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Physics NCERT Exemplar Solutions Class 12th Chapter Nine

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Nine Mechanical Properties of Solids

A closely wounded circular coil of radius 5cm produces a magnetic field of 37.68 * 104 T at its centre. The current through the coil is_ A. [Given, number of turns in the coil is 100 and = 3.14]

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Vishal Baghel

Contributor-Level 10

$B_{c e n t r e} = N \frac{μ_{0} i}{2 r}$

$\Rightarrow \frac{100 * 4 π * 1 0^{- 7} * i}{2 * 5 * 1 0^{- 2}} = 37.68 * 1 0^{- 4}$

$∴ i = \frac{37.68}{4 * 3.14} = 3 A$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

The current I flowing through the given circuit will be_________A.

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Vishal Baghel

Contributor-Level 10

$i = \frac{v}{R_{N e t}} = \frac{6}{3} = 2 A$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics NCERT Exemplar Solutions Class 12th Chapter Ten

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture of constant volume is $\frac{α^{2}}{4} R J / m o l K;$ then the value of will be________.

(Assume that the given diatomic gas has no vibrational mode.)

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Vishal Baghel

Contributor-Level 10

$C_{v} = \frac{α^{2} R}{4} J / m o l^{- k}$

As, Cv (mix) = $\frac{1 * \frac{3}{2} R + 3 * \frac{5}{2} R}{4} = \frac{9 R}{4} = \frac{α^{2} R}{4}$

$∴ α = 3$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Mechanical Properties of Solids

The pressure P1 and density d1 of diatomic gas $(γ = \frac{7}{5})$ changes suddenly to P2(>P1) and d2 respectively during an adiabatic process. The temperature of the gas increases and becomes ________times of its initial temperature. (given $\frac{d_{2}}{d_{1}} = 32)$

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Vishal Baghel

Contributor-Level 10

For adiabatic process – PVY = const

$T_{1} {V_{1}}^{Y - 1} = T_{2} V_{2}^{Y - 1}$

$\frac{T_{2}}{T_{1}} = {(\frac{v_{1}}{v_{2}})}^{Y - 1} = {(\frac{d_{2}}{d_{1}})}^{Y - 1} = {(32)}^{(\frac{7}{5} - 1)} = {(32)}^{2 / 5}$

$= {(2)}^{2} = 4$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics Mechanical Properties of Solids

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics NCERT Exemplar Solutions Class 12th Chapter Eight

If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a dept ah(h <e) from the earth surface. The value of a will be_______.

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Vishal Baghel

Contributor-Level 10

$g_{(a t h)} = g_{(a t d e p t h α h)}$ h << R

$\Rightarrow g (1 - \frac{2 h}{R}) = g (1 - \frac{α h}{R})$

$\begin{array}{l} 1 - \frac{2 h}{R} = 1 - \frac{α h}{R} \Rightarrow \frac{2 h}{R} = α \frac{h}{R} \\ α = 2 \end{array}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An object is projected in the air with initial velocity u at an angle q. The projectile motion is such that the horizontal range R, in maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be_______degree.

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Vishal Baghel

Contributor-Level 10

$R = \frac{u^{2} s i n (2 * 45 °)}{g} = \frac{u^{2}}{g}$

$\frac{R}{2} = \frac{u^{2}}{2 g} = \frac{u^{2} s i n 20}{g}$

$s i n 2 θ = \frac{1}{2}$

$\Rightarrow 2 θ = 30 ° \Rightarrow θ = 15 °$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine 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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Physics NCERT Exemplar Solutions Class 12th Chapter Nine 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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Physics NCERT Exemplar Solutions Class 12th Chapter Nine 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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