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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.
| 5251. |
Distinguish between sky wave and space wave propagation. Give a brief description with the help of suitable diagrams indicating how these waves are propagated. |
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Answer» Solution :SKY wave : A radio wave transmitted TOWARDS the sky and reflected by the ionosphere towards the desired location of the earth is called a sky wave. Space wave : A radiowave that travels directly from a HIGH transmitting antenna to the receiving station is called a space wave. Diagram : Various propagation MODES of Electromagnetic waves (EMW) are shown in diagram. 
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| 5252. |
Two positive charges separated by a distance 2 m each other with a force of 0.36 N. If the combined charge is 26 mu C, the charges are |
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Answer» `20 MU C, 6 mu C` |
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| 5253. |
In the circuit shown the resistance of voltmeter is 10,000 ohm and that of ammeter is 20 ohm. The ammeter reading is 0.10 Amp and voltmeter reading is 12 volt. Then R is equal to (##MOT_CON_NEET_PHY_C22_E01_003_Q01.png" width="80%"> |
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Answer» ` 122 Omega` |
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| 5254. |
For a spherical shell |
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Answer» If potential insideit is ZERO then it necessarily ELECTRICALLY neutral |
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| 5255. |
The distance between two consecutive maxima and minima is called ___. |
| Answer» SOLUTION :FRINGE WIDTH | |
| 5256. |
A body experiences acceleration for 6 seconds after starting from rest. If it travels a distance x_1 in first 2 s, x_2 in the next two s, and x_3 in the last two s. then : |
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Answer» `x_1 : x_2 : x_3 : =1:1:1` `impliesx_1 =2a` `x_1+x_2=(1)/(2)a(4)^(2)implies x_2+x_1=8a` `x_1+x_2+x_3=(1)/(2)a(6)^(2)=18a` |
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| 5257. |
The width of an aperture is 4 mm and wavelength is 5000 Å. Calculate the distance upto which ray optics is valid. |
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Answer» |
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| 5258. |
Assertion (A): For identical charges of magnitude g each are placed at the vertices of a square of side'a'. The electric field at the centre point O is zero but electric potential is non-zero and finite. Reason (R) : Electric field is a vector quantity but electric potential is a D scalar. |
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Answer» If both assertion and REASON are true and the reason is the CORRECT explanation of the assertion. |
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| 5259. |
Starting with a sample of pure .^(66)Cu, 1/8 of it decaysinto Zn in 15 minutes. The corresponding half life is |
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Answer» `(7)1/2` minutes `1/8 = (1/3)^(3) = (1/2)^(15//t_(1)//2)therefore15/(T_(1//2)) =3 or T_(1//2) = 5` minutes |
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| 5260. |
Apparent dip at a place is delta_(1). When the dip circle is rotated by 90^(@), the apparent dip becomes delta_(2). Declination a at the place (prior to 90^(@) rotation) is given by |
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Answer» `tan ALPHA= tan delta_(1), tan delta_(2) ` |
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| 5261. |
For hydrogen atom, energy of nth level is given by |
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Answer» `E_n=-(13.6/n)EV` |
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| 5262. |
A 10 muFcapacitor and a 15 Omega resistor are connected in series and are connected across a DC supply of 100V. Calculate the impedance of circuit. |
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Answer» |
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| 5263. |
What is the colour of fresh mango leaves? |
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Answer» REDDISH brown |
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| 5264. |
In the given circuit the ammeter A_(1)" and "A_(2) are ideal and the ammeter A_(3) has a resistance of 1.9xx10^(-3)Omega.If sum of readings of all three meters is given by ((1n)/27) Ampere the value of n will be. |
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| 5265. |
A train is moving with a uniform speed of 33 m/s and an observer is aproaching the train with the same speed.If the train blows a whistle of frequency 1000 Hz and the velocity of sound is 333 m/s then the apparent of sound is 333m/s then the apparent frequency of the sound the observer hears is |
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Answer» 1220 HZ |
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| 5266. |
Calculate the radius of capillary tube if water rises to a height 4.5cm. Surface tension of water is 72 dyne/cm. Take the angle of contact of water as 0^@ and g = 980 cm//s^2. |
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Answer» 0.032 `cm |
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| 5267. |
Magnetic field produced at the centre of circular coil having radius 0.1 m and 2 turns is _____ if 1/(4pi)A current passes through it. |
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Answer» `10xx10^(-6)T` `thereforeB=(4PIXX10^(-7)xx2xx1)/(2xx0.1xx4pi)" "thereforeB=1xx10^(-6)T` |
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| 5268. |
The kinetic energy K of a particle moving along a circle of radius R depends upon the distance s as K=as^2. The force acting on the particle is |
| Answer» ANSWER :A | |
| 5269. |
Nitin and Rajeec were studying the effect of certain radiations on flower plants. Nitin expossed his plants to ultraviolet rays, found that his plants got damaged after a few days. Rajeev exposed his plants to infrared rays, found that his plants had a beautifulbloom, after a few days. (i) What is the difference between ultraviolet rays and infrared rays? Why were the plants exposed to ultraviolet rays damaged and the plants exposed to infrared rays had a beautiful bloom? (iii) What are the basic values you have learnt from this study? |
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Answer» Solution :(i) The frequency of ultraviolet rays `(v_(UV))` is `8xx10^(14) Hz` to `5xx10^(16)Hz`. The frequency of infrared rays `(v_(IR))` is `3xx10^11Hz 4xx10^14 Hz`. As energy, `E=hv`, so ultraviolet rays are much more energetic than infrared rays. (ii) A flower plant is very delicate. It can not tolerate the exposure of HIGH energy rays. As ultraviolet rays are of high energy than infrared rays therefore, the plants EXPOSED to ultraviolet rays were diamaged. And the plants exposed to infrared rays had a beutiful bloom. (iii) This study IMPLIES that small children are like flower plants. They require soft and gentle care by their mothers. Exposure of young kids to harsher treatment is dangerousand it must be avoided. |
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| 5270. |
What is an oscillator ? Discuss the use of a junction transistor as an oscillator. |
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Answer» Solution :Oscillator. It is an electronic device which provides a.c. output waveform of desired from d.c. power withour input signal. It consists of a tank circuit consisting of inductor L and capacitor C connected in parallel. The frequency of the tank circuit is given by `v = 1/(2pi sqrt(LC))` Due to the resistance of inductive coil, there OCCURS al small but constant energy loss and oscillations thus  produced are damped. To TRANSMIT speech or music, we require UNDAMPED electromagnetic waves called carrier waves. To do so L - C circuit is coupled with transistor in such a way that there is a proper feedback to the L-C circuit at the proper timings so that the energy of L-C circuit remains the same throughout oscillations. When key K is pressed the collector attains positive potential due to which a weak collector current will start rising in `L_1`. The increasing magnetic flux is linked with `L_1` and hence with L. The e.m.f. induced in L will charge upper plate of the capacitor positively and hence support the forward biasing of the E-B circuit. This results in an increase in `I_e` and hence `I_c`. Due to this more increasing magnetic flux is linked with L and hence with L. The process continues till the collector current becomes maximum i.e. saturated. Capacitor C gets discharged through inductor L and support to forward biasing of E-B circuit is withdrawn. Thus `I_e` and hence `I_c` DECREASES. There is a decreasing magnetic field in `L_1` and hence with L. The e.m.f. induced in L charges the lower plate positively. This results in opposition to forward biasing and further decrease of `I_e`. and hence Is This process repeats till `I_c = 0`. Thus oscillations set up in the circuit. |
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| 5271. |
Four charges of +q, +q, +q and +q are placed at the corners A, B, C and D of a square of side a. Find the resultant force on the charge at D |
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Answer» SOLUTION :Let SIDE `=a, BD = sqrt2a` Along AD FORCE `F_(1) = (1)/(4pi in_(0)) .(q^(2))/(a^(2))` Along CD force `F_(2) = (1)/(4pi in_(0)).(q^(2))/(a^(2)) therefore F_(1) = F_(2)` Resultant of `F_(1) and F_(2) = SQRT(F_(1)^(2) + F_(2)^(2))= sqrt2.F_(1)` Force along BD `F_(3) = (1)/(4pi in_(0)) .(q^(2))/((sqrt2a)^(2)) = (1)/(2) (1)/(4pi in_(0)) (q^(2))/(a^(2))` Resultant force at `D= sqrt2F_(1) + F_(3)` `= sqrt2 (1)/(4pi in_(0)) .(q^(2))/(a^(2)) + (1)/(2) (1)/(4pi in_(0)).(q^(2))/(a^(2))` `=(1)/(4pi in_(0))(q^(2))/(a^(2)) [sqrt2 + (1)/(2)] = (q^(2))/(8pi in_(0)a^(2)).[1 + 2 sqrt2]` 
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| 5272. |
A :Electric field lines intersect each other. R :Electric field lines are parallel in uniform electric field. |
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Answer» Both Assertion and Reason are TRUE and the Reason is CORRECT explanation of the Assertion. |
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| 5273. |
Four perfect polarising plates are stacked so that the axis of each is turned 30^@ clockwise to the preceding plate, the last plate therefore being crossed with the first. A beam of unpolarised light of intensity 1 passes through the stack perpendicularly. The transmitted beam has intensity |
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Answer» `27/128`I |
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| 5274. |
The variation of electirc potentia with distance d from a fixed point is as shown in the figure. The electirc intensity at d = 5m is |
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Answer» `2.5 VM^(-1)` |
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| 5275. |
In Young's double slit experiment is performed with blue and with green light of wavelengths 4360 Å and 5460 A respectively. If x is the distance of 4^(th) maxima from the central one, then..... |
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Answer» X (BLUE) = x (GREEN) `:.xprop lambda` `:. Lambda_("blue") lt lambda_("green") "" :. X_("blue") le x_("green")` |
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| 5276. |
Tempareture and volume of one mole of an ideal momatomic gas in a process are related as TV^(2/3)=K ,where k is constant.The molar specific heat capacity for the process is |
| Answer» ANSWER :A | |
| 5277. |
Consider a zener diode with the breakdown voltage 6.2V.What is the significance of the breakdown volt age for a zener diode? |
| Answer» Solution :It is the CONSTANT voltage ACROSS zener diode in reverse bias at which the CURRENT increases rapidly. | |
| 5278. |
Two blocks A and B having masses 100 kg and 200 kg are placed as shown in the figure. The block A is connected to a wall with a light inextensible string. The coefficient of friction between the blocks is 0.3 and that between the block B and the floor is 0.03. Then the minimum force required to move the block B will be (take g=10m//s^(2)) |
| Answer» ANSWER :D | |
| 5279. |
Current 'i' is flowing in hexagonal coil of side a. The magnetic induction at the centre of the coil will be |
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Answer» `( 3 SQRT(3)mu_0i) /( PIA)` |
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| 5280. |
In Bohr's model of hydrogen atom, the radius of the first electron orbit is 0.53 A. What will be the radius of the third orbit? |
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Answer» `4.77 Å` |
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| 5281. |
A converging lens forms a five fold magnified image of an object. The screen is moved towards the object by a distance 0.5m, and the lens is shifted so that the image has the same size as the object. Find the lens power and the initial distance between the object and the screen. |
| Answer» SOLUTION :6.4D, 1.125m | |
| 5282. |
As shown in figure, a ray of light is incident on glass cube. If it experiences total internal reflection in vertical plane, then what is the refractive index of glass? |
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Answer» `sqrt(3/2)` `therefore" "sinr=(1)/(sqrt2a^(mu)g)` For point B,sin(`90^@`-r)=`""_gmu_a` where, (`90^@-r)` becomes critical ANGLE `therefore COS r=""_gmu_g=(1)/(COSR)` `therefore =(1)/(1-sin^2r)` `=(1)/(sqrt(1-(1)/(2_amu_g^2)))`  `therefore ""_amu_g^2=(1)/(1-(1)/(2_amu_g^2))=(2_amu_g^2)/(2_amu_g^2-1)` `therefore 2_amu_g^2-1=2` `therefore ""_amu_g=sqrt(3/2)` |
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| 5283. |
At which temperature, a pure semiconductor behave slightly as a conductor? |
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Answer» LOW TEMPERATURE |
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| 5284. |
You walk slowly toward a large concave mirror.At first, you see your inverted image moving toward you. After you pass a certain point, you no longer see your image clearly.Moving still closer, you see a clear, enlarged and erect image of yourself behind the mirror.During the time when you cannot see a clear image |
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Answer» you are closer to the mirrro than the focal point and the IMAGE is now virtual and invisible |
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| 5285. |
A moving electron has energy 728 eV. It's velocity is: |
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Answer» 728 m/sec |
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| 5286. |
The impurity levels of a doped semiconductor are 30 me V below the conduction band. Determine whether the semiconductor is n-type or p-type. At the room temperature thermal collisons occur as a result of which the extra electron loosely bound to the impurity ion gets an amount of emergy kT [where k is Boltzmann constant =8.62xx10^(-5) eV//K] and hence this electron can jump into conduction band. What is the value of T? |
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Answer» Solution :The separation of impurity energy LEVEL from conduction band is less in case of n-type semiconductor and more in case of p-type semiconductor. As energy separation of impurity level `=30xx10^(-3)EV` is much smaller than energy GAP of pure semiconductor `(~~1EV)`, therefore, the doped semiconductor is n-type. Here, `E_(g)=30xx10^(-3)eV=kT` or `T=(30xx10^(-3))/k=(30xx10^(-3))/(8.62xx10^(-5))=348K` |
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| 5287. |
The surface tension of water is 7 xx 10^-2 N/m. What is the height at which water rises in a capillary tube of bore diameter 0.048 cm will be and angle of cotact is zero(Density of water = 1000 kg//m^3, g = 9.8 m//s^2) |
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Answer» 5.142 cm |
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| 5288. |
The phase and orientation of the electric field vector linked with electromagnetic wave differ from those of the corresponding magnetic field vector, respectively by: |
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Answer» ZERO and zero |
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| 5289. |
A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of 10 Omega and a resistance R to a 100 V line as shown in figure. What should be the value of R so that the heater operates with a power of 62.5 W? |
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Answer» `10 OMEGA` |
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| 5290. |
The mutual inductance of an induction coil is 5H. In the primary coil the current reduces from 5A to 0 in 10^(-3)s. What is the induced emf in the secondary coil? |
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Answer» 2500V |
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| 5291. |
A coil has an area of 0.05 m^2 and has 800 turns. After placing the coil in a magnetic field of strength 4 xx 10^(-5) "Wb/m"^2 perpendicular to the field the coil is rotated through 90^@ in 0.1 s. The average emf induced is ____ |
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Answer» zero `=N(AB cosphi_2 -AB COS phi_1)` `=NAB (cos theta_2 - cos theta_1)` `=800xx5xx10^(-2)xx4xx10^(-5)("cos"pi/2- cos0^@)` `=160xx10^(-5)[0-1]` `=-160xx10^(-5)` Wb `therefore epsilon=-(N Deltaphi)/(DELTAT)=(160xx10^(-5))/0.1 =160xx10^(-4)` `therefore epsilon` =0.016 V |
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| 5292. |
If the Earth were a homogenous sphere of wood of density 800 kg/m^(3). What would be (i)the acceleration due to gravity on the Earth's surface ? (ii) the critical velocity of a satellite orbiting close to its surface ? |
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Answer» <P> Solution :Data : `p = 800 kg//m^(3) , G = 6.67 xx 10^(-11) N.m^(2)//Kg^(2)R_(E) = 6.4 xx 10^(6)` m(i) ` g = GM_(E)//R_(E)^(2)` ASSUMING the Earth to be a HOMOGENOUS sphere of density p. ` M_(E) = 4/3 piR_(E)^(3)p` Then , the surface gravity on a wooden Earth . ` g = (G (4/3 piR_(E)^(3)p))/(R_(E)^(2)) = 4/3piGR_(E)p` ` 4/3 (3.142) (6.67 xx 10^(-110 ) (6.4 xx 10^(6)) (800) ` ` = 1.431 m//s^(2)` (ii) The critical VELOCITY close the the surface of the wooden Earth. ` v_(c0= sqrt(gR_(E)) = sqrt(1.431 xx 6.4xx10^(6))` `3.026 xx 10^(3) m//s` |
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| 5293. |
The co-ordinates of moving particle at any time t are given by x = at^(2) and y = bt^(2). The velocity magnitude of the particle : |
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Answer» `2tsqrt(a^(2)-b^(2))` `V_(y)=(DY)/(dt)=2bt` `:. V=sqrt(V_(x)^(2) + V_(y)^(2))=sqrt(4a^(2)t^(2) + 4B^(2)t^(2))` `=2tsqrt(a^(2)+ b^(2))` |
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| 5294. |
Two capacitors C_1 = 4muF and C_2 = 2muF are charged to same potential V = 500 Volt, but with opposite polarity as shown in the figure. The switches S_1 and S_2 are closed. |
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Answer» The potential difference across the two capacitors are same and is GIVEN by 500/3V |
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| 5295. |
The earth's magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably? |
| Answer» SOLUTION :YES, it does change with time. Time SCALE for appreciable change is roughly a few hundred years. But even on a much smaller scale of a few years, its variations are not completely NEGLIGIBLE. | |
| 5296. |
A radio isotope .X. with a half life 1.4 xx 10^(9)years decays to .Y. which is stable. A sample of the rock from a cave was found to contain .X. and .Y. in the ratio 1:7. The age of the rock is |
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Answer» `3.92 XX 10^(9)` years |
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| 5297. |
In X-ray tube the accelerating potential applied at the anode is V volt. The minimum wavelength of the emitted X-rays will be : |
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Answer» `E V // h` `:. lambda=(hc)/(eV)` |
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| 5298. |
Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on 'r' for r/a gt gt 1, and contrast your result with that due to an electric dipole, and an electric monopole (i.e, a single charge) |
Answer» Solution : Potential at P, `V=V_(1)+V_(2)+V_(3)+V_(4)` `=(Q)/(4pi epsi_(0)) ((1)/((R+a))-1/r-1/r+(1)/((r-a)))=q/(4pi epsi_(0)) ((1)/((r+a)) -2/r+(1)/((r-a)))` `V=(q)/(4pi epsi_(0)) ((r(r-a)-2(r^(2)-a^(2))+r(r+a))/(r(r-a)(r+a)))=(q)/(4pi epsi_(0)) ((r^(2)-ra-2r^(2)+2a^(2)+r^(2)+ra)/(r(r^(2)-a^(2))))` `=(2q.a^(2))/(4pi epsi_(0) r(r^(2)-a^(2)))=(2q.a^(2))/(4pi epsi_(0) r.a^(2) (r^(2)/a^(2)-1))=(2q)/(4pi epsi_(0) r xx r^(2)/a^(2))=(2q a^(2))/(4pi epsi_(0) r^(3))` V varies for QUADRUPOLE as `1/r^(3)` V varies for DIPOLE as `1/r^(2)` V varies for monopole as `1/r` |
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| 5299. |
Two balls are thrown from an inclined plane at angle of projection with the plane, one up the incline and the other down the incline as shown. Which one is not correct? |
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Answer» `h_(1)-h_(2)=(v_(0)^(2)sin^(2)alpha)/(2g cos theta)` `R_(2)=V_(0) cos alphaT_(2)+(1)/(2)(g sin theta)T_(2)^(2)` `T_(1)=T_(2)R_(1)cancel=R_(2)` |
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| 5300. |
In figure , there are two patterns of electric field lines absent of dielectric and in presence of dielectric slab. The relative permittivity of the dielectric slab is , |
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Answer» `5/3` `K=(E_("ext"))/(E_("net"))=(5)/(3)` |
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