In given potentiometer, e.m.f of primary cell is 40 V and its internal resistance is 2 Ω resistance of wire AB is 4 Ω. In primary circuit, there is also a rheostat whose resistance varies as Rh = (2x + 2)Ω. If balance length obtained for x = 0 is l1 and x = 2 is l2 respectively, then l1l2.
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a
32
b
34
c
23
d
43
answer is C.
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Detailed Solution
Let resistance per unit length of wire‘AB’ be kCase : (i)x = 0 → balance length: l1Rh=2Ωi=404+2+2⇒i=5A∴ At balance lengthe = 5kl1 …(1)Case : (ii)x = 2 → balance length l2Rh=6Ωi=404+2+6⇒i=103A∴ At balance length,e = (10/3)kl2 …. (2)From (1) and (2)5xl1 = (10/3)kl2⇒l1l2=23
In given potentiometer, e.m.f of primary cell is 40 V and its internal resistance is 2 Ω resistance of wire AB is 4 Ω. In primary circuit, there is also a rheostat whose resistance varies as Rh = (2x + 2)Ω. If balance length obtained for x = 0 is l1 and x = 2 is l2 respectively, then l1l2.