JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 11)
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 :
_2nd_September_Morning_Slot_en_11_1.png)
$$Mg\sqrt {1 - {{\left( {{{R - a} \over R}} \right)}^2}} $$
$$Mg\sqrt {1 - {{{a^2}} \over {{R^2}}}} $$
$$Mg{a \over R}$$
$$Mg\sqrt {{{\left( {{R \over {R - a}}} \right)}^2} - 1} $$
Explanation
_2nd_September_Morning_Slot_en_11_2.png)
to pull up, $$\tau $$F $$ \ge $$ $$\tau $$mg
$$ \Rightarrow $$ F $$ \times $$ R $$ \ge $$ Mg $$ \times $$ x
$$ \Rightarrow $$ Fmin = $${{Mg} \over R} \times \sqrt {{R^2} - {{\left( {R - a} \right)}^2}} $$
= $$Mg\sqrt {1 - {{\left( {{{R - a} \over R}} \right)}^2}} $$
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