TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Given that, resistance of wire, R1=40Ω
Length of wire, l=10m
Resistance of resistance box =R Potential gradient of wire,
=‌

dV

dx

=0.1mV∕cm=‌

0.1×10−3

10−2

V∕m=0.01V∕m
Let V1 and V2 be the voltage drop across R1 and R.
Then, V1=IK=‌

R1

R1+R2

V
10×0.01=(‌

40

40+R

)2⇒R=760Ω