Question

A $10$ cm long rod $AB$ moves with its ends on two mutually perpendicular straight lines $OX$ and $OY$. If the end $A$ be moving at the rate of $2$ cm/sec, then when the distance of $A$ from $O$ is $8$cm, the rate at which the end $B$ is moving, is

• A. $\frac{8}{3}$cm / sec
• B. $\frac{4}{3}$cm / sec
• C. $\frac{2}{9}$cm / sec
• D. None of these

Answer 1

The correct option is A

$\frac{8}{3}$cm / sec

Explanation for the correct option:

Step1. Draw the diagram:

Here it is given that the rod $AB$ moves with its ends on two mutually perpendicular straight lines $OX$ and $OY$

Let the end $A$ of rod move on line $OX$ with respect to time $t$ and end $B$ of rod move on line $OY$ with respect to time $t$ .

Let the distance from point $A$ to point $O$ be $x$ and the distance from point $B$ to point $O$ be $y$.

Step2. To find the rate at which the end $B$ is moving i.e. $\frac{dy}{dt}$:

From the figure, we know that ΔBOA$\mathrm{\Delta }BOA$ is a right-angled triangle.

Therefore, from Pythagoras theorem we get,

$x^2+y^2=100.....\left(1\right)$

Differentiating on both side we will get,

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.....\left(2\right)$

It is given that end $A$ is moving at the rate of $2$ ${\displaystyle \frac{cm}{s}}$ i.e. ${\displaystyle \frac{dx}{dt}}=2$ cm

and the distance of $A$ from $O$ is $8$cm i.e. $x=8$cm

Substituting the value of $x$ in equation )$\left(1\right)$ , we get $8^2+y^2={10}^2$

$\Rightarrow y^2=100-64$

$\Rightarrow y=\sqrt{36}$
So, $y=6$ cm

Hence the distance from point $B$ to point $O$ is $6$cm.

From equation $\left(1\right)$ and $\left(2\right)$,

$$\begin{gathered} \frac{dy}{dt}=-\frac{16}{6} \\ =-\frac{8}{3} \end{gathered}$$

Thus, the rate at which B is moving is $\frac{8}{3}$cm / sec

Hence, Option(A) is the correct answer.