A 10m long potentiometer wire has a resistance of 20 ohm . It is connected in series with a battery of emf 3 V and a resistance of 10 Ω . The internal resistance of cell is negligible. If the length can be read accurately up to 1 mm , the potentiometer can read voltage with an accuracy of (k/10) (in milli volt). Then the value of k is

Question

A 10m long potentiometer wire has a resistance of 20 ohm . It is connected in series with a battery of emf 3 V and a resistance of 10 Ω . The internal resistance of cell is negligible. If the length can be read accurately up to 1 mm , the potentiometer can read voltage with an accuracy of (k/10) (in milli volt). Then the value of k is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-7ra07jbdjhsj7lfb.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Current in wire =(3/30)=0.1A V=IR=2 V maximum voltage can read =2 V λ =(V/l)=(2/10 × 103)=0.2 m Vmm- 1 least account of scale =1 mm accuracy =0.2 mV

Current in wire =303​=0.1A V=IR=2V maximum voltage can read =2V λ=lV​=10×1032​=0.2mVmm−1 least account of scale =1mm accuracy =0.2mV

Current in wire $= \frac{3}{30} = 0.1 A$
$V = I R = 2 V$
maximum voltage can read $= 2 V$
$λ = \frac{V}{l} = \frac{2}{10 \times 1 0 ^{3}} = 0.2 m Vm m^{- 1}$
least account of scale $= 1 mm$
accuracy $= 0.2 mV$