### Neet Solved Paper 2015 Question 10

##### Question: A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

#### Options:

A) $ \frac{wx}{d} $

B) $ \frac{wd}{x} $

C) $ \frac{w(d-x)}{x} $

D) $ \frac{w(d-x)}{d} $

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

As the weight w balances the normal reactions. So, $ w=N _1+N _2 $ ..(i)

Now balancing torque about the COM, i.e. anti-clockwise momentum = clockwise momentum

$ \Rightarrow ,N _1x=N _2(d-x) $ Putting the value of $ N _2 $ from Eq. (i),

we get $ N _1x=(w-N _1)(d-x) $

$ \Rightarrow N _1x=wd-wx-N _1d+N _1x $

$ \Rightarrow ,N _1d=w(d-x) $

$ \Rightarrow N _1=\frac{w(d-x)}{d} $