### What should be the lengths of an aluminium and a brass rods at STP.so that brass rod is longer than the aluminium rod by 4cm at all conditions of temperature.(alpha alumn 24exp -6 brass 20 expo -6c

**Data :**

Let

Length of aluminum = La

Length of brass = L

_{b}=La + 0.04 (In meters)
α for aluminum= 24x10

^{-6 }c^{-1}
α for brass= 20 x10

^{-6}c^{-1}
Lengths at 0

^{0}C La’ = ? and L_{b}’=?**Solution:**

Length of aluminum is given by

La’=La ( 1 + α Δt)----------------1

L

_{b}’=L_{b}( 1 + α Δt)-----------------2
Subtracting eq(1) from eq(2), we get

La’ - L

_{b}’ =La ( 1 + α Δt) - L_{b}( 1 + α Δt)
= La + La
α Δt) - L

_{b}+ L_{b}α Δt
=(La –
Lb )+ Δt (La α - L

_{b}α)
0.04 =0.04 + Δt
(La α - L

_{b}α)
So Δt (La α - L

_{b}α) =0
Since La α = L

_{b}α
La x 24x10

^{-6}= L_{b}x20 x10^{-6}
La =(20 /24) x Lb =20 Lb / 24 ------------3

Since L

_{b}=La + 0.04
Putting value of La from eq (3) , we get

Lb =(20 Lb / 24) +0.04
or 24 x (Lb -0.04)=20 Lb

24 Lb – 20 Lb =0.96
or Lb= 0.96/4 =

**Lb=0.24 m Answer**
Now La= 0.24 -0.04 =

**0.02 m Answer**

My brother recommended I would possibly like

ReplyDeletethis website. He was entirely right. This put up truly made my day.

You can not imagine simply how so much time I had spent for

this information! Thank you!

Look at my web-site - transformers age of extinction hack tool