TG EAMCET 2 May 2025 Shift 2 Paper

The mass of a wire is m=ρAL
Since, the material is the same, the density ρ is constant for both wires.
‌

mA

mB

‌=‌

ρAALA

ρABLB

⇒‌

mA

mB

=‌

AALA

ABLB

‌

2

3

‌=‌

1

2

×‌

LA

LB

‌

LA

LB

‌=‌

2

3

×2=‌

4

3

The ratio of elongations is
‌

∆LA

∆LB

=‌

‌ FLA AAY

FLB ABY

=‌

LA

AA

×‌

AB

LB

Substitute ‌

LA

LB

=‌

4

3

and ‌

AB

AA

=‌

2

1

‌

∆LA

∆LB

=‌

4

3

×‌

2

1

=‌

8

3

The ratio of the elongations of the wires A and B is ‌

8

3

.