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Current Electricity question

2010 · Q140
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Current Electricity question

2010 · Q140

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at length lll1 cm and lll2 cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to

AIPMT 2010 Prelims Physics - Current Electricity Question 80 English
  1. A
    k(l2−l1)k({l_2} - {l_1})k(l2​−l1​) and kl2k{l_2}kl2​
  2. B
    kl1k{l_1}kl1​ and k(l2−l1)k({l_2} - {l_1})k(l2​−l1​)
  3. C
    k(l2−l1)k\left( {{l_2} - {l_1}} \right)k(l2​−l1​) and kl1k{l_1}kl1​
  4. D
    kl1k{l_1}kl1​ and kl2k{l_2}kl2​
View written solutionFree

Correct answer: B

(i) When key between the terminals 1 and 2 is plugged in,

P.D. across R = IR = kl1k l_1kl1​

⇒\Rightarrow⇒ R =kl1 k l_1kl1​ as I = 1A

(ii) When key between terminals 1 and 3 is plugged in,

P.D. across (X + R) = I(X + R) = kl2k l_2kl2​

⇒\Rightarrow⇒ X + R = kl2k l_2kl2​

∴\therefore∴ X = k(l2–l1)k (l_2 – l_1)k(l2​–l1​)

∴\therefore∴ R = kl1kl_1kl1​ and X = k(l2–l1)k (l_2 – l_1)k(l2​–l1​)

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