Physics

9.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

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Hydraulic pressure exerted on glass slab, p = 10 atm = 10 $* 1.1013 * 10^{5}$ Pa

Bulk modulus of glass, k = 37 $* 10^{9}$ N $m^{- 2}$

From the relation k = $V \frac{Δ P}{Δ V}$ , we get $\frac{Δ V}{V}$ = $\frac{Δ P}{k}$ = $\frac{10 * 1.1013 * 10^{5}}{37 * 10^{9}}$ = 2.976 $* 10^{- 5}$

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11.25 Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never 'count photons', even in barely detectable light.

(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.

(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive ( 10^–10 W m^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white lig

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11.25 The power of the medium wave transmitter, P = 10 kW = 10 $* 10^{3}$ W = $10^{4}$ J/s

Hence energy emitted by the transmitter per second, E = $10^{4}$ J

Wavelength of the radio wave, $λ$ = 500 m

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

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

Energy of the wave is given as :

$E_{w}$ = $\frac{h c}{λ}$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{500}$ = 3.98 $* 10^{- 28}$ J

Let n be number of photons emitted by the transmitter. Hence, total energy transmitted is given by:

n $E_{w}$ = E

n = $\frac{E}{E_{w}}$ = $\frac{10^{4}}{3.98 * 10^{- 28}}$ = 2.52 $* 10^{31}$

Intensity of light perceived by the human eye, I = $10^{- 10}$ W $m^{- 2}$

Area of the pupil, A = 0.4 ${c m}^{2}$ = 0.4 $* 10^{- 4} m^{2}$

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11.25 The power of the medium wave transmitter, P = 10 kW = 10 $* 10^{3}$ W = $10^{4}$ J/s

Hence energy emitted by the transmitter per second, E = $10^{4}$ J

Wavelength of the radio wave, $λ$ = 500 m

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

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

Energy of the wave is given as :

$E_{w}$ = $\frac{h c}{λ}$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{500}$ = 3.98 $* 10^{- 28}$ J

Let n be number of photons emitted by the transmitter. Hence, total energy transmitted is given by:

n $E_{w}$ = E

n = $\frac{E}{E_{w}}$ = $\frac{10^{4}}{3.98 * 10^{- 28}}$ = 2.52 $* 10^{31}$

Intensity of light perceived by the human eye, I = $10^{- 10}$ W $m^{- 2}$

Area of the pupil, A = 0.4 ${c m}^{2}$ = 0.4 $* 10^{- 4} m^{2}$

Frequency of white light, $ν$ = 6 $* 10^{14}$ Hz

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

Energy emitted by photon is given as:

$E = h ν$ = 6.626 $* 10^{- 34} * 6 * 10^{14}$ = 3.9756 $* 10^{- 19}$ J

If n is total number of photons falling per second, per unit area of the pupil,

Total energy per unit area of n photons is = n $*$ 3.9756 $* 10^{- 19}$ J = Intensity of light

n $*$ 3.9756 $* 10^{- 19}$ = $10^{- 10}$

n = 2.52 $* 10^{8}$

Total number of photons entering the pupil per second is given as:

$n_{A}$ = n $* A$ = 2.52 $* 10^{8} *$ 0.4 $* 10^{- 4}$ = 10.06 $* 10^{3}$ /s

The number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.

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9.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 * 10³ kg m^–3?

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Let us assume the depth = h, pressure at depth, $P_{h}$ = 80 atm = 80 $* 1.013 * 10^{5}$ Pa

Density of water at the surface, $ρ_{s}$ = 1.03 $* 10^{3}$ kg/ $m^{3}$

Let density of water at depth h be $ρ_{h}$

Let $V_{s}$ be the volume at the surface and $V_{h}$ be the volume at depth h and ΔV be the change in volume. Let m be the mass of water.

ΔV = $V_{s} - V_{h}$ From the relation m = $ρ V,$ we get

ΔV = m ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ ) = $ρ_{s} V_{s}$ ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ )

$\frac{Δ V}{V_{s}}$ = 1- $\frac{ρ_{s}}{ρ_{h}}$ ……(i)

Bulk modulus of water, k = $V_{1} \frac{Δ P}{Δ V}$ = $V_{s} \frac{Δ P}{Δ V}$

$\frac{Δ V}{V_{s}}$ = $\frac{Δ P}{k}$ …….(ii)

Bulk modulus of water, k = 2.1 $* 10^{9}$ Pa

Hence $\frac{Δ V}{V_{s}} = \frac{80 * 1.013 * 10^{5}}{2.1 * 10^{9}}$ = 3

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Let us assume the depth = h, pressure at depth, $P_{h}$ = 80 atm = 80 $* 1.013 * 10^{5}$ Pa

Density of water at the surface, $ρ_{s}$ = 1.03 $* 10^{3}$ kg/ $m^{3}$

Let density of water at depth h be $ρ_{h}$

Let $V_{s}$ be the volume at the surface and $V_{h}$ be the volume at depth h and ΔV be the change in volume. Let m be the mass of water.

ΔV = $V_{s} - V_{h}$ From the relation m = $ρ V,$ we get

ΔV = m ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ ) = $ρ_{s} V_{s}$ ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ )

$\frac{Δ V}{V_{s}}$ = 1- $\frac{ρ_{s}}{ρ_{h}}$ ……(i)

Bulk modulus of water, k = $V_{1} \frac{Δ P}{Δ V}$ = $V_{s} \frac{Δ P}{Δ V}$

$\frac{Δ V}{V_{s}}$ = $\frac{Δ P}{k}$ …….(ii)

Bulk modulus of water, k = 2.1 $* 10^{9}$ Pa

Hence $\frac{Δ V}{V_{s}} = \frac{80 * 1.013 * 10^{5}}{2.1 * 10^{9}}$ = 3.86 $* 10^{- 3}$

From equation (i) $\frac{ρ_{s}}{ρ_{h}}$ = 0.99614

$ρ_{h} = 1.03 * 10^{3} / 0.99614$ = 1.034 $* 10^{3}$ kg/ $m^{3}$

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9.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 liter, Pressure increase = 100.0 atm (1 atm = 1.013 * 10⁵ Pa), Final volume = 100.5 liter. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

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Initial volume, $V_{1}$ = 100 lit = 100 $* 10^{- 3}$ $m^{3}$

Final volume, $V_{2}$ = 100.5 lit = 100.5 $* 10^{- 3}$ $m^{3}$

Increase in volume, ΔV = $V_{2} - V_{1}$ = 0.5 $* 10^{- 3}$ $m^{3}$

Increase in pressure, ΔP = 100 atm = 100 $*$ 1.013 $* 10^{5} P a$

Bulk modulus, k = $V_{1} \frac{Δ P}{Δ V}$ = $\frac{100 * 10^{- 3} * 100 * 1.013 * 10^{5}}{0.5 * 10^{- 3}}$ Pa = 2.026 $* 10^{9}$ Pa

Bulk modulus of air = 1.0 $* 10^{5} P a$

(Bulk modulus of water / Bulk modulus of air) = (2.026 $* 10^{9}) /$ 1.0 $* 10^{5}$ = 2.026 $* 10^{4}$

This higher ratio is attributed to the higher compressibility of air than water.

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11.24 In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two -rays of equal energy. What is the wavelength associated with each -ray? (1BeV = 10⁹ eV)

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11.24 The total energy of two X-rays = 10.2 BeV = 10.2 $* 10^{9} e V$ = 10.2 $* 10^{9} * 1.6 * 10^{- 19}$ J

Hence energy of each X-ray $E$ = $\frac{10.2 * 10^{9} * 1.6 * 10^{- 19}}{2}$ = 8.16 $* 10^{- 10}$ J

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

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

From the relation of energy and wavelength, we get

$E$ = $\frac{h c}{λ}$ or

$λ = \frac{h c}{E}$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{8.16 * 10^{- 10}}$ = 2.436 $* 10^{- 16}$ m

Therefore the wavelength associated with each X-ray is 2.436 $* 10^{- 16}$ m

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9.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.

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Mass, m = 14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity, $ω$ = 2 rev/s

Cross sectional area of the wire, A = 0.065 ${c m}^{2}$ = 0.065 $* 10^{- 4} m^{2}$

Let Δl be the elongation of the wire

When the mass is placed at the position of the vertical circle, the total force on the mass is

F = mg + ml $ω^{2}$ = 14.5 $* (9.81 + 1 * 2^{2})$ = 200.25 N

Young's modulus for steel, $Y_{s}$ = Stress / Strain = 2 $* 10^{11}$ Pa

Stress = F/A = 200.25/0.065 $* 10^{- 4}$

Strain = (Δl/l) = (Δl/1) = Δl

Δl = (200.25/0.065 $* 10^{- 4}) /$ 2 $* 10^{11}$ = 1.54 $* 10^{- 4}$ m

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11.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?

(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

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11.23 Wavelength produced by X-ray, $λ =$ 0.45 Å = 0.45 $* 10^{- 10}$ m

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

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

The maximum energy of a photon is given as:

$E$ = $\frac{h c}{λ}$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{0.45 * 10^{- 10}}$ = 4.417 $* 10^{- 15}$ J = $\frac{4.417 * 10^{- 15}}{1.6 * 10^{19}}$ eV = 27.6 $* 10^{3}$ eV = 27.6 keV

Therefore, the maximum energy of an X-ray photon is 27.6 keV

To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic energy. Hence an accelerating voltage in the order of 30 keV will be required.

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9.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

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It is given that the tension, F in each wire is same. Since the wire is of same length, Strain also will be same.

If $d_{c}$ is the diameter of copper wire and Young's modulus of copper $Y_{c}$ = 110 $* 10^{9} P a$ and strain is s, then $Y_{c}$ = $\frac{F}{\frac{π}{4} {d_{c}}^{2}}$ , $d_{c} = \sqrt{\frac{4 F}{π * Y_{c}}}$

Similarly, if $d_{i}$ is the diameter of iron wire and $Y_{i}$ is the Young's modulus of iron = 190 $* 10^{9} P a$ , then $d_{i}$ = $\sqrt{\frac{4 F}{π * Y_{i}}}$

$\frac{d_{c}}{d_{i}}$ = $\sqrt{\frac{Y_{i}}{Y_{c}}}$ = $\sqrt{\frac{190}{110}}$ = 1.314

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11.22 An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure ( mm of Hg). A magnetic field of 2.83 * 10^–4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the 'fine beam tube' method.) Determine e/m from the data.

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11.22 Potential, V = 100 V

Magnetic field experienced by electron, B = 2.83 $* 10^{- 4}$ T

Radius of the circular orbit, r = 12.0 cm = 12 $* 10^{- 2}$ m

Mass of each electron = m

Charge on each electron = e

Velocity of each electron= v

The energy of each electron is equal to its kinetic energy, i.e.

$\frac{1}{2}$ m $v^{2}$ = $e V$

$v^{2} = \frac{2 e V}{m}$ …….(1)

Since centripetal force ( $\frac{m v^{2}}{r})$ = Magnetic force (evB), we can write

$\frac{m v^{2}}{r} = e v B$

$v$ = $\frac{e B r}{m}$ ………………(2)

Equating equations (1) and (2) we get

$\frac{2 e V}{m} = \frac{e^{2} B^{2} r^{2}}{m^{2}}$

$\frac{e}{m}$ = $\frac{2 V}{B^{2} r^{2}}$ = $\frac{2 * 100}{({2.83 * 10^{- 4})}^{2} * ({12 * 10^{- 2})}^{2}}$ = 1.734 $* 10^{11}$ C/kg

Therefore, the specific charge ratio (e/m) is 1.734 $* 10^{11}$ C/kg

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9.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ N m–2, what is the maximum load the cable can support?

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