Tags Physics

Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question

5.6k

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

Physics physics ncert exemplar solutions class 12th chapter two

In an experiment of find out diameter of wire using screw gauge, the following observations were noted:

(A) Screw moves 0.5mm on main scale in one complete rotation

(B) Total divisions on circular scale = 50

(C) Main scale reading is 2.5 mm

(D) 45th division of circular scale is in the pitch line

(E) Instrument has 0.03 mm negative error

Then the diameter of wire is:

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

MS â€“ Reading = 2.5 mm.

50 division on CS = 0.5 mm on MS.

$∴ 4 5^{t h} d i v i s i o n o n C S = \frac{0.5}{50} * 45 m m M . S .$

= .45 mm on M.S.

So, diameter = 2.5 mm

+0.45 mm

+0.03 mm

= 2.98 mm

0 0

New answer posted

a year ago

Physics physics ncert exemplar solutions class 12th chapter two

A traveling microscope has 20 divisions per cm on the main scale while its vernier scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

MSD = 20 divisions per cm 1 MSD = $\frac{1}{20} c m .$

VSD = 50 divisions

As, 25 VSD = 24 MSD

$∴ 1 V S D = \frac{24}{25} M S D$

L.C. = 1 MSD â€“ 1 VSD

= $= \frac{1}{500} c m = 0.002 c m$

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

Find the modulation index of an AM wave having 8V variation where maximum amplitude to the wave is 9V.

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

Modulation Index, $μ = \frac{A_{m}}{A_{C}}$

Variation = $2 A_{m} = 8 \Rightarrow A_{m} = 4 v$

$A_{m} + A_{c} = 9$

$A_{C} = 9 - A_{m} = 5 v$

$∴ μ = \frac{4}{5} = 0.8$

0 0

New question posted

a year ago

Physics Communication Systems

Find the modulation index of an AM wave having 8V variation where maximum amplitude to the wave is 9V.

0 Follower

New answer posted

a year ago

Physics Chemistry NCERT Exemplar Solutions Class 12th Chapter Twelve

Find the ratio of energies of photos product due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level. and (ii) the highest permitted energy level to the first permitted level.

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

$Δ E_{1} = - \frac{E_{0}}{4} + \frac{E_{0}}{1} = \frac{3}{4} E_{0}$

$Δ E_{2} = 0 - (- E_{0}) = E_{0}$

$\frac{Δ E_{1}}{Δ E_{2}} = \frac{3}{4}$

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

The kinetic energy of emitted electron is E when the right incident on the metal has wavelength $λ$ . To double the kinetic energy, the incident light must have wavelength:

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

$\frac{h c}{λ} - ? = E$ -(i)

$\frac{h c}{λ'} - ? = 2 E$ -(ii)

$h c (\frac{1}{λ'} - \frac{1}{λ}) = E$

$λ' = \frac{h c λ}{E λ + h c}$

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Match List – I with List – II:

List – I List – II

(A) UV rays (i) Diagnostic tool in medicine

(B) X-rays(ii) Water purification

(C) Microwave(iii) Communication, Radar

(D) Infrared wave(iv) Improving visibility in foggy daya

Choose the correct answer from the options given below:

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

Based an theoretical data.

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

A coil of inductance 1H and resistance 100 $Ω$ is connected to a battery of 6.V. Determine approximately:

(A)The time elapsed before the current acquires half of its steady – state value.

(B)The energy stored in the magnetic field associated with the coil at instant 15ms after the circuit is switched on. (Given In 2 = 0.693, e3/2 = 0.25)

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

L = 1H, R = 100 $Ω$

$A s, i = i_{0} e^{- t / τ}$

$F o r i = \frac{i_{0}}{2} \Rightarrow \frac{i_{0}}{2} = i_{0} e^{- t / τ}$

$\Rightarrow - l n 2 = - t / τ$

$i = \frac{6}{100} e^{- \frac{15}{1 / 100}} = . 06 e^{- 1500 * 1 0^{- 3}} = 0.06 e^{- 1.5} = 0.06 * 0.25 = . 0.015 - A$

So, $U = \frac{1}{2} L i^{2} = \frac{1}{2} * 1 * {(15 * 1 0^{- 3})}^{2} = \frac{1}{2} (225) * 1 0^{- 6} = 112.5 * 1 0^{- 6} J = 0.1125 m J$

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Orange light of wavelength $6000 * 10^{- 10} m$ illuminates a single slit of width $0.6 * 10^{- 4} m$ . the maximum possible number of diffraction minima produced on both sides of the central maximum is

0 Follower 14 Views

alok kumar singh

Contributor-Level 10

Sol. Angular momentum conservation:

$\begin{array}{r} \Rightarrow I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{f} \\ \Rightarrow \frac{{M R}^{2}}{2} ω_{o} = (\frac{{M R}^{2}}{2} + \frac{{M R}^{2}}{8}) ω_{f} \\ \Rightarrow ω_{f} = \frac{4}{5} ω_{o} \\ \Rightarrow {K E}_{final} = \frac{1}{2} (I_{1} + I_{2}) ω_{f}^{2} = \frac{{M R}^{2} ω^{2}}{5} \\ \Rightarrow K E_{initial} = \frac{1}{2} I, ω_{0}^{2} = \frac{{M R}^{2} ω^{2}}{4} \\ \Rightarrow % loss \Rightarrow 20 % . \end{array}$

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

Four resistance $40 Ω, 60 Ω, 90 Ω$ , and $110 Ω$ make the arms of a quadrilateral ABCD. Across $A C$ is a battery of emf $40 V$ and internal resistance negligible. The potential difference across $B D$ in $V$ is

0 Follower 15 Views

alok kumar singh

Contributor-Level 10

10553.33

Sol. $\frac{1}{λ_{min}} = R [\frac{1}{1} - \frac{1}{\infty}] = R [n = \infty \to n = 1]$

$\frac{1}{λ_{m a x}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4} [n = 2 \to n = 1]$

$\Rightarrow Δ λ \Rightarrow \frac{4}{3 R} - \frac{1}{R} \Rightarrow \frac{1}{3 R} = 340$

For Paschan $\Rightarrow \frac{1}{λ_{min}} = R [\frac{1}{9}] [n = \infty \to h = 3]$

$\Rightarrow \frac{1}{λ_{m a x}} = R [\frac{1}{9} - \frac{1}{16}] = \frac{7 R}{144}$

$\Rightarrow Δ λ = \frac{81}{7 R}$

0 0

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

On Shiksha, get access to

67k Colleges
1.2k Exams
718k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Physics

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience