1.

A rod of leng th 'l' rests at a point A against a smooth vertial wall while end B is on the floor as shown in the fig. If the end Amoves uniformly downwards, what will be the velocity of the end B if x is the distance of point B from wall.

Answer»

`v_(B)=sqrt(l^(2)/x^(2) -1.v_(a))`
`sqrt(x^(2)/l^(2)-1.v_(a))`
`(l^(2)-x^(2))/x.v_(a)`
NONE of these

Solution :Let the position of POINT B be x at any INSTANT such that dx OB = x and OA = y. The velocity of B. `v_(b)=(dx)/(dt)` is positive along x-axis while velocity of A `v_(a) =(dy)/(dt)`is constant and is down wards. There is a friction between rod floor at B. Now
`x^(2)+y^(2)=l^(2)`
or `2xdx/dt+2y(dy)/(dt)=0`
or `(dx)/(dt)=v=-y//x((dy)/(dt))=y//x.v_(a)`
As `v_(a)` is constant magnitude of the given DOWNWARD velocity as .y. decreases with time.
Then `v_(b)=y//x.v_(a)`
But `y=sqrt(l^(2)-x^(2))`
`:.v_(b)=sqrt((l^(2)-x^(2))/x).v_(a)`


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