_"max"=m_1r_1(omega.)^2 or m_2r_2 (omega.)^2+f_"max"=m_1r_1(omega.)^2`
`(omega.)^2 xx [m_1r_1-m_2r_2]=f_"max"`
`ie, omega.=sqrt(f_"max")/sqrt(m_1r_1-m_2r_2)= sqrt(49)/sqrt(10 xx 0.124-5 xx 0.176)=11.67 "red"//sec`