1
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
Consider two uniform discs of the same thickness and different radii R1 = R and
R2 = $$\alpha$$R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of $$\alpha$$ is :
A
$$\sqrt 2$$
B
2
C
$$2\sqrt 2$$
D
4
2
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :
A
$${1 \over 2}$$
B
$${1 \over 4}$$
C
$${1 \over 8}$$
D
$${2 \over 3}$$
3
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
$$\cos \theta = {g \over {l{\omega ^2}}}$$
4
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
$${{2 \over 3}}$$
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