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3-2.Motion in Plane
hard
A block of $200\, g$ mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius $20\, cm$. If the block takes $40\, s$ to complete one round, the normal force by the side walls of the groove is
A$0.0314\, N$
B$9.859 \times 10^{-2}\, N$
C$6.28 \times 10^{-3}\, N$
D$9.859 \times 10^{-4}\, N$
(JEE MAIN-2021)
Solution
$N = m \omega^{2} R$
$N = m \left[\frac{4 \pi^{2}}{ T ^{2}}\right] R$
Given $m =0.2 kg , T =40 S , R =0.2 m$
$N =9.859 \times 10^{-4} N$
$N = m \left[\frac{4 \pi^{2}}{ T ^{2}}\right] R$
Given $m =0.2 kg , T =40 S , R =0.2 m$
$N =9.859 \times 10^{-4} N$
Standard 11
Physics
