A small ball of mass m starts from rest from point A(b,c) on a smooth slope which is a parabola. The normal force that the ground exerts at the instant, the ball arrives at lowest point (0,0) is
A
mg(b2+4c2b2)
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B
4mgc2b2
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C
mg
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D
3mg
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Solution
The correct option is Amg(b2+4c2b2) From the figure, mgc=12mv2 ⇛v2=2gc As x2=4Ay ∴b2=4Ac ⇒A=b24c ∴x2=4b24c×y i.e., x2=b2cy ∴y=cb2x2 ∴1r=d2ydx2[1+(dydx)2]3/2=2cb2[1+(2cb2x)2]3/2 At x=0,y=0 ∴1r=2cb2 or r=b22c Now, N−mg=m×2gcb2×2c ∴N=4mgc2b2+mg ∴N=mg(4c2+b2b2)