VITEEE 2009 Paper

Given,

l1

l2

=a,

r1

r2

=b,

Y1

Y2

=c
Free body diagram of the two blocks brass and steel are

Let young's modulus of steel is Y1 and of brass is Y2.
∴ Y1=

F1.l1

A1.Δl1

...(i)
and Y2=

F2.l2

A2.Δl2

...(ii)
Dividing Eq.(i) by (ii),

Y1

Y2

=

F1.l1 A1.Δl1

F2.l2 A2.Δl2

or ‌‌

Y1

Y2

=

F1.A2.l1.Δl2

F2.A1.l2.Δl1

...(iii)
Force on steel wire from free body diagram
T=F1=(2g) newton
Force on brass wire from free body diagram
F2=T′=T+2g=(4g) newton
Now, putting the value of F1,F2 in Eq. (iii), we get
‌‌

Y1

Y2

=(

2g

4g

).(

πr22

πr12

).[

l1

l2

].(

Δl2

Δl1

)
or ‌‌c=

1

2

(

1

b2

).a(

Δl2

Δl1

)
or ‌‌

Δl1

Δl2

=(

a

2b2c

)