circular motion.... easy doubtsss...............

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for B part

net acceleration = \sqrt{\frac{v^{2}}{r}+ a }

thus f_s > Mv²/r

@Anirudh,
the net acceleration is not \sqrt{\frac{v^2}{r}+a}.
Its actually, \sqrt{\left(\frac{v^2}{r}\right)^2+a^2}

The idea is that the centripetal acceleration and the tangential acceleration both must be provided by the friction. So the net frictional force is
f=m\sqrt{\left(\frac{v^2}{r}\right)^2+a^2}
Since f\le \mu N
so we must have
m\sqrt{\left(\frac{v^2}{r}\right)^2+a^2}\le \mu mg giving
\sqrt{\left(\frac{v^2}{rg}\right)^2+\left(\dfrac{a}{g}\right)^2}\le \mu

so in wich direction will d car skid if i increase v more than the limiting value ???

the tangential fric. force is constant and it wont increase on increasing v so it does not affect direction of skidding.

but kaymant sir earlier told that friction itself is providing tangential acceleration [7]

and how is the tangential frictional force = \mumg ???the net frictional force shud be \mumg , if its a limiting case

friction did not produces any accln.it acts against direction of motion.the centripetal accln is also not produced by friction.it acts radially ouwards. friction acts towards centre and is the force which counters the radially outwards accln.

what kayamant sir meant is that friction neutralises the tangential accln.the magnitude remains same either way.

sorry kaymant sir .