JEE Mains · Physics · STD 12 - 3. current electricity
On interchanging the resistances, the balance point of a meter bridge shifts to the left by \(10\ cm\). The resistance of their series combination is \(1\ k\Omega\). How much was the resistance on the left slot before interchanging the resistances? .................. \(\Omega\)
- A \(505 \)
- B \(550\)
- C \(910\)
- D \(990\)
Answer & Solution
Correct Answer
(B) \(550\)
Step-by-step Solution
Detailed explanation
\(\mathrm{R}_{1}+\mathrm{R}_{2}=1000\) \(\Rightarrow \mathrm{R}_{2}=1000-\mathrm{R}_{1}\) On balancing condition \(\mathrm{R}_{1}(100-I)=\left(1000-\mathrm{R}_{1}\right) l\) ...\((i)\) On Interchaanging resistance balance point shifts left by \(10 \,cm\) On balancing condition…
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