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ThePN junctionregion of aJunction Diodehas the following important characteristics: Semiconductors contain two types of mobile charge carriers, “Holes” and “Electrons”. ... A semiconductor may be doped with acceptor impurities such as Boron (P-type doping), so that it contains mobile charges which are mainly holes.

What is Rectification?

A direct current flows from the positive to negative terminal of a source supply when connected by a circuit. The current flows continuously in a single direction. However, in the case of alternating current, the direction of the flow keeps on changing on the basis of the frequency of the alternator producing it. This means that the current first flows in one direction in the circuit, rises to the peak value and comes down, and then changes the direction, reaches the peak value and again comes back to normal. The rise and fall of the current follows a sine wave pattern and is the characteristic of alternating current.

In a situation wherein direct current is required from a source supplying alternative current, the later needs to be rectified in order to make it usable for the D.C requirement. This process of rectifying the AC current is known as rectification. Thus, rectification can be defined as a process of converting alternating current (A.C) to Direct current (D.C). In this process the flow of current in only one direction is permitted while in the other direction is resisted. This means that in an alternating current (A.C) the positive component of the current is only permitted to pass whereas the negative is resisted. The equipment used for this rectification process is known as rectifier.

Downwardforce on water drop = mass * acceleration due togravity

let the electric field strength be applied vertically upwards to oppose gravity.

electric force = charge * electric field strength 0.0000001 kg* 9.81 m/s² = 1.6 * 10⁻¹⁹ C* E E = 6.13 * 10¹² Newton/Coul

Height of object ho = 4 cm

Focal lenght of convex mirror = ? cm

Object distance u = ? cm

Image distance v = 60 cm

Image height hi = 2 cm

Magnification m = hi/ho = 2/4=1/2

v/u = 1/2

60cm/u = 1/2

u = 120 cm

mirror formula: 1/v + 1/u = 1/-f

1/-f = 1/v + 1/u

= 1/60 + 1/120

= 3/120 = 1/40

⇒ f= -40 cm

Given O = 2cmu = 16cm(-ve)I = 3cm(-ve){Since image is real it is inverted)⇒ -v/u = I / O⇒ v = -1(I * u/o) = (-3cm * -16cm/2 cm) = 24 cm (-ve)

By relation 1/f = 1/u + 1/v⇒1/f = 1/-16 + 1/-24= 3 + 2/-48=-5/48

⇒f = 48/5 (-ve) = 9.6 cm(-ve)∵ C > u > f⇒ Image formed is beyond C

h = 4 cm h' = - 6 cm m = -6/4 = - 1.5 Since image is magnified, inverted real, it is a convex lens.

| u | + v = 20 cm v - u = 20 cm as u is -ve.

m = v / u => v = - 1.5 u

- 1.5 u - 1 u = 20 => u = - 20/2.5 = - 8 cm

v = +12 cm

1/v - 1/u = 1/f => 1/12 + 1/8 => f = 24/5 = 4.8 cm

The addition of impurity atoms to a pure-semi-conductor is known as doping. The doping is done to increase the conductivity of a pure semi-conductor. It must be noted that the impurity atom is about 1 in every 1010atoms.

Dopinga semiconductor in a good crystal introduces allowed energy states within the band gap, but very close to the energy band that corresponds to thedopanttype. In other words, electron donor impurities create states near the conduction band while electron acceptor impurities create states near the valence band.

(i) Collector current decreases slightly.(ii) Base current increases slightly.

According to this law when unpolarised light is incident at polarising angle (i) on an interface separating a rarer medium from a denser medium, of refractive index μ such thatμ = tan i,then light reflected in the rarer medium is completely polarised. Reflected and refractive rays are perpendicular to each other.

Application One general example for theapplicationof Brewster's lawis polarized sunglasses. These glasses uses the principle ofBrewster's angle. The polarized glasses reduce glare that is reflecting directly from the sun and also from horizontal surface like road and water.

The wedge is tangent to the sphere. Using that it touches at height R/5 you can easily work out the slope of the wedge. The velocity of the sphere follows directly (20m/s times slope).

Brewster’s law is a relationship of light waves at the maximumpolarization angle of light. This law is named after Sir David Brewster, a Scottish physicist, who proposed the law in the year 1811. The law states that the p-polarized rays vanish completely on different glasses at a particular angle.

One general example for the application of Brewster’s law is polarized sunglasses. These glasses uses the principle of Brewster’s angle. The polarized glasses reduce glare that is reflecting directly from the sun and also from horizontal surface like road and water. Photographers also use the same law to reduce the reflection from reflective surfaces by using polarizing filter for the lens.

sonar = sound navigation and ranging ultrasound . error.

Sonar(originally anacronymforsoundnavigationranging) is a technique that usessoundpropagation (usually underwater, as insubmarine navigation) tonavigate, communicate with or detect objects on or under the surface of the water, such as other vessels.[2]Two types of technology share the name "sonar":passivesonar is essentially listening for the sound made by vessels;activesonar is emitting pulses of sounds and listening for echoes. Sonar may be used as a means ofacoustic locationand of measurement of the echo characteristics of "targets" in the water. Acoustic location in air was used before the introduction ofradar. Sonar may also be used in air for robot navigation, andSODAR(an upward-looking in-air sonar) is used for atmospheric investigations. The termsonaris also used for the equipment used to generate and receive the sound. The acoustic frequencies used in sonar systems vary from very low (infrasonic) to extremely high (ultrasonic). The study of underwater sound is known asunderwater acousticsorhydroacoustics.The first recorded use of the technique was byLeonardo da Vinciin 1490 who used a tube inserted into the water to detect vessels by ear.[3]It was developed during World War I to counter the growing threat ofsubmarine warfare, with an operationalpassive sonarsystem in use by 1918.[2]Modern active sonar systems use an acoustictransponderto generate a sound wave which is reflected back from target objects.[2]

The SI unit of modulus of elasticity (E, or less commonly Y) is the pascal; N/m^2option b

here f=100/2.5

f=40cm

m=f/(u-f)

m=40/(20

m=2

hi/ho=2

hi=ho×3

hi=24

The expression of velocity in S.H.M is

V² = ω²(a² -x²) here.. a = maximum amplitude and ω is the frequency..

so, compraing V = √144-16x² = √(12)²-(4x)² , we get ω (as coefficients of x)= 4 and a as √9 = 3m

so, maximum acceleration is aω² = 3*4*4 = 48m/s²

so option D.

MOI of solid sphere is 2/5mr² ...(i1)MOI of hollow sphere is 2/3mr² .....(i2)MOI of Ring is mr² ......(i3)

so, i3 > i2> i1

When length changes area also changes... R is directly proportional to square of length... Hence R=9R

R/3 is the right answer because disc is increased by three times.

thank you sir

Given, m =1/8. u = -40 cmNow we know , m = -v/u1/8 = - v/-(40)1/8 = v/40thus v = +5 cm , Here positive sign indicates that image is formed at a distance 5 cm behind the mirror.Now from mirror formula,1/V+1/u=1/f1/5+(1/-40)=1/f8-1/40=1/ff=40/7

Hence focal length = 5.71 cm

u = -20 cm

f = -10 cm

position = v

1 / f = 1 / v + 1 / u

1 / -10 = 1 / v + 1 / -20

1 / v = -1 / 10 + 1 / 20

= -2 / 20 + 1 / 20

= -1 / 20

v = -20cm

Nature : Real and inverted same size

Power = 1/f where f is focal length in meternow p1 = 100/25= 4 (for convex lens)p2 = -100/10 = -10(power is negative because focla length of concave lens is negative)now resultant power = p1+p2 = 4+(-10)=-6

U= 50m/sg= -10m/s^2

The velocity at the highest point will be 0 m/sThus v=0

Height can be calculated by

2gs=v^2-u^2-20s= -2500s= 125m

Time :v= u+gt0=50-10t-50=-10tt= 5sec

the mirror is likely to be concave mirror

convex lens correct answer

convex lens is best answer

it must be convex lens

The mirror is concave mirror

I think convex lens is the correct answer

if the spherical mirror has focal length negative then convex mirror will be formed.

If spherical mirror has a focal length negative then Convex mirror will be formed.

if the spherical mirror has focal length 10cm then convex mirror will be formed

Convex mirror is right answer

concave mirror is the best answer

Height of object = 2cm

Object distance = 15cm

Focal length = 10cm

1/f = 1/v +1/u

-1/10 = -1/15 +1/v

1/v = -3 +2 /30

v = -30 cm

Height of image/Height of object = -v /u

Height of image / 2 = 30/-15

Height of image = -4cm

Nature - Real and inverted

Position - 30 cm from concave mirror

Size = 4 cm

flux is zero inside the electrice field

F=Kq1q2/r^2k = 9*10^9q1=q2= 2*10^-6 CF = 9*10^9 * (2*10^-6)^2/(0.5)^2F=0.144N