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 = Lb =La + 0.04 (In meters)
α for aluminum= 24x10 -6 c-1
α for brass= 20 x10 -6 c-1
Lengths at 0 0C La’ = ? and Lb’=?

Solution:

 Length of aluminum is given by
La’=La ( 1 + α Δt)----------------1
Lb’=Lb ( 1 + α Δt)-----------------2
Subtracting eq(1) from eq(2), we get
La’ - Lb’ =La ( 1 + α Δt) - Lb ( 1 + α Δt)
              = La + La α Δt) - Lb + Lb α Δt
              =(La – Lb )+ Δt (La α - Lb α)
       0.04 =0.04 + Δt (La α - Lb α)
So Δt (La α - Lb α) =0
Since La α = Lb α
La x 24x10 -6 = Lb x20 x10 -6
La =(20 /24) x Lb =20 Lb / 24 ------------3
Since Lb =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
 

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