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 frictional force om `m_1` is

Options

A. 12 N
B. 6 N
C. 24 N
D. 36 N

Answer

D

Explanation

The cantripetal force is provided by the tension in the string
`:. T=m_2r_2 omega^2 or T=5(0.176)(10)^2=88N`
For circular motion of `m_1`,
`"Centripetal force"=m_1r_1 omega^2=10(0.124)(10)^2=124N`
`"But centripetal force"=T+"force of friction on" m_1`
`ie, 124=88+f or f=36N`