Two wires A and B are of the same materials. Their lengths are in the ratio 1 :2 and diameters are in the ratio 2 :1 . When stretched by force F_A and…

Question

Two wires A and B are of the same materials. Their lengths are in the ratio 1 :2 and diameters are in the ratio 2 :1 . When stretched by force F_A and F_B respectively, they get an equal increase in their lengths. Then the ratio F_AF_B should be

MARKS Content Team · Accepted Answer

Correct answer: (D) 8 :1. Force F=(YA∆l)/l Substituting area A=πd^22=(πd^2)/4 So, we have F=(Yπd^2∆l)/4l F∝(d^2∆l)/l ∵(Yπ)/4=constant Hence, using equation (i), we have F_A∝d/(A2∆l_Al_A) ...(i) F_B∝d/(B2∆l_Bl_B) ...(ii) Form equations (i) and (ii), we get &there4;F_A/F_B=d/(A2∆l_Al_A)×l_B/(dB2∆l_B) As ∆l_A= ∆l_B=∆l Therefore F_A/F_B= d/A2dB2×l_B/l_A Putting d_A/d_B=2/1 And l_B/l_A=2/1 We get F_A :F_B=8 :1

Two wires $A$ and $B$ are of the same materials. Their lengths are in the ratio $1 : 2$ and diameters are in the ratio $2 : 1$ . When stretched by force $F_{A}$ and $F_{B}$ respectively, they get an equal increase in their lengths. Then the ratio $\frac{F_{A}}{F_{B}}$ should be

Solution

More Mechanical Properties of Solids Questions

Two wires A and B are of the same materials. Their lengths are in the ratio 1 :2 and diameters are in the ratio 2 :1 . When stretched by force FA and FB respectively, they get an equal increase in their lengths. Then the ratio FAFB should be

Solution

More Mechanical Properties of Solids Questions

Two wires $A$ and $B$ are of the same materials. Their lengths are in the ratio $1 : 2$ and diameters are in the ratio $2 : 1$ . When stretched by force $F_{A}$ and $F_{B}$ respectively, they get an equal increase in their lengths. Then the ratio $\frac{F_{A}}{F_{B}}$ should be