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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Dual Nature of Radiation and Matter

An ideal fluid of density 800kgm^-3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2} .$ The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt[]{x}}{6} m s^{- 1}$ for x =________. (Given g = 10 ms^-²)

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Payal Gupta

Contributor-Level 10

$ρ = 800 k g / m^{3}$

P₁ – P₂ = 4100 Pa

Bernoulli's question b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

$\frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{v_{2}^{2}}{2 g} + 0$ -(1)

Equation of continuity

A₁v₁ = A₂v₂

v₂ = 2v₁-(2)

from (1) & (2)

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{{(2 v_{1})}^{2}}{2 g}$

$\frac{4100}{800 * 10} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{4 v_{1}^{2}}{2 g}$

$\frac{12100}{8000} = \frac{3 v_{1}^{2}}{2 g}$

$\Rightarrow v_{1}^{2} = \frac{2 * 10}{3} * \frac{12100}{8000}$

$= \frac{2}{3} * \frac{121}{8}$

x = 363

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Dual Nature of Radiation and Matter

An electron with speed $υ$ and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are E_e and p_e and that of photon are E_ph and p_ph respectively. Which of the following is correct?

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Payal Gupta

Contributor-Level 10

$λ_{e} = \frac{h}{m v} = \frac{h}{p_{e}}$ - (1)

$λ_{P h} = \frac{h}{P_{P h}}$ - (2)

A/C to question

$λ_{e} = λ_{P h}$

$P_{e} = P_{P h} \Rightarrow \frac{P_{P}}{P_{P h}} = 1$ - (3)

$k . E_{e} = E_{e} = \frac{1}{2} m v^{2}$

$= \frac{1}{2} P e v$ - (4)

$k . E_{P h} = E_{P h} = m c^{2}$

= mc c

q = P_Ph c- (5)

$(4) \div (5)$

$\frac{E_{e}}{E_{P h}} = \frac{v}{2 c}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Physics Thermal Properties of Matter

As per the given figure, two plates A and B of thermal conductivity K and 2K are joined together of form a compound plate. The thickness of plates are 4.0cm and 2.5cm respectively and the area the area of cross-section is 120 cm² for each plate. The equivalent thermal is 120 cm² for each plate. The equivalent thermal conductivity of the compound plat is $(1 + \frac{5}{α}) K,$ then the value of a will be________.

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alok kumar singh

Contributor-Level 10

k_{eq=
$\frac{L_{1} + L_{2}}{\frac{L_{1}}{k_{1}} + \frac{L_{2}}{k_{2}}} = \frac{4 + 2.5}{\frac{4}{k} + \frac{2.5}{2 k}}$}

_{keq=
$\frac{6.5}{\frac{8 + 2.5}{2 k}} = \frac{6.5}{10.5} * 2 k = \frac{13 * 2}{21} k$}

_{=
$(1 + \frac{5}{51}) k$}

_{=
$(1 + \frac{5}{α})$
=21}

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Thermal Properties of Matter

In van dar Wall equation $[P + \frac{a}{V^{2}}] [V - b] = R T; P$ is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants $\frac{a}{b}$ is dimensionally equal to:

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alok kumar singh

Contributor-Level 10

dimension of a Pv²

$\frac{a}{b} = p v$ dimension of b = v

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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Laws of Motion

A block of metal weighing 2kg is resting on a frictionless plane (as shown in figure). It is struck by a jet relasing water at a rate of 1 kgs^-1 and at a speed of 10 ms^-1. Then, the initial acceleration of the block, in ms^-2, will be:

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alok kumar singh

Contributor-Level 10

$a = \frac{F}{m} = \frac{m v}{m} = \frac{1 * 10}{2} = 5 m / s e c^{2}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Dual Nature of Radiation and Matter

Given below are two statements : one is labelled as Assertion A and other is labelled as Reason R.

Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal.

Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal.

In the light of the above statements, chose the most appropriate answer from t

...more

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alok kumar singh

Contributor-Level 10

To free the electron from metal surface minimum energy required, is equal to the work function of that metal.

So Assertion A, is correct

have = w₀ + K.E_max

If have = w₀

Þ K.E_max = 0

Hence reas

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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Physics Thermal Properties of Matter

An ice cube of dimensions 60cm * 50cm * 20cm is placed in an insulation box of wall thickness 1cm. The box keeping the ice cube at $0 ° C$ 898

of temperature is brought to a room of temperature $40 ° C .$ The rate of melting of ice is approximately:

(Latent heat of fusion of ice is 3.4 * 105 J kg1 and thermal conducting of insulation wall is 0.05 $W m^{- 1} ° C^{- 1}$ )

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alok kumar singh

Contributor-Level 10

$A r e a = 2 * [l b + b h + h l]$

$\Rightarrow A = 2 * [0.6 * 0.5 + 0.5 * 0.2 + 0.2 * 0.6]$

= 1.04 M2

$R_{t h e r m a l} = \frac{t}{K A} = \frac{1 * 1 0^{- 2}}{0.05 * 1.04}$

$\Rightarrow m = 61 * 1 0^{- 5} k g / s$

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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Dual Nature of Radiation and Matter

Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is

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Payal Gupta

Contributor-Level 10

Case I :- $2 ? - ? = \frac{1}{2} m v_{1}^{2} . . . . . . . . (i)$

Case II :- $10 ? - ? = \frac{1}{2} m v_{2}^{2} . . . . . . . . (i i)$

$\Rightarrow \frac{1}{9} = \frac{v_{1}^{2}}{v_{2}^{2}}$

$\Rightarrow v_{1} : v_{2} = 1 : 3$

x = 1

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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Dual Nature of Radiation and Matter

A parallel beam of light of wavelength 900nm and intensity 100Wm^-2 is incident on a surface perpendicular to the beam. The number of photons crossing 1cm² area perpendicular to the beam in one second is:

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Vishal Baghel

Contributor-Level 10

$E = I A$

$= 100 * 1 * 1 0^{- 4}$

$= 1 0^{- 2} W$

$E = n h \frac{c}{λ}$

$1 0^{- 2} = \overset{?}{n} * \frac{6.64 * 1 0^{- 34} * 3 * 1 0^{8}}{900 * 1 0^{- 9}}$

$\overset{?}{n} = \frac{1 0^{- 2} * 9 * 1 0^{- 7}}{6.64 * 1 0^{- 34} * 3 * 1 0^{8}}$

$\overset{?}{n} = 4.5 * 1 0^{16}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Dual Nature of Radiation and Matter

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

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Vishal Baghel

Contributor-Level 10

$E_{1} = 5 ? - ? = 4 ? = \frac{1}{2} m v_{1}^{2}$

$E_{2} = 10 ? - ? = \frac{1}{2} m V_{2}^{2}$

$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{\sqrt[]{\frac{8 ?}{m}}}{\sqrt[]{\frac{18 ?}{m}}} = \sqrt[]{\frac{8}{18}} = \frac{\sqrt[]{4}}{9} = \frac{2}{3}$

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