JEE Mains · Physics · STD 12 - 3. current electricity

The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as \(R_1\) has the colour code (Orange, Red, Brown). The resistors \(R_2\) and \(R_4\) are \(80\, \Omega \) and \(40\,\Omega \), respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as \(R_3\) would be

A Brown, Blue, Brown
B Brown, Blue, Black
C Red, Green, Brown
D Grey, Black, Brown

Verified Solution

Answer & Solution

Correct Answer

(A) Brown, Blue, Brown

Step-by-step Solution

Detailed explanation

\(R_{1}=32 \times 10=320 \,\Omega\) \(\mathrm{R}_{3}=\frac{\mathrm{R}_{\mathrm{y}}}{\mathrm{R}_{2}} \times \mathrm{R}_{1}=\frac{40 \times 320}{80}=160 \,\Omega\) \(\therefore \) Colour code of \(\mathrm{R}_{3}\) be Brown, Blue, Brown.

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.

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Match the LIST-I with LIST-II

LIST-I	LIST-II
(A) Boltzmann constant	I. \(\mathrm{ML}^2 \mathrm{~T}^{-1}\)
(B) Coefficient of viscosity	II. \(\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\)
(C) Planck's constant	III. \(\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\)
(D) Thermal conductivity	IV. \(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)

Choose the correct answer from the options given below :

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