NTSE Stage 2 SAT 2018 Exam Question Paper

Question : 38 of 100

Marks: +1, -0

Two nichrome wires A and B, each of length 5 cm and of radius 1 cm and 3 cm respectively are connected to each other in series. If a current of 5 A flows through the combination of wires, the ratio of potential difference across wire A to that across wire B will be:

$1:3$
$3:1$
$9:1$
$1:9$

Validate

Solution:

From Ohm’s law

{V}_{{A}}={IR}_{{A}}=\;\;{{I}ρℓ_{A}}/{π(r_{A})^{2}}(∵R=ρ\;\;{ℓ}/{A})

and

{V}_{{B}}=\;\;{ρℓ_{B}I}/{π(r_{B})^{2}}

\;\;{V_{A}}/{V_{B}}=\;\;{Iρℓ_{A}}/{π(r_{A})^{2}}×\;\;{π(r_{B})^{2}}/{Iρℓ_{B}}(_{{A}}=_{{B}}=5{cm})

=\;\;{(r_{B})^{2}}/{(r_{A})^{2}}=\;\;{9}/{1}({r}_{{B}}=3{cm}\&{r}_{{A}}=1{cm})

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