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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.
| 43701. |
When a coil joined to a cell, the current through the coil grows with a time constant tau . After what time, the current will reach 10% of its steady-state value ? |
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Answer» Solution :Given by `i=i_0 (1-E^(t/TAU)), (i_0)/(10) = i_0 (1-e^(t/tau)) , 9/10 = e^(-t/tau)` `10/9 = e^(t/tau) ` , applying log on both sides `ln(10/9) = t/tau , t = tau ln (10/9)` |
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| 43702. |
Which one of the following property does not support wave theory of light ? |
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Answer» Light obeys LAWS of REFLECTION and REFRACTION |
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| 43703. |
For the working of potentiometer, the emf of cell in the primary circuit (E) compared to the emf of the cell in the secondary circuit (E^(1)) is |
| Answer» Answer :A | |
| 43704. |
The image formed by the objective of a compound microscope is |
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Answer» VIRTUAL and magnified |
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| 43705. |
When the magnetic flux is said to be neutral? |
| Answer» Solution :When normal to the AREA, POINTS perpendicular to `vec(B), theta= 90^(@), "i.e." phi= BA COS 90^(@)`, magnetic FLUX is NEUTRAL. | |
| 43706. |
What is isotopes? |
| Answer» Solution :These are the NUCLEI having the same atomic NUMBER but different MASS number. | |
| 43707. |
भारत की एकमात्र सक्रिय ज्वालामुखी किस द्वीप पर स्थित है? |
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Answer» बैरन द्वीप |
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| 43708. |
The radius of earth is about 6400 km and that of mars is 3200 km. The mass of the earth is about 10 times mass of mars. An object weighs 200 N on the surface of earth. Its weight on the surface of mars will be |
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Answer» 20 N Radius of mars `(R_(m))=3200" km"`, Mass of earth `(M_(e ))=10 M_(m)` and weight of the object on earth `(W_(e ))=200" N"`. `(W_(m))/(W_(e ))=(mg_(m))/(mg_(e ))=(M_(m))/(M_(e ))xx((R_(e ))/(R_(m)))^(2)=(1)/(10)xx(2)^(2)=(4)/(10)=(2)/(5)` or`W_(m)=W_(e )xx(2)/(5)=200xx0.4=80" N"`. |
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| 43709. |
Ground/ surface wave propagation is a mode of propagation in which the signal wave glides over thesurface of earth, while going from transmitter to receiver. In this mode, there is a loss of power (i.e., attenuation) of signal wave due to diffraction and absorption of signal wave energy by ground. The attenuation of ground / surface wave signal increases very rapidly with the increase in its frequency. Therefore, the ground/ surface wave propagation is not suitable for the propagation of high frequency signal wave and for very long range communication. Read the above passage and answer the following questions: (i) What is the frequency range of a signal wave employed for the ground wave propagation? (ii) For what purpose the ground wave propagation is used? (iii) What does the study of ground wave propagation imply in day to day life? |
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Answer» SOLUTION :(i) For ground wave propagation, the frequency range of signal wave is `530 kHz "to" 1710 kHz.` (ii) The ground wave propagation is used for local BROADCASTING as a medium wave broadcast service. (iii) In day to day life, the ground wave propagation can be compared to the movement of a small child ALONG the surface of ground. As child can go only upto certain distance on ground and GETS exhausted, similarly the ground wave frequency signal loses its energy while gliding over the surface, of earth and HENCE can go up to limited distance. As ground wave propagation is for local broadcasting, similarly, the small child movement is confined with in home under the supervision of some elders. |
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| 43710. |
A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of ""_(29)^(63)Cu atoms (of mass 62.92960 u). |
| Answer» SOLUTION :`1.584 XX 1025` MEV or `2.535xx 10^12 J` | |
| 43711. |
In molecular crystals , the force between molecules is |
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Answer» Van der WAAL's FORCE ARISING out of electric polarization: |
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| 43712. |
In a p-type semiconductor, germanium is doped with |
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Answer» gallium |
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| 43713. |
Two forces vec p and vec Q have a resultant perpendicular to vec p. The angle between the forces |
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Answer» `TAN ^(-1) (-p/Q)` |
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| 43714. |
" "_(2)^(3)He and " "_(1)^(3)H nuclei have the same mass number. Do they have the same binding energy ? |
| Answer» SOLUTION :No, the binding ENERGY of `" "_(1)^(3)H` is more than that of `" "_(2)^(3)He`. | |
| 43715. |
A thin double convex lens of focal length f is broken into two equal halves at the axis. The two halves are combined as shown in figure. What is the focal length of combination in (ii) and (iii). |
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Answer» Solution :FOCAL LENGTH of COMBINATION is infinite, (ii) `f//2` |
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| 43716. |
The contral rods used in a nuclear can be made up of |
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Answer» Graphite |
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| 43717. |
A spring - block system undergoes vertical oscillations above a large horizontal metal sheet with uniform positive charge. The time period of the oscillation is T. If the block is given a charge Q, its time period of oscillation will be |
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Answer» `T` |
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| 43718. |
Find the degeneracy of the states .^(2)p,.^(3)D, and .^(4)F possesing the greatest possible values of the total angular momentum. |
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Answer» <P> Solution :The state with GREATEST possible total angular momentum areFor a `.^(2)P` state `J=(1)/(2)+1=(3)/(2)i.e.,.^(2)P_(3//2)` Its degeneracy is `4`. For a `.^(3)D` state `J=1+2=3 i.e.,^(2)P_(3//2)` Its degeneracy is `2xx3+1=7` For a `.^(4)F` state `J=(3)/(2)+3=(9)/(2)i.e., .^(4)F_(4/2)`. Its degeneracy is `2xx(9)/(2)+1=10`. |
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| 43719. |
Lassigne's test for the detection of nitrogen will fail in the case of : |
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Answer» `NH_(2)CONH_(2)` |
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| 43720. |
A parallel beam of light emerges from the opposite surface of the sphere when a point source of light lies at the surface of the sphere. The refractive index of the sphere is |
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Answer» `(3)/(2)` |
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| 43721. |
प्रेरित आवेश की प्रकृति सदैव प्रेरित करने वाले आवेश के - |
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Answer» समान होती है |
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| 43722. |
A body is droped from a height 122.5 m.If its stopped after 3 seconds and again released the further time of descent is (g=9.8 m//s^(2)) |
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Answer» 2 s |
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| 43723. |
Two identical capacitors of plate dimensions l xx b and plate separation 'd' have dielectric slabs filled in between the space of the plates as shown in Fig Obtain the relation between the dielectric constants K , K_1 and K_2. |
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Answer» Solution :For capacitor shown in FIG area of each plate of capacitor `A_1 = lb` and capacitance `C_1 = (K in_(0) A_(1))/(d)` The capacitor shown in Fig is equivalent to two capacitors each of plate area `A_(2) = (l)/(2).b = (lb)/(2) = (A_(1))/(2)` but FILLED with dielectrics of dielectric constants `K_1` and `K_2` respectively JOINED in series . Therefore , net capacitance of this arrangement is `C_(2) = (K_(1) in_(0) (A_(1)//2))/(d) + (K_(2) in_(0) (A_(1)//2))/(d) = (in_0 A_(1))/(d) [ (K_(1) + K_(2))/(2)]` Since `C_(1) = C_(2)` hence we CONCLUDE that `(K in_(0) A_(1))/(d) = (in_(0) A_(1))/(d) [ (K_(1) + K_(2))/(2)] implies K= (K_(1) + K_(2))/(2)` |
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| 43724. |
The dimension of (1)/(2) epsilon_(0)E^(2) . Where epsilon_(0) is permittivity of free space and E is electric field is |
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Answer» `[ML^(2) T^(-2)]` |
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| 43725. |
Which of the following happens when a monochromatic light wave passes from air to glass? |
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Answer» Both, the FREQUENCY and WAVELENGTH decrease. |
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| 43726. |
Coloured spectrum is seen when we look through a muslim cloth. Why? |
| Answer» Solution :The fine threads in MUSLIM cloth from NARROW SLITS. When WHITE light is passed through these slits DIFFRACTION occurs and forms the coloured spectrum. | |
| 43727. |
In experiment of photoelectric effect width of slit is equal to de-Broglie wavelength.Beam of light is incident normal to plane and by using detector D it is detected.Which of the following best represent number of electron (N) and position of detector (y)? |
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Answer»
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| 43728. |
Explain n and p-type semiconductor based on band theory. |
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Answer» Solution :In EXTRINSIC semiconductor because of the abundance of majority current carriers the minority carriers produced thermally have more chance of meeting majority carriers and thus getting destroyed and indirectly helps to REDUCE the intrinsic concentration of minority carriers. The semiconductor.senergy band structureis affected by doping. In extrinsic semiconductor, additionalenergy states due to donor impurities `E_(D)` and acceptor impurities `E_(A)` also exist. In n-type of Si semiconductor, the donor energy level `E_(D)` is slightly below the bottom `E_(C )` of the CONDUCTION band and electrons from this level move into the conduction band with very small of energy. At room temperature most of donor atoms get ionised but very few `(~10^(12))` atoms of Si get ionised so the conduction band will have most electrons coming from the donor impurities as shown in figure (a).  For p-type semiconductor the acceptor energy level `E_(A)` is slightly above the top `E_(V)` of the valence band. With very small supply of energy an electron from the valence band can jump to the level `E_(A)` and ionise the acceptor negatively. In other words,withvery small supply of energy the hole from level `E_(A)` sinks down into the valence band `E_(V)` and electron comes up. This is shown in figure (b).  At room temperature, most of the acceptor atoms get ionised leaving holes in the valence band `E_(V)`. Thus, at room temperaturethe density of holes in the valence band is mostly due to impurity in the extrinsic semiconductor. The above description is grossly approximate and hypothetical it helps in understanding the difference between metals, insulators and semiconductors (extrinsic and intrinsic) in a simple manner. The energy difference between `E_(V)` and `E_(C )` for C (diamond) is 5.4 eV, for Si is 1.1 eV and for Ge it is 0.7 eV and Sn is also a GROUP IV element but it is a metal because the energygap in its case is 0 eV. |
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| 43729. |
A horizontal wire carries 200A current below which another wire of linear density 20 xx 10^(-3) Kg/m carrying a current is kept at 2cm distance. If the wire kept below hangs in air, then the current in the wire is |
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Answer» 9.8 A |
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| 43730. |
The half life of a radioactive substance is 30 days. What is the time taken to disintegrate to 3//4^(th) of its original mass ? |
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Answer» 30 days `therefore` No of half LIVES, n=2 Time TAKEN to disintegrate = Half LIFE X No. of half lives = 30 x 2 =60 days |
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| 43731. |
When unpolarised light beam is incident from air into glass (n =1.5) at the polarised angle ....... |
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Answer» REFLECTED BEAM is polarised 100 % |
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| 43732. |
A man walks east with certain relocity. A car is travelling along a road which is 30^(@) West of North. While a bus is travelling in another road which is 60^(@) South of West. Find the angle between velocity vector of (a) man and car (b) car and buc) bus and man. |
Answer» SOLUTION :  From the DIAGRAM the angle between velocity vector of man and car is `90^(@)+30^(@)=120^(@)=` From the diagram the angle between velocity vector of man and car is `60^(@)+60^(@)=120^(@)` The angle between velocity vector of bus and man is `30^(@)+90^(@)=120^(@)` |
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| 43733. |
What kinetic energy must a proton possess to split a deuteron H^(2) whose binding energy is E_(b)=2.2 MeV? |
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Answer» Solution :The result of the previous problem applies and we find that ENERGY required to SPLIT a DEUTERON is `Tge(1+(M_(p))/(M_(d)))E_(B)= 3.3MeV` |
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| 43734. |
A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter (d)/(2) is central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively. |
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Answer» `f and (3I)/(4)` |
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| 43735. |
According to uncertainty principal for an electron, time measurement will become uncertain if following is measured with high certainty |
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Answer» ENERGY |
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| 43736. |
Where on the surface of Earth is the angle of dip zero? |
| Answer» SOLUTION :The LOCUS of the point having zero dip is CALLED the magnetic equator. | |
| 43737. |
The direction of the electric field in p-n junction diode is ……. |
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Answer» from p-side to n-side The direction of electric field in p-n junction diode is from n-side to p-side. |
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| 43738. |
A tiny particle of mass4 mugis kept over a large horizontal sheet of charge density4 xx 10 ^(-6)Cm ^(-1)What charge should be given to the particle so that if released it does not fall down? |
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Answer» Solution :Let charge q be GIVEN to the particle. As particle close not fall, hence downward weight of particle mg = upward force due to electric field of charged plate ` q_E =q . (SIGMA)/( 2 in _0) ` ` therefore "" q= ( 2 mg in_0)/( sigma ) ` Here , m ` mug= 5 XX 10 ^(-6) g = 5 xx 10 ^(-9) kg , sigma= 4 xx 10 ^(-6)C m^(-1) ` ` therefore"" q= ( 2xx 5 xx 10 ^(-9)xx 9.81 xx 8.85 xx 10 ^(-12)) /( 4 xx 10 ^(-6)) = 2.17 xx 10 ^(-13)C . ` |
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| 43739. |
Figure 30-50 shows a rod of length L = 12.0 cm that is forced to move at constant speed v = 5.00 m/s along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance 0.400 Omega the rest of the loop has negligible resistance. A current i = 100 A through the long straight wire at distance a= 5.00 mm from the loop sets up a (nonuniform) magnetic field through the loop. Find the (a) emf and (b) current induced in the loop. (c) At what rate is thermal energy generated in the rod? (d) What is the magnitude of the force that must be applied to the rod to make it move at constant speed? (e) At what rate does this force do work on the rod? |
| Answer» SOLUTION :(a) 322 μV, (B) 0.805 mA, (C) 0.259 μW, (d) 51.8 nN, (e) 0.259 μW | |
| 43740. |
The current in c coil of self - inductance 10 mH changes from 0 to 1.5 amp in 1 milli-sec. The emf induced in the coil is |
| Answer» Answer :B | |
| 43741. |
Mention any three properties of a photon. |
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Answer» Solution :(i) They are electrically NEUTRAL. (ii) They are not deflected by electric and magnetic field. (III) They travel with the speed of light. (iv) The energy of each photon is given by, E = hv (Where h = Plank.s constant and v = Frequency of photon) (v) MOMENTUM of the photon is given by, `p= (h)/(lambda)=(h)/(c)` (vi) The energy and MOMENTUIN of the photons are conserved during photon-particle collision. |
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| 43742. |
Diameter of spherical mirror is 10 cm, what will be aperture of mirror ? |
| Answer» ANSWER :A::B::C | |
| 43743. |
Find the tension needed to produce stationary waves with 4 loops in a string lm long and 0.5 gram in weight, fixed to a tunig fork of frequency 200Hfc, when the prongs of the fork are vibrating perpendicular to the string, |
| Answer» ANSWER :A | |
| 43744. |
First law of thermodynamics can be explained on the basis of |
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Answer» MAXWELL’s law |
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| 43745. |
Four point charges 1muC,3muC, - 2muC and -2muC are arranged on the four vertices of a square of side 1 cm. The dipole moment of this charge assembly is |
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Answer» zero |
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| 43747. |
A sample radio active subsistence has 10^6 radio active radio active nuclei. It.s half-life is 20 sec. How many nuclei will remain after 10 seconds? |
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Answer» `7xx10^5` |
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| 43748. |
Find out the current flowing through 2Omega resistance in the given circuit. |
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Answer» 5A |
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| 43749. |
Whatdo you mean by Internet of Things? |
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Answer» Solution :Internet of Things (IoT), it is made possible to CONTROL various DEVICES from a SINGLE device. Example: home automation USING a MOBILE phone. |
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| 43750. |
Calculate the mutual inductance between two coils when a current of 4A changing to 8A in 0.5s in one coil, induces an emf of 50 mV in the other coil. Data : I_(1) = 4 A, I_(2) = 8 A, dt = 0.5 s, e=50 mV = 50 xx 10^(-3) V, M = ? |
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Answer» Solution :`E=M.(dI)/(dt)` `:.M=-(e)/(((dI)/(dt)))= -(e)/(((I_(2)-I_(1))/(dt)))` ` = - (50 XX 10^(-3))/(((8-4)/(0.5)))= 6.25 xx 10^(-3)` `:.M= 6.025` MH |
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