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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

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⇆/ 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