CUET · PHYSICS · PYQ PAPER 2023

8. An ammeter having internal resistance of \( 0.81 \, \Omega \) can read upto 1A. To convert it into an ammeter reading 0 - 10 A we have to connect :

A a \( 90 \, \Omega \) resistor in series
B a \( 9.0 \, \Omega \) resistor in parallel
C a \( 0.09 \, \Omega \) resistor in parallel
D a \( 0.9 \, \Omega \) resistor in parallel

Verified Solution

Answer & Solution

Correct Answer

(C) a \( 0.09 \, \Omega \) resistor in parallel

Step-by-step Solution

Detailed explanation

\( R_{sh} = \frac{I_g R_g}{I - I_g} \) \( R_{sh} = \frac{1 \, A \times 0.81 \, \Omega}{10 \, A - 1 \, A} \) \( R_{sh} = \frac{0.81}{9} \, \Omega = 0.09 \, \Omega \) Connect a \( 0.09 \, \Omega \) resistor in parallel.

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) L.C. Oscillation	(I) Inductor supplied with AC
(B) \( Q = \frac{\omega_o L}{R} \)	(II) \( \frac{d^2q}{dt^2} + \frac{1}{LC}q = 0 \)
(C) Voltage leads current	(III) \( \frac{i_p}{i_s} = \frac{V_s}{V_p} \)
(D) Transformer	(IV) Sharpness of resonance curve

Choose the correct answer from the options given below :

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