In a Wheatstone bridge (see fige ) Resitances P and Q are approximately equal .When `R= 400 Omega` the bridge is blanced .on is inerchanging P and Q ,the value of R, for balance is 405 `Omega ` .The value of X is close to :
A
401.5 ohm
B
402.5 ohm
C
403.5 ohm
D
404.5 ohm
Text Solution
Verified by Experts
The correct Answer is:
B
Let the ratio `P/Q` be `1+delta` Where `delta` < < 1 So, `400/x = 1+ delta` and `405/x = 1/(1+delta) 1-delta` Adding `400/x + 405/x =2 ` Solving x= 402.5 `Omega`
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VMC MODULES ENGLISH-JEE Main Revision Test-6 | JEE-2020 -PHYSICS
