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 ⁰C 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