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200 Omega resistor is connected in one of the gaps of the meter bridge. Series combination of X Omega and 50 Omega resistors is connected in the second gap. Here unknown resistance X Omegais kept in a heat bath at a certain temperature. The unknown resistance and its temperature is....... and.......respectively if the balance point is obtained at 50 cm. The total length of the wire of the meter bridge is equal to 1 meter. The resistance of the unknown resistance at 0" "^(@)Ctemperature is equal to 100 Omega . alpha = 0.5 xx 10^(-3) ""^(@) C^(-1) for the material of the X Omega resistors. |
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Answer» Solution :`R_(1)= 200 OMEGA, "" R_(2) = (X + 50) Omega` `l_(1) = 50 cm , "" l_(2) = 100 - 50 = 50 cm ` Here , we have`(R_(1))/(R_(2)) = (l_(1))/(l_(2))` `therefore (200)/(X + 50) = (50)/(50) ` `therefore X +50 = 200 ` `therefore X= 150Omega` Now, X = `X_(0) [ 1 + ALPHA (theta - 0 ) ] ` `therefore 150 = 100 [ 1 + 5 xx 10^(-3) theta ]` `therefore 1.5 = 1+ 5 xx 10^(-3) theta` `therefore theta = 100^(@) `C |
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