Q87 / 180 · COMEDK 2015 · Mathematics
A ladders $5 \mathrm{~m}$ long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall, at the rate of $2 \mathrm{~m} / \mathrm{sec}$. The speed at which its height on the wall decreases when the foot of the ladder is $4 \mathrm{~m}$ away from the wall is
- A$\frac{3}{8} \mathrm{~m} / \mathrm{sec}$
- Correct answerB$\frac{8}{3} \mathrm{~m} / \mathrm{sec}$
- C$\frac{5}{3} \mathrm{~m} / \mathrm{sec}$
- D$\frac{2}{3} \mathrm{~m} / \mathrm{sec}$
Solution
Let $A B$ be the ladder of length $5 \mathrm{~m}$. We are given, $\frac{d x}{d t}=2 \mathrm{~m} / \mathrm{sec}$
In $\triangle A B C$ $$ \begin{aligned} & A B^{2}=A C^{2}+B C^{2} \\ \Rightarrow \quad &(5)^{2}=x^{2}+h^{2} \quad \text{...(i)} \end{aligned} $$ Differentiating Eq. (i) w.r.t. $t$, we get $$ \begin{aligned} 0 &=2 x \frac{d x}{d t}+2 h \frac{d h}{d t} \\ \Rightarrow \frac{d h}{d t} &=\frac{-x}{h} \frac{d x}{d t} \quad \text{...(ii)} \end{aligned} $$ Now, from Eq. (i), when $x=4$ $\quad h^{2}=25-16$ $\Rightarrow h^{2}=9 \Rightarrow h=3 \mathrm{~m}$ From Eq. (ii), we get $\left[\frac{d h}{d t}\right]=\frac{-4}{3} \times 2=\frac{-8}{3}$ $\because$ The negative sign shows the height decreases and decreasing rate is $\frac{8}{3} \mathrm{~m} / \mathrm{sec}$.
In $\triangle A B C$ $$ \begin{aligned} & A B^{2}=A C^{2}+B C^{2} \\ \Rightarrow \quad &(5)^{2}=x^{2}+h^{2} \quad \text{...(i)} \end{aligned} $$ Differentiating Eq. (i) w.r.t. $t$, we get $$ \begin{aligned} 0 &=2 x \frac{d x}{d t}+2 h \frac{d h}{d t} \\ \Rightarrow \frac{d h}{d t} &=\frac{-x}{h} \frac{d x}{d t} \quad \text{...(ii)} \end{aligned} $$ Now, from Eq. (i), when $x=4$ $\quad h^{2}=25-16$ $\Rightarrow h^{2}=9 \Rightarrow h=3 \mathrm{~m}$ From Eq. (ii), we get $\left[\frac{d h}{d t}\right]=\frac{-4}{3} \times 2=\frac{-8}{3}$ $\because$ The negative sign shows the height decreases and decreasing rate is $\frac{8}{3} \mathrm{~m} / \mathrm{sec}$.
Hence, the correct answer is option: 'B'.