By QB365 on 31 Dec, 2022
QB365 provides a detailed and simple solution for every Possible Questions in Class 12 Physics Subject - Important 5 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
Calculate the electric field due to a dipole on its axial line and equatorial plane.
Answer all the following questions.
Obtain the expression for energy stored in the parallel plate capacitor.
Explain the equivalent resistance of a series and parallel resistor network.
Explain the determination of unknown resistance using meter bridge.
Show that for a straight conductor, the magnetic field
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } (cos\varphi _{ 1 }-cos\varphi _{ 2 })\hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (sin{ \theta }_{ 1 }+sin{ \theta }_{ 2 })\hat { n } \)
What is tangent law? Discuss in detail.
Define self-inductance of a coil interms of
(i) magnetic flux and
(ii) induced emf.
Mention the various energy losses in a transformer.
Explain the Maxwell’s modification of Ampere’s circuital law.
Explain the types of absorption spectrum.
Describe the Fizeau’s method to determine the speed of light.
Derive the equation for angle of deviation produced by a prism and thus obtain the equation for refractive index of material of the prism.
Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
Derive an expression for de Broglie wavelength of electrons.
Explain the J.J. Thomson experiment to determine the specific charge of electron.
Discuss the alpha decay process with example.
Discuss the applications of Nanomaterials in various fields.
Discuss the experiment to determine the wavelength of different colours using diffraction grating.
Mention different parts of spectrometer and explain the preliminary adjustments.
Give circuit symbol, logical operation, truth table, and Boolean expression of
i) AND gate
ii) OR gate
iii) NOT gate
iv) NAND gate
v) NOR gate and
vi) EX-OR gate.
Describe the function of a transistor as an amplifier with the neat circuit diagram. Sketch the input and output wave forms.
A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge \(\frac{Q}{2}\) is placed at the centre C and another is placed at the centre C and another a distance x from the centre as shown in the figure.
(i) Find the electric flux through the shell.
(ii) Find the force on the charges at C and A.
Two charges each of +q coulomb are placed along a line. A third charge -q is placed between them. At what position will the system be in equilibrium?
The current through an element is shown in the figure. Determine the total charge that pass through the element at a) t = 0 s, b) t = 2 s, c) t = 5s
Let I1 and I2 be the steady currents passing through a long horizontal wire XY and PQ respectively. The wire PQ is fixed in horizontal plane and the wire XY be is allowed to move freely in a vertical plane. Let the wire XY is in equilibrium at a height d over the parallel wire PQ as shown in figure.
Show that if the wire XY is slightly displaced and released, it executes Simple Harmonic Motion (SHM). Also, compute the time period of oscillations.
A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere paralled to the line AB. Find the angle of refraction at A if refraction index of the material of the sphere is \(\sqrt { 3 } \).
For photo electronic effect in sodium, the figure shows the plot of cut-off voltage versus frequency of incident radiation. Calculate
(i) threshold frequency
(ii) work function for sodium.
Write the advantages of Robotics than the human.
A long solenoid with 20 turns per cm has a small loop of area 2cm2 placed inside the solenoid normal to its axis. If the circuit carried by the solenorid changes steadily from lA to 3A in 0.2s. What is the-induced emf in the loop while the current is changing.
A plane Electromagnetic wave travels in vacuum along z - direction. What can you say about the directions of electric and magnetic field vectors? If the frequency of the wave is 30 MHz. What is its wavelength?
Answers
Case (i): Electric field due to an electric dipole at points on the axial Iine:
Consider an electric dipole placed on the x-axis as shown in Figure. A point C is located at a distance of r from the midpoint O (of the dipole) along the axial line.
The electric field at a point C due to +q is
\({ \vec { E } }_{ + }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r-a) }^{ 2 } } \) along BC
Since the electric dipole moment vector \(\vec { p } \) is from -q to +q and is directed along BC, the above equation is rewritten as
\({ \vec { E } }_{ + }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r-a) }^{ 2 } } \hat { p } \) ....(1)
When \(\vec { p } \) is the electric dipole moment unit vector from -q to +q. The electric field at a point C due to -q is
\({ \vec { E } }_{ - }=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r+a) }^{ 2 } } \hat { p } \) ....(2)
Since +q is located closer to the point C than -q, \({ \vec { E } }_{ + }\) is stronger than \({ \vec { E } }_{ - }\). Therefore, the length of the \({ \vec { E } }_{ + }\) vector is drawn larger than that of \({ \vec { E } }_{ - }\) vector.
The total electric field at point C is calculated using the superposition principle of the electric field.
\({ \vec { E } }_{ tot }={ \vec { E } }_{ + }+{ \vec { E } }_{ - }\)
\(=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r-a) }^{ 2 } } \hat { p } -\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r+a) }^{ 2 } } \hat { p } \)
\({ \vec { E } }_{ tot }=\frac { q }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { 1 }{ { (r-a) }^{ 2 } } -\frac { 1 }{ { (r+a) }^{ 2 } } \right) \hat { p } \) ....(3)
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } q\left( \frac { 4ra }{( { r }^{ 2 }-{ a }^{ 2 })^2 } \right) \hat { p } \) ...(4)
Note that the total electric field is along \({ \vec { E } }_{ + }\), since +q is closer to C than -q.
If the point C is very far away from the dipole then (r >> a). Under this limit the term \(({ r }^{ 2 }-{ a }^{ 2 })\approx { r }^{ 2 }\).
Substituting this into equation (4), we get
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { 4aq }{ { r }^{ 3 } } \right) \hat { p } (r>>a)\)
\(since\quad 2aq\hat { p } =\vec { p } \)
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2\vec { p } }{ { r }^{ 3 } } (r>>a)\) ...(5)
The direction \({ \vec { E } }_{ tot }\) is shown in Figure.
NOTE: If the point C is chosen on the left side of the dipole, the total electric field is still in the direction of \(\vec { p } \).
Case (ii) Electic field due to an electric dipole at a point on the equatorial plane:
Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane. Since the point C is equidistant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of \({ \vec { E } }_{ + }\) is along BC and the direction of \({ \vec { E } }_{ - }\) is along CA. \({ \vec { E } }_{ + }\) and \({ \vec { E } }_{ - }\) are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(|{ \vec { E } }_{ + }|\) sinθ and \(|{ \vec { E } }_{ -}|\) sinθ are equal in magnitude and oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of \({ \vec { E } }_{ + }\) and \({ \vec { E } }_{ - }\) and its direction is along \(-\hat{p}\) as shown in the Figure.
\({ \vec { E } }_{ tot }=-|{ \vec { E } }_{ + }|cos\theta \hat { p } -|{ \vec { E } }_{ - }|cos\theta \hat { p } \) ...(6)
The magnitudes \({ \vec { E } }_{ + }\) and \({ \vec { E } }_{ - }\) are the same and given by,
\(|{ \vec { E } }_{ + }|=|{ \vec { E } }_{ - }|=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ (r^2+a^2) } \) ...(7)
By substituting equation (7) into equation (6), we get
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2qcos\theta }{ ({ r }^{ 2 }+{ a }^{ 2 }) } \hat { p } ....(8)\)
\(=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2qa }{ ({ r }^{ 2 }+{ a }^{ 2 })^{ \frac { 3 }{ 2 } } } \hat { p } \)
Since \(cos \theta =\frac { a }{ \sqrt { { r }^{ 2 }+{ a }^{ 2 } } } \)
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { \vec { p } }{ ({ r }^{ 2 }+{ a }^{ 2 })^{ \frac { 3 }{ 2 } } } \)
Since \(\vec { p } \) = 2qa\(\hat { p } \) ...(9)
At very large distances (r >> a), the equation (9) becomes
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { \vec { p } }{ { r }^{ 3 } } (r>>a)\) ...(10)
Negative sign shows that direction of Electric field is opposite to the direction of dipole moment vector.
Energy stored in the capacitor
i) Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge - Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
ii) To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = V dQ
Where \(V=\frac { Q }{ C } \) .....(1)
iii) The total work done to charge a capacitor is
\(W=\int _{ 0 }^{ Q }{ \frac { Q }{ C } } dQ=\frac { { Q }^{ 2 } }{ 2C } \quad \quad ....(2)\)
This work done is stored as electrostatic potential energy (UE) in the capacitor.
\({ U }_{ E }=\frac { { Q }^{ 2 } }{ 2C } =\frac { 1 }{ 2 } { CV }^{ 2 },\quad (\therefore Q=CV)\quad ....(3)\)
(iv) This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.Substituting \(C=\frac { { \varepsilon }_{ 0 }A }{ d } \) and V = Ed.
\(U=\frac { 1 }{ 2 } \left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) { (Ed) }^{ 2 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }(Ad){ E }^{ 2 }\quad \quad \quad \quad \quad ...(4)\)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \({ u }_{ E }=\frac { U }{ Volume } \)
Equation (4) ⇒ \({ u }_{ E }=\frac{1}{2}{ \varepsilon }_{ 0 }{ E }^{ 2 }\).....(5)
(v) From equation (5),
(a) We infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.
(b) The energy density depends only on the electric field and not on the size of the plates of the capacitor.
(c) This is true for the electric field due to any type of charge configuration.
Resistors in series:
(i) When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Figure (a) shows three resistors R1, R2 and R3 connected in series.
(ii) The amount of charge passing through resistor R1 must also pass through resistors R2 and R3 since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors are the same.
(iii) According to Ohm's law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V1, V2 and V3 be the potential difference (voltage)across each of the resistors R1, R2 and R3 respectively, then we can write V1 = IR1, V2= RI2 and V3 = IR3. But the total voltage V is equal to the sum of voltages across each resistor.
V = V1 + V2 + V3 = IR1+ IR2 + IR3
V = I (R1 + R2 + R3)
V = IRS
where Rs is the equivalent resistance,
RS = R1 + R2 + R3
(iv) When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances.
Note: The value of equivalent resistance in series connection will be greater than each individual resistance.
Resistors in parallel:
(i) Resistors are in parallel when they are connected across the same potential difference as shown in figure (a).
(ii) In this case, the total current I that leave the battery split into three separate components.
Let I1, I2 and I3 be the current through the resistors R1, R2, and R3 respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I1 + I2 + I3 .....(1)
(iii) Since the voltage across each resistor is the same, applying Ohm's law to each resistor, we have
\({ I }_{ 1 }=\cfrac { V }{ { R }_{ 1 } } ,{ I }_{ 2 }=\cfrac { V }{ { R }_{ 2 } } ,{ I }_{ 3 }=\cfrac { V }{ { R }_{ 3 } } \)
Substituting these values in equation (1), we get,
\({ I }_{ 1 }=\cfrac { V }{ { R }_{ 1 } } +\cfrac { V }{ { R }_{ 2 } } +\cfrac { V }{ { R }_{ 3 } } =V\left[ \cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R }_{ 2 } } +\cfrac { 1 }{ { R }_{ 3 } } \right] \)
\(I=\cfrac { V }{ { R }_{ p } } \)
\(\cfrac { 1 }{ R_{ P } } =\cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R_{ 2 } } } +\cfrac { 1 }{ R_{ 3 } } \)
Here RP is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination.
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.
(i) The meter bridge is another form of Wheatstone's bridge. It consists of a uniform manganin wire AB of one meter length.
(ii) This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip E is mounted to enclose two gaps G1 and G2.
(iii) An unknown resistance P is connected in G1 and a standard resistance Q is connected in G2. A jockey (conducting wire) is connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance (HR).
(iv) The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are connected across the ends of the bridge wire.
(v) The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the position of jockey at the wire be at J.
(vi) The resistances corresponding to AJ and JB of the bridge wire now form the resistance R and S of the Wheatstone's bridge. Then for the bridge balance.
\(\cfrac { P }{ Q } =\cfrac { R }{ S } =\cfrac { { r }.AJ }{ { r }.JB } \)
where r' is the resistance per unit length of wire
\(\cfrac { P }{ Q } =\cfrac { AJ }{ JB } =\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \)
\(P=Q\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \)
(vii) By interchanging P and Q, another set of readings are taken and the average value of P is value of unknown resistance.
In a right angle triangle OPN let the angle \(\angle\)OPN = \(\theta \)1 which implies, \({ \varphi }_{ 1 }=\frac { \pi }{ 2 } -{ \theta }_{ 1 }\) and also in a right angle triangle OPM,
\(\angle\)OPN = \(\theta \)2 which implies, \({ \varphi }_{ 2 }=\frac { \pi }{ 2 } +{ \theta }_{ 2 }\)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)
(i) When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.
(ii) Let B be the magnetic field produced by passing current through the coil of the tangent galvanometer and BH be the horizontal component of earth's magnetic field.
(iii) Under the action of two magnetic fields, the needle comes to rest making angle with BH , such that
B = BH tan \(\theta\) .........(1)
Where B ⇒ magnetic field produced by current
BH ⇒ horizontal component of earth's magnetic field
Construction:
(i) Copper coil of wire wound on a non-magnetic circular frame such as brass or wood. Compass box is kept at centre.
(ii) This compass box consists of pivoted magnet and aluminum pointer.
(iii) This compass box is having circular scale graduated with four quadrants.
Working:
(i) Two magnetic fields are perpendicular to each other.
(ii) Magnetic induction due to the current in the coil acting to normal to the plane of the coil.
(iii) Magnetic induction at the centre of the coil,
\(B=μ_o\frac{NI}{2R}\) .....(2)
Sub. eqn.(1) in eqn. (2)
\(B_H tan \theta =μ_o\frac{NI}{2R}\)
\(B_H =μ_o\frac{NI}{2R}\frac{1}{tan\theta}\)
(i) Self inductance of a coil is desined as the flux linkage with the coil, when a current of 1 A flows through it.
\(L=\frac{{ N\Phi }_{ B }}{i}\)
\(L={ N\Phi }_{ B }\)
(ii) Self inductance of a coil is defined as the opposing emf induced in the coil when the rate of change of current through the coil is 1 A s-1
\(\varepsilon =\frac { d{ (N\Phi }_{ B }) }{ dt } \)
\(=-\frac { d(Li) }{ dt }\)
\(\varepsilon =-L\frac { di }{ dt } \)
L = -e
S.No | Name of the losses | Source of losses | Method to minimise |
(i) | (a) Core loss (or) Iron loss (or) Hysteresis loss | Transformer core is magnetised and demagnetised repeatedly |
Using steel of high silicon content in making transformer core |
(b) Eddy current loss | Alternating magnetic flux in the core induces eddy currents in it. |
Using very thin laminations of transformer core. | |
(ii) | Copper loss | When the electric current flows through windings of transformers, some amount of energy is dissipated due to Joule heating |
Using wires of larger diameter |
(iii) | Flux leakage | The magnetic lines of primary coil are not completely linked with secondary coil. |
Windings the coils one over the other. |
(i) We have stated Ampere's law as \(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_oi\)
(ii) Where, i is the electric current crossing a surface bounded by a closed curve and the line integral of \(\vec{B}\) is calculated along that closed curve. This equation is valid only when the electric field at the surface does not change with time.
(iii) Maxwell strongly believed that when the time varying magnetic field produces an electric field, the time varying electric field must produce a magnetic field.
(iv) To understand how a varying electric field produces magnetic field, let us consider a situation of charging a parallel plate capacitor.
(v) Let ic be the conduction current. To calculate the magnetic field at P (fig. 1 ) an amperian loop. S1 is drawn. Applying Ampere circuital law for the surface S1, we get
\(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 i_c\) Where, \(\mu_0\) is permeability of free space.
(vi) Applying the same for the surface S2, we get \(\oint \vec{B} \cdot \overrightarrow{d l}=0.\)
Because the surface S2 nowhere touches the wire carrying conduction current. Therefore for the point P at one surface (S1) it has some value and at another surface (S2) it has zero value.
(vii) So, Maxwell believed that there must be a current associated with the changing electric field in between the capacitor and he called that current as displacement current.
(viii) Applying Gauss law to the electric flux between the plates of the capacitor \(\phi_E=\oint \vec{E} \cdot \overrightarrow{\mathrm{dA}}=E A=\frac{q}{\varepsilon_0}\) where, A is the area of the plate.
The change in electric flux is \(\frac{d \phi_F}{d t}=\frac{1}{\varepsilon_0} \frac{d q}{d t} (or) \frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{i}_{\mathrm{d}}=\varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\), where id is the displacement current.
(ix) The displacement current can be defined as the current which comes into play in the region in which the electric field and electric flux are changing with time.
(x) So, Maxwell modified Ampere's law \(\oint_{l} \vec{B} \cdot d \vec{l}=\mu_{0} i_c+\mu_{0}-i_d\) which means the total current enclosed by the surface is sum of conduction current and displacement current.
Absorption spectra:
When light is allowed to pass through a medium or an absorbing substance then the spectrum obtained is known as absorption spectrum. It is the characteristic of absorbing substances. Absorption spectrum is classified into three types:
(i) Continuous absorption spectrum:
When we pass white light through a blue glass plate, it absorbs all the colours except blue and gives continuous absorption spectrum.
(ii) Line absorption spectrum:
When light from the incandescent lamp is passed through cold gas (medium), the spectrum obtained through the dispersion due to prism is line absorption spectrum . Similarly, if the light from the carbon are is made to pass through sodium vapour, a continuous spectrum of carbon are with two dark lines in the yellow region are obtained.
(iii) Band absorption spectrum:
When the white light is passed through the iodine vapour, dark bands on continuous bright background is obtained. This type of band is also obtained when white light is passed through diluted solution of blood or chlorophyll or through certain solutions of organic and inorganic compounds.
Apparatus:
The apparatus used by Fizeau for determining speed of light in air is shown in Figure
(i) The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light.
(ii) The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism.
(iii) The light passing through one cut in the wheel will get reflected by a mirror M kept at a long distance d, about 8 km from the toothed wheel.
(iv) The toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer, through the partially silvered glass plate.
Working:
(i) The angular speed of rotation of the toothed wheel was increased from zero to a value w until light passing through one cut would completely be blocked by the adjacent tooth.
(ii) This is ensured by the disappearance of light while looking through the partially silvered glass plate.
Expression for speed of light:
The speed of light in air 'v' is equal to the ratio of the distance 2d (the distance light travelled from the toothed wheel to the mirror and back) to the time taken 't'.
\(v=\cfrac { 2d }{ t } \) ...(i)
(v) The distance d is a known value from the arrangement. The time taken 't' for the light to travel the distance (2d) is calculated from the angular speed 'ω' of the toothed wheel.
(vi) The angular speed 'ω' of the toothed wheel when the light disappeared for the first time is,
\(\omega =\cfrac { \theta }{ t } \) ...(2)
(vii) Here, θ is the angle between the tooth and the next slot which is turned within that time t.
\(\theta =\cfrac { total \ angle\ of \ the\ circle \ in \ radian }{ number\ of\ teeth+number\ of\ cuts } \)
\(\theta =\cfrac { 2\pi }{ 2N } =\cfrac { \pi }{ N } \)
Substituting for θ in the equation for ω,
\(\omega =\cfrac { \pi /N }{ t } =\cfrac { \pi }{ Nt } \)
Rewriting the above equation for t,
\(t=\cfrac { \pi }{ N\omega } \) ....(3)
Substituting 't' in equation (1)
\(V=\cfrac { 2d }{ \pi /N\omega } \)
After rearranging,
\(v=\cfrac { 2dN\omega }{ \pi } \)
From this, the speed of light in air was determined as, v = 2.997 x 108 m/s.
Angle of deviation Produced by Prism:
(i) Let light ray PQ is incident on one of the refracting faces of the prism.
(ii) The angles of incidence and refraction at the first face AB are i1 and rl. The path of the light inside the prism is QR.
(iii) The angle of incidence and refraction at the second face AC is r2 and i2 respectively.
(iv) RS is the ray emerging from the second face. Angle i2 is also caned angle of emergence.
(v) The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
(vi) The two normals drawn at the point of incidence Q and emergence R meet at point N. They meet at point N.
(vii) The extended incident ray and the emergent ray meet at a point M.
The angle of deviation d1 at the surface AB is,
ㄥRQM = d = i1 - r1 ...(1)
The angle of deviation d2 at the surface AC is
ㄥQRM = d2 = i2 - r2 .......(2)
Total angle of deviation d produced is,
d = d1 + d2 .....(3)
Substituting for d1 and d2 in equation (3)
d = (i1 - r1) + (i2 - r2)
After rearranging,
d = (i1 - r1) + (i2 - r2) ........(4)
In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
\(\angle A+\angle QNR={ 180 }^{ 0 }\) .........(5)
From the triangle ΔQNR
\({ r }_{ 1 }+{ r }_{ 2 }+\angle QNR={ 180 }^{ o }\) ......(6)
Comparing these two equations (5) and (6) we get,
r1 + r2 = A .......(7)
Substituting this in equation (4) for angle of deviation,
d = i1+ i2 - A .............(8)
(viii) Thus, the angle of deviation depends on the angle of incidence i1, angle of emergence i2 and the angle for the prism A.
(ix) For a given angle of incidence the angle of emergence is decided by the refractive index of the material of the prism. Hence the angle of deviation depends on these following factors.
(i) the angle of incidence
(ii) the angle of the prism.
(iii) the refractive index of the material of the prism (which decides the angle of emergence).
Refractive index of the material of the prism:
At minimum deviation, i1 = i2 = i and r1 = r2 = r
Now, the equation (8) becomes,
D - i1 + i2 - A = 2i - A (or) \(i=\cfrac { \left( A+D \right) }{ 2 } \)
The equation (7) becomes
r1 + r2 = A ⇒ 2r = A (or) \(r=\cfrac { A }{ 2 } \)
Substituting i and r in Snell's law
\(n=\cfrac { sini }{ sinr } \)
\(n=\cfrac{\cfrac{sin(A+D)}{2}}{sin(A/2)}\)
From Maxwell's theory we learnt that light is an electromagnetic wave consisting of coupled electric and magnetic oscillations that move with the speed of light and exhibit typical wave behaviour. Let us try to explain the experimental observations of photoelectric effect using wave picture of light.
(i) When light is incident on the target, there is a continuous supply of energy to the electrons in the metal surface.
(ii) According to wave theory, light of greater intensity should impart greater kinetic energy to the liberated electrons (Here, intensity of light is the energy delivered per unit area per unit time). But this does not happen. The experiments show that maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light.
(iii) According to wave theory, if a sufficiently intense beam of light is incident on the surface, electrons will be liberated from the surface of the target, however low the frequency of the radiation is. From the experiments, we know that photoelectric emission is not possible below a certain minimum frequency. Therefore, the wave theory fails to explain the existence of threshold frequency.
(iv) Since the energy of light is spread across the wavefront, the electrons which receive energy from it are large in number. Each electron needs considerable amount time (a few hours) to get energy sufficient to overcome the work function and to get liberated from the surface. But experiments show that photoelectric emission is almost instantaneous process (the time lag is less than 10-9 s after the surface is illuminated) which could not be explained by wave theory.
Thus, the experimental observations of photoelectric emission could not be explained on the basis of the wave theory of light.
(i) An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
\(\cfrac { 1 }{ 2 } { mv }^{ 2 }=ev\)
(ii) Therefore, the speed v of the electron is
\(v=\sqrt { \cfrac { 2ev }{ m } } \)
Hence, the de Broglie wavelength of the matter waves associated with electron is
\(\lambda =\cfrac { h }{ mv } =\cfrac { h }{ \sqrt { 2mev } } \)
(iii) Substituting the known values in the above equation, we get
\(\lambda =\cfrac { 6.26\times { 10 }^{ -34 } }{ \sqrt { 2V\times 1.6\times { 10 }^{ -19 }\times 9.11\times { 10 }^{ -31 } } } \)
= \(\cfrac { 12.27\times { 10 }^{ -10 } }{ \sqrt { V } } m\)
\(\lambda =\cfrac { 12.27 }{ \sqrt { V } } \overset { o }{ A } \)
(iv) Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
\(\lambda =\cfrac { h }{ \sqrt { 2mK } } \)
Principle:
Cathode rays are deflected in electric and magnetic fields.
By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.
Construction and Working:
(i) Cathode rays (electron beam) produced at cathode of a highly evacuated discharge tube. Cathode rays are attracted towards anode disc A.
(ii) Pin hole in the anode disc allows only a narrow beam of cathode rays.
(iii) These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage.
(iv) Further, discharge tube is kept in between pole pieces of magnet.
(v) Electric and magnetic fields are perpendicular to each other.
(vi) When the cathode rays strike the zinc sulphide coated screen (O), produces scintillation and hence bright spot is observed.
(i) Determination of velocity of cathode rays:
(a) For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O.
(b) This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field.
(ie) Ee = Bev
\(\Rightarrow v=\frac { E }{ B } \) ...(1)
(ii) Determination of specific charge:
(a) Since accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode.
(b) Let V be the potential difference between anode and cathode, then the potential energy is eV.
Then from law of conservation of energy,
\(eV=\frac { 1 }{ 2 } { mv }^{ 2 } \)
\(\frac { e }{ m } =\frac { { v }^{ 2 } }{ 2V } \) ....(2)
Substituting (1) in (2),
\(\frac { e }{ m } =\frac { 1 }{ 2V } \frac { { E }^{ 2 } }{ { B }^{ 2 } } \)
By substituting known values, we get
\(\frac { e }{ m } =1.7\times { 10 }^{ 11 }{ CKg }^{ -1 }\)
The specific charge \(\frac{\mathrm{e}}{\mathrm{m}}\) is independent of (a) gas used (b) nature of the electrodes
(i) When unstable nuclei decay by emitting an \(\alpha\) - particle (\(_{ 4 }^{ 2 }{ He }\) nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4.
\(_{ Z }^{ A }{ X }\rightarrow _{ Z-2 }^{ A-4 }{ Y+ }_{ 2 }^{ 4 }{ He }\)
(ii) X is called the parent nucleus and Y is called the daughter nucleus.
Example : when uranium \(_{ 92 }^{ 238 }U\) emit (α - particle) it is converted into thorium \(_{ 90 }^{ 234 }{ Th }\) .
\(_{ 92 }^{ 238 }{ U\rightarrow }_{ 90 }^{ 234 }{ Th }+_{ 2 }^{ 4 }{ He }\)
(iii) Total mass of the daughter nucleus and \({ }_{2}^{4} \mathrm{He}\) nucleus is always less than that of parent nucleus.
(iv) The difference in mass \(\left(\Delta m=m_{X}-m_{Y}-m_{\alpha}\right)\) is released as energy called disintegration energy Q
\(\mathrm{Q}=\Delta \mathrm{m} \times \mathrm{c}^{2}=\left(\mathrm{m}_{\mathrm{X}}-\mathrm{m}_{\mathrm{Y}}-\mathrm{m}_{\alpha}\right) \mathrm{c}^{2}\)
(v) For spontaneous decay (natural radioactivity) Q > 0. In alpha decay process.
(a) The disintegration energy is positive (Q > 0).
(i) if the parent nucleus is not at rest, Q is equal to the net kinetic energy gained in the decay process.
(ii) if the parent nucleus is at rest, Q is equal to the total kinetic energy of daughter nucleus and the \({ }_{2}^{4} \mathrm{He}\) nucleus.
(b) The disintegration energy is negative (Q < 0).
The decay process cannot occur spontaneously and energy must be supplied to induce the decay.
Materials made up of particles sizes of 1-100 made up nm differ in their properties from the materials made up of bulk particles. Thus, nano materials find wide range of applications
Automotive industry :
(i) Lightweight construction
(ii) Painting (fillers, base coat, clear coat)
(iii) Catalysts
(iv) Tires (fillers)
(v) Sensors
(vi) Coatings for wind-screen and car bodies.
Chemical industry :
(i) Fillers for paint systems
(ii) Coating systems based on nanocomposites
(iii) Impregnation of papers
(iv) Switchable adhesives
(v) Magnetic fluids
Engineering :
(i) Wear protection for tools and machines (anti blocking coatings, scratch resistant coatings on plastic parts etc.)
(ii) Lubricant-free bearings.
Electronic industry :
(i) Data memory
(ii) Displays
(iii) Laser diodes
(iv ) Glass fibres
(v) Optical switches
(vi) Filters (lR-blocking)
(vii) Conductive, antistatic coatings.
Construction :
(i) Construction materials
(ii) Thermal insulation
(iii) Flame retardants
(iv) Surface-functionalised building materials for wood, floors, stone, facades, tiles, roof tiles, etc.
(v) Facade coating
(v) Groove mortar.
Medicine :
(i) Drug delivery systems
(ii) Active agents
(iii) Contrast medium
(iv) Medical rapid tests
(v) Prostheses and implants
(vi) Antimicrobial agents and coatings
(vii) Agents in cancer therapy.
Textile/fabrics/non-wovens :
(i) Surface-processed textiles
(ii) smart clothes.
Energy :
(i) Fuel cells
(il) Solar cells
(iii) Batteries
(iv) Capacitors.
Cosmetics :
(i) Sun protection
(ii) Lipsticks
(iii) Skin creams
(iv) Tooth paste.
Food and drinks :
(i) Package materials
(ii) Storage life sensors
(iii) Additives
(iv) Clarification of fruit juices.
Household :
(i) Ceramic coatings for irons
(ii) Orders catalyst
(iii) Cleaner for glass, ceramic, floor, windows.
Sports/Outdoor :
(i) Ski wax
(ii) Antifogging of glasses /goggles
(iii) Antifouling coating for ships /boats
(iv) Reinforced tennis rackets and balls.
(i) Thediffraction pattern for white light consists of a white central maximum and on both side continuous coloured diffraction pattenrs are formed.
(ii) The central maximum is white as all the colours constructively meet at centre with no path difference. As \(\theta\) increases, the path difference fuifills the condition for maxima of different orders for all colours from violet to red.
(iii) It produces a spectrum of diffraction pattern from violet to red on either side of central maximum as shown in Figure.
(iv) By measuring the angle at which these colours appear for various orders of diffraction, the wavelength of different colours could be calculated using the formula.
\(\lambda =\cfrac { sin\theta }{ Nm } \)
(v) Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.
i) The spectrometer is an optical instrument used to analyse the spectra of different sources of light, to measure the wavelength of different colours and to measure the refractive indices of materials of prisms.
ii) It basically consists of three parts namely. They are (i) collimator, (ii) prism table and (iii) Telescope
Adjustments of the spectrometer
(i) The following adjustments must be done in a spectrometer before doing the experiment.
(a) Adjustment of the eyepiece:
The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.
(b) Adjustment of the telescope:
The telescope is adjusted to' receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.
(c) Adjustment of the collimator:
The telescope is brought in line with the collimator. The distance between the illuminated slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire.
(d) Levelling the prism table:
The prism table is brought to the horizontal level by adjusting the levelling screws and it is ensured by using sprit level.
i) AND gate
a) Circuit Symbol:
The circuit symbol of a two input AND gate is shown in Figure (a). A and B are inputs and Y is the output. It is a logic gate and hence A, B, and Y can have the value of either 1 or 0
Two input AND gate
Inputs | outputs | |
A | B | Y = A + B |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
Truth table
b) Boolean equation:
Y = A.B
It performs logical multiplication and is different from arithmetic multiplication.
c) Logic operation:
The output of AND gate is high only when all the inputs are high. In the rest of the cases, the output is low. It is represented in the truth table (Figure (b).
ii) OR gate
a) Circuit Symbol:
The circuit symbol of a two input OR gate is shown in Figure (a). A and B are inputs and Y is the output.
The input OR gate
Inputs | outputs | |
A | B | Y = A + B |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 1 |
Truth table
a) Boolean equation:
A + B = Y
It performs logical addition and is different from arithmetic addition.
b) Logic operation:
The output of OR gate is high (logic 1 state) when either of the inputs or both are high. The truth table of OR gate is shown in Figure (a).
iii) NOT gate
a) Circuit Symbol:
The circuit symbol of NOT gate is shown in Figure (a). A and B are inputs and Y is the output.
NOT gate
Inputs | Output |
A | Y = Ā |
0 | 1 |
1 | 0 |
Truth table
a) Boolean equation:
Y = Ā
b) Logic operation:
The output is the complement of the input. It is represented with an overbar. It is also called as inverter. The truth table infers that the output Y is I when input A is 0 and vice versa. The truth table of NOT is shown in Figure (b).
iv) NAND gate
a) Circuit Symbol:
The circuit symbol of NAND gate is shown in Figure (a). A and B are inputs and Y is the output.
Two input NAND gate
Inputs | Output (AND) |
outputs (NAND) |
|
A | B | Z = A.B | Y = \(\overline { A.B } \) |
0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 |
1 | 1 | 1 | 0 |
Truth table
b) Boolean equation:
Y = \(\overline { A.B } \)
Logic operation:
The output Y equals, the complement of AND operation. The circuit is an AND gate followed by a NOT gate. Therefore, it is summarized as NAND. The output is at logic zero only when all the inputs are high. The rest of the cases, the output is high (Logic I state). The truth table of NAND gate is shown in Figure (b).
v) NOR gate
a) Circuit Symbol:
The circuit symbol of NOR gate is shown in Figure (a). A and B are inputs and Y is the output.
Two input NANS gate
Inputs | Output (OR) |
outputs (NOR) |
|
A | B | Z = A + B | Y = \(\overline { A+B } \) |
0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 |
Truth table
Boolean equation:
Y = \(\overline { A+B } \)
Logic operation:
The output Y equals the complement of OR operation (A OR B). The circuit is an OR gate followed by a NOT gate and is summarized as NOR. The output is high when all the inputs are low. The output is low for all other combinations of inputs. The truth table of NOR gate is shown in Figure (b).
vi) Ex-OR gate
a) Circuit Symbol:
The circuit symbol of Ex-OR gate is shown in Figure (a). A and B are inputs and Y is the output. The Ex-OR operation is denoted as ⊕
Ex-OR gate
Inputs | outputs (Ex-OR) |
|
A | B | Y = A ⊕ B |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
Truth table
b) Boolean equation
Y = \(A.\overline { B } \) + \(\overline { A }.B \)
Y = A ⊕ B
Logic operation:
The output is high only when either of the two inputs is high. In the case of an Ex-OR gate with more than two inputs, the output will be high when odd number of inputs are high. The truth table of Ex-OR gate is shown in Figure (b).
Construction:
(a) The amplification of an electrical signal is explained with a single-stage transistor amplifier as shown in figure.
(b) Single stage indicate that the circuit consists of one transistor with the allied components.
(i) An NPN transistor is connected in the common-emitter configuration
(ii) To start with, the Q point or the operating point of the transistor is fixed, so as to get the maximum signal swing at the output (neither towards saturation point nor towards cut-off).
(iii) A load resistance, RC is connected in series with the collector circuit to measure the output voltage.
The resistance R1, R2, and RE, form the biasing and stabilization circuit.
(iv) The capacitor C, allows only the AC signal to pass through.
(v) The emitter by pass capacitor CE provides a low reactance path to the amplified AC signal
(vi) The coupling capacitor CC is used to couple one stage of the amplifier with the next stage, while constructing multistage amplifiers
Vs is the sinusoidal input signal source applied across the base-emitter. The output is taken across the collector-emitter.
Collector current IC = βIB [∵β = IC/IB]
Applying Kirchhoff's voltage law to the output loop, the collector-emitter voltage is given by
VCE = VCC - ICRC
Working of the amplifier :
During the positive half cycle :
(i) Input signal (Vs) increases the forward voltage across the emitter base. As a result, the base current (IB in μA) increases. consequently the collector current (ICin mA) increases β times.
(ii) This increase the voltage drop across RC(ICRC) which in turn decreases the collector-emitter voltage (vCE). Therefore, the input signal in the positive direction produces an amplified signal in the negative direction at the output. Hence the output signal is reversed by 1800 as shown in figure
During the negative half cycle:
(i) Input signal (Vs) decreases the forward voltage across the emitter base. As a result base current (IB in μA) decreases and in tum increases the collector current (IB in μA).
(ii) The increase in collector current (IC) decreases the potential drop across RC and increases the collector - emitter voltage (VCE).
(iii) Thus the input signal in the negative. direction produces an amplified signal in the positive direction at the output.
(iv) Therefore, 180 phase reverse is observed during the negative half cycle of the input signal as well as shown in figure.
(i) Electric flux \(\phi =\frac { Total\ enclosed\ charge }{ { \varepsilon }_{ 0 } } \)
Net charge enclosed inside the shell q = 0
∴ electric flux through the shell \(\frac { q }{ { \varepsilon }_{ 0 } } =0\)
(ii) The electric field or net charge inside the spherical conducting shell is zero.
Hence the force on charge \(\frac{Q}{2}\)is zero.
Force on charge at \(A,{ F }_{ A }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { 2Q\left( Q+\frac { Q }{ 2 } \right) }{ { x }^{ 2 } } \)
For charge -q to be in equilibrium, for an -q due to +Q at point A should be equal and opposite to that due to +Q at point B.
i.e \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Qq }{ { x }^{ 2 } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q.q }{ { (r-x) }^{ 2 } } \)
r (r - x)2
x r - x
x = \(\frac{r}{2}\)
∴ for equilibrium, -q must be kept at the middle of the line joining the A and B.
Charge Q = Current x Time interval
= I x t
At t = 0 s,
dq = dI x t
= 5 x 0
dq = 0 C
At t = 2 s,
dg = dI x t
=5 x 2
dq = 10 C
At t = 5 s,
dg = dl x t
=0 x 5
dq = 0 C
At t= 0 s, dg = 0 C; At t = 2 s, dg = 10 C; At t= 5 s, dg = 0 C.
Let the currents flowing through wires XY and PQ be I1 and I2
Magnetic field along PQ is \(\mathrm{B}_{1}=\frac{\mu_{o} I_{2}}{2 \pi r}\)
Force per unit length on PQ is \(\frac{F_{2}}{l}=\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\)
If the wire XY is slightly displaced and released, it executes simple harmonic motion with the condition that acceleration is directly proportional to the displacement y
\(\therefore a=-\omega^{2} y\) .....(1)
The distance between two wires = d
Time period
\(T=\frac{2 \pi}{\omega} \)
\(a=\frac{g}{d} y \) .....(2)
By comparing the equations (1) and (2) we get
\(\omega^{2} =-\frac{g}{d} \ \therefore \omega=\sqrt{\frac{g}{d}} \)
\(\text { Time period } =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{d}{g}} \)
\(\therefore T =2 \pi \sqrt{\frac{d}{g}} \)
Formula:
Refractive index,\(\mu =\cfrac { sini }{ sinr } \)
\(\sqrt { 3 } =\cfrac { sin{ 60 }^{ o } }{ sinr } \)
\(sinr=\cfrac { \sqrt { 3 } }{ 2 } \times \cfrac { 1 }{ \sqrt { 3 } } =\cfrac { 1 }{ 2 } \)
sin r = sin30o
⇒ Angle of refraction = 30o
(i) The threshold frequency is the frequency of incident light at which kinetic energy of ejected photoelectron is zero.
∴ From fig. threshold frequency,
v0 = 4.5 x 1014 Hz
(ii) Work function, W = hv0
= 6.6 x 10-34 x 4.5 x 1014 joule
= \(\frac { 6.6\times { 10 }^{ -34 }\times 4.5\times 10^{ 14 } }{ 1.6\times { 10 }^{ -19 } } \) eV
= 1.85 eV
(i) The robots are much cheaper than human.
(ii) It can work for 24 x 7. Robots never get tired like human.
(iii) Robots are more precise and error free on performing the task.
(iv) Stronger and faster then humans.
(v) Robots can work in extreme environmental conditions like extremely hot or cold, space or underwater. In dangerous situations like bomb detection and bomb deactivation.
(vi) In warfare, robots can save human lives.
(vii) Robots are significantly used in handling material in chemical industries especially in nuclear plants which can lead to health hazards in human.
Give: No. of turns of long solenoid = 20 = N
No. of turns of small loop N2= 1
Length l = 1 = 1 crn = 10-2m2
To find:
Induced emf e = ?
\(e=\mu \frac { dI }{ dt } \)
Formula:
Mahal inductance of a solenoid
\(\mu =\frac { { \mu }_{ o }{ N }_{ 1 }{ N }_{ 2 }{ A }_{ 2 } }{ l } \)
Solution:
\(\mu =\frac { 4\pi \times { 10 }^{ -7 }\times 20\times { 10 }^{ -4 }\times 2\times 1 }{ { 10 }^{ -2 } } \)
=160π x 10-9 x \(\frac { (3-1) }{ 0.2 } \)
= 160π x 10-8
= 502 x 10-8 = 5.02 x 10-6 V
E and B vectors must be in x and y directions.
Formula: We know \(\lambda =\frac { v }{ \gamma } =\frac { 3\times { 10 }^{ 8 } }{ 30\times { 10 }^{ 6 } } \)
λ = 10m.