In a meter bridge experiment null point is obtained at \(40\, cm\) from one end of the wire when resistance \(X\) is balanced against another resistance \(Y\). If \(X < Y\), then the new position of the null point from the same end, if one decides to balance a resistance of \(3X\) against \(Y\), will be close to .............. \(cm\)

A \(80\)
B \(75\)
C \(67\)
D \(50\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(67\)

Step-by-step Solution

Detailed explanation

From question, \(\frac{x}{y}=\frac{40}{100-40}=\frac{2}{3}\) \(\Rightarrow x=\frac{2}{3} y\) Again, \(\frac{3 x}{y}=\frac{z}{100-z}\) or \(\frac{3 \times \frac{2 y}{3}}{y}=\frac{z}{100-z}\) Solving we get \(Z=67\, \mathrm{cm}\) Therefore new position of null point…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List\(-I\) with List\(-II:\)

List\(-I\)	List\(-II\)
\((A)\) \(3\) Translational degrees of freedom	\((I)\) Monoatomic gases
\((B)\) \(3\) Translational,\(2\) rotational degrees of freedoms	\((II)\) Polyatomic gases
\((C)\) \(3\) Translational,\(2\) rotational and \(1\) vibrational degrees of freedom	\((III)\) Rigid diatomic gases
\((D)\) \(3\) Translational,\(3\) rotational and more than one vibrational degrees of freedom	\((IV)\) Nonrigid diatomic gases

Choose the correct answer from the options given below:

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