Physics Circle home Student Login

Home›Chapters›Current Electricity PYQs›Ohm's Law, Resistivity & Temperature›Question 30

JEE Main 2022 (29 Jun, Shift 1) · Ohm's Law, Resistivity & Temperature

Ohm's Law, Resistivity & Temperature question from JEE Main 2022 (29 Jun, Shift 1)

Q30JEE Main 2022 · 29 JUN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature

The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x ×\times 10−-3 Ω\Omega cm. The value of x is ____________. [Take π\pi = 3.14]

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 144

  1. Slope of VV–II graph =R=tan⁡45∘=1 Ω= R = \tan45^\circ = 1\ \Omega.
  2. A=πd2/4=3.14×(2.4)2/4=4.5216 cm2A = \pi d^2/4 = 3.14\times(2.4)^2/4 = 4.5216\ \text{cm}^2.
  3. ρ=RA/l=4.5216/31.4=0.144 Ω cm=144×10−3 Ω cm\rho = RA/l = 4.5216/31.4 = 0.144\ \Omega\,\text{cm} = 144\times10^{-3}\ \Omega\,\text{cm}
  4. So x=144x = 144.
Report a mistake

More questions on Ohm's Law, Resistivity & Temperature