Question

Two rods, one made of brass and the other of lead, have the same dimensions. When a particular stress is applied to the brass rod, it stretches by 0.18 mm. How much does the lead rod stretch under the same stress? Use a value of 9.0×10 10 Pa for the Young’s modulus of brass and use a value of 1.6×10 10 Pa for the Young’s modulus of lead.

Answer 1

Explanation:

The diameter of a brass rod is 4 mm and Young's modulus of brass is 9×10

10

N/m

2

. The force required to stretch by 0.1% of its length is :-

ANSWER

ANSWERY=

ANSWERY= AΔl

ANSWERY= AΔlFL

ANSWERY= AΔlFL

ANSWERY= AΔlFL

ANSWERY= AΔlFL F=

ANSWERY= AΔlFL F= L

ANSWERY= AΔlFL F= LYΔlA

ANSWERY= AΔlFL F= LYΔlA

ANSWERY= AΔlFL F= LYΔlA =

ANSWERY= AΔlFL F= LYΔlA = L

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×(

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 2

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3 )

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3 ) 2

ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3 ) 2 =360π

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