Class 12th

The circumference C of the circle with radius r is C = 2πr.

Then, rate of change in circumference is $\frac{\mathrm{dC}}{dt}=\frac{d}{dt}2\pi r$ = $2\pi \frac{dx}{df}.$

Q Radius of circle increases at rate 0.7 cm/s we get,

$\frac{dy}{dt}=0.7$ cm/s

So,  $\frac{\mathrm{dC}}{dt}$ = 2.* 0.7 cm/s = 1.4 * cm/s.

The area A of the circle with radius π is A = πr²

Then, rate of change in area $\frac{\mathrm{dA}}{dt}=\frac{d\stackrel{?}{x}{r}^2}{dt}$ = $\frac{dx{r}^2}{dr}·\frac{dr}{dt}$

= 2πr $\mathrm{dr}$
${\frac{}{dt}}_{.}$

Q The wave moves at a rate 5cm/s we have,

$\frac{dr}{\mathrm{dF}}$ = 5cm/s

So,  $\frac{\mathrm{dA}}{dt}$ =r. 5 = 10πr. ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When r = 8 cm.

$\frac{d}{dt}$ = 10.π.8 cm ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ = 80 * cm ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

Let 'x' cm be the length of edge of the cube which is a fxⁿ of time t then,

$\frac{dx}{dt}$ = 3cm/s as it is increasing.

Now, volume v of the cube is v = x³

Ø Rate of change of volume of the cube $\frac{dv}{dt}=\frac{d{q}^3}{dt}.$

$=\frac{d{x}^3}{dx}\frac{dx}{dt}$

= 3x².3

= 9x² ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When x = 10cm.

$\frac{dv}{dt}$ = 9 x (10)²= 900 ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

Let 'r' cm be the radius of the circle which is afxn of time.

Then,  $\frac{dy}{dt}$ = 8 3cm/s as it is increasing.

Now, the area A of the circle is A = πr2.

So, the rate at which the area of the circle change $\frac{d}{dt}=\frac{d}{dt}$ πr2.

$=\frac{d}{dr}\pi r^2·\frac{dr}{dt}$

= 2πr 3

= 6πr. ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When r = 10cm,

$\frac{d}{dt}$ = 6.π * 10 = 60π ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

Let x be the length of edge,v be the value and s be the surface area of the cube then,

y = x3.

and S = 6x2, where x is a fxn of time.

Now, $\frac{dY}{dF}=8c{m}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dt}$ (x3) = 8

$\frac{d{x}^3}{dx}·\frac{dx}{dt}=8$ (by chain rule)

$\Rightarrow$ 3x2 $\frac{dx}{dx}=8.$

$\frac{dx}{dt}=\frac{8}{3x^2.}$

Now, $\frac{dS}{dt}=\frac{d}{dt}$ (bx2) = $\frac{d\left(6x^2\right)}{dx}-\frac{dx}{dx}$ = 12x $\frac{8}{3x^2}=\frac{32}{x}$ $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

When x = 12 cm,

$\frac{dS}{dt}=\frac{32}{12}=\frac{8}{3}$ $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

(a) r = 3cm

When r = 3 cm,

$\frac{d}{dr}$ = 2 * π * 3 cm = 6π.

Thus, the area of the circle is changing at the rate of 6π cm $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

(b) r = 4cm

whenr = 4cm,

$\frac{d}{dr}$ = 2 * π * 4cm = 8π.

Thus, the area of the circle is changing at the rate of 8 $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

If one solves all these questions then it is great but it is not something mandatory. Practicing these questions helps students develop a strong understanding of Coulomb's law, electric charges, the superposition principle, and electric field lines. Even practicing a significant portion helps in boosting exam confidence and helps students to score high in the examination.

The NCERT textbook is good to start with as it introduces students to the basic concepts and explains these topics to provide conceptual clarity. However, the exemplar focuses on reasoning, challenging multiple-choice questions and numerical problems. The exemplar is like the supplement that boosts students' grasp on these concepts given in the NCERT textbook through reasoning, multiple-choice and numerical problems.

Chapter 1 Electric Charges and Fields NCERT Exemplar goes beyond the basic NCERT textbook and contains application-based and advanced-level questions. It is designed to enhance students' problem-solving skills, improve their understanding of electrostatics, and prepare them for their CBSE board exams and competitive exams like NEET and JEE.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, d)

Explanation- In reverse biasing, the minority charge carriers will be accelerated due to reverse biasing, which on striking with atoms cause ionization resulting secondary electrons and thus more number of charge carriers.

When doping concentration is large, there will be large number of ions in the depletion

region, which will give rise to a strong electric field.