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## Mechanical Properties of Solids

Question : 1

Total: 21

**Solution:**

Here, for steel wire,

Area of cross-section,A 1 = 3.0 × 10 − 5 m 2

Stretching,Δ l 1 = Δ l (say)

Stretching force on steel,F 1 = F

For copper wire, length of wire,l 2 = 3.5 m

Area of cross-section,A 2 = 4.0 × 10 − 5 m 2

Stretching,Δ l 2 = Δ l (given);

Stretching force on copper,F 2 = F

LetY 1 and Y 2 be the Young's modulus of steel and copper wire respectively.

∴ Y 1 =

=

. . . ( i ) and Y 2 =

=

=

. . . ( i i )

Dividing (i) by (ii), we get

=

×

=

= 1.79 = 1.8

⇒ Y 1 : Y 2 = 1.8 : 1

Area of cross-section,

Stretching,

Stretching force on steel,

For copper wire, length of wire,

Area of cross-section,

Stretching,

Stretching force on copper,

Let

Dividing (i) by (ii), we get

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