Question

A rigid rod is placed against the wall as shown in figure. When its velocity of lower end is $10m/s$ and its base makes an angle $\alpha ={60}^o$ with horizontal, then the vertical velocity of its end $B$ will be:

• A. $10\sqrt{3}$
• B. $\frac{10}{\sqrt{3}}$
• C. $5\sqrt{3}$
• D. $\frac{5}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{10}{\sqrt{3}}$

let the $x$ intercept be $lcos\alpha$ and $y$ intercept be $lsin\alpha$

let the length of the string be $l$ which will always remain the same

hence the constraint is $(lcos\alpha {)}^2+(lsin\alpha {)}^2=l^2$

now the decrement in $x$ intercept is $10m$ in a sec so we can calculate $y$ intercept

$(lcos\alpha -10{)}^2+(lsin\alpha +y{)}^2=l^2$

$l^{2}co{s}^2\alpha +100-20lcos\alpha +l^{2}si{n}^2\alpha +y^2+2ylsin\alpha =l^2$

cancelling $l^2$ on both sides

$y^2+2ylsin\alpha +100-20lcos\alpha =0$

putting $\alpha ={60}^o$

$y^2+\sqrt{3}yl+100-10l=0$

roots for above equation are

$y=\frac{-\sqrt{3}l+-\sqrt{3l^2+40l-400}}{2}$

$y=\frac{-\sqrt{3}l+-\sqrt{(l+20)(3l-20)}}{2}$

for $l=\frac{20}{3}$ we get $y$ as

$y=\frac{\sqrt{3}\times \frac{20}{3}}{2}=\frac{10}{\sqrt{3}}$