KCET · Physics · Current Electricity
When a metal conductor connected to left gap of a meter bridge is heated, the balancing point
- A shifts towards left
- B remains unchanged
- C shifts to the centre
- D shifts towards right
Answer & Solution
Correct Answer
(D) shifts towards right
Step-by-step Solution
Detailed explanation
We know that, resistance of unknown wire can be determined by using meter bridge as
\(\frac{R}{S}=\frac{l}{100-l} \Rightarrow S=\frac{R(100-l)}{l}\)
where \(R=\) known resistance, \(l=\) balancing length and \(S=\) unknown resistance.
When a metal conductor of resistance \(R\) connected to left gap is heated, then its resistance increases i.e. \(R\) increases. Hence, from Eq. (i), we see that when \(R\) will increase, then \((100-l)\) will decrease also, i.e \(l\) will increase. Therefore, balancing point will shift towards right.
\(\frac{R}{S}=\frac{l}{100-l} \Rightarrow S=\frac{R(100-l)}{l}\)
where \(R=\) known resistance, \(l=\) balancing length and \(S=\) unknown resistance.
When a metal conductor of resistance \(R\) connected to left gap is heated, then its resistance increases i.e. \(R\) increases. Hence, from Eq. (i), we see that when \(R\) will increase, then \((100-l)\) will decrease also, i.e \(l\) will increase. Therefore, balancing point will shift towards right.
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