1thanks virang n ankit
a wire of length l and 3 identical cells of negligible internal resistances are connected in series .due to the current the temp of wire is raised by Δτ in time t .N number of similar cells is now connected in series with a wire of same material and cross section but of length 2l the temp of change by same amountΔτ in same t.value of N
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17 Answers
11The perfect Solution
Part 1
Q = 9t/R
Part 2
H = N2t/2R
Now since the mass is doubled therefore twice the energy
H = 2Q
N2/2 = 2*9
N2 =36
N = 6
Therefore the answer is 6
1hey what r u saying ... wats in .. the new potential difference is nv and the exterlnal resistance is twice the previous so it is 2R
.. please point out where is my error .. clearly
1hw d hell yaar
lenght double kara toh woh uska cross section area kam hoga ........
mass remains same
1heat is V2/r
Let total V in first case = 3
Resistance R ...
power generated = 92/R
new V is = n
resistance 2*R
9v2/R * 2 = n2/2R / /rate of heat generated should be 2times the previous because 2times new mass ...
n = √36 = 6 cell ..
11Guys i think you forgot to consider that when the length of wire is Doubled . Its Mass is also doubled . Therefore it will require twice the Energy.
Twice the energy and half the resistance . Therefore N = 6
Richa do you have the answer?
1heat is =v^2/R *tym
nw itz inversely proportional to R
so nw R is dirctly proportional to lenght(l) and inversely proportional to area (A)
nw u can equate d heats wich r same n u egt d value
sinc ethey r in series R will b NR