Physics Ncert Solutions Class 12th

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1.15 (a) Electric field intensity, E = 3 $* 10^{3}$ î N/C

Magnitude of electric field intensity, $|E|$ $=3* 10^{3}$ =
N/C

Side of the square, s = 10 cm = 0.1 m

Area of the square, A = $s^{2}$ = 0.01 $m^{2}$

The plane of the square is parallel to the y-z plane, hence the angle between the unit vector normal to the plane and electric field, $θ$ = 0 $°$

Flux ( $φ)$ through the plane is given by the relation, $φ$ = $|E| A \cos ? θ$ = $3 * 10^{3} * 0.01 * \cos ? 0 ° = 30$
${Nm}^{2} /$ C

(b) When the normal to its plane make a 60 $°$ angle with x-axis, $θ$ = 60
. From the equation $φ$ = $|E| A \cos ? θ$ we get = $φ$ $3 * 10^{3} * 0.01 * \cos ? 60 ° = 15$ ${Nm}^{2} /$ C

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10.4 In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

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10.4 Distance between the slits, d = 0.28 mm = 0.28 $* 10^{- 3}$ m

Distance between the slits and the screen, D = 1.4 m

Distance between the central bright fringe and the fourth ( n = 4) fringe, u = 1.2 cm = 1.2 $* 10^{- 2}$ m

In case of a constructive interference, we have the relation for the distance between two fringes as : u = n $? \frac{D}{d},$ where n = order of fringes = 4 and $?$ = wavelength of the light used

Hence, $?$ = $\frac{u d}{n D}$ = $\frac{1.2 * 10^{- 2} * 0.28 * 10^{- 3}}{4 * 1.4}$ = 6 $* 10^{- 7}$ m = 600 $* 10^{- 9}$ m = 600 nm

Hence, wavelength of the light is 600 nm.

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10.3 (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 * 108 m s–1)

(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

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Physics Ncert Solutions Class 12th Electric Charges and Fields

Contributor-Level 10

10.3 The refractive index of glass, $?$ = 1.5

Speed of light in vacuum, c = 3.0 $* 10^{8}$ m/s

Speed of light in glass is given by the relation, $v$ = $\frac{c}{?}$ = $\frac{3.0 * 10^{8}}{1.5}$ = 2 $* 10^{8}$ m/s

The speed of light in glass is not independent of the colours of light. The refractive index of a violet component of white light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

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1.14 Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

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1.14 Since the charges 1 & 2 are attracted towards + ve, their charges will be – ve. The charge 3 is attracted towards – ve, hence its charge will be +ve.

The charge to mass ratio (emf) is directly proportional to the displacement, charge 3 will have the highest charge to mass ratio.

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10.2 What is the shape of the wave front in each of the following cases:

(a) Light diverging from a point source.

(b) Light emerging out of a convex lens when a point source is placed at its focus.

(c) The portion of the wave front of light from a distant star intercepted by the Earth.

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Ans.10.2 The shape of the wave front in case of a light diverging from a point source is spherical.

The shape of the wave front in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid.

The portion of the wave front of light from a distant star intercepted by the Earth is a plane.

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10.1 Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.

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Physics Ncert Solutions Class 12th Electric Charges and Fields

Contributor-Level 10

10.1 Wavelength of incident monochromatic light, $?$ = 589 nm = 589 $* 10^{- 9}$ m

Speed of light in air, c = 3 $* 10^{8}$ m/s

Refractive index of water, $?$ = 1.33

In case of reflection, the ray goes back to the same medium. Hence wavelength, frequency and speed of reflected beam will be same as incident beam.

Frequency of light beam is given by the relation, $?$ = $\frac{c}{?}$ = $\frac{3 * 10^{8}}{589 * 10^{- 9}}$ = 5.09 $* 10^{14}$ Hz.

Hence speed = 3 $* 10^{8}$ m/s, Wavelength = 589 $* 10^{- 9}$ m, Frequency = 5.09 $* 10^{14}$ Hz of incident ray and reflected ray will remain unchanged.

(i) Frequency = does not depend on the property of the medium in

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10.1 Wavelength of incident monochromatic light, $?$ = 589 nm = 589 $* 10^{- 9}$ m

Speed of light in air, c = 3 $* 10^{8}$ m/s

Refractive index of water, $?$ = 1.33

In case of reflection, the ray goes back to the same medium. Hence wavelength, frequency and speed of reflected beam will be same as incident beam.

Frequency of light beam is given by the relation, $?$ = $\frac{c}{?}$ = $\frac{3 * 10^{8}}{589 * 10^{- 9}}$ = 5.09 $* 10^{14}$ Hz.

Hence speed = 3 $* 10^{8}$ m/s, Wavelength = 589 $* 10^{- 9}$ m, Frequency = 5.09 $* 10^{14}$ Hz of incident ray and reflected ray will remain unchanged.

(i) Frequency = does not depend on the property of the medium in which it is travelling, hence frequency will remain same, i.e. 5.09 $* 10^{14}$ Hz

(ii) Speed of light in water depends upon the refractive index of water, hence speed of light, v = $\frac{c}{?}$ = $\frac{3 * 10^{8}}{1.33}$ = 0.226 $* 10^{9}$ m/s

Wavelength in water, $?$ = $\frac{v}{?}$ = $\frac{0.226 * 10^{9}}{5.09 * 10^{14}}$ = 4.432 $* 10^{- 7}$ m = 443.2 nm

Hence the speed, frequency and wavelength of refracted light are 0.226 $* 10^{9}$ m/s, 5.09 $* 10^{14}$ Hz and 443.2 nm

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1.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

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Physics Ncert Solutions Class 12th Electric Charges and Fields

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1.13 When the third uncharged sphere C is brought in contact with the sphere A, then the charge is shared and becomes half. Then

$q_{A}$ = $\frac{q}{2}$ and = $q_{C}$ = $\frac{q}{2}$

When the charged sphere C is brought in contact with charged sphere B, the charge between both the sphere is shared and becomes half

$q_{B}, q_{C} = \frac{1}{2}$ (q + = $\frac{q}{2})$ $\frac{3 q}{4}$

Hence the force of repulsion between sphere A and B can be given as

F =
$\frac{1}{4 π ?_{0}}$ $* \frac{q_{1} q_{2}}{r^{2}}$ , where $?_{0}$ = Permittivity of free space = 8.854 $* 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$ = = $\frac{1}{4 * π * 8.854 * 10^{- 12}}$ = $* \frac{(\frac{q}{2} * \frac{3 q}{4})}{{0.5}^{2}}$ = $\frac{1}{4 * π * 8.854 * 10^{- 12}}$ $* \frac{3 * q^{2}}{{8 * 0.5}^{2}}$
= $\frac{1}{4 * π * 8.854 * 10^{- 12}}$ $* \frac{3 * {(6.5 * 10^{- 7})}^{2}}{{8 * 0.5}^{2}}$ = 5.695 $* 10^{- 3}$ N

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(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 $* 10^{- 7}$ C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

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Physics Ncert Solutions Class 12th Nuclei

Contributor-Level 10

1.12 (a) Charge on sphere A, $q_{A}$ = 6.5 $* 10^{- 7}$ C

Charge on sphere B, $q_{B}$ = 6.5 $* 10^{- 7}$ C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between two spheres

F = $\frac{1}{4 π ?_{0}}$ $* \frac{q_{1} q_{2}}{r^{2}}$ , where $?_{0}$ = Permittivity of free space = 8.854 $* 10^{- 12}$
$C^{2} N^{- 1}$ $m^{- 2}$ Therefore, F = $\frac{1}{4 * π * {8.854 * 10^{- 12}}_{}}$ $* \frac{6.5 * 10^{- 7} * 6.5 * 10^{- 7}}{{0.5}^{2}}$ N = 0.0152 N = 1.52 $* 10^{- 2}$ N

(b) Charge on sphere A, $q_{A}$ = 2 $*$ 6.5 $* 10^{- 7}$ C = 1.3 $* 10^{- 6} C$

Charge on sphere B, $q_{B}$ = 2 $*$ 6.5 $* 10^{- 7}$ C = 1.3 $* 10^{- 6} C$

Distance between the spheres, r =
$\frac{50}{2}$ = 25 cm = 0.25 m

Force of repulsion between two spheres

F = $\frac{1}{4 π ?_{0}}$ $* \frac{q_{1} q_{2}}{r^{2}}$ , where $?_{0}$ = Permittivity of free space = 8.854 $* 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$ Therefore, F =
$\frac{1}{4 * π * {8.854 * 10^{- 12}}_{}}$ $* \frac{1.3 * 10^{- 6} * 1.3 * 10^{- 6}}{{0.25}^{2}}$ N = 0.243 N

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13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ${}^{235}U$ to be about 200MeV.

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Physics Ncert Solutions Class 12th Nuclei

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13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $* 10^{5}$ MW = 2 $* 10^{4}$ MW = 2 $* 10^{4} * 10^{6}$ J/s

= 2 $* 10^{4} * 10^{6} * 60 * 60 * 24 * 365$ J/y = 6.3072 $* 10^{17}$ J/y

Heat energy release per fission of a ${}^{235}U$ nucleus, E = 200 MeV

Efficiency of a reactor = 25%

So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV

= 50 $* 10^{6}$ eV = 50 $* 10^{6} * 1.6 * 10^{- 19}$ J = 8 $* 10^{- 12}$ J

Therefore, number of atoms required per year = $\frac{6.3072 * 10^{17}}{8 * 10^{- 12}}$ = 7.884 $* 10^{28}$

1 mole of ${}^{235}U$ = 235 gm of ${}^{235}U$ contains 6.023 $* 10^{23}$ atoms

Hence the mass of 7.884 $* 10^{28} a t o m s$ = $\frac{235 * 10^{-3}}{6.023 * 10^{23}}$ $*$ 7.884 $* 10^{28}$ = 30.76 $* 10^{3}$ kg

Hence, the Urani

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13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $* 10^{5}$ MW = 2 $* 10^{4}$ MW = 2 $* 10^{4} * 10^{6}$ J/s

= 2 $* 10^{4} * 10^{6} * 60 * 60 * 24 * 365$ J/y = 6.3072 $* 10^{17}$ J/y

Heat energy release per fission of a ${}^{235}U$ nucleus, E = 200 MeV

Efficiency of a reactor = 25%

So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV

= 50 $* 10^{6}$ eV = 50 $* 10^{6} * 1.6 * 10^{- 19}$ J = 8 $* 10^{- 12}$ J

Therefore, number of atoms required per year = $\frac{6.3072 * 10^{17}}{8 * 10^{- 12}}$ = 7.884 $* 10^{28}$

1 mole of ${}^{235}U$ = 235 gm of ${}^{235}U$ contains 6.023 $* 10^{23}$ atoms

Hence the mass of 7.884 $* 10^{28} a t o m s$ = $\frac{235 * 10^{-3}}{6.023 * 10^{23}}$ $*$ 7.884 $* 10^{28}$ = 30.76 $* 10^{3}$ kg

Hence, the Uranium needed per year is 30.76 $* 10^{3}$ kg

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13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of ${}_{92}^{235}U$ in a fission reactor.

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