Question

If wire is stretched such that its length increases by 0.5% then find change in resistance

Answer 1

Answer:

0.5×10∧-2

Explanation:

R = kL/A

ΔR/R = ΔL/L - ΔA/A = 0.5×10∧-2

Answer 2

Answer

Δx = stretch in the length of wire = 0.5 cm = 0.005 m

m = mass attached to the wire = 5 kg

g = acceleration due to gravity = 9.8 m/s²

W = weight attached to the wire

weight of mass hanged is given as

W = mg

W = 5 x 9.8

W = 49 N

k = force constant of the wire

the spring force by the wire balances the weight here

hence  , Spring force = weight

k Δx = mg

k (0.005) = 49

k = 9800 N/m