Resistance of 100 cm long potentiometer wire is 10 Ω. it is connected to a battery of 2 volt and resistance R is in series. A source of 10 mV gives null point at 40cm length, then external resistance R is

Question

Resistance of 100 cm long potentiometer wire is 10 Ω. it is connected to a battery of 2 volt and resistance R is in series. A source of 10 mV gives null point at 40cm length, then external resistance R is (A) 490 Ω (B) 790 Ω (C) 590Ω (D) 990Ω

Tardigrade Admin · Accepted Answer

V=((E/R+Rn+r)) (r/L) × ℓ ⇒ 10 × 10-3 =(2/(10+R+0)) × (10/1) × 0.4 R =790 Ω

Q. Resistance of $100$ cm long potentiometer wire is $10Ω$ . it is connected to a battery of 2 volt and resistance R is in series. A source of 10 mV gives null point at 40cm length, then external resistance R is

$490Ω$

$790Ω$

$590Ω$

$990Ω$

$V = (\frac{E}{R + R _{n} + r}) \frac{r}{L} \times ℓ$
$\Rightarrow 10 \times 1 0^{- 3}$
$= \frac{2}{( 10 + R + 0 )} \times \frac{10}{1} \times 0.4$
$R = 790Ω$

Q. Resistance of 100 cm long potentiometer wire is 10Ω. it is connected to a battery of 2 volt and resistance R is in series. A source of 10 mV gives null point at 40cm length, then external resistance R is

490Ω

790Ω

590Ω

990Ω