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NEET-XII-Physics
neet 2020 questions with solutions year:2020
- #44A resistance wire connected in the left gap of a metre bridge balances a `` 10\, \Omega `` resistance in the right gap at a point which divides the bridge wire in the ratio `` 3 : 2 `` . If the length of the resistance wire is `` 1.5\,m `` , then the length of 1 `` \Omega `` of the resistance wire is:
(A) `` 1.0 \times 10^{-2} m ``
(B) `` 1.0 \times 10^{-1} m ``
(C) `` 1.5 \times 10^{-1} m ``
(D) `` 1.5 \times 10^{-2} m ``
digAnsr: BAns :
Initially, `` \frac{P}{10}=\frac{l_{1}}{l_{2}}=\frac{3}{2} ``
`` \Rightarrow P=\frac{30}{2}=15\Omega ``
Now resistance, `` R=\frac{\rho\,l}{A} ``
`` \frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}} `` : Length of `` 15\,\Omega `` resistance wire is 1.5 m
`` \Rightarrow \frac{15}{1}=\frac{1.5}{l_{2}} ``
`` \Rightarrow l_{2}=0.1\,m ``
`` =1.0 \times 10^{-1}\,m ``
`` \therefore `` Length of `` 1\,\Omega `` resistance wire is `` 1.0\times 10^{-1}\,m ``