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With an unknown resistance X i...

Updated On: 27-06-2022

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Solution : Data: R=30 Omega, L_(X) = 40 cm, <br> L_(R) = 100- L_(X)= 60 cm <br> (i) With the resistance X in the left gap and R= 30 Omega in the right gap <br> X/R= L_(X)/L_(R) <br> therefore The unknown resistance is, <br> X=R(L_(X))/L_(R) = 30 xx 40/60 = 20 Omega <br> (ii) When the resistance in each gap is increased by 15Omega, the resistance in the left gap is <br> X'=X +15=20+15 = 35 Omega, and the resistance in the right gap is <br> R'=R+ 15= 30 + 15 = 45 Omega <br> therefore X^(')/R^(') = L_(X)^(')/L_(R)^(') <br> therefore (L_(X)^('))/(100 - L_(X)^(')) = (35/45) = 7/9 <br> therefore 9 L_(X)^(') = 700- 7 L_(X)^(') therefore 16 L_(X)^(') = 700 <br> therefore L_(X)^(') = 700/16=43.75 cm <br> Since, L_(X)^(') gt L_(X), the null point shifts to the right by <br> L_(X)^(') - L_(X)= 43.75 cm - 40 cm = 3.75 cm
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