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Physics Numerical Problems

Electromagnetic Waves · Tamil Nadu Board STD 12 (Tamil - English Medium). 6 questions.

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Question 15 Marks
Light With an energy flux of $25 \times 10^4 Wm ^{-2}$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15 cm ^2$, calculate the average force exerted on the surface.
Answer
Average force $=$ momentum transferred per second$F _{ av }=\frac{P}{T}=\frac{2 u}{C}$Where $U$ is the energy falling on th surface per second.
$F_{a v}=\frac{2 \times 25 \times 10^4 \times 15 \times 10^{-4}}{3 \times 10^8}=250 \times 10^{-8}$
$F_{a v}=2.5 \times 10^{-6} N .$
Common Errors And Its Rectifications
Common Errors:
  1. Students do mistakes most of the time in the unit of frequency. They write the units in improper ways. Eg. Hertz (or) H. This is the wrong way.
  2. They may confuse the frequency range of radiations and wavelength range of radiation.
Rectifications:
  1. The correct way of the unit is hertz (or) Hz . The unit of frequency is hertz (or) Hz (or) $s ^{-1}$.
  2. The easy way to understand the frequency and wavelength range of radiations are, Frequency increases the order of gamma, X-ray, UV, visible, IR, microwave, radio wave.
  3. Wavelength increases the order of the reverse of frequency order.
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Question 25 Marks
The energy of the EM wave is of the order of $15\ KeV$. To which part of the spectrum does it belong?
Answer
$E = hv =\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E}$
$h =6.626 \times 10^{-34} JS ; c =3 \times 10^8 ms ^{-1} $
$E =15 \times 10^3 eV =15 \times 10^3 \times 1.6 \times 10^{-19} V $
$\lambda=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{15 \times 10^3 \times 1.6 \times 10^{-19}}=\frac{19.878 \times 10^{-26}}{24 \times 10^{-16}}=0.8282 \times 10^{10} m $
$\lambda=0.8282 \times \mathring A$
Hence X-rays
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Question 35 Marks
Radiation of energy $E$ falls normally on a perfectly reflecting surface. Find the momentum transferred to the surface.
Answer
Momentum of radiation of energy $E$ is $P =\frac{E}{C}$Since the radiation is completely reflected, its momentum changes by $\frac{2 E}{C}$ Therefore, by the law of conservation of momentum the momentum transferred to the surface is $\frac{2 E}{C}$.
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Question 45 Marks
Electromagnetic waves travel in a medium at a speed of $2 \times 10^8 ms ^{-1}$. The relative permeability of the medium is 1 . Find the relative permittivity.
Answer
The speed of an em wave in a medium is given by$\begin{aligned}V & =\sqrt{\frac{1}{\mu \varepsilon}}=\frac{1}{\sqrt{\mu_r \mu_0 \varepsilon_r \varepsilon_0}}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \frac{1}{\sqrt{\mu_r \varepsilon_r}} \\V & = C \frac{1}{\sqrt{\mu_r \varepsilon_r}} \\V ^2 & =\frac{ C ^2}{\mu_r \varepsilon_r}\end{aligned}$Relative permitivity,$\begin{array}{c}\varepsilon_r=\frac{ C ^2}{ V ^2 \mu_r}=\frac{\left(3 \times 10^8\right)^2}{1 \times\left(2 \times 10^8\right)^2}=\frac{9 \times 10^{16}}{4 \times 10^{16}} \\\varepsilon_r=2.25 \text { (no unit) }\end{array}
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Question 55 Marks
The voltage between the plates of a parallel - plate capacitor of capacitance 1 $\mu F$ is changing at the rate of $5 Vs ^{-1}$. What is the displacement current in the capacitor?
Answer
Capacitance of parallel plate capacitor, $C =1 \mu F$$C =1 \times 10^{-6} F$The rate of voltage $b / n$ the plate, $\frac{d v}{d t}=5 Vs ^{-1}$
Displacement current,
$I _{ d }=\varepsilon_0 \frac{d \varphi_F}{d t}=\varepsilon_0 \frac{d}{d t}( EA )$
$\varepsilon_0= A \frac{d}{d t}\left(\frac{V}{d}\right) E =\frac{V}{d}$
$=\frac{\varepsilon_{ o } A }{d} \frac{d V}{d t}= C \cdot \frac{d V}{d t}$
$I _{ d }=1 \times 10^{-6} \times 5$
$Id =5 \mu A .$
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Question 65 Marks
A parallel plate capacitor has circular plates, each of radius $5.0 cm$. It is being charged so that electric field in the gap between its plates rises steadily at the rate of $10^{12} V m ^{-1} s ^{-1}$. What is the displacement current?
Answer
Radius, $r=5 cm =5 \times 10^{-2} m$The rate okf electric frield, $\frac{d E}{d t}=10^{12} V m ^{-1} s ^{-1}$Displacement current, $I _{ d }=\varepsilon_0 \frac{d \varphi_E}{d t}=\varepsilon_0 \frac{d}{d t}( EA )=\varepsilon_0\left(\pi r ^2\right) \frac{d E}{d t}$$=8.85 \times 10^{-12} \times 3.14 \times\left(5 \times 10^{-2}\right)^2 \times 10^{12}$$I _{ d }=0.069$$I _{ d }=0.07$ (or) $70 mA$
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Numerical Problems - Physics STD 12 Questions - Vidyadip