A) 15 cm to the right
B) 15 cm to the left
C) 20 cm to the right
D) 20 cm to the left
Correct Answer: B
Solution :
(b) 15 cm to the left Using, \[\frac{{{R}_{1}}}{l}=\frac{{{R}_{2}}}{100-l}\] (1) Put \[{{R}_{1}}=20\text{ }\Omega \] and \[{{R}_{2}}=30\text{ }\Omega \] in eq. (1), we get \[\frac{20}{l}=\frac{30}{100-l}\] \[\Rightarrow 2000-20l=30l\] \[\Rightarrow 50l=2000\] \[\Rightarrow l=40\,\,cm\] Let the new balance point lies at a distance l Put \[{{R}_{1}}=10\,\Omega \left( \frac{20}{2} \right)\] and \[{{R}_{2}}=30\,\Omega \]in eq.(1),we get \[\frac{10}{l'}=\frac{30}{100-l'}\] \[\Rightarrow 1000-10l'=30l'\] \[\Rightarrow 40l'=1000\] \[\Rightarrow l'=25\,\,cm\] As \[l'<l\], so balance point shift towards left. Shift in balance point \[=l-l'=40-25=15\,cm\]
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