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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.
| 38151. |
How we define self-inductance? |
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Answer» Solution :It is defined as the magnetic FLUX LINKED through a coil due to the flow of ONE ampere current in it and is given by L = `phi/I` where `phi` is the magnetic flux linked through a coil due to flow of current I in it. |
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| 38152. |
Two charges of dipole +-10 mu C are placed 1mm apart. Determine the electric field at a point on the axis of the dipole 10 cm away from its centre and at 10 cm away from centre on the equator of the dipole. |
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Answer» (2) `4.5 xx 10^(-2) NC^(-1)` on equator |
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| 38153. |
n resistors each of resistance r are connected to a battery of emf epsilonand internal resistance r. Then the ratio of terminal voltage to emf of battery = ...... . |
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Answer» N `epsilon = I XX ` total RESISTANCE of circuit  `epsilon = I xx (n + 1) r "" `...... (1) and V= I (r + r + r + ... N TIMES) V = Inr...... (2) `therefore (V)/(epsilon) = (n)/(n+ 1)` |
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| 38154. |
The isotopes of U^(238) and U^(235) occur in nature in the ratio 128 : 1. Assuming that at time of earth's formation, they were equal in ratio, make an estimate of age of earth. The half lives of U^(238) and U^(235) are 4.5 xx 10^9 yrs and 7.13 xx 10^8 yrs respectively: |
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Answer» `2.2 xx 10^(7) yrs` `N_(2)=N_(0) e^(-(ln2)/(T_(2))t` `N_(1)/N_(2) approx e^(ln 2) (1/T_(2)-1/T_(2))t` `128 =e^(ln 2) (1)/(7.13 xx 10^(8))-(1)/(4.5 xx 10^(9))t` `or ln 2^(7) =ln 2 xx 1.18 xx 10^(-9)t` `RARR 7 ln 2=ln 2 xx 1.18 xx 10^(-9)t` `rArr t=(7)/(1.18) xx 10^(9) =5.9 xx 10^(9)yrs.` |
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| 38156. |
What is a thermo-nuclear reaction ? |
| Answer» Solution :Nuclear FUSION reaction TAKING PLACE under extreme high temperature `(approx10^(9)K)` is called a THERMONUCLEAR reaction. | |
| 38157. |
What is particle physics ? |
| Answer» Solution :Particle physics deals with the theory of fundamental PARTICLES of nature and it is one of the ACTIVE research areas in physics. Initially it was THOUGHT that atom is the fundamental ENTITY of matter. | |
| 38158. |
Which of the following of these are called "optoelectronic junction devices"? |
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Answer» SOLAR cell |
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| 38159. |
What is the opposite of 'natural? |
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Answer» innatural |
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| 38160. |
A radioactive nuclide with Z protons and N neutrons emit an alpha-particle, two beta-particles and two gamma-rays. The number of protons and neutrons in the nuclide formed after the decay are |
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Answer» Z-3,N-1 |
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| 38161. |
A thin equiconvex lens is made of glass of refractive index 1.5 and its focal length is 0.2m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the refractive index of liquid is |
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Answer» `17/8` |
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| 38162. |
4muFcapacitor and a resistance 2.5MOmegaare in series with 12V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given In ( 2 ) = 0.693] |
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Answer» Solution :CHARGING current `i=(V_0)/(R ) e^((t)/(RC))` `therefore` Potential difference across R is `V_R = IR = V_0e^(-(t)/(RC))` `therefore` Potential difference across .C. is `V_C = V_0 - V_R` `= V_0 - V_0e^((t)/(RC)) = V_0 (1-e^((t)/(RC)) )` but given `V_C = 3V_R` , we get `1-e^(-t//RC) = 3e^(-t//RC) or 1 = 4E^(-t//RC)` `e^((t)/(RC)) = 4 rArr (t)/(RC) = ln 4 rArr t= 2RC ln 2` `t = 2.5 xx 10^6 xx 4 xx 10^(-6) xx 2 xx 0.693` or t = 13.86 sec |
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| 38163. |
A circuit containing a 80 mH inductor and a 60 mu F capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values. |
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Answer» Solution :(a) For `V=V_(0)sin omega t` `I=(V_(0))/(|omegaL-(1)/(OMEGAC)|) sin (omega t +(pi)/(2)), " if "R=0` where - sign appears if `omegaL gt 1//omegaC`, and + sign appears if `omegaL lt 1//omegaC.` `I_(0)=11.6A, I_("rms")=8.24A` (b) `V_("Lrms")=207V, V_("Crms")=437V` (c) Whatever be the current I in L, actual voltage leads current by `pi//2`. Therefore, AVERAGE POWER CONSUMED by L is zero. (d) For C, voltage lags by `pi//2`. Again, average power consumed by C is zero. (E) Total average power absorbed is zero. |
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| 38164. |
A message signal of frequecy 10 kHz and peak voltage 10V is used to modulate a carrier wave of frequency 1 MHz and peak voltage 20V. Determine : (i) The modulation index, (ii) The side bands produced . |
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Answer» Solution :`f_m=10 kHz` `V_m = 10V` `F_C= 1 MHz` `V_C= 20 V ` (i) Modulation index = `(V_C-V_m)/(V_C + V_m)` `=(20 -10)/(20 + 10)` `=(10)/(30)` `=33%` (II) Side bands `=f_c pm f_m` ` =1000 pm 10KHZ` ` = 1010 kHz & 990 kHz` |
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| 38165. |
The relation between the intensity of t electric field of an electric dipole at a distan r from its centre on its axis and the distance is....... (where r » 2a) |
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Answer» `E PROP 1/r^(4)` |
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| 38166. |
A non - conducting sphere has mass of100 gand radius 20 cm . Aflat compact coil of wire with turns 5 iswrapped tightly around it with each turns concentric with the sphere. This sphere is placed on an inclined plane such that plane of coil is parallel to the inclined plane. A uniform magnetic field of 0.5 T exists in the region in vertically upwards direction. Compute the current I required to rest the sphere in equilibrium.(##SUR_PHY_XII_V01_C03_E04_005_Q01##) |
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Answer» Solution :The sphere is in translational equilibrium , thus`f_(s)- mgsin theta ` = 0 ….(1) The sphere is in rotational equilibrium . If torques are taken about the centre of the sphere, the magnetic field produces a clockwise torque of magnitude i.e`tau = mBsinq [mu = NIA]` The frictional force `(f_(s))` produces a anticlockwise torque of magnitude `tau = f_(s)R`, where R is theradius of the sphere. Thus ` f_(s) R - mB SIN theta = 0 ` .....(2) From (1) and (2)[i.e `f_(s)= mgsin theta ` substituting in (2)]mg `sin theta. R - mu B sin theta mgR= mu B ` Substituting ` mu ` mgR = NIAB ` I = "mgR"/"NBA" ` [where A is the area of the sphere ` A = pi R^(2) ` ] ` :.I = "mg"/(pi"RBN")` Given : mass of thesphere` mu= 100 g = 100xx 10^(-3) kg ` Radius of the sphere` R = 20 cm= 20XX 10 ^(-2) m ` No. of turns of wire wrapped` N = 5` Magnetic field ` B = 0.5 ` T Current required to rest the sphere in equilibrium` I = ? ` `I = (100 xx 10^(-3) xx cancel10^(2))/(pi xx cancel5 xx 20 xx 10^(-2) xx 0.5) ` ` I = 2/pi A ` |
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| 38167. |
(A): If a heavy nucleus is split into two medium sized parts, each of the new nuclei will have more binding energy per nucleon than the original nucleus. (R): Joining two light nuclei together to give a single nucleus of medium size means more B.E per nucleon in the new nucleus. |
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Answer» Both .A. and .R. are true and .R. is the correct EXPLANATION of .A. |
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| 38168. |
If the ground state ionisation energy of H-atom is 13.6 eV, the energy required to ionize a H-atom froms second, excited state is: |
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Answer» 1.51 eV |
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| 38169. |
The plane of the coil of tangent galvanometer is kept parallel to magnetic meridian to |
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Answer» avoid the influence of EARTH's MAGNETIC FIELD |
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| 38170. |
हेनरी के नियम के अनुसार किसी गेस का आंशिक दाब निम्न के समानुपाती होता है |
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Answer» मोल प्रभाज |
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| 38171. |
Give de-Broglie's explanationof Bohr's second postulate. |
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Answer» SOLUTION :According to Bohr the ANGULAR momentum of the electron is an integral multiple of `h//2pi` i.e., `mvr=(nh)/(2pi)` de-Broglie argued that these ELECTRONS EXHIBIT wave propertyhaving stationary wave patternhaving nodes at the ends. The total distance travelled by a wave is an integral multiple of the wave length . i.e., `2pir=nlambda` `lambda=h/P=h/(mv)` `therefore 2pir=(nh)/(mv)` `mvr=(nh)/(2pi)` But angular momentum of the ELECTRONIS L=mvr Hence `L=(nh)/(2pi)` |
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| 38172. |
Consider the shown arrangement where each capacitor (without dielectric) has capacitance of 2 muF. However, one of the capacitor has got a dielectric of dielectric constant 2. When the system is in equilibrium, the dielectric is suddenly removed. How much charge would flow along the indicated direction |
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Answer» `(10)(3)muC` |
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| 38173. |
Define unit of current. |
| Answer» Solution :The current through a CONDUCTOR is called one AMPERE, if one COULOMB of charge FLOWS the conductor in one second. | |
| 38174. |
In certain Young's double slit experiment, the slit separation is 0.05 cm. The slit to screen distance is 100 cm. When blue light is used the distance from central fringe to the fourth order fringe is 0.36 cm. What is the wavelength of blue light ? |
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Answer» 4000 Ã… |
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| 38175. |
(a)1 mW of light of wavelength 456 nm is incident on a cesium surface. Calculate the photoelectric current produced, if the quantum efficiency of the surface for photoelectric emission is only 0.5 %. (b) In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches its saturation value. Assuming that on the average one out of every 10^(6) photons is able to eject a photoelectron, find the photocurrent in the circuit. |
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Answer» Solution :(a) `P =n(hc)/(lambda)`, n: NUMBER of photons/sec `n=(P lambda)/(h c)` Quantum efficiency `eta =("number of electrons ejected"//"sec")/("number of photons incident"//sec") =(n')/(n)` `n' eta n` Photoelectric CURRENT `I =n' E=eta n e` `=eta(P lambda)/(hc)e` `=((0.5)/(100)) ((10^(-3))(456xx10^(-9)))/((6.6xx10^(-34))(3xx10^(8)))xx1.6xx10^(-19)` `=1.84xx10^(-6) A=1.84 mu A` (b) Here `eta=(1)/(10^(6))` `i=eta(P lambda)/(hc)e =(1)/(10^(6))xx(5xx400xx10^(-9))/(6.6xx10^(-34)xx3xx10^(8))xx1.6xx10^(-19)` `=1.6xx10^(-6) A=1.6 mu A` |
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| 38176. |
Diameter of objective of a telescope is 10 cm Its distance from two different objects is 1 km If wavelength of light is 5000 Å, then at wha minimum distance, these objects are clearl seen by telescope ? |
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Answer» 5 cm or `x=(1.22xx5xx10^3xx10^(-10)xx10^3)/(10xx10^(-2))` `=1.22xx5xx10^(-3)`m =6.1 mm which is order of 5 mm for x. |
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| 38177. |
Give the expression for energy stored in an inductance coil carrying current. |
| Answer» SOLUTION :`U=1/(2)LI^2` | |
| 38178. |
The half life of ""_(92)^(238)U undergoing alpha decay is 4.5 xx 10^9 years what is the activity of 1g. Sample of ""_(92)^(238)U. |
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Answer» Solution :`T_(1//2)=4.5 xx 10^9 y` `=4.5 xx10^9xx3.16 xx10^7s` `=1.42xx10^(17)s` `1g "of" ""_(92)^(238)U "contains" =1/(238)xx6.025 xx10^(23)"ATOM"` `=25.3 xx10^(20) "atom"` `THEREFORE "decay RATE " =R =LAMBDA N=(0.693)/TxxN` `=(0.693 xx25.3 xx10^(20))/(1.42xx10^(17))s^(-1)` `=1.23 xx 10^4 `bq. |
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| 38179. |
An audio signal V_(m)=5sin 6 pixx10^(3)t is to be modulated on a carrier wave given by V_(c)=15sin 2pi xx 10^(5)t The frequencies of side bands and band width |
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Answer» 103 KHZ , 97 KHz , 6 KHz |
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| 38180. |
How in early days, long distance messages were sent? |
| Answer» Solution :(i) By TRAINED PIGEONS, (II) through messenger, (iii) MESSAGE through horse/cart. | |
| 38181. |
A body projected at 45^@ with a velocity of 20 m/s has a range of 10m. The decrease in range due to air resistance is (g = 10ms^(-1)) |
| Answer» ANSWER :D | |
| 38182. |
When a car is started, the intensitiy of light of it's head light undergoes a momentary decreases. Why is it so? |
| Answer» SOLUTION :When starter is PUSHED, motor gets started. In the beginning the speed of ARMATURE is small. Therefore, there is a small back emf. Thus, there is a GREATER of potential across it. So, the intensity of light decreases. When the armature has acquired sufficient speed, there is appreciate back emf which reduces the fall of potential across it, therby resorting the brightness of headlights. | |
| 38183. |
Total energy of an electron in n^th Bohr's orbit of hydrogen atom E_n is given by: |
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Answer» `E PROP`N |
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| 38184. |
If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object'is put 1.25 cm away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is |
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Answer» 150 |
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| 38185. |
What will happen if both , emitter and collector of a transistor are reverse biased ? |
| Answer» SOLUTION :No CONDUCTION TAKES PLACE. | |
| 38186. |
A 5muF capacitor is fully charged by a 12V battery and then disconnected. If it is connected now parallel to an uncharged capacitor, the voltage across it is 3V. Then the capacitance of the uncharged capacitor is |
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Answer» a. `5muF` |
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| 38188. |
Two coherent radio point sources that are separated by 2.0 m are radiating in phase with a wavelength of 0.25m. If a detector moves in a large circle around their midpoint, at how many points will the detector show a maximum signal ? |
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Answer» 10 |
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| 38189. |
An unpolarised light is incident on a surface separating two transparent media of different optical densities at the polarizing angle. Then the reflected ray and refracted ray are |
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Answer» parallel to each other |
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| 38190. |
Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N//C) cos [(1.8 rad//m) y + (5.4 xx 10^(6) rad//s)t]} hat i. (a) What is the direction of propagation? (b) What is the wavelength lamda ? (c) What is the frequency ν ? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave. |
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Answer» SOLUTION :(a) `-HAT j` (b) 3.5m, (c ) 86 MHZ, (d) 100 nT, `{(100 nT) cos [(1.8" rad/m")y+(5.4xx10^(6)" rad/s")t]} hat K` |
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| 38191. |
The force F is given in terms of time t and displacement x by the equation F=A cos Bx+Csin Dt. The dimensions of (D)/(B) are |
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Answer» `M^(@)L^(@)T^(@)` |
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| 38192. |
A small air bubble of radius 0.1 mm is formed just below the surface of water. If the surface tension of water is 0.07 N/m and atmospheric pressure is 10^5 N//m^2, find the pressure of air in the air bubble |
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Answer» `1.0014 XX 10^5 N/m^2` |
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| 38193. |
The work done to get 'n' smaller equal size spherical drops from a bigger size spherical drop of water is proportional to : |
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Answer» `1//n^(2//3)`-1 `(4piR^(3))/3=n(4pir^(3))/3rArrR^(3)=NR^(3)rArrr=n^(-1//3)R` Now INCREASE in area=`n4pir^(2)-4piR^(2)=4pi[nr^(2)-R^(2)]` `=4pi[n.n^(-2//3)R^(2)-R^(2)]` `=4piR^(2)[n^(1//3)-1]` `therefore` Work DONE `= T xx` increase in area `W=T.4piR^(2)[n^(1//3)-1]` `rArrWprop(n^(1//3)-1)` Hence the correct choice is (c). |
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| 38194. |
T_(1//2 ) for a radioactive nuclei is 12.5 hr. and its mass is 256 gm. The time after which an amount of the remaining substance is 1 gm is _ hr. |
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Answer» 75 |
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| 38195. |
Assertion : A body falling freely under the force of gravity has constant acceleration (9.81 m//sec^(2)) Reason : Earth attracts every body towards its centre by the same force. |
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Answer» Both Assertion & Reason are TRUE & the Reason is a CORRECT explanation of the Assertion. |
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| 38196. |
There is a beam of electrons and protons in whichboth the particles are moving with the same velocity, entering a thin region, where a uniform magnetic field is applied perpendicular to the motion of beam. The protons and the electrons |
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Answer» will be deviated in the same direction |
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| 38197. |
A particle of mass m and charge q is placed at rest in uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is |
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Answer» QEY `qEy = K,` As E isuniform `K=qEy` |
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| 38198. |
The IUPAC name of CH_(3)-underset(COOC_(2)H_(5))underset(|)(C)=CH-CH_(2)-overset(O)overset(||)underset(OH)underset(|)C is |
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Answer» 4-ethoxycarbonylpent-3-enoic ACID  4-Ethoxy carbonylpent-3-enoic acid |
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| 38199. |
A square platform of side length 8m is is situated in x-z plane sucha that it is at 16 m form the x-axis and 8 m from the z-axis as shown in figure. A particle is projected with velocity overset(vec)(v)=(v_(2)overset(hat)(i)+25hat(j)) m//s relative to wind form origin and at the same instant the platform starts with acceleration overset(vec)=(2hat(i)+2.5hat(j)) m//s^(2). Wind is blowing with velocity v_(1)hat(k).g=10m//s^(2) {:(,,"ListI",,"List II"),((P),,"Least possible values of " V_(2) "(in M/s) so that particle hits the platform or edge of platfrom is",,(1) 4),((Q),,"Least possible values of " v_(1) "(in m/s) so that particle hits the platform or edge or platform is ",,(2) 6),((Q),,"If t is the time (in seconds) after particle hits the platform then 2t is equal to ",,(3) 8),((S),,"Value of displacement with respect to ground (in m) of the particle in y-direction , when " v_(2) " has its minimum possible value is (till particle hits the platform or edge of platform)",,(4) 20):} |
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Answer» <P>`{:(,P,Q,R,S),((A),2,4,3,1):}` `veca_(p.p)=-2hati-12.5hatj` `vecV_(P.P)=` Velocity of particle relative to platform Time `=(2xx25)/(12.5)=4` SEC. `8leV_(2)xx4-(1)/(4)xx2xx4^(2)le16` `6leV_(2)le8` `16leV_(1)xx4le24` `4leV_(1)le6` `Y=25xx4-(1)/(2)xx10xx4^(2)=100-80=20m` |
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| 38200. |
Referring to previous Wustration, what are the magnitudes of the displacement and distance covered when the car returns to A? |
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Answer» Solution :Displacement =`|vecr_(2)-vecr_(1)|=0,(because|vecr_(2)|-|vecr_(1)|=0)` DISTANCE =|vec(AB)|+|vec(BA)|=20m` |
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