Question

Two wires of the same material and same length but diameters in the ratio 1 : 2 are stretched by the same force. The potential energy per unit volume of the two wires will be in the ratio

Answer 1

Answer:

Explanation:

According to the Young's modulus -  

Y =  the tensile stress or the compressive stress / tensile strain or compressive strain.  

Y=  F/A)/(dL /L)  

where F is the force per unit normal area and L is the original length.

Y= (F/A)/(dL/L)= (FL)/(A dL)…….(1)

Since, both wires are of the same material , therefore Y will be same for them. Also, the radii of the wires are same, so their normal cross sections will also be same in area,A.  Likewise, same stretching force F is also same for them.

Length L1:L2 = 1:2.

Stress = F/a

a = πr²

a ∝ r²

Strain in first wire = dL1/L1 = (F/A)/Y  

Strain in second = dL2/L2=(F/A)/Y

a1/a2 = (r1/r2)²=(2/1)²

= 4

S1/S2 = F/a1/ F/a2 = (a2/a1) = 1/4

Thus, the extensions in them will be in the ratio  1 : 4.

Answer 2

Answer:

Explanation:

According to the Young's modulus -

Y = the tensile stress or the compressive stress / tensile strain or compressive strain.

Y= F/A)/(dL /L)

where F is the force per unit normal area and L is the original length.

Y= (F/A)/(dL/L)= (FL)/(A dL)…….(1)

Since, both wires are of the same material , therefore Y will be same for them. Also, the radii of the wires are same, so their normal cross sections will also be same in area,A. Likewise, same stretching force F is also same for them.

Length L1:L2 = 1:2.

Stress = F/a

a = πr²

a ∝ r²

Strain in first wire = dL1/L1 = (F/A)/Y

Strain in second = dL2/L2=(F/A)/Y

a1/a2 = (r1/r2)²=(2/1)²

= 4

S1/S2 = F/a1/ F/a2 = (a2/a1) = 1/4

Thus, the extensions in them will be in the ratio 1 : 4.