Trigonometrical Ratios And Identities Ques 3
- A $2 \mathrm{m}$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $25$ $\mathrm{cm} / \mathrm{s}$, then the rate (in $\mathrm{cm} / \mathrm{s}$ ) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $1 \mathrm{m}$ above the ground is
(2019 Main, 12 April I)
(a) $25 \sqrt{3}$
(b) $\frac{25}{\sqrt{3}}$
(c) $\frac{25}{3}$
(d) $25$
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Answer:
Correct Answer: 3.(b)
Solution: (b) Given a ladder of length $l=2 \mathrm{m}$ leans against a vertical wall.
Now, the top of ladder begins to slide down the wall at the rate $25 \mathrm{cm} / \mathrm{s}$.
Let the rate at which bottom of the ladder slides away from the wall on the horizontal ground is $\frac{d x}{d t} \mathrm{cm} / \mathrm{s}$.
$ \because \quad x^2+y^2=l^2 \quad $ [by Pythagoras theorem]
$\Rightarrow \quad x^2+y^2=4$ $\quad [\because l=2 \mathrm{m}]$ $\quad$ ……..(i)
On differentiating both sides of Eq. (i) w.r.t. ’ $t$ ‘, we get
$2 x \frac{d x}{d t}+2 y \frac{d y}{d t} =0 $
$\frac{d x}{d t} = - \left(\frac{y}{x}\right) \frac{d y}{d t}$ $\quad$ ……..(ii)
From Eq. (i), when $y=1 \mathrm{m}$, then
$ x^2+1^2=4 \Rightarrow x^2=3 \Rightarrow x=\sqrt{3} \mathrm{m}\quad \quad[\because x>0] $
On substituting $x=\sqrt{3} \mathrm{m}$ and $y=1 \mathrm{m}$ in Eq. (ii), we get
$ \begin{array}{rlr} \frac{d x}{d t} & =-\frac{1}{\sqrt{3}}\left(-\frac{25}{100}\right) \mathrm{m} / \mathrm{s} & {\left[\text { given } \frac{d y}{d t}=-25 \mathrm{cm} / \mathrm{sec}\right]} \\ & =\frac{25}{\sqrt{3}} \mathrm{cm} / \mathrm{s} & \end{array} $
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