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Two diametrically opposite points of a metal ring are connected to two terminals of the left gap of a metre bridge. A resistance of 22 Omega is connected in the right gap. If the null point is obtained at 45 cm from the left end, find the resistance of the metal ring. |
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Answer» Solution :Data: `R= 22 Omega, L_(X) = 45 cm` `therefore L_(R) =100 - L_(X) = 100-45= 55 cm,` Let X be the resistance of the metal ring. `therefore` The resistance of each half of the metal ring `=X//2`. Therefore, the resistance in the left gap is the effective resisance of the parallel combination of `X//2` and `X//2`. `R_(p) = ((X//2)(X//2))/(X//2 + X//2) = (X^(2)//4)/X = X/4` [ OR `1/R_(P) = 1/(X//2) + 1/(X//2) = 2/(X//2) = 4/Xtherefore R_(P) = X/4`] At balance, `R_(P)/R = L_(X)/L_(R)therefore (X//4)/(22) = 45/55=9/11therefore X/4 = 18 Omega` `therefore X= 4 xx 18 = 72 Omega` The resistance of the metal ring is `72 Omega`. |
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