EAMCET Engineering 2009 Solved Paper

Free body diagram of the two blocks areGiven, ‌‌‌

l1

l2

=a,‌

r1

r2

=b,‌

Y1

Y2

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

F1⋅l1

A1⋅∆l1

. . . (i)
and Y2=‌

F2⋅l2

A2∆l2

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

Y1

Y2

=‌‌

‌ F1⋅l1 A1⋅∆l1

F2⋅l2 A2⋅∆l2

or ‌

Y1

Y2

‌=‌

F1⋅A2⋅l1⋅∆l2

F2⋅A1⋅l2⋅∆l1

. . . (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

]⋅(‌

∆l2

∆l1

)
or ‌=‌

1

2

(‌

1

b2

)⋅a(‌

∆l2

∆l1

)
or ‌

∆l1

∆l2

‌=(‌

a

2b2c

)