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6 months 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:

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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

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6 months 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?

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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

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

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6 months 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.

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Vishal Baghel

Contributor-Level 10

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

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

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

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

$∴ μ = \frac{4}{5} = 0.8$

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6 months ago

Physics Communication Systems

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

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6 months 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.

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Vishal Baghel

Contributor-Level 10

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

$Δ E_{2} = 0 - (- E_{0}) = E_{0}$

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

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6 months 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:

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Vishal Baghel

Contributor-Level 10

$\frac{h c}{λ} - ? = E$ -(i)

$\frac{h c}{λ'} - ? = 2 E$ -(ii)

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

$λ' = \frac{h c λ}{E λ + h c}$

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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:

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Vishal Baghel

Contributor-Level 10

Based an theoretical data.

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6 months 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)

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

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6 months 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

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alok kumar singh

Contributor-Level 10

Sol. Angular momentum conservation:

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

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6 months 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

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

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

$\Rightarrow Δ λ = \frac{81}{7 R}$

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