Physics

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.

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.

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

Fundamental harmonic frequency open pipe

$=\frac{\mathrm{V}}{2\mathrm{L}}={\mathrm{v}}_1$

(say)

Fundamental harmonic frequency of closed pipe $=\frac{\mathrm{V}}{4\mathrm{L}}$

$={\mathrm{v}}_2$ (say)

$\Rightarrow \frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{\frac{\mathrm{V}}{2\mathrm{L}}}{\frac{\mathrm{V}}{4\mathrm{L}}}\mathrm{}2:1$

Diameter = main scale reading + (circular scale reading * least count)
Diameter = 0 + (52 * 0.01 mm) = 0.52 mm = 0.052 cm.
In the RLC circuit:
Given I? = 10√2 A, so I? = I? /√2 = 10 A.
V? = √ [V? ² + (V? - V? )²] = √ [40² + (40 - 10)²] = √ [1600 + 30²] = √ [1600 + 900] = √2500 = 50 V.
Impedance Z = V? / I? = 50 V / 10 A = 5 Ω.
For Hindi: I? = 10√2 A, V? = 50V, Z = V? /I? = 50/ (10√2) = 5/√2 Ω.

Least count = (1 mm) / 100 = 0.01 mm.

$\mathrm{T}\mathrm{i}=-{50}^{?}\mathrm{C}$

$\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$

As $\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$  increased by 3 times

So ${\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\mathrm{f}}=4{\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\text{initial}}$

${\mathrm{T}}_{\mathrm{f}}=16{\mathrm{T}}_{\mathrm{i}}$

$=3568\mathrm{K}$

${\mathrm{T}}_{\mathrm{f}}=(3568-273{)}^{?}\mathrm{C}$

$={3295}^{?}\mathrm{C}$

$\eta =1-\frac{T_{\text{sin}k}}{T_{\text{source}}}=0.5$

$\begin{array}{rr}& \frac{{\mathrm{T}}_{\text{sin}}}{{\mathrm{T}}_{\text{source}}}=0.5\\ & \Rightarrow {\mathrm{T}}_{\text{sin}}=\frac{1}{2}*(273+327)\\ & =\frac{1}{2}*600\\ & =300\mathrm{K}\\ & ={27}^{?}\mathrm{C}\end{array}$

Factual (theory based)