Question

Two pieces of wire A & B of same material have their lengths in the ratio 1:2 and their
diameters in the ratio 2:1, if they are stretched by same force, their elongation will be in
the ratio of
(A) 8:1
(B) 2:1
19 1:8/1 (D) 1:4
b​

Answer 1

Answer:

$\sqrt{124}$

Answer 2

Answer:

8:1

Explanation:

Using Hooke's law                  Y=  

ΔL/L

F/A

​  

       where    A=  

4

πD  

2

 

​  

 

⟹     F=Y  

4

πD  

2

 

​  

 

L

ΔL

​  

 

As  Y and  ΔL are constant, thus         F∝  

L

D  

2

 

​  

 

⟹      

F  

B

​  

 

F  

A

​  

 

​  

=  

D  

B

2

​  

 

D  

A

2

​  

 

​  

×  

L  

A

​  

 

L  

B

​  

 

​  

 

Given :           D  

A

​  

:D  

B

​  

=2:1     and     L  

A

​  

:L  

B

​  

=1:2

∴          

F  

B

​  

 

F  

A

​  

 

​  

=  

1

4

​  

×  

 1=  1

2   8

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