Exams › NEET › Physics

A carbon resistor of (47 ± 4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be:

1. Violet – Yellow – Orange – Silver
2. Yellow – Violet – Orange – Silver
3. Green – Orange – Violet – Gold
4. Yellow – Green – Violet – Gold

Correct answer: Yellow – Violet – Orange – Silver

Solution

The resistance value is 47 kΩ with a tolerance of ±4.7 kΩ. Using the resistor color code, 47 corresponds to Yellow (4) and Violet (7), with Orange (×10³) as the multiplier. Silver represents a 10% tolerance. Thus, the correct sequence is Yellow – Violet – Orange – Silver.

Related NEET Physics questions

• \( \mathbf{1}^{o} \boldsymbol{A} \) is
• Which of the following is different from the rest?
• Assertion nm is not same as mN Reason \( \ln m=10^{-9} m \operatorname{and} 1 m N=10^{-3} N \)
• If the dimension of a physical quantity is given by \( \left[M^{a} L^{b} T^{c}\right] \) then the physical quantity will be
• In a new system, unit of mass is \( 10 \mathrm{kg} \) unit of length is \( 5 \mathrm{m} \) and unit of time is 10s. Then in the new system 5N is equal to :
• A cube has a side of length \( 2.342 m \). The surface area and volume in correct significant figures are respectively:

⚔️ Practice NEET Physics free + battle 1v1 →