1
JEE Main 2020 (Online) 3rd September Evening Slot
+4
-1
A uniform rod of length ‘$$l$$’ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $$\omega$$ the rod makes an angle $$\theta$$ with it (see figure). To find $$\theta$$ equate the rate of change of angular momentum (direction going into the paper) $${{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta$$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of $$\theta$$ is then such that :
A
$$\cos \theta = {{2g} \over {3l{\omega ^2}}}$$
B
$$\cos \theta = {{3g} \over {2l{\omega ^2}}}$$
C
$$\cos \theta = {g \over {2l{\omega ^2}}}$$
D
$$\cos \theta = {g \over {l{\omega ^2}}}$$
2
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is
I = $$M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$$. If such a cylinder is to be made for a given mass of a material, the ratio $${L \over R}$$ for it to have minimum possible I is
A
$${3 \over 2}$$
B
$$\sqrt {{3 \over 2}}$$
C
$$\sqrt {{2 \over 3}}$$
D
$${{2 \over 3}}$$
3
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are 0.1 kg-m2 and 10 rad s–1 respectively while those for the second one are 0.2 kg-m2 and 5 rad s–1 respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is :
A
$${{20} \over 3}J$$
B
$${{5} \over 3}J$$
C
$${{10} \over 3}J$$
D
$${{2} \over 3}J$$
4
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its centre ‘O’ perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is :
A
$$Mg\sqrt {1 - {{\left( {{{R - a} \over R}} \right)}^2}}$$
B
$$Mg\sqrt {1 - {{{a^2}} \over {{R^2}}}}$$
C
$$Mg{a \over R}$$
D
$$Mg\sqrt {{{\left( {{R \over {R - a}}} \right)}^2} - 1}$$
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