The resistivity of potentiometer wire is 40 × 10-8 ohm - metre and its area of cross-section is 8 × 10-6 m2. If 0.2 ampere current is flowing through the wire, the potential gradient of the wire is

Question

The resistivity of potentiometer wire is 40 × 10-8 ohm - metre and its area of cross-section is 8 × 10-6 m2. If 0.2 ampere current is flowing through the wire, the potential gradient of the wire is (A) 10-1 V /m  (B) 10-2 V /m  (C) 10-3 V /m  (D) 10-4 V /m

Tardigrade Admin · Accepted Answer

Given: ρ=40 × 10-8 ohm - m I=0.2 A A=8 × 10-6 m 2 Resistance of the wire R=ρ (L/A) Potential difference across the wire V = IR = I ρ ( L / A ) Thus potential gradient of the wire ( V / L )=( I ρ/ A ) ∴ ( V / L )=(0.2 × 40 × 10-8/8 × 10-6) ⇒ ( V / L )=10-2 V / m

Q. The resistivity of potentiometer wire is $40 \times 1 0^{- 8}$ ohm - metre and its area of cross-section is $8 \times 1 0^{- 6} m^{2}$ . If $0.2$ ampere current is flowing through the wire, the potential gradient of the wire is

$1 0^{- 1} V / m$

$1 0^{- 2} V / m$

$1 0^{- 3} V / m$

$1 0^{- 4} V / m$

Q. The resistivity of potentiometer wire is 40×10−8 ohm - metre and its area of cross-section is 8×10−6m2. If 0.2 ampere current is flowing through the wire, the potential gradient of the wire is

10−1V/m

10−2V/m

10−3V/m

10−4V/m

Given : ρ=40×10−8 ohm −mI=0.2AA=8×10−6m2 Resistance of the wire R=ρAL​ Potential difference across the wire V=IR=IρAL​ Thus potential gradient of the wire LV​=AIρ​ ∴LV​=8×10−60.2×40×10−8​ ⇒LV​=10−2V/m

Q. The resistivity of potentiometer wire is $40 \times 1 0^{- 8}$ ohm - metre and its area of cross-section is $8 \times 1 0^{- 6} m^{2}$ . If $0.2$ ampere current is flowing through the wire, the potential gradient of the wire is

$1 0^{- 1} V / m$

$1 0^{- 2} V / m$

$1 0^{- 3} V / m$

$1 0^{- 4} V / m$