By QB365 on 31 Dec, 2022
QB365 provides a detailed and simple solution for every Possible Questions in Class 12 Physics Subject - Important 3 Mark English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
12th Standard
Physics
Answer all the following questions.
Consider two point charges q1 and q2 at rest as shown in the figure.
They are separated by a distance of 1m. Calculate the force experienced by the two charges for the following cases:
(a) q1 = +2μC and q2 = +3μC
(b) q1 = +2μC and q2 = -3μC
(c) q1= +2μC and q2 = -3μC kept in water (εr = 80)
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.
To create the spark, an electric field of magnitude 3 x 106 Vm-1 is required.
(a) What potential difference must be applied to produce the spark?
(b) If the gap is increased, does the potential difference increase, decrease or remains the same?
(c) find the potential difference if the gap is 1 mm.
A battery of voltage V is connected to 30 W bulb and 60 W bulb as shown in the figure.
(a) Identify brightest bulb
(b) which bulb has greater resistance?
(c) Suppose the two bulbs are connected in series, which bulb will glow brighter?
Three identical lamps each having a resistance R are connected to the battery of emf ε as shown in the figure.
Suddenly the switch S is closed.
(a) Calculate the current in the circuit when S is open and closed
(b) What happens to the intensities of the bulbs A, B, and C.
(c) Calculate the voltage across the three bulbs when S is open and closed
(d) Calculate the power delivered to the circuit when S is opened and closed
(e) Does the power delivered to the circuit decrease, increase or remain same?
A proton moves in a uniform magnetic field of strength 0.500 T magnetic field is directed along the x - axis. At initial time, t = 0s, the proton has velocity
\(\hat { v } =(1.95\times { 10 }^{ -5 }\hat { i } +2.00\times { 10 }^{ 5 }\hat { k } )m{ s }^{ -1 }\). Find
(a) At initial time, what is the acceleration of the proton.
(b) Is the path circular or helical? If helical, calculate the radius of helical trajectory and also calculate the pitch of the helix (Note: Pitch of the helix is the distance travelled along the helix axis per revolution).
Calculate the magnetic field at the centre of a square loop which carries a current of 1.5 A, length of each side being 50 cm.
A circular loop of area 5 x 10–2 m2 rotates in a uniform magnetic field of 0.2T. If the loop rotates about its diameter which is perpendicular to the magnetic field as shown in figure. Find the magnetic flux linked with the loop when its plane is
(i) normal to the field
(ii) inclined 60o to the field and
(iii) parallel to the field.
A 500 μH inductor, \(\\ \frac { 80 }{ { \pi }^{ 2 } } pF\) capacitor and a 628 Ω resistor are connected to form a series RLC circuit. Calculate the resonant frequency and Q-factor of this circuit at resonance.
Write short notes on
(a) microwaves
(b) X - rays
(c) Radio waves
(d) Visible spectrum
Give two uses each of
(i) IR radiation,
(ii) Microwaves and
(iii) UV radiation.
A compound microscope has a magnification of 30. The focal length of eye piece is 5 cm Assuming the final image to be at least distance of distinct vision, find the magnification produced by the objective.
Find the position of the image of a point object O in the two cases given. Take the radius of curvature of the surface R as 15 cm, n1 = 1 and n2 = 2.
Case i) O is located 10 cm to the left of the surface.
Case ii) O is located 30 cm to the left of the surface.
When a 6000Å light falls on the cathode of a photo cell, photoemission takes place. If a potential of 0.8 V is required to stop emission of electron, then determine the
(i) frequency of the light
(ii) energy of the incident photon
(iii) work function of the cathode material
(iv) threshold frequency and
(v) net energy of the electron after it leaves the surface.
Calculate the momentum and the de Broglie wavelength in the following cases:
i) an electron with kinetic energy 2 eV.
ii) a bullet of 50 g fired from rifle with a speed of 200 m/s
iii) a 4000 kg car moving along the highways at 50 m/s
Hence show that the wave nature of matter is important at the atomic level but is not really relevant at macroscopic level.
On your birthday, you measure the activity of the sample 210Bi which has a half-life of 5.01 days. The initial activity that you measure is 1µCi.
(a) What is the approximate activity of the sample on your next birthday? Calculate
(b) the decay constant
(c) the mean life
(d) initial number of atoms.
(a) Calculate the disintegration energy when stationary \(_{ 92 }^{ 232 }{ U }\) nucleus decays to thorium \(_{ 90 }^{ 228 }{ Th }\) with the emission of α particle. The atomic masses are of \(_{ 92 }^{ 232 }{ U }\) = 232.037156 u, \(_{ 90 }^{ 228 }{ Th }\) = 228.028741u and \(_{ 2 }^{ 4 }{ He }\) = 4.002603 u
(b) Calculate kinetic energies of \(_{ 90 }^{ 228 }{ Th }\) and α-particle and their ratio.
What is the difference between Nano materials and Bulk materials?
Elaborate any two types of Robots with relevant examples.
What are polariser and analyser?
A monochromatic light of wavelength 5000 Å passes through a single slit producing diffraction pattern for the central maximum as shown in the figure. Determine the width of the slit.
Four silicon diodes and a 10 Ω resistor are connected as shown in figure below. Each diode has a resistance of 1Ω. Find the current flows through the 10Ω resistor.
The output characteristics of a transistor connected in common emitter mode is shown in the figure. Determine the value of IC when VCE = 15 V. Also determine the value of IC when VCE is changed to 10 V
Define and derive an expression for the energy density in parallel plate capacitor.
The mass of an electron in hydrogen atom is 9.1 x 10-31 Kg. It revolves around the nucleus in a circular orbit of radius 0.53 \(\stackrel {o} A\) calculate its angular velocity [e = 1.6 x 10-19C]
Derive a relation between internal resisance and emf of a cell.
State seebeck effect and give two applications.
Calculate the magnetic field at a point P which is perpendicular bisector to current carrying straight wire as shown in figure.
Can a galvanometer be used for measuring the current? Explain.
State the principle of a Transformers.
The frequency of a ratio station is in the range from 7.5 MHz to 12 MHz band. Calculate the corresponding wavelength band.
Answers
(a) q1 = +2 μC, q2 = +3 μC, and r = 1m. Both are positive charges. so the force will be repulsive.
Force experienced by the charge q2 due to q1 is given by
\(\vec { { F }_{ 21 } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } \)
Here \(\hat { r_{ 12 } } \) is the unit vector from q1 to q2. Since q2 is located on the right of q1, we have
\(\hat { r_{ 12 } } =\hat { i } \), and \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } }=9 \times10^9\) so that
\(\vec { { F }_{ 21 } } =\frac { 9\times 10^{ 9 }\times 2\times 10^{ -6 }\times 3\times 10^{ -6 } }{ 1^{ 2 } } \hat { i } \)
= 54 x 10-3 N\(\hat { i } \)
According to Newton’s third law, the force experienced by the charge q1 due to q2 is \(\vec { { F }_{ 12 } } =-\vec { { F }_{ 21 } } \) Therefore,
\(\vec { F_{ 12 } } \)= 54 x 10-3 N\(\hat { i } \)
The directions of \(\vec { { F }_{ 21 } } \) and \(\vec { { F }_{ 12 } } \) are shown in the figure (case (b)).
(b) q1 = +2 μC, q2 = –3 μC, and r = 1m. They are unlike charges. So the force will be attractive.
Force experienced by the charge q2 due to q1 is given by
\( \vec{F}_{21} =\frac{9 \times 10^{9} \times\left(2 \times 10^{-6}\right) \times\left(-3 \times 10^{-6}\right)}{1^{2}} \hat{r}_{12} \)
\(=-54 \times 10^{-3} \mathrm{~N} \hat{i}\left(\mathrm{Using} \hat{r}_{12}=\hat{i}\right)\)
The charge q2 will experience an attractive force towards q1 which is in the negative x direction.
According to Newton’s third law, the force experienced by the charge q1 due to q2 is \(\vec{F}_{12}=-\vec{F}_{21}\) Therefore,
\(\vec{F}_{12}=54 \times 10^{-3} \widehat{i} \mathrm{~N}\)
The directions of \(\vec{F}_{21} \text { and } \vec{F}_{12}\) are shown in the figure (case (b)).
(c) If these two charges are kept inside the water, then the force experienced by q2 due to q1
\(\vec { { F }_{ 21 }^{ W } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } \)
since ε = εrε0,
we have \(\vec { { F }_{ 21 }^{ W } } =\frac { 1 }{ 4\pi { { \varepsilon }_{ r }\varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } =\frac { \vec { F_{ 21 } } }{ { \varepsilon }_{ r } } \)
Therefore,
\(\vec { { F }_{ 21 }^{ W } } =\frac { 54\times { 10 }^{ -3 }N }{ 80 } \hat { i } \) = -0.675 x 10-3 N\(\hat { i } \)
Distance of separation d = 0.6 x 10-3 m
Electric field E = 3 x 106 Vm-1
\(E=\frac{V}{d}\)
(a) Potential difference V = E x d
V = 3 x 106 x 0.6 x 10-3
V = 1800 V
(b) If the gap (distance d) is increased then potential difference would be increased, since V = E x d
(c) If E = 3 x 106 Vm-1 and distance of separation d = 1 x 10-3 m
Potential difference V = E x d
V = 3 x 106 x 10-3
V = 3 x 103 = 3000 V
(a) The power delivered by the battery P = VI. Since the bulbs are connected in parallel, the voltage drop across each bulb is the same. If the voltage is kept fixed, then the power is directly proportional to current (P ∝ I). So 60 W bulb draws twice as much as current as 30 W and it will glow brighter than 30 W bulb.
(b) To calculate the resistance of the bulbs, we use the relation \(P=\frac { { v }^{ 2 } }{ R } \) In both the bulbs, the voltage drop is the same, so the power is inversely proportional to the resistance or resistance is inversely proportional to the power \(\left( R∝ \frac { 1 }{ P } \right) \). It implies that, the 30W has twice as much as resistance as 60 W bulb.
(c) When these two bulbs are connected in series, the current passing through each bulb is the same. It is equivalent to two resistors connected in series. The bulb which has higher resistance has higher voltage drop. So 30W bulb will glow brighter than 60W bulb. So the higher power rating does not always imply more brightness and it depends whether bulbs are connected in series or parallel.
Electrical quantities | Switch S is open | Switch S is closed |
(a) Current | \(\frac { \varepsilon}{ 3R } \) | \(\frac { \varepsilon }{ 2R } \) |
(b) Intensity | All the bulbs glow with equal intensity. | The intensities of the bulbs A and B equally increase. Bulb C will not glow since no current pass through it. |
(c) Voltage | \({ V }_{ A }=\frac {\varepsilon }{ 3 } ,\) \({ V }_{ B }=\frac { \varepsilon }{ 3 } ,\) \({ V }_{ C }=\frac { \varepsilon }{ 3 } \) |
\({ V }_{ A }=\frac { \varepsilon }{ 2 } ,\) \({ V }_{ B }=\frac { \varepsilon }{ 2 } ,\) Vc= 0 |
(d) Power | \(P_{ A }=\frac { {\varepsilon }^{ 2 } }{ 9R } ,\) \({ P }_{ B }=\frac { { \varepsilon }^{ 2 } }{ 9R } ,\) \({ P }_{ C }=\frac { { \varepsilon }^{ 2 } }{ 9R } \) |
\(P_{ A }=\frac { { \varepsilon }^{ 2 } }{ 4R } ,\) \({ P }_{ B }=\frac { {\varepsilon }^{ 2 } }{ 4R } ,\) Pc=0 |
(e) Total power delivered to the circuit increases. |
Magnetic field \(\overset { \rightarrow }{ B } ={ 0.500\hat { i } T } \)
Velocity of the particle
\(\hat { v } \) = (1.95 x 105\(\hat { i } \) + 2.00 x 105\(\hat { k } \)) ms-1
Charge of the proton q = 1.67 x 10-19 C
Mass of the proton m = 1.67 x 10-27kg
(a) The force experienced by the proton is \(\overset { \rightarrow }{ F } \) = q(\(\overset { \rightarrow }{ v } \) x \(\overset { \rightarrow }{ B } \) )
= 160 x 10-19 x ((1.95 x 105\(\hat { i } \) + 2.00 x 105\(\hat { k } \)) x (0.500 \(\hat { i } \)))
\(\overset { \rightarrow }{ F } \)= 1.60 x 10-14 \(\hat { j } \) N
Therefore, from Newton’s second law,
\(\overset { \rightarrow }{ a } =\frac { 1 }{ m } \overset { \rightarrow }{ F } =\frac { 1 }{ 1.67\times { 10 }^{ -27 } } (1.60\times { 10 }^{ -14 })\hat j\)
\(=9.58\times { 10 }^{ 12 }\hat jm{ s }^{ -2 }\)
(b) Trajectory is helical Radius of helical path is
\(R=\frac { { mv }_{ z } }{ \left| q \right| B } =\frac { 1.67\times { 10 }^{ -27 }\times 2.00\times { 10 }^{ 5 } }{ 1.60\times { 10 }^{ -19 }\times 0.500 } \)
= 4.175 x 10-3m = 4.18mm
Pitch of the helix is the distance travelled along x-axis in a time T, which is P = vx T
But time, \(T=\frac { 2\pi }{ \omega } =\frac { 2\pi m }{ \left| q \right| B } =\frac { 2\times 3.14\times 1.67\times { 10 }^{ -27 } }{ 1.60\times 1{ 0 }^{ -19 }\times 0.500 } =13.1\times { 10 }^{ -8 }s\)
Hence, pitch of the helix is
\(p={ v }_{ x }T=(1.95\times { 10 }^{ 5 })(13.1\times { 10 }^{ -8 })=25.5\times { 10 }^{ -3 }m=25.5mm\)
The proton experiences appreciable acceleration in the magnetic field, hence the pitch of the helix is almost six times greater than the radius of the helix.
The magnetic field at the centre of a square loop due to either arm is
\( \mathrm{B}_{1}=\frac{\mu_{o}}{4 \pi}\left(\frac{I}{L / 2}\right)\left(\sin \phi_{1}+\sin \phi_{2}\right) \)
\(Here, \ \phi_{1}=\phi_{2}=45^{\circ}\)
\(\therefore B_{1} =\frac{\mu_{o}}{4 \pi} \times \frac{2 I}{L}\left[\sin 45^{\circ}+\sin 45^{\circ}\right] \)
\(=\frac{\mu_{o}}{4 \pi} \times \frac{2 I}{L}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right] \)
\(B_{1} =\frac{\mu_{o}}{4 \pi} \times \frac{2 I}{L} \times\left(\frac{2}{\sqrt{2}}\right)=\frac{\mu_{0}}{4 \pi} \times 2 \frac{\sqrt{2} I}{L} \)
\(\therefore B_{1} =\frac{\mu_{o}}{4 \pi L} \times(2 \sqrt{2}) I \) .....(1)
Magnetic field at the centre of a square loop carrying a current I
\({B}=4 B_{1}=4 \times \frac{\mu_{o}}{4 \pi} \times \frac{2 \sqrt{2} I}{L}\)
\({B}=\frac{\mu_{o}}{\pi} \times \frac{2 \sqrt{2} I}{L}\) .........(2)
Substituting I = 1.5 and l = 50 x 10-2 m in the equations (2) we get
\(B =\frac{4 . \pi \times 10^{-7}}{\pi} \times 2 \sqrt{2} \times \frac{1.5}{50 \times 10^{-2}} \)
\(=8 \sqrt{2} \times 10^{-7} \times 0.03 \times 10^{2} \)
\(=8 \sqrt{2} \times 10^{-7} \times 3 \times 10^{-2} \times 10^{2} \)
\(=24 \sqrt{2} \times 10^{-7} \)
\(=24 \times 1.414 \times 10^{-7}=33.936 \times 10^{-7} \mathrm{~T} \)
\(\mathrm{~B} =3.3936 \times 10^{-6} \mathrm{~T} \)
∴ Magnetic field B = 3.4 x 10-6 T
A = 5 x 10-2 m2; B = 0.2 T
(i) θ = 0°;
\({ \Phi }_{ B }=BAcos\theta =0.2\times 5\times { 10 }^{ -2 }\times { cos }0^{ o }\)
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }Wb\)
(ii) θ = 90° – 60° = 30°;
\(\Phi_B\) = BAcosθ = 0.2 x 5 x 10-2 x cos 30o
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }\times \frac { \sqrt { 3 } }{ 2 } =8.66\times { 10 }^{ -3 }Wb\)
(iii) θ = 90°;
\({ \Phi }_{ B }\) = BA cos90o = 0
L = 500 x 10-6H; C\(=\frac { 80 }{ { \pi }^{ 2 } } \times { 10 }^{ -12 }F;R=628\Omega \)
(i) Resonant frequency is
\({ f }_{ r }=\frac { 1 }{ 2\pi \sqrt { LC } } =\frac { 1 }{ 2\pi \sqrt { 500\times { 10 }^{ -6 }\times \frac { 80 }{ { \pi }^{ 2 } } \times { 10 }^{ -12 } } } \)
\(=\frac { 1 }{ 2\sqrt { 40,000\times { 10 }^{ -18 } } } \)
\(=\frac { 10,000\times { 10 }^{ 3 } }{ 4 } = 2500\)
fr = 2500 KHz
(ii) Q–factor
\(=\frac { { \omega }_{ r }L }{ R } =\frac { 2\times 3.14\times 2500\times { 10 }^{ 3 }\times 500\times { 10 }^{ -6 } }{ 628 } \)
Q = 12.5
(a) Microwaves:
It is produced by special vacuum tubes such as klystron, magnetron and gunn diode. The frequency range of microwaves is 109 Hz to 1011 Hz. These waves undergo reflection and can be polarised.
Uses:
It is used in radar system for aircraft navigation, speed of the vehicle, microwave oven for cooking and very long distance wireless communication through satellites.
(b) X-rays:
lt is produced when there is sudden stopping of high speed electrons at high-atomic number target, and also by electronic transitions among the innermost orbits of atoms. The frequency range of X-rays is from 1017 Hz to 1019 Hz. X-rays have more penetrating power than ultraviolet radiation.
Uses:
X-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.
(c) Radio waves:
It is produced by accelerated motion of charges in conducting wires. The frequency range is from few Hz to 109 Hz. It obeys reflection and diffraction.
Uses:
It is uses in radio and television communication systems and also in cellular phones to transmit voice communication in the ultra high frequency band.
(d) Visible light:
Visible light is produced by incandescent bodies and also it is radiated by excited atoms in gases. The frequency range is from 4 x 1014 Hz to 8 x 1014 Hz. It obeys the laws of interference, diffraction and can be polarised. It exhibits photo-electric effect also.
Uses:
It can be used to study the structure of molecules, arrangement of electrons in external shells of atoms and it causes sensation of vision.
(i) IR radiation:
(i) It is used to produce dehydrated fruits, in green houses to keep the plants warm, heat therapy for muscular pain or sprain, TV remote as a signal carrier, to look through haze fog or mist.
(ii) It is used in night vision or infrared photography.
(ii) Microwaves:
It is used in radar systems for aircraft navigation, speed of the vehicle microwave oven for cooking and very long distance wireless communication through satellites.
(iii) UV radiation:
It is used to destroy bacteria in sterilizing the surgical instruments, burglar alarms, to detect the invisible writing, finger prints and also in the study of atomic structure.
Given data :
Magnification of compound microscope m = 30
Focal length of eye piece f. = 5 cm
Image distance D = 25 cm
To find:
Magnification of objective mo = ?
Formula :
Magnification m = mome
\(m={ m }_{ 0 }\left( 1+\cfrac { D }{ { f }_{ e } } \right) \)
\(30={ m }_{ 0 }\left( 1+\cfrac { 25 }{ 5 } \right) \)
⇒ mo = 5
Case i) \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
applying sign convention, u = –10 cm, R = 15 cm
\(\cfrac { 2 }{ v } -\cfrac { 1 }{ -10 } =\cfrac { \left( 2-1 \right) }{ 15 } ;\cfrac { 2 }{ v } +\cfrac { 1 }{ 15 } =\cfrac { \left( 1 \right) }{ 10 } \)
∴ =− 60 cm
[a virtual image is formed 60 cm, to the left of the surface]
Case ii) \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
applying sign convention, u = –30 cm, R = 15 cm
\(\cfrac { 2 }{ v } -\cfrac { 1 }{ -30 } =\cfrac { \left( 2-1 \right) }{ 15 } ;\cfrac { 2 }{ v } +\cfrac { 1 }{ 30 } =\cfrac { \left( 1 \right) }{ 15 } \)
∴ = 60 cm [a real image is formed 60 cm, to the right of the surface]
\(\lambda=6000Å=6000 \times 10^{-10} \mathrm{~m} ; \mathrm{V}=0.8 \mathrm{v} \)
\(\mathrm{k} \cdot \mathrm{E}=\mathrm{hv}-\phi \)
\(\mathrm{eV}_o=\mathrm{hv}-\phi=\frac{\mathrm{hc}}{\lambda}-\phi \)
\((i) v=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6000 \times 10^{-10}}=5 \times 10^{14} \mathrm{~Hz} \)
\((ii)\ \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10}}=3.313 \times 10^{-19} J\)
\(\mathrm{E}=\frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}}=2.07 \mathrm{eV} \)
\((iii) \ \mathrm{E}=\mathrm{hv}-\mathrm{W}\Rightarrow\mathrm{W}=\mathrm{h} v-\mathrm{E} \)
\(\mathrm{E}=\mathrm{eV_o}=1.6 \times 10^{-19} \times 0.8=1.2 8 \times10^{-19}J\)
\(\mathrm{hv}=6.626 \times 10^{-34} \times 5 \times 10^{14}=3.313 \times 10^{-19}J \)
\(\mathrm{~W}=\frac{(3.313-1.28) \times 10^{-19}}{1.6 \times 10^{-19}}=1.270 \mathrm{eV} \)
W = 1.27 eV
\((iv) \ \mathrm{W}=\mathrm{h} \mathrm{v}_{0} \)
\(v_{0}=\frac{W}{h}=\frac{2.033 \times 10^{-19}}{6.626 \times 10^{-34}}=3.07 \times 10^{14} \mathrm{~Hz} \)
\((v)\ \mathrm{E}=\mathrm{eV}_o=\frac{0.8 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=\mathbf{0 . 8} \mathrm{eV}\)
i) Momentum of the electron is
p = \(\sqrt { 2mK } =\sqrt { 2\times 9.1\times { 10 }^{ -31 }\times 2\times 1.6\times 10^{ -19 } } \)
= 7.63 x 10-25 kg ms-1
Its de Broglie wavelength is
\(\lambda=\frac { h }{ p } =\frac { 6.626\times { 10 }^{ -34 } }{ 7.63\times { 10 }^{ -25 } } \) = 0.868 x 10-9 m
= 8.68 \(\mathring { A }\)
ii) Momentum of the bullet is
p = m\({ \upsilon }\) = 0.050 x 200 = 10 kg ms-1
It's de Broglie wavelength is
\(\lambda=\frac { h }{ p } =\frac { 6.626\times { 10 }^{ -34 } }{ 10 } \) = 6.626 x 10-35 m
iii) Momentum of the car is
p = mv = 4000 x 50 = 2 x 105 kg ms-1
Its de Broglie wavelength is
\(\lambda=\frac { h }{ p } =\frac { 6.626\times { 10 }^{ -34 } }{ 2\times { 10 }^{ 5 } } \) = 3.313 x 10-39 m
From these calculations, we notice that electron has a significant value of de Broglie wavelength (≈10-9m which can be measured from diffraction studies) but the bullet and car have negligibly small de Broglie wavelengths associated with them (≈10-33m and 10-39m respectively, which are not measurable by any experiment). This implies that the wave nature of matter is important at the atomic level but it is not really relevant at the macroscopic level.
\(\mathrm{T}_{\frac{1}{2}}=5.01 \text { days} \), \(\mathrm{R}_{o}=1 μ \mathrm{Ci}=3.7 \times 10^{10} \times 10^{-6} \) decays per second
\(if\ \mathrm{t}=1\ year\ \mathrm{R}= ? \), \(\tau=? \), \(\lambda=? \), \(\mathrm{~N}_{o}=? \)
a) \(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\lambda t} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\frac{0.6931}{\mathrm{~T}_{1 / 2} }t} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\frac{0.6931}{5.01} \times 365} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-50.5}=\mathrm{R}_{o} \times 1.17 \times 10^{-22} \)
\(\mathrm{R}=1 \mu \mathrm{Ci} \times 1.17 \times 10^{-22}=1.17 \times 10^{-22} \mu \mathrm{Ci} \)
\(\mathrm{R}=1.17 \times 10^{-22} \mu \mathrm{Ci} \)
b) \(\lambda=\frac{0.6931}{T_{\frac{1}{2}}}=\frac{0.6931}{5.01}=0.1383 \text { day }^{-1} \)
\(\lambda=\frac{0.6931}{T_{\frac{1}{2}}}=\frac{0.6931}{5.01 \times 24 \times 60 \times 60}=1.6 \times 10^{-6} \mathrm{~s}^{-1} \)
\(\lambda=1.6 \times 10^{-6} \mathrm{~s}^{-1} \)
c) \(\tau=\frac{1}{\lambda}=\frac{1}{0.1383}=7.23 \text { days } \)
\(\tau=7.23 \text { days } \)
\(R_{o}=\lambda N_{o} \)
\(N_{o}=\frac{R_{o}}{\lambda}=\frac{3.7 \times 10^{10} \times 10^{-6}}{1.6 \times 10^{-6}}=2.3 \times 10^{10} \)
\(N_{o}=2.3 \times 10^{10} \)
The difference in masses
Δm = (mU - mTh - mα)
= (232.037156–228.028741 – 4.002603)u
The mass lost in this decay = 0.005812 u
Since 1u = 931MeV, the energy Q released is
Q = (0.005812 u) x (931 MeV / u)
= 5.41 MeV
This disintegration energy Q appears as the kinetic energy of α particle and the daughter nucleus. In any decay, the total linear momentum must be conserved.
Total linear momentum of the parent nucleus = total linear momentum of the daughter nucleus and α particle. Since before decay, the uranium nucleus is at rest, its momentum is zero. By applying conservation of momentum, we get
0 = \({ m }_{ Th }{ \overrightarrow { \upsilon } }_{ Th }+{ m }_{ \alpha }\overrightarrow { \upsilon } _{ \alpha }\)
\({ m }_{ \alpha }\overrightarrow { \upsilon } _{ \alpha }\) = - \({ m }_{ Th }{ \overrightarrow { \upsilon } }_{ Th }\)
It implies that the alpha particle and daughter nucleus move in opposite directions.
In magnitude mα ሀα = mTh ሀTh
The velocity of α particle ሀα = \(\frac { { m }_{ Th } }{ { m }_{ \alpha } } { \upsilon }_{ Th }\)
Since mTh > mα , ሀα > ሀTh. The ratio of the kinetic energy of α particle to that the daughter nucleus,
\(\frac { K.{ E }_{ \alpha } }{ K.{ E }_{ Th } } =\frac { 1/2{ m }_{ \alpha }{ { \upsilon }_{ \alpha } }^{ 2 } }{ 1/2{ m }_{ Th }{ { \upsilon }_{ Th } }^{ 2 } } \)
By substituting, the value of ሀα into the above equation, we get \(\frac { K.{ E }_{ \alpha } }{ K.{ E }_{ Th } } =\frac { { m }_{ Th } }{ { m }_{ \alpha } } =\frac { 228.02871 }{ 4.002603 } =57\)
The kinetic energy of α particle is 57 times greater than the kinetic energy of the daughter nucleus (\(_{ 90 }^{ 228 }{ Th }\))
The disintegration energy Q = total kinetic energy of products
K.Eα + K.ETh = 5.41 MeV
57K.ETh + K.E Th = 5.41 MeV
K.ETh = \(\frac{5.41}{58}\) MeV = 0.0093 MeV
K.Eα = 57K.ETh = 57 x 0.093 = 5.301 MeV
In fact, 98% of total kinetic energy is taken by the α particle.
(i) If the particle of a solid is of size less than 100 nm, it is said to be a nano solid, if the particle size exceeds 100 nm, it is a bulk solid.
(ii) Both have same chemical properties. But their mechanical, electrical, optical, magnetic properties differ.
a) Human Robot :
Robots are made to resemble humans in appearance and replicate the human activities like walking, lifting, and sensing, etc.
The important part of human Robots are :
(i) Power conversion unit : Robots are powered by batteries, solar power, and hydraulics.
(ii) Actuators: It converts energy into movement. The majority of the actuators produce rotational or linear motion.
(iii) Pneumatic Air Muscles : They are devices that can contract and expand, when air is pumped inside.
(iv) Muscle wires : They are thin strands of wire made of shape memory alloys. They can contract by 5% when electric current is passed through them.
(v) Sensors: Generally used in task environments as it provides information of real-time knowledge.
(vi) Electric motors : Different types of motors are used. The most often used ones are AC motor, Brushed DC motor, Brushless DC motor, Geared DC motor etc,. They are used to actuate the parts of the robots like wheels, arms, fingers, legs, sensors, camera, weopons, systems, etc.
(vii) Robot locomotion : It provides different types of movement to the Robot. They are,
(a) Legged.
(b) Wheeled.
(c) Combination of Legged and Wheeled Locomotion.
(d) Tracked slip/skid.
(viii) Controller : lt is known as the "brain" which is run by a computer program. It gives commands for the moving parts to perform the job.
(vi) Piezo Motors and Ultrasonic Motors : It is used for industrial robots.
(x) Artificial Intelligence : The artificial intelligence is introduced into robots to bring in human like behavior in robots. Now they work on,
(a) Face recognition.
(b) Providing response to player's actions in computer games.
(c) Taking decisions based on previous actions.
(d) To regulate the traffic by analyzing the density of traffic on roads.
(e) Translate words from one language to another.
b) Industrial Robots :
There are six types of industrial robots.
(i) Cartesian.
(ii) Selective Compliance Assembly Robot Arm (SCARA).
(iii) Cylindrical.
(iv) Delta.
(v) Polar.
(vi) Vertically articulated.
Again six-axis robots are ideal for Are Welding, Spot Welding, Material Handling and Machine Tending and other applications. They can work for 24 x 7, stronger and faster than humans. However, humans cannot be replaced by robots in decision making.
(i) The polaroid which polarises the light passing through it is called a polariser.
(ii) The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.
λ = 5000 Å = 5000 x 10 -10 m; sin 30o = 0.5; n = 1; a =?
Equation for diffraction minimum is,
asin θ = nλ
The central maximum is spread up to the first minimum. Hence, n = 1
Rewriting, \(a=\cfrac { \lambda }{ sin\theta } \)
Substituting, \(a=\cfrac { 5000\times { 10 }^{ -10 } }{ 0.5 } \)
a = 1 x 10-6m = 0.001 x 10-3 m = 0.001 mm.
Diode D1 and D4 is reverse biased [open]
Diode D1 and D3 are forward biased.
The resistances are in series
R = 1 + 10 + 1 - 12 Ω
Barier Potential, V = 0.7 + 0.7 = 1.4 V (Silicon diode)
Applying Kirchhoff's voltage Law,
0.7 + I(1) + I(10) + 0.7 + I(1) = 3V
12 I = 3 - 1.4
12 I = 1.6
\(I=\frac{1.6}{12}=\mathbf{0 . 1 3 3 A}\)
When VCE = 15 V, IC = 1.5 μA
When VCE is changed to 10 V, IC = 1.4 μA
Energy stored in the capacitor
\(U=\frac { 1 }{ 2 } { Cv }^{ 2 }\quad \quad ...(1)\)
This is rewritten as using \(C=\frac { { \varepsilon }_{ 0 }A }{ d } \& Ed=V\)
\(U=\frac { 1 }{ 2 } \left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) { (Ed })^{ 2 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }(Ad)\quad { E }^{ 2 }...(2)\)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \({ U }_{ E }=\frac { U }{ Volume } \) From equation (4),
We get
\({ u }_{ E }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }{ E }^{ 2 }\quad \quad \quad \quad \quad ...(3)\)
(iv) The energy density depends only on the electric field and not on the size of the plates of the capacitor.
Force acting on the election \(F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{d^{2}} \quad A=10^{-10} \mathrm{~m}\)
\(F=9 \times 10^{9} \times \frac{\left(1.6 \times 10^{-19}\right)^{2}}{\left(0.53 \times 10^{-10}\right)^{2}}\)
= 8.1 x 10-8N
Centripetal force on the electron is \(F=\frac{m v^{2}}{r}=m r \omega^{2} \quad \therefore \omega^{2}=\frac{F}{m r}\)
\(\omega=\sqrt{\frac{F}{m r}}=\sqrt{\frac{8.1 \times 10^{-8}}{9.31 \times 10^{-31} \times 0.53 \times 10^{-10}}}\)
= 4 x 1016 rad s-1
(i) The emf of cell \(\xi \) is measured by connecting a high resistance voltmeter across it without connecting the external resistance R.
(ii) Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence the voltmeter reading gives the emf of the cell.
(iii) Then, external resistance R is included in the circuit, and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).
(iv) The potential drop across the resistor R is V + IR
(v) Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell . It is because a certain amount of voltage (Ir) has dropped across the internal resistance r.
Then \(V=\xi -Ir\)
\(Ir=\xi -V\)
(vi) Dividing equation (2) by equation (1) we get
\(\cfrac { Ir }{ IR } =\cfrac { \xi -V }{ V } \)
\(r=\left| \cfrac { \xi -V }{ V } \right| R\)
Since \(\xi \) V and R are known, internal resistance r can be determined.
Drift velocity | Mobility | |
i | It is the average velocity acquired by the electrons inside the conductor when it is subject to an electric field |
Mobility is defined as the magnitude of the drift velocity per unit electric field. |
ii | Its unit is m/s | Its unit is m2/Vs |
Let the length MN = y and the point P is on its perpendicular bisector. Let O be the point on the conductor as shown in figure. Therefore,
\(OM=ON=\frac { y }{ 2 } ,then\)
\(cos\varphi _{ 1 }=\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ d }^{ 2 } } } =\frac { adjacent \ length }{ hypotenuse \ length } \)
\(=\frac { ON }{ PH } =-\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ a }^{ 2 } } } =-\frac { y }{ \sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \)
\(cos\varphi _{ 1 }=\frac { adjacent \ length }{ hypotenuse \ length } =\frac { OM }{ PM } \)
\(=-\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ a }^{ 2 } } } =-\frac { y }{ \sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a\sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \hat { n } \)
For long straight wire, Y\(\rightarrow \infty ,\)
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 2\pi a } \hat { n } \)
The result obtained is the same as we obtained in equation (3.39).
A galvanometer cannot be used for measuring the current
(i) A galvanometer has a finite large resistance and is connected in series in the circuit. So it will increase the resistance of circuit and hence change the value of current in the circuit.
(ii) A galvanometer is a very sensitive device, it gives a full scale deflection for the current of the order of microampere hence if connected, it will not measure current of the order of ampere.
(i) principle of transformer is the mutual induction between two coils.
(ii) That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.
Wavelength \(\lambda=\frac{c}{f}\)
when \(\mathrm{f}=7.5 \mathrm{MHz}=7.5 \times 10^{6} \mathrm{~Hz}\)
wavelength \(\lambda=\frac{3 \times 10^{8}}{7.5 \times 10^{6}}=40 \mathrm{~m}\)
when \(\mathrm{f}=12 \mathrm{MHz}=12 \times 10^{6} \mathrm{~Hz}\)
wavelength \(\lambda=\frac{3 \times 10^{8}}{12 \times 10^{6}}=25 \mathrm{~m}\)
\(\therefore \text { The wavelength band is from } 25 \mathrm{~m} \text { to } 40 \mathrm{~m} \text {. }\)