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JEE Main 2 April 2017 (offline), Physics Q6

Question 6 of 90 in this shift, Physics question 6 of 30.

A slender uniform rod of mass M and length

l

is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle

\theta

with the vertical is :

The figure, in words

Rod pivoted at the origin of axes (z vertical, x horizontal), making angle theta with the vertical z-axis; double-headed arc arrow shows rotation.

(1) $\frac{3g}{2l}\sin\theta$ Official answer
(2) $\frac{2g}{3l}\sin\theta$
(3) $\frac{3g}{2l}\cos\theta$
(4) $\frac{2g}{3l}\cos\theta$

Official answer

Option 1

CBSE answer key (25/04/2017, used for result).

Chapter: System of Particles and Rotational Motion
Topic: Torque and Equilibrium
Idea tested: Torque about an Axis

JEE Main 2 April 2017 (offline), Physics Q6

Same idea in other shifts