Class 12th

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New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

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

Then,  dydt = 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 ddt=ddt πr2.

=ddrπr2·drdt

= 2πr 3

= 6πr. cm25

When r = 10cm,

ddt = 6.π * 10 = 60π cm25

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

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, dYdF=8cm35

ddt (x3) = 8

dx3dx·dxdt=8 (by chain rule)

 3x2 dxdx=8.

dxdt=83x2.

Now, dSdt=ddt (bx2) = d(6x2)dxdxdx = 12x 83x2=32x cm25.

When x = 12 cm,

dSdt=3212=83 cm25.

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

(a) r = 3cm

When r = 3 cm,

ddr = 2 * π * 3 cm = 6π.

Thus, the area of the circle is changing at the rate of 6π cm cm25.

(b) r = 4cm

whenr = 4cm,

ddr = 2 * π * 4cm = 8π.

Thus, the area of the circle is changing at the rate of 8 cm25.

New answer posted

a year ago

0 Follower 2 Views

P
Pallavi Pathak

Contributor-Level 10

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.

New answer posted

a year ago

0 Follower 11 Views

P
Pallavi Pathak

Contributor-Level 10

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.

New answer posted

a year ago

0 Follower 5 Views

P
Pallavi Pathak

Contributor-Level 10

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.

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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.

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c, d

Explanation-ripple factor is inversely proportional to RL, C  and frequency. so to reduce ripple factor all these should be increase.

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b, d

Explanation- During regulation action of a Zener diode, the current through the Rs changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant.

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, d

Explanation- The space-charge regions on both the sides of p-n junction which has immobile ions and entirely lacking of any charge carriers will form a region called depletion region of a diode.

The number of ionized acceptors on the p -side equals the number of ionized donors on the n-side.

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