# When a wire is drawn until its radius decrease by 3%. then the percentage of increase in resistance is? (ans -12%) how 2 proceed with the sum?

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above eqn can be also written dq/dt = (t1-t2)÷ l/Ka.(the L/KA is known as Resistance).

Now according to above ques

radius is dec by 3 %.But be know that mass remains constant .let r be radius,l be initial length be length and d be density and x we new lehngt we can write

d*volume intial = d *volume final

volume initial = volume final

we can write

pi r² . l = pi (97r/100)² . x (since R decreased by 3%)

therfore x = (100/97)² .l

now resistance = L/KA

initial resistance = l÷(K × pi r²)

new resitance =l×(100/97)² ÷ (K× pi (97r/100)²)

now new percentage can be cacluted by (new resistance ×100)÷initial resistance

which you will get 112 approx

therefor % increased = 112-100 =12

mark it as brainliest answer