Tags Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Get insights from 70 questions on Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Oscillations

A particle executes S.H.M. with amplitude â€˜aâ€™ and time period â€˜Tâ€™. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt[]{x} a}{2} .$ The value of x is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$v = ω \sqrt[]{A^{2} - s^{2}} \Rightarrow \frac{A ω}{2} = ω \sqrt[]{A^{2} - s^{2}} \Rightarrow s = \frac{\sqrt[]{3} A}{2} \Rightarrow x = 3$

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Oscillations

Time period of a simple pendulum is T. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{α}{β} T .$ The value of a is

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

Total oscillation = $\frac{1}{2} + \frac{1}{8} = \frac{5}{8}$

$\Rightarrow t i m e t a k e n = \frac{T}{2} + \frac{T}{12} = \frac{7 T}{12}$

= 7

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Oscillations

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

x = A sin $ω t$

$\Rightarrow v = A ω c o s ω t$

$\Rightarrow v = \pm ω \sqrt[]{A^{2} - x^{2}}$

$\frac{v^{2}}{ω^{2}} + x^{2} = A^{2}$

Elliptical

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Oscillations

Given below are two statements:

Statement I : A second's pendulum has a time period of 1 second.

Statement II : It takes precisely one second to move between the two extreme positions.

In the light of the above statements, choose the correct answer from the options given below

0 Follower 1 View

Payal Gupta

Contributor-Level 10

Time period of second pendulum is 2 seconds.

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Oscillations

A particle executes S.H.M., the graph of velocity as a function of displacement is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$v = ω \sqrt[]{A^{2} - x^{2}} \Rightarrow \frac{v^{2}}{A^{2} ω^{2}} + \frac{x^{2}}{A^{2}} = 1$ Path is ellipse.

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductor Electronics

Take the breakdown voltage of the zener diode used in the given circuit as $6 V$ . For the input voltage shown in figure below, the time variation of the output voltage is: (Graphs drawn are schematic and not to scale)

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

Kindly go through the solution

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Oscillations

The potential energy of a particle of mass 4 kg in motion along the x-axis is given by $U = 4 (1 - c o s 4 x) J .$ The time period of the particle for small oscillation (sin = ) is $(\frac{π}{K}) s .$ The value of K is_______.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

m = 4kg

U = 4 (1 – cos 4x) J

$F = - \frac{\partial υ}{\partial x} = - 4 * 4 s i n 4 x$

$∴ a = - \frac{16}{4} s i n 4 x = - 4 s i n 4 x = - 16 x$

$\Rightarrow a = - 16 x$

$ω^{2} = 16 \Rightarrow ω = 4 = \frac{2 π}{T} \Rightarrow T = \frac{2 π}{4} = \frac{π}{2} s e c$

k = 2

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductors

An n.p.n transistor with current gain = 100in common emitter configuration is shown in figure. The output voltage of the amplifier will be

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

n.p.m. transistor current gain B = 100

As, B = $\frac{I_{C}}{I_{B}} V o l t a g e g a i n, \frac{V_{o u t}}{V_{i n}} = (\frac{I_{C}}{I_{B}}) \frac{R_{2}}{R_{i}}$

$V_{o u t} = \frac{B R_{2}}{R_{i n}} V i n$

$\frac{1 0^{3}}{1 0^{3}} V = 1 v$

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductor Electronics

A small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06 T$ experiences a torque of $0.018 N m$ . The minimum work required to rotate it from its stable to unstable equilibrium position is:

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

$\overset{?}{τ} = \overset{?}{μ} * \overset{?}{B}$

$\begin{array}{l} \Rightarrow & 0.018 = μ (0.06) (s i n ? 30^{?}) \\ \Rightarrow & μ = 0.6 \\ \Rightarrow & Work = U_{f} - U_{i} \\ \Rightarrow & 2 μ B \\ \Rightarrow & 7.2 * 10^{- 2} J . \end{array}$

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductors

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : n-p-n transistor permits more current that a p-n-p transistor.

Reason R : Electrons have greater mobility as a change carrier.

Choose the correct answer form the options given below

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

Using factual data.

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
686k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience