JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 4)
A $$\sqrt {34} $$ m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If Ef and Fw are the reaction forces of the floor and the wall, then ratio of $${F_w}/{F_f}$$ will be :
(Use g = 10 m/s2.)
_28th_June_Evening_Shift_en_4_1.png)
$${6 \over {\sqrt {110} }}$$
$${3 \over {\sqrt {113} }}$$
$${3 \over {\sqrt {109} }}$$
$${2 \over {\sqrt {109} }}$$
Explanation
_28th_June_Evening_Shift_en_4_2.png)
Taking torque from B
$${F_w} \times 5 = {3 \over 2}mg$$
$$ \Rightarrow {F_w} = {3 \over {10}} \times 10 \times 10$$
$$ = 30\,N$$
$$N = mg = 100\,N$$
and $${f_r} = {F_w} = 30\,N$$
so $${F_f} = \sqrt {{N^2} + f_r^2} = \sqrt {10900} = 10\sqrt {109} \,N$$
so $${{{F_w}} \over {{F_f}}} = {3 \over {\sqrt {109} }}$$
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