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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?



We know that rate of of flow dq/dt is directly propertional to Area ,Temperature difference,and inversly proportional to length of wire so the eqation is⇆/dq/dt = KA(t1 -t2)/l
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

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