Physics Ncert Solutions Class 12th

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

E = ½CV²
= ½ (ε? A/d)Ed²
= ½ε? E²Ad

New answer posted

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Least count = 1 mm / 100 = 0.01 mm
Diameter = main scale reading + circular scale reading
Diameter = 0 + 52 * 0.01 mm
= 0.52 mm = 0.052 cm

New answer posted

a month ago

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V
Vikash Kumar Vishwakarma

Contributor-Level 7

If you want the Current Electricity Class 12 important questions, then check the Class 12 Physics previous year question paper pdf. Also, focusing on the Class 12 Physics past 5 years of exam questions to identify the current electricity class 12 important questions. Well, we have framed all the important questions in the NCERT Physics Class 12 Current Electricity solution pdf, which you can easily download through this article.

New answer posted

a month ago

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V
Vikash Kumar Vishwakarma

Contributor-Level 7

Before attempting the NCERT solutions PDF, solve the textbook exercise questions. Hereafter, verify your answer using the NCERT Class 12 Physics Ch 3 solutions pdf. Following a systematic approach to solve a problem will help to score good marks in the CBSE Class 12 board exam.

New answer posted

a month ago

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V
Vikash Kumar Vishwakarma

Contributor-Level 7

Class 12 Physics Chapter 3 Current Electricity is important for the board exam. The weightage of the chapter is 8-10%. When you practice the Current Electricity class 12 NCERT solutions, you get an overview of your preparation. If you are lacking in any concept, then it helps to build the conceptual knowledge.

New answer posted

a month ago

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V
Vikash Kumar Vishwakarma

Contributor-Level 7

The topics focused on while preparing the Current Electricity NCERT solutions are:

  • Ohm's law

  • Drift velocity

  • Resistivity vs temperature

  • Combination of cells

  • Kirchhoff's rules and 

  • Wheatstone bridge

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

| τ ? | = P E s i n ? θ 4 = q * 2 a * E s i n ? 30 ?

q = 4 2 * 10 - 2 * 2 * 10 5 * 1 2

= 2 * 10 - 3 C

= 2 m C

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Fundamental harmonic frequency open pipe

= V 2 L = v 1

(say)

Fundamental harmonic frequency of closed pipe = V 4 L

= v 2 (say)

v 1 v 2 = V 2 L V 4 L 2 : 1

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

A reverse-biased Zener diode is used as a voltage regulator.
The potential barrier for Germanium (Ge) is approximately 0.3 V.
The potential barrier for Silicon (Si) is approximately 0.7 V.

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