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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.
| 1201. |
The reading of ammeter in the adjoining diagram will be |
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Answer» `(2)/(17)A`  `I-I_(1)3V,3Omega` `Ixx2-2+21+Ixx1+2I_(1)=0` `3I+_2I_(1)=2""(1)` `-1-(I-I_(1))xx4+3+2I_(1)=0` `-4I+4I_(1)+2+2I_(1)=0` `-41+61^(1)=-1` `6I+4I_(1)=4""...(2)` `-6I+9I_(1)=-3""...(1)` `13I_(1)=1""thereforeI_(1)=1//13A""...(2)` |
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| 1202. |
An observer starts moving with uniform acceleration .a. towards a stationary sound source of frequency f. As the observer approaches the source, the apparent frequency f heard by the observer varies with time t as: |
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Answer»
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| 1203. |
Explain the experimental deterimental of material of the prism usingspectrometer. Determination of refractive index of material of the prism. |
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Answer» Solution :The preliminary adjustments of the telescope, colimator and the prism table of the sepectometer are MADE. The refractive index of the prism can be determined by knowing the angle of the prism and the angle of minimum devition. (i) Angle of the prism (A) The prism is PLACED on the prism table with its refracting edge facing the colimator. The slit is illuminated by a sodium light (monochromotic light). The parallel rays coming from the colimator fall on the two faces AB and AC. The telescope is rotated to the position `T_(1)` unlit the image of the slit formed by thereflection at the face AB is made to coincide with the vertical cross wire of the telescope.  The readings of the vermiers are noted. The telescope is then rotated to the position ` T_(2)` where the image of the slit formed by the reflection at the face AC coinides with the vertical cross wire. The readings are again noted. (ii) Angle of minimum deviation (D) The prism is placed on the prism table so that the light from the colimator falls on a refracting face, and the refracted image a observed through the telescope. The prism table is now rotated so that angle of deviaton decreases. A stage comes when the image stops for a moment and if we rotate the prism table further in the same DIRECTION, the image is seento reede and the angle of deviation increases. The vertical cross wire of the telescope is made to coincide with the image of the slit where it turns back. This gives the minimum deviation position.  The readings of the verniers are noted. Now, the prism is removed and teh telescope is turned to receive the direct ray and the vertical cross wire is made to coinside with the image. The readings of the verniers are noted. The difference between the two readings gives the angle of minimum deviation D. The refractive index of the material of the prism n is calculated using the formula. `n=(sin(A+d)/(2))/(sin((A)/(2)))` The refractive index of a liquid may be determined in the same way using a hollow glass prism filled with the given liquid. |
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| 1204. |
The simple Bohr model can not be directly applied to calculate the energy levels of an atom with many electrons. This is because |
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Answer» of the electrons not being subjected to a central force |
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| 1205. |
Device that converts one form of energy into another is called |
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Answer» transmitter |
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| 1206. |
Ball pen functions on the principles of |
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Answer» a)VISCOSITY |
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| 1207. |
A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the vernier scale, which match with 16 main scale divisions. For this vernier calipers the least count,is |
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Answer» 0.02 MM |
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| 1208. |
Assertion: Electric conduction in gases is possible at normal pressure. Reason: The electric conduction in gases depends only upon the potential difference between the electrodes. |
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Answer» If both assertion and reason are TRUE and the reason is the correct explanation of the ASSERT ion. |
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| 1209. |
What is relation between lambda_(K alpha), lambda_(K beta) and lambda_(Lalpha) |
| Answer» SOLUTION :`(1)/(lambda_(K ALPHA))=(1)/(lambda_(k BET))+(1)/(lambda_(L alpha))` | |
| 1210. |
Why did the author decide to go for a round the-world voyage? |
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Answer» to duplicate the voyage made 200 years ago by Captain JAMES Cook |
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| 1211. |
What do you understand by self-inductance of a coil? Given its physical singificance. |
Answer» SOLUTION :i. When a circuit is switched on, the increasing current induces an emf which opposes the growth of current in a circuit (Figure (a)). Likewise, when circuit is broken, the reverse direction. This emf now opposes the decay of current (Figure (B)).  ii. Thus, inductance of the coil opposes any change in current and TRIES to maintain the original state. III. When the current i CHANGES with time, an of electromagnetic induction, this self-induced emf is given by `epsilon=(d(NPhi_(B)))/(dt)=-(d(Li))/(dt)""...(1)` `("using equation 1")` `thereforeepsilon=-L(di)/(dt)(or)L=(-epsilon)/(di//dt)` |
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| 1212. |
Calculate the wavelength associated with an electron accelerated by p.d. of 100 volts |
| Answer» SOLUTION :`LAMBA=(1.227xx10^-9)/sqrtV=(1.227xx10^-9)/sqrt100=1.227xx10^-10m` | |
| 1213. |
Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is: |
| Answer» ANSWER :A | |
| 1214. |
In the figure given below, find the (a) equivalent capacitance of the network between points A and B. Given: C_(1)=C_(5)=8muF, C_(2)=C_(3)=C_(4)=4muF. (b) maximum charge supplied by the battery, and (c ) total energystored in the network. |
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Answer» Solution :(a) In the network capacitors `C_(1) and C_(2)` as well as `C_(4) and C_(5)` are short CIRCUITED and do not contribute to equivalent capacitance of the circuit. `therefore` Equivalent capacitance `C=C_(3)=4MUF`. (b) `therefore` Charge supplied by the battery `Q=CV=(4muF)XX 7V=28muC` (c ) Total energy stored in the network `U=(1)/(2) CV^(2)=(1)/(2)(4F)xx(7V)^(2)=98muJ` |
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| 1215. |
A curved surface of radius R separates two medium of refractive indices mu_(1) "and" mu_(2) as shown in figures A and B . Choose the correct statement(s) related to the real image formed by the object O placed at a distance x, as shown in figure A. |
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Answer» Real IMAGE is always FORMED irrespective of the position of OBJECT `mu_(2) gt mu_(1)` _E01_077_S01.png) Real image can't be formed always virtual. |
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| 1216. |
A paramagnetic sample shows a new magnetisation of 8 "Am"^(-1) when placed in and external magnetic field of 0.6 T at a temperature of 4 K. When the same sample is placed in an external magnetic field of 0.2 T at temperature of 16 K, the magnetisation will be |
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Answer» `(32)/(3) Am^(-1)` So magnetisation, `I prop (B)/( T)` `therefore I_(1) prop (B_1)/( T_1) and I_(2) prop (B_2)/( T_(2))` `therefore (I_2)/( I_1) = (B_2)/( B_1) xx (T_1)/( T_2) ` `I_1= 8 "Am"^(-1) , "Taking" B_(1) = 0.6 T, T_(1) = 4K` `B_(2) = 0.2 T , T_(2) = 16` K `(I_2)/( 8) = ((0.2)/( 0.6)) ((4)/(16))` `I_(2) = (8)/(12) = (2)/(3) "Am"^(-1)` |
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| 1217. |
A non-polar molecule is located at the axis of a thin uniformly charged ring of rasius R. At what distancex fromthe ring's centre is the magnitude of the force F acting on the given molecule (a) equal tozero, (b) maximum ? Draw the appoximate plot F_(x) (x). |
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Answer» Solution :The electric field`E`at distance `x` from the centre of the ring is, `E(x)= (qx)/(4pi epsilon_(0) (R^(2) + x^(2))^(3//2))` The induced DIPOLE moment is `p = beta epsilon_(0) E = (q beta x)/(4pi (R^(2) + x^(2))^(3//2))` The force on this molecule is The force on this molecule is `F = p (del)/(del x) E = (q beta x)/(4pi (R^(2) + x^(2))^(3//2))(q)/(4pi epsilon_(0)) (del)/(dx) (x)/((R^(2) + x^(2))^(3//2)) = (q^(2) beta)/(16 pi^(2) epsilon_(0)) (x (R^(2) - 2x^(2)))/((R^(2) + x^(2))^(4))` Thisvanishes for `x = (+-R)/(sqrt(2))` (apart from `x = 0, x = oo`) It is maximum when `(del)/(del x) (x (R^(2) - x^(2) xx 2))/((R^(2) + x^(2))^(4)) = 0` or, `(R^(2) - 2x^(2)) (R^(2) + x^(2)) - 4x^(2) (R^(2) + x^(2)) - 8X^(2) (R^(2) - 2x^(2)) = 0` or, `R^(4) - 13 x^(2) R^(2) + 10 x^(4) = 0`, or `x^(2) = (R^(2))/(20) (13 +- sqrt(129))` or, `x = (R )/(sqrt(20)) sqrt(13 +- sqrt(129))` (on either side), Plot of `F_(n) (x)` is as shown in the ansersheet. |
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| 1218. |
The figure (i) shows the graphical representation of the air molecules in a tube of air ( length =L) at atmosphric pressure on the absolute pressure P(x) graph.which of following picture corresponds to absolute pressure P(x) of figure (ii)? (##TRG_PHY_MCQ_XII_C07_E04_022_Q01.png" width="80%"> |
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Answer» ` (##TRG_PHY_MCQ_XII_C07_E04_022_O01.png" WIDTH="30%"> |
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| 1219. |
In single slit experiment , the width of the slit is reduced . Then , the linear width of the principal maxima… |
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Answer» INCREASES but BECOMES LESS bright |
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| 1220. |
Name the electromagenticreadiations used for (a) water puification, and (b) eye surgery. |
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Answer» Solution :(a) U.V RAYS are used for WATER purification. (b) INFRARED Raysare used for eye surgrey. |
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| 1221. |
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn fromthe standard cell, a very high resistance of 600 k Omega is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf e and the balance point found similarly, turns out to be at 82.3 cm length of the wire (a) What is the value epsilon ? (b) What purpose does the high resistance of 600 k Omega have ? (c) Is the balance point affected by this high resistance? (d) Is the balance point affected by the internal resistance of thedriver cell?(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V? (f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit? |
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Answer» Solution :(a) Constant emf of the given standard cell, `E_(1)=1.02 V` Balance point on the wire, `l_(1) = 67.3` CM A cell of unknown emf, `epsilon`, replaced the standard cell. Therefore, new balance point on the wire, `l = 82.3` cm The relation CONNECTING emf and balance point is, `(E_(1))/(l_(1))=(epsilon)/(l)` `epsilon = (l)/(l_(1))xxE_(1)` `= (82.3)/(67.3) xx1.02 = 1.247` V The value of unknown emfits 1.247 V (b) The purpose of using the high resistance of `600 k Omega` is to reduce the current through the galvanometer when the movable contact is far from the balance point (c) The balance point is not affected by the presence of high resistance (d) The point is not affected by the internal resistance of the driver cell (e) The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire. (F) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error. The given circuit can be MODIFIED if a series resistances is connected with the wire AB. the potential drop acorss AB is slightly greater than the emf measured. The percentage error would be small. |
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| 1222. |
The current (in A) flowing through 6Omegaresistor in the following circuit is: |
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Answer» `1.2` `R_(eq)=2+2+2"",""R_(eq)=6OMEGA` `I=(30)/(6)"",""I=5A` Current through `6Omega` resistor `=5XX(3)/(6+3)=(15)/(9)A=1.7A` |
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| 1223. |
SI unit of permittivity of free space is |
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Answer» Farad |
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| 1224. |
In Fig. 4-28, particle A moves along the line y = 30 m with a constant velocity vecv of magnitude 3.0m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with a zero initial speed and a constant acceleration veca of magnitude 0.40m//s^(2). What angle theta between veca and the positive direction of the y axis would result in a collision? |
| Answer» SOLUTION :`60^(@)` | |
| 1225. |
(a) Write the factors by which the resolving power of a telescope can be increased. (b) Estimate the angular separation between first-order maximum and third-order minimum of the diffraction pattern due to a single slit of width 1mm, when light of wavelength 600 nm is incident normal on it. |
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Answer» SOLUTION :Resolving power = `D/(1.22 lambda) = (Dv)/(1.22 c)` For LIGHT of given frequecy, resolving power can be increased by increasing diameter of objective LENS. |
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| 1226. |
Three large identical conducting plates of area A are closely placed parallel to each other as shown (the area A is perpendicular to plane of diagram). The net charge on left, middle and right plates are Q_(L), Q_(M) and Q_(R ) respectively. Three infinitely large parallel surface S_(L), S_(M) and S_(R ) are drawn passing through middle of each plate such that surface are perpendicular to plane of diagram as shown. Then pick up the correct option(s). |
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Answer» The net charge on LEFT side of surface `S_(L)` is equal to net charge on RIGHT side of surface `S_(R )` `:.` Charge on left most surface `q_(1)` is equal to charge on rigth most surface `q_(6),` that is, `q_(1)=q_(6)` hence all STATEMENTS are true. 
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| 1227. |
Express the combined power of two lenses, one of focallength +f_(1) and of the other-f_(2) in contact with each other. |
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Answer» <P> Solution :`(1)/(F)=(1)/(f_(1))+(1)/(-f_(2))`where .f. is in metre. i.e., `P=(P_(1)-P_(2))` and .P. is in DIOPTRE. |
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| 1228. |
A vehicle of mass m is moving on a rough horizontal road with momentum p. If coefficient of Motion between the tyres and the road is mu calculate the stopping distance. |
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Answer» `P//2MU MG` |
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| 1229. |
The earth's core is known to contain tron. Yet geologists do not regard this as a source of the earth's magnetism. Why? |
| Answer» Solution :(b) Because, temperature at the core of Earth is very high, iron and MAGNETIC ores of iron are in molten state where they do not possess domain structure and hence they are not FERROMAGNETIC. Hence, they are not RESPONSIBLE for terestrial MAGNETISM. (i.e. geo-magnetism). | |
| 1230. |
A constant couple of 500 Nm turns a wheel of moment of inertia 100 kg m^(2) about an axis through its centre, the angular velocity gained in two second is : |
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Answer» 5 rad `s^(-1)` `omega=omega_(0)+alphat.=0+5xx2=10" rad/s"` |
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| 1231. |
In the previous Q.No. 6-23, all the particles: |
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Answer» Of the MEDIUM VIBRATE in the same phase |
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| 1232. |
Three identical capacitors A,B and C are charged to the same potential and then made to discharge thorugh three resistances R_(A),R_(B) and R_(C), where R_(A)gtR_(B)gtR_(C). Their potential differences (V) are plotted agains time t giving the curves 1,2, and 3. Find the correlations between A,B,C and 1,2,3 |
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Answer» `1toA` |
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| 1233. |
Define nuclear fission. Write the fission reaction of a neutron with uranium isotope ""_(92)U^(235). Write one use of nuclear fission reaction. |
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Answer» `""_(92)U^(235) +""_(0)n^(1) rarr ""_(36)Kr^(92)+""_(56)Ba^(141)+3 ""_(0)n^(1)+Q` Used in nuclear reactor. |
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| 1234. |
For a simple microscope would you prefer a lens of higher focal length or smaller focal length ? Why? |
| Answer» SOLUTION :A convex LENS of least POSSIBLE focal length so that its MAGNIFYING power is high. | |
| 1235. |
A circular copper loop is placed perpendicular to a uniform magnetic field of 0.50 T. Due to external forces, the area of the loop decreases at a rate of 1.26 xx 10^(-3) m^(2)/s. Determine the induced emf in the loop. |
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Answer» `3.1xx10^(-4)V` |
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| 1236. |
A cannon of mass 1000 kg located at the base of an inclined plane fires a shell of mass 100 kg in ahorizontal direction with a velocity 180 km/hr. The angle of inclination of inclined plane with thehorizontal is 45^@. The coefficient of friction between the cannon and inclined plane is 0.5. The height inmeter to which the cannon ascends the inclined plane as a result of recoil is : (g = 10 m//s^2): |
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Answer» SOLUTION :Momentum of cannon after recoil = Momentum of SHELL `=100xx180xx5/18` =5000 kg-m/s Velocity of recoil of cannon =`5000/1000`=5 m/s LET the cannon ascends a distance s on the rough inclined surface. The effective deceleration is  `a=(muR+Mgsin45^@)/M` `a=(muMgcos 45^@ + Mgsin 45^@)/M` `a=g/sqrt2[mu+1]` `v^2=u^2-2as` `0=(5)^2-2xxg/sqrt2(mu+1)s` or `s=25/(sqrt2xx10(0.5+1))=25/(sqrt2xx15) = 5/(3sqrt2)` m Now, `H = s.sin45^@=5/(3sqrt2)xx1/sqrt2=5/6`m |
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| 1237. |
In Bainbridge's mass spectroscope the singly ionised ion of N_(e)^(20) strike the photographic plate at a distance of 20 cm from the slit. Then the distance at which singly ionised ions of N_(e)^(22) strike the photographic plate is : |
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Answer» 10cm `q_(1) and q_(2)` are same single ionised `N_(e)^(200) and N_(e)^(22)` but `(m_(1))/(m_(2))=(20)/(22)` `:.(r_(1))/(r_(2))=(20)/(22)` When `r_(1)=20 cm` and `r_(2)= 22 cm` `r_(2)=22 cm` |
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| 1238. |
A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV. |
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Answer» Solution :Mass NUMBER of heavy nucleus X, A = 240 Binding energy PER nucleon of nucleus X, B = 7.6 MeV `therefore` Total binding energy of heavy nucleus` = AB = 240 XX 7.6 MeV = 1824 MeV` Mass numbers of fragments Y and Z are `A_(1) = 110` and `A_(2) = 130` Binding energy per nucleon of each fragment `B_(1)=B_(2)=8.5MeV` `therefore` Total binding energy of fission fragments=`A_(1)B_(1)+A_(2)B_(2)=110xx8.5+130xx8.5=2400MeV` `therefore` Total energy Q released per fission=Binding energy of fission fragments-Binding energy of heavy nucleus=2040-1824=216MeV. |
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| 1239. |
Draw a schematic diagram of the experimental arrangement used by Davisson and Germer to establish the wave nature of electrons. Explain how the de-Broglie relation was experimentally verified in case of electrons. |
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Answer» Solution :A labelled diagram of the apparatus used by Davisson and Germer in their famous experiment to demonstrate WAVE NATURE of electron beam using diffraction effect is shown in figure. In their experiment, a fine collimated beam of mono-energetic electrons was allowed to fall on the surface of a nickel crystal. the intensity of the electron beam, scattered in a given direction, was measured by an electron detector (movable collector). it could be rotated on a circular scale and was connected to a sensitive galvanometer. galvanometer reading i.e., deflection was directly proportional to the intensity of the electron beam entering the collector. variation of the intensity (I) of the scattered electrons with the angle of SCATTERING `theta` is obtained for any value of ACCELERATING voltage. figure shows one such graph for accelerating voltage V=54 V. in this graph, a sharp peak is obtained at `theta=50^(@)`. the appearance of the peak is due to constructive interference of electrons scaterred from different layers of the regularly spaces atoms of Nickel crystal. From electron diffraction measurements, the wavelength of electron waves was found to be 1.67 Å, whereas theoretical value `LAMDA=(12.27)/(sqrt(54))Å=1.66Å`  As the two values are in an excellent agreement, it proves the wave nature of electrons and the formula for de-Broglie wavelength. |
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| 1240. |
The decay constant (lambda) of radium is 0.0356 per year (1.13 xx 10^(-9) per sec). How much time is taken to reduce a. to (1//10)^(th) of the original value? b. by (1//10)^(th) of the original value? |
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Answer» Solution :Data supplied, a. `N = (N_0)/(N) , "" lambda = 0.0356` PER year `t = 1/(lambda) XX 2.303 LOG ((N_0)/(N)) = (1 xx 2.303 xx log 10)/(0.0356) = 64.69 ` years b. `N = (9N_0)/(10) "" :. (N_0)/N = 10/9` `t = 1/lambda xx 2.303 log ((N_0)/(N)) = (2.303 xx log (10/9))/(0.0356)` `= (2.303)/(0.0356) xx (log 10 - log 9) = (2.303)/(0.0356) xx 0.0458 = 2.96` years . |
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| 1241. |
The galaxies are moving away from us as indicated by |
| Answer» Answer :C | |
| 1242. |
A spherical conductor A lies inside a hollow spherical conductor B. Charges AQ_(1)and Q_(2)are given to A and B respectively . |
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Answer» Charges `Q_(1) ` will appear on the outer surface of A . |
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| 1243. |
In a region ofspace having a spherical symmetic distribution of charge , the electric flux enclosed by a concentric spherical Gaussion surface, varies with radius r" as "phi={(phi_(0)r^(2))/R^(3), r le R phi_(0),ltR where R and phi_(0) are theconstants. |
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Answer» `E=(phi_(0))/(4pir^(2))`,for `rleR` |
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| 1244. |
How does the self inductance of a coil change when the number of turns in the coil is decreased? |
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Answer» Solution :Self-inductance of a COIL, L= `mu_0n^2lA` When number of turns is DECREASED, the self-inductance will INCREASE. It because `Lpropn^2` |
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| 1245. |
A power transmission line feeds input power a 3300 V to a step down transformer with it primary windings having 2000 turns. Wha should be the number of turns in th secondary in order to get output power a 330 V ? |
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Answer» 400 `N_P`= 2000 `V_S`=330 V Transformation RATIO , `gamma=V_S/V_P` and `gamma=N_S/N_P` `therefore V_S/V_P=N_S/N_P` `therefore N_S=(V_SN_P)/V_P`= 200 |
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| 1246. |
Three capacitors of 3 muF, 10 muF and 15 muF are connected in series to a voltage supply of 100 V. The charge on 15 muF capacitor is |
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Answer» `50 mu C ` `:.` Charge on each capacitor `q = CV = (2MUF) XX (100V) = 200 muC` |
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| 1247. |
In a full wave rectifier operating at 50 Hz, the fundamental frequency in ripples would be : |
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Answer» Solution :`f=Hz, T = 1/f = 1/50` For FULL wave rectifier , `T.=T/2=1/50xx1/2=1/100` then f. = 100 Hz |
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| 1248. |
Assuming the nuclei to be spherical in shape, how does the surface area of a nucleus of mass number A_(1) compare with that of a nucleus of mass number A_(2). |
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Answer» Solution :Since radius of a nucleus `R propA^(1/3)` , HENCE SURFACE area `SpropA^(2/3)` `IMPLIES S_(1)/S_(2)=(A_(1)/A_(2))^(2/3)`. |
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| 1249. |
12 cells each having same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two cells which are in series. Current is 3 A when cells and battery aid each other and is 2 A when cells and battery oppose each other. The number of cells wrongly connected is |
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Answer» 4 |
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| 1250. |
The speed of light in air is 3 xx 18^8 ms^-1. What will be its speed in diamond whose refractive index is 2.4 ? |
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Answer» `3 XX 10^8 m//s` |
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