JEE Main 2015 · 11 Apr · Shift 1 · Physics Q6

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PhysicsSingle correct+4 / -1Question 6 of 90

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad s $^{-1}$ , the magnitude of its angular momentum about a point on the ground right under the centre of the circle is :

8.64 kg m

^2

^{-1}

11.52 kg m

^2

^{-1}

14.4 kg m

^2

^{-1}

20.16 kg m

^2

^{-1}

Answer and worked solution

Official answer: C

$p=mv=m\omega R=2\times12\times0.6=14.4$ kg m/s, tangential. The position vector from the ground point (0.6 m horizontal, 0.8 m vertical, length 1.0 m) is perpendicular to the velocity, so
$|L|=1.0\times14.4=14.4\text{ kg m}^2\text{s}^{-1}.$

Topics:Angular momentum and its conservation

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 6