JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 13)
A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate
25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the
horizontal ground when the top of the ladder is 1 m above the ground is :
$${{25} \over 3}$$
25
25$$\sqrt 3 $$
$${{25} \over {\sqrt 3 }}$$
Explanation
_12th_April_Morning_Slot_en_13_1.png)
x2 + y2 = 4 $$ \Rightarrow $$ $$2x{{dx} \over {dt}} + 2y{{dy} \over {dt}} = 0$$
$$ \Rightarrow {{dx} \over {dt}} = - {y \over x}.{{dy} \over {dt}}$$
When upper end is 1m above the ground, $${{dx} \over {dt}} = - {1 \over {\sqrt 3 }}.25 = - {{25} \over {\sqrt 3 }}$$ cm/sec.
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