JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 4)
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 :
9.859 $$\times$$ 10$$-$$2 N
0.0314 N
9.859 $$\times$$ 10$$-$$4 N
6.28 $$\times$$ 10$$-$$3 N
Explanation
Normal force will provide the necessary centripetal force.
$$ \Rightarrow $$ N = m$$\omega$$2R
Also, $$\omega$$ = $${{2\pi } \over t}$$
N = (0.2)$$\left( {{{4{\pi ^2}} \over {{T^2}}}} \right)$$(0.2)
$$ \Rightarrow $$ N = 0.2 $$\times$$ $${{4 \times {{(3.14)}^2}} \over {{{(40)}^2}}}$$ $$\times$$ 0.2
$$ \therefore $$ N = 9.859 $$\times$$ 10$$-$$4 N
$$ \Rightarrow $$ N = m$$\omega$$2R
Also, $$\omega$$ = $${{2\pi } \over t}$$
N = (0.2)$$\left( {{{4{\pi ^2}} \over {{T^2}}}} \right)$$(0.2)
$$ \Rightarrow $$ N = 0.2 $$\times$$ $${{4 \times {{(3.14)}^2}} \over {{{(40)}^2}}}$$ $$\times$$ 0.2
$$ \therefore $$ N = 9.859 $$\times$$ 10$$-$$4 N
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