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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2251. |
If theta is the angle between the transmission axes ofthe polarizer and the analyser and l_0 be the intensity of the polarized light incident on the analyser, then Intensity of transmitted light through analyser would be |
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Answer» `(I_0 COSTHETA)^2` |
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| 2252. |
Three cells are connected in parallel, with their like poles connected together, with wires of negligible resistance. If the emf of the cell are 2V, 1V and 4V and if their internal resistance are4 Omega, 3 Omegaand2 Omegarespectively,find thecurrentthrougheach cell. |
| Answer» SOLUTION : ` I _1 = (-2)/(13) A, I _2= (-7)/(13) A, I_3 = (9)/(13) A ` | |
| 2253. |
The binding energies per nucleon for deuteron (""_1^(2)H)and helium ("'_(2)^(4)He)are 1.1 MeV and 7.0 MeV respectively. Calculate the energy released, when two deuterons fuse to form a helium nucleus. |
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Answer» |
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| 2254. |
The distnace between the plates of a parallel plate capacitor is 3 mm the potential difference applied is 3xx10^(5) V. If an electron travels from one plate to another the change in its potential energy is |
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Answer» `3xx10^(5)` eV An ELECTRON volt (eV) is defined as the energy acquired by an electron when it MOVES through a potential DIFFERENCE of one volt . |
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| 2255. |
Charges of +2 +2, and-2 muCrespectively are placed at the vertices of an equilateral triangle of side0.3meach in Find the net forces experienced by each charges. |
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Answer» Solution :Here AB = AC =BC =0.3 and `q_A =q_B =+ 2mu C = 2xx 10 ^(-6) C and q_C =- 2muC =-2 xx 10 ^(-6)C ` Thus , it is clear that mutual force of attraction/repulsion between any two charges have equal MAGNITUDE GIVEN by ` ""F= |oversetto (F_(AB))|=|oversetto (F_(CB)) |=(9xx 10 ^(9) xx (2xx 10 ^(-6))^(2)) /( (0.3)^(2)) =0.4 N ` Direction of these forces have been SHOWN in (a)As angle between `oversetto (F_(CA)) and oversetto (F_(CB)) is 60^(@) ` , hence `|oversetto (F_C) |=2F _(CA) cos ""( (60)/( 2))^(@) ` ` ""=2 xx0.4 xx (sqrt( 3))/(2)` ` = 0.69 N and ` the force is `TT `to AB. (b)As angle between `oversetto (F_(AB))and oversetto (F_(AC))is 120 ^(@) `, hence `|oversetto (F_A) |2F_(AB) cos 60^(@) = 2xx 0.4 xx (1)/(2)= 0.4 N ` and the force makes an angle ` 60^(@) ` from line AC. (c)As in (b)force `|oversetto (F_B)|=0.4 N `inclined at an angle of ` 60^(@) `from line BC. ` (##U_LIK_SP_PHY_XII_C01_E10_022_S01.png" width="80%"> |
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| 2256. |
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope ? If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48xx10^(6)m, and radius of lunar orbit is 3.8xx10^(8)m. |
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Answer» Solution :if d is the DIAMETER of the IMAGE (in cm) `(d)/(1500)=(3.48xx10^(6))/(3.8xx10^(8))` `d=13.7` |
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| 2257. |
An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and eye piece is 36 cm and the final image is formed at infinity. Determine the focal length of objective and eye piece. |
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Answer» SOLUTION :For final image at INFINITY, `M_(oo)=(f_(0))/(f_(e))andL_(oo)=f_(o)+f_(e)` `therefore5=f_(0)/(f_(e))""...(i)` and `36=f_(0)+f_(e)" "...(ii)` Solving these two equations, we have `f_(0)=30cmandf_(e)=6cm` |
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| 2258. |
The energy in an electromagnetic wave is ..... |
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Answer» WHOLLY SHARED only by ELECTRIC field vector |
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| 2259. |
X-rays can be used |
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Answer» to detect FLAWS in CASTING and welding |
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| 2260. |
Radio waves lie in upper frequency range of the electromagnetic spectrum. |
| Answer» Solution :Gamma RAYS LIE in the upper FREQUENCY range of the electromagnetic spectrum. | |
| 2261. |
Due to which reasons the energy losses do occurs in actual transformer ? |
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Answer» Solution :(i) Flux LEAKAGE `:` Not all of the flux due to primary PASSES through the secondary due to poor design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other. (ii) Resistance of the windings `:` The wire used for the winding has some resistance and so, energy is lost due to heat PRODUCED in the wire `I^(2) R`. In high current, low voltage windings theses are minimised by using thick wire. (iii) Eddy Currents `:` The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by having a LAMINATED coil. (iv) Hysterests `:` The magnetisation of the core is repeated reversed by the alternating magnetic field. The resulting expenditure ofenergy in the coreappears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss. (v) TRANSFORMER work on AC voltage hence in each cycle, the length of ocre increases and decreases. As a result transformer produces buzz noise. So, some electric energy dissipated in buzz noise. |
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| 2262. |
The coil of a moving coil galvanometer has 5 turns and each turn has an effective area of 2 xx 10^(-2) m^(2) . It is suspended in a magnetic field whose strength is 4 xx 10^(-2) Wb m^(-2) . If the torsional constant K of the suspension fibre is4xx 10^(9) N m deg^(-1) . (a) Find its current sensitivity in degree per micro - ampere (b) Calculate the voltage sensitivity of the galvanometer for it to havefull scale deflection of 50 divisions for 25 mV. (c ) Compute the resistance of the galvanometer . |
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Answer» Solution :The number of turns of coil is 5 turns The area of each coil is ` 2xx 10^(-2) m^(2)` Strength of the magnetic FIELD is 4 `xx 10^(-2) ` Wh `m^(-2)` Torsional constant is 4 `xx 10^(-9) ` N m `deg^(-10` `I_(x) = ("N A B")/( K)= (5 xx 2 xx 10^(-2) xx 4 xx 10^(-2))/( 4 xx 10^(-9)) = 10^(@)` divisions per ampere `I mu`A = 1 microampere = `10^(-6)` ampere THEREFORE, ` I_(x)= 10^(6) ("div")/(A) = 1 ("div")/(10^(-6) A) = 1 ("div")/(mu A)` `I_(x) ` = 1div `(mu A)^(-1)` (b) Voltage sensitivity `V_(x)= (theta)/(V) = (50"div")/(25 m V) = 2xx 10^(3) "div" V^(-1)` (c) The resistance of the galvanometer is `R_(g) = (I_(s))/(V_(s)) = (10^(6) ("div")/(A))/( 2 xx 10^(3) ("div")/(A)) = 0.5 xx 10^(3) (V)/(A) = 0.5 k Omega` |
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| 2263. |
Explain declination and dip. |
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Answer» Solution :Declination : It is the angle between the geographic NORTH south and magnetic north-south of EARTH,denoted by `theta`. It is the angle made by a magnet suspended FREELY from its CENTRE of gravity with the HORIZONAL, denoted by `delta` |
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| 2264. |
For a bar magnet the magnetic length is 10cm. Then ii's geometric length is : |
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Answer» 8 cm |
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| 2265. |
AB is a uniform wire of length L = 100 cm. A cell of emf V_(0) = 12 volt is connected across AB. A resistance R, cell of emf V and a milliammeter (which can show deflection in both directions] is connected to the circuit as shown. Contact C can be slid on the wire AB. Distance AC = x. The current (I) through the milliammeter is taken positive when the cell of emf V is discharging. A graph of I Vs x has been shown. Neglect internal resistance of the cells. (a) Find V (b) Find R (c) find I when x = 100 cm |
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Answer» (B) `80 Omega ` (c) `-120 mA` |
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| 2266. |
जनन छिद्र का कार्य है |
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Answer» प्रांकुर का निर्माण करना |
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| 2267. |
In an A.C. generator, a coil with N turns, all the same area A and total resistance R, rotate with frequency omega in a magnetic field B. The maximum value of emf generated in the coil is ….. |
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Answer» NABR If sin `omegat=1` then `epsilon=epsilon_"MAX"` `therefore epsilon_"max" =NABomega` |
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| 2268. |
In the above question the torque on the coil is |
| Answer» Answer :D | |
| 2269. |
A wire of resistance 15 Omega is bent to form a regular hexagon ABCDEFA. Find the equivalent resistance of the loop between the points (a) A and B (b) A and C and (c) A and D |
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Answer» SOLUTION :`2.08Omega` `3.33Omega` 3.75Omega` |
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| 2270. |
A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacement x and y vary with time t in second as : x=10sqrt(3t) and y=10t-t^(2) The maximum height attained by the ball is : |
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Answer» 100 m |
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| 2271. |
Electromagnetic waves used in the telecommunication are _____. |
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Answer» VISIBLE |
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| 2272. |
A point source of sound is vibrating with a frequency of 256 vibrations per second in air and propagating energy uniformly in all directions at the rate of 5 joules per second. Calculate the amplitude of the wave at a distance of 25 m from the source. Assume non-dissipative medium [c=330m//s & density of air =1.29kg//m^(3)] |
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Answer» `0.8xx10^(-2)m` `thereforeA=[(W)/(8pi^(3)rhovupsilon^(2))]^(1//2).(1)/(R)cong10^(-6)m` |
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| 2273. |
In the below figure, an electron moves at speed v=100m//s along an x-axis through uniform electric and magnetic fields. The magnetic field is directed into the page and has magnitude 5.00 T. In unit-vector notation, what is the electric field? |
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Answer» |
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| 2274. |
A man is standing on a spring platform Reading of spring balance is 60Kg. Wt. If man jumps outside platform, then reading of the spring balance |
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Answer» FIRST increase and then DECREASES to zero |
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| 2275. |
प्रसिद्ध चिपको आंदोलन संबंधित है |
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Answer» मेनका गांधी |
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| 2276. |
A piece of copper and piece of germanium and cooled upto 80 K temperature so …… |
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Answer» resistance of both INCREASES |
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| 2277. |
Charges Q_0 and 2Q_0 are given to parallel plates A and B, respectively, and they are separated by a small distance. The capacitance of the given arrangement is C. Now, plates A and B are connected to positive and negative terminals of battery of potential difference V = 2Q_0//C respectively, as shown, then the work done by the battery is |
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Answer» `(2Q_(0)^(2))/C` The final charge in INNER plate of `A` will be `2Q_(0)`. HENCE charge supplied by BATTERY `DeltaQ=(2Q-((-Q_(0))/(2)))=(5Q_(0))/(2)` Hence work done by battery `W=(DeltaQ)V=(5Q_(0))/(2)xx(2Q_(0))/(C)=(5Q_(0)^(2))/(C)` |
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| 2278. |
The rectangular coil is moving-coil galvanometer has 100 turns, each of length 5 cm and breadth 3 cm, and is suspended in a radial magnetic field of intuction 0.05 Wb//m^(2). The twist constant of the suspension fibre is 2 xx 10^(-9) N.m/degree. Calculate the deflecting torque and the current through the coil which will deflect it through 30^(@). |
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Answer» Solution :Data: `N=100, A=lb =5 cm xx 3cm = 15 cm^(2) = 15 xx 10^(-4) m^(2)`, `B=0.05 Wb//m^(2), C=2 xx 10^(-9) N.m//"DEGREE", theta=30^(@)` `tau_(p) = tau_(R)therefore NIAB=Ctheta` `therefore` The current through the coil, `I=(Ctheta)/(NAB)` `=(6 xx 10^(-8))/(100 xx 15 xx 10^(-4) xx 0.05) = (600 xx 10^(-10))/(15 xx 5 xx 10^(-4)) = 8 xx 10^(-6)` A |
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| 2279. |
Mass of mars is (1)/(9)and radius (1)/(2) that of earth. A body weighs54 kg on surface of the earth. Its wt. on the surface of mars will be: |
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Answer» 50 kg `W_(2)=(4)/(3)W_(1)=(4)/(9)xx54=24 kg`. Thus correct choice is (b). |
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| 2280. |
निम्नलिखित फलनों का x के सापेक्ष अवकलन कीजिए: y=(x-1)(x-2)(x-3) |
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Answer» `(DY)/(DX)=(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)` |
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| 2281. |
बीज बचाओ आंदोलन संबंधित है |
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Answer» गढ़वाल क्षेत्र से |
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| 2282. |
When is a wheatstone's bridge is said to be balanced ? |
| Answer» Solution :If no current flows through the GALVANOMETER. In BALANCED condition P/Q=R/S. | |
| 2283. |
Answer the following questions : (a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band. (b) In what way is diffraction from each slit related to interference pattern in a double slit experiment ? ( c) When a tiny circular obstacle is placed in the path if light from a distant source, a bright sound seen at the centre of the shadow of the obstacle. Explain why ? ltbr. (d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend round corners, how is it that the students are unable to see each other even through they can converse easily. (e) Ray optics is based on the assumptionthat light travels in a straight line. Diffraction effects (observed when light propagates through small apertures//slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so sommonly used in understanding location and several other properties of images in optical instruments. What is the justification ? |
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Answer» Solution :(a) when width (a) of single slit is made double, the half angular width of central maximum which is `lambda//a`, reduces to half. The intensity of central maximum will become `4 times`. This is because area of central diffraction band would become `1//4th`. (b) If width of each slit is of the order of `lambda`, then INTERFERENCEPATTERN in the double slit experiment is modified by the diffraction pattern from each of the two slits. ( c) This is becasue waves diffracted from the edges of circular obstacle INTERFERE constructively at the centre of the shadow resulting in the formation of a bright spot. (d) For diffraction of waves by obstacle//aperture, through a large angle, the size of obstacle//aperture should be comparable to wavelengt. This follows from `sin theta = lambda//a`. For light, `lambda ~~ 10^(-7)m` and sizze of wall `a ~~ 10m.:.sin theta = (lambda)/(a) = (10^(-7))/(10) = 10^(-8):. theta RARR 0`. i.e light GOES almost unbent. The students are thus unable to see each other. For sound waves of FREQUENCY `~~ 1000 Hz, lambda = (v)/(n) = (330)/(1000) = 0.33m` `sin theta = (lambda)/(a) = (0.33)/(10) = 0.033` `:. theta` has a definite values i.e. sound waves bend around the partition. Hence students can cinverse easily. (e) The ray optics assumption is used in understanding loaction and several other properties of image in optical intstruments. This is because typical sizes of paertures invloved in ordinary optical instruments are much larger than the wavelength of light. Therefore, diffraction or bending of waves is of no significance. |
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| 2284. |
Electric charge of 1 mu c, -1 mu C andmu c are placed in air at the corners A,B and C respectively of an equilateral triangle ABC having length of each side 10 cm .The resultant force on the charge at C is |
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Answer» `0.9 N` |
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| 2285. |
The distance between a slit and a biprism of acute angle 2^(@) is 10 cm. Find (a) the fringe width and (b) the width of the entire fringe pattern when observation is made on screen at a distance of 90 cm from the biprism. The refractive index of the biprism is 1.5 and l = 5890 Å. |
| Answer» SOLUTION :(a) 0.168 MM, (B) 31.5 mm | |
| 2286. |
A circular coil, having 100 turns of wire of radius (nearly) 20 cm each, lies in the X-Y plane with its centre at the origin of coordinates. Find the magnetic field at the point (0, 0, 20sqrt(3)cm), when the coil carries a current of (2/pi) A. |
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Answer» Solution :As per QUESTION the given POINT lies at the AXIAL line of the circular coil at a distance `x = 20sqrt(3)` CM `=0.2 XX sqrt(3)m` from the centre of coil. Moreover, `N = 100, R = 20 cm = 0.2 m and I = (2/pi)A` `:.` Magnetic field `B = (mu_0 NIR^2)/(2[R^2 + x^2]^(3/2)) = ((4 pi xx 10^(-7)) xx 100 xx (2/pi) xx (0.2)^(2))/(2[(0.2)^(2) + (0.2 xx sqrt(3))^(2)]^(3//2)) = 6.3 xx 10^(-5) T` |
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| 2287. |
A point of application of a force vec F= 5hati - 3hatj +2hatk is moved from vec r_1= 2hati + 7hatj +4hatk to vec r_2 =-5hati + 2hatj +3hatk. The work done is |
| Answer» Answer :A | |
| 2288. |
Explain the modes of vibration of an air column in an open pipe . |
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Answer» Solution :Modes of vibratioin of an AIR COLUMN in an open pipe `:` (1) For a open pipe both the ends are open. So antinodes will be formed at both the ends. But two antinodes cannot exist without a node between them. (2) The possible harmonics in vibrating air column of a open pipe is given by `v = ( n v)/( 2l)` Where n =1,2,3, (`1^(st)`HARMONIC or fundamental frequence ) (3) In first normal Mode of vibrating air column in a open pipe`v_(1) =( v)/( 2l) =v ` (4) In second normal Mode of vibrating air column in a openpipe,`v_(2) = (2v)/(2l)= 2v ` (5) In third, normal Mode of vibrating air column in a open pipe, `v_(3)=( 3v)/(2l) =3v ` (6) In open pipe the ratio of frequencies of harmonics is `v_(1): v_(2) : v_(3) =v,2v: 3v= 1:2:3` 
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| 2289. |
Ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is |
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Answer» `1:2:3` |
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| 2290. |
Explain the polarization by a thin plastic like sheet. |
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Answer» Solution :Phenomenon of acquiring plane polarized light from unpolarized light is called polarization. A sheet by which the polarized light that can be OBTAINED from unpolarized light is called polaroid. For example : Thin plastic sheet and tourmaline plate. A polaroid consists of long CHAIN molecules alinged in a particular direction. Such electric vectors along the direction of the aligned molecules get absorbed and molecules arranged in a perpendicular direction, this direction is called pass axis of polaroid. It passes components of light vectors parallel to the pass axis and perpendicular components are absorbs. So the waves that come out of it will get linearly plane polarized If the light from an ordinary source passes through a polaroid sheet `P_(1)` its INTENSITY is reduced by half. Rotating `P_(1)` has no effect on the transmitted beam and transmitted intensity remains constant but the intensity on sheet `P_(1)` is less than the intensity of transmitted light.  Now LET an identical piece of polaroid `P_(2)` be placed before `P_(1)` and now rotating `P_(1)` the intensity of light coming from `P_(2)` reduces. When `P_(2)` and `P_(1)` are parallel, the light vectors passes from `P_(2)` also pass through `P_(1)` so the intensity of light is equal to the light transmitted from a `P_(1)` sheet. When the `P_(2)` sheet is kept in the same position, the `P_(1)` sheet is given `90^(@)` rotation then intensity from `P_(1)` becomes zero. This is shown in figure.  It is shown in figure (a) that light passes through `P_(2)` and `P_(1)` polaroid and by keeping one fixed the relative intensity of the EMERGING light becomes 1 to 0 as the other polaroid rotates from `0^(@)` to `90^(@)`.  Figure (b) shows that the light (electric) component (polarized light) passing through both polaroids is parallel to the polaroid axis. |
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| 2291. |
Each question contains statements given in two Columns, which are to be a matched. Statements in Column-I are labelled as B A, B, C and D, whereas statements in Column-II C are labelled as p, q, r and s. Match the entries of D Column-I with appropriate entries in Column-II. Each entry in Column-I may have one or more than one correct option from Column-II. The answers to these questions have to be appropriately bubbled as illustrated in the given example, if the correct matches are A to (q,r), B to (p,s), C to (r,s) and D to (q) |
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Answer» Gamma rays are ELECTROMAGNETIC waves which have highest penetrating power and cause nuclear reactions. `B to ( r)` `gamma` rays also have penetrating power but LESS than gamma rays and can cause nuclear reactions. `C to (q)`: When a radioactive element emits `BETA^(-)` rays, then the mass number of the element remains unchanged and only atomic number increases. `D to (q)` For a positron emission by a radioactive element, mass number remains same and atomic number of daughter nucleus DECREASES. |
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| 2292. |
Each question contains statements given in two Columns, which are to be a matched. Statements in Column-I are labelled as B A, B, C and D, whereas statements in Column-II C are labelled as p, q, r and s. Match the entries of D Column-I with appropriate entries in Column-II. Each entry in Column-I may have one or more than one correct option from Column-II. The answers to these questions have to be appropriately bubbled as illustrated in the given example, if the correct matches are A to (q,r), B to (p,s), C to (r,s) and D to (q) |
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Answer» |
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| 2293. |
Define critical temperature, critical pressure and critical volume. |
| Answer» SOLUTION :Critical temperature. It is the temperature of a substance in gaseous state, below which the GAS can be liquified by PRESSURE only and above which the gas can not be liquified. So a gas is SIMPLY a vapour below its critical temperature. Critical pressure. It is a pressure required to liquify a gas at its critical temperature.Critical VOLUME. It is the volume occupied by one gm of gaseous substance at critical temperature and critical pressure | |
| 2294. |
A telescope consisting of two convex lenses of focal lengths 0.20m and 0.01m is focussed on a distant object for normal vision. What is the distance between the glasses? If the same telescope is now focussed on an object which is 2 m away from the object glass without altering the accommodation of the eye, by how much does the eye-piece have to be shifted? |
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Answer» |
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| 2295. |
Consider the followingA and Band identifythe correctchoicein the given answers. (A ) For a bodyresting on a roughhorizontaltable , it is easierto pull at angle that pushat the same angle th cause motion (B) A body sliding downa rough inclined planeof inclination equal to angle offriction hac nonzeroacceleration |
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Answer» both A and BARE true |
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| 2296. |
A charged ball hangs from silk thread which makes an angle 'theta ' with large charged conducting sheet 'P' as shown. The surface charge density sigma ( sigma ) of the sheet is proportional to |
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Answer» `cos THETA ` |
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| 2297. |
If the ratio of the concentration of electrons to that of holes in a semiconductor is 7//5 and the ratio of currents is 7//4,then what is the ratio of their drift velocities ? |
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Answer» `5/8` |
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| 2298. |
Three plane sources of sound of frequency n_(1) = 400 Hz, 401 Hz and n_(3) = 402 Hzof equal amplitude a each, are sounded together. A detector receives waves from all the three sources simultaneously.It can detect signals of amplitude gt A. Calculate (a) period ofone complete cycle of intensity received by the detector and (b) time for which the detector remains idle in each cycle of intensity. |
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Answer» Solution :We know that `y = A sin omega t = A sin (2 PI n t)` The displacements of the medium particles CAUSED by these waves are given as `y_(1) = A sin (800 pi t)` `y_(2) = A sin(802pi t)` `y_(3) = A sin (804 pi t)` The resultant displacement of medium particle at time t is given by `y = y_(1) + y_(2) + y_(3)` = `A[sin (800 pi t) + sin(802 pi t ) + sin (804 pi t)]` `A[sin (800 pi t) + sin(802 pi t ) + sin (804 pi t)]` `A [ 1 + COS 2 pi t ] sin 802 pi t` The resultant is also a plane wave. Let its amplitude be R. The, `y = R sin omega t` Here `R = A ( 1 + cos 2 pi t)` Equation (6.115) shows that the resultant amplitude (or resultant intensity, I infty`R^(2))` varies with time. The resultant amplitude is maxiumum, when ` `2PI t = 0, 2 pi, 4 pi`, ... `t = 0, 1, 2, 3 ....` Hence period of one complete cycle of intensity is one second. (b ) Given that signal is detected when A ge a Thus `cos 2 pi t ge 0` Thus cos `2pi t` should lie either in first quadrant or in fourth quadrant I,e.. Between either 0 to `pi/2 or 3pi/2 and 2pi`. So, during first cycle of intensity, the signal is detected when `0 le t le (1)/(4)` and `(3)/(4) le t le 1` This shows that detector remains ideal from `t = (1)/(4)s` to `t = (3)/(4)s` Therefore, in each cycle of intensity, the detector remains ideal for`(3)/(4) - (1)/(4) = (2)/(4)` = 0.5 s. |
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| 2299. |
A 4muF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 2muF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electrostatic radiation? |
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Answer» Solution :`C_(1)=4 xx 10^(-6)F"V=200V"C_(2)=2 xx 10^(-6)F` `E_(1)=1/2 C_(1)V^(2)=1/2 xx 4 xx 10^(-6)xx 200 x 200=8 xx 10^(-2)J` `Q=C_(1)V=4 xx 10^(-6) xx 200 =8 xx 10^(-4)C` `E_(2)=1/2 (Q_(2))/(C_(1)+C_(2))=1/2 xx (8 xx 10^(-4) xx 8 xx 10^(-2))/(6 xx 10^(-6))=(16)/(3) xx 10^(-2)` `triangleE=8 xx 10^(-2) -(16)/(3) xx 10^(-2) =8/3 x 10^(-2)J` |
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| 2300. |
The ovaries are located one on each side of the lower abdomen and is connected to the pelvic wall and uterus by |
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Answer» bone |
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