Physics

Mass, m = 14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity,  $\omega$ = 2 rev/s

Cross sectional area of the wire, A = 0.065 ${cm}^2$ = 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 ${\omega }^2$ = 14.5 $*\left(9.81+1*2^2\right)$ = 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

11.23 Wavelength produced by X-ray, $\lambda =$ 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:

$E$ = $\frac{hc}{\lambda }$ = $\frac{6.626*{10}^{-34}*3*{10}^8}{0.45\mathrm{}*{10}^{-10}}$ = 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.

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 $*{10}^{9}Pa$ and strain is s, then $Y_c$ = $\frac{F}{\frac{\pi }{4}{d_c}^2}$ , $d_c=\sqrt{\frac{4F}{\pi *Y_c}}$

Similarly, if $d_i$ is the diameter of iron wire and $Y_i$ is the Young's modulus of iron = 190 $*{10}^{9}Pa$ , then $d_i$ = $\sqrt{\frac{4F}{\pi *Y_i}}$

$\frac{d_c}{d_i}$ = $\sqrt{\frac{Y_i}{Y_c}}$ = $\sqrt{\frac{190}{110}}$ = 1.314

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 $v^2$ = $eV$

$v^2=\frac{2eV}{m}$ …….(1)

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

$\frac{m{v}^2}{r}=evB$

$v$ = $\frac{eBr}{m}$ ………………(2)

Equating equations (1) and (2) we get

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

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

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

Area of cross-section, A = $\pi {1.5}^2$ ${cm}^2$ = 7.07 $*{10}^{-4}{m}^2$

Maximum stress = Maximum load / cross sectional area

Maximum load = Maximum stress $*$ cross sectional area = 108 $*$ 7.07 $*{10}^{-4}$ = 7.07 $*{10}^4$ N

Area of cross-section, A = 15.2 mm $*19.1\mathrm{m}\mathrm{m}$ = 15.2 $*19.1*{10}^{-6}{\mathrm{m}}^2=2.9\mathrm{}*{10}^{-4}{\mathrm{m}}^2$

Force, F = 44500 N

Stress, F/A = (44500/$2.9\mathrm{}*{10}^{-4})$ N/ ${\mathrm{m}}^2$

Modulus of elasticity,  $\eta$ = Stress / Strains, Strains = Stress / $\eta$

For copper,  $\eta$ = 42$*{10}^9$ $\mathrm{N}/{\mathrm{m}}^2$

Strains = (44500/$2.9\mathrm{}*{10}^{-4})/($ 42 $*{10}^9)$ = 3.65 $*{10}^{-3}$

Mass of the big structure, M = 50000 kg = 50000 $*9.8N$ = 4.9 $*{10}^5$ N

Inner radius of the column, r = 30 cm = 0.3 m

Outer radius of the column, R = 60 cm = 0.6 m

Young's modulus of steel, Y = 2 $*{10}^{11}$ Pa

Total force exerted on 4 columns, F = 4.9 $*{10}^5$ N

Force exerted on single column, f = F/4 = 1.225 $*{10}^5$ N

Cross sectional area of each column, A = $\pi (R^2-r^2)$ = $\pi ({0.6}^2-{0.3}^2)$ = 0.848 $m^2$

Stress in each column = f/A

Young's modulus, Y = Stress / Strain, Strain = Stress / Y = f/ (A $*Y)$ = $\frac{1.225\mathrm{}*{10}^5}{0.848*2*{10}^{11}}=7.22*{10}^{-7\mathrm{}}$ m

Edge of the aluminium cube, L = 10 cm = 0.1 m, Area A = 0.01 $m^2$

Mass attached, m = 100 kg = 100 $*$ 9.8 = 980 N = Applied force F

Shear modulus $\eta$ = 25 GPa = 25 $*{10}^{9}Pa$

Shear modulus $\eta$ = Shear stress / Shear strain = $\frac{\frac{F}{A}}{\frac{?L}{L}}$ ,  $?L=\frac{F*L}{\eta *A}$ = $\frac{980*0.1}{25\mathrm{}*{10}^9*0.01}$ = 3.92 $*{10}^{-7}m$

Diameter of wires, d = 0.25 cm, radius, r = 0.125 cm

Cross-sectional area, $A_1$ = $A_2$ = $\pi *r^2=0.049{cm}^2$ = 4.908 $*{10}^{-6}$ $m^2$

Length of the steel wire, $L_1=1.5\mathrm{}\mathrm{m}$ , length of the brass wire, $L_2=1.0\mathrm{}\mathrm{m}$

Change in length of the steel wire $=?L_1,$ Change in length of the copper wire $=?L_2$

Total force exerted on the steel wire, $F_1$ = ( 4+6) kg = 10 kg = 98 N

Young's modulus of steel , $Y_1$ = $\frac{\frac{F_1}{A_1}}{\frac{?L_1}{L_1}}$ = 2.0 $*{10}^{11}$ Pa

$?L_1$ = $\frac{F_1}{A_1}*\frac{L_1}{Y_1}=\mathrm{}\frac{98*1.5}{4.908*{10}^{-6}*2.0*{10}^{11}}$ = 1.497 $*{10}^{-4}$ m

Similarly for brass wire, $F_2$ = 6 kg = 58.8 N, $Y_2$ = $0.91*{10}^{11}\mathrm{}\mathrm{P}\mathrm{a}$

$?L_2$ = $\frac{F_2}{A_2}*\frac{L_2}{Y_2}=\mathrm{}\frac{58.8*1.0}{4.908*{10}^{-6}*0.91*{10}^{11}}$ = 1.316 $*{10}^{-4}$ m