InterviewSolution
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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.
| 1701. |
A `0.5 kg` ball moving with a speed of `12 m//s` strikes a hard wall at an angle of `30^(@)` with the wall. It is reflected with the same speed and at the same angle . If the ball is in contact with the wall for `0.25 s`, the average force acting on the wall is A. 48 NB. 24 NC. 12 ND. 96 N |
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Answer» Correct Answer - B F=`(Deltap)/(Deltat)=(2mvsin30^(@))/(0.25)=24N` |
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| 1702. |
A coil of inductive reactance `31 Omega` has a resistance of `8 ohm`. It is placed in series with a condenser of capacitive reactance `25 Omega`. The combination is connected to an `ac` source of `110 V`. The power factor of the circuit isA. 0.56B. 0.64C. 0.8D. 0.33 |
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Answer» Correct Answer - C Power factor =cos`phi=(R)/(|Z|)=(8)/sqrt(8^(2)+(31-25)^(2)` `(8)/sqrt(8^(2)+6^(2))=(8)/(10)=0.8` |
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| 1703. |
In a face centred cubic lattice unit cell is shared equally by how many unit cells?A. 4B. 2C. 6D. 8 |
| Answer» Correct Answer - C | |
| 1704. |
The wattability of a surface by a liquid depends primarily onA. viscosityB. surface tensionC. densityD. angle of contact between the surface and the liquid |
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Answer» Correct Answer - D The wettability of a surface by a liquid depends on angle of contact between the surface and the liquid. |
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| 1705. |
Which one of the following statement is false ?A. Pure SI doped with trivalent impurties gives a p type semiconductorB. Majority carriers in a n - type semiconductor are holesC. Minority carriers in a p - type semiconductor are electronsD. The resistance of intrinsic semiconducotor decreases with increase of temperature |
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Answer» Correct Answer - B Electric and magnetic field vectors are perpendicular to each other in electromagnetic wave. |
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| 1706. |
A circular disc of moment of inertia `I_(t)` is rotating in a horizontal plane about its symmetry axis with a constant angular velocity `omega_(i)`. Another disc of moment of inertia `I_(b)` is dropped co-axially onto the rotating disc. Initially, the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed `omega_(f)`. Calculate the energy lost by the initially rotating disc due to friction.A. `(1)/(2) (I_(b)^(2))/((I_(t)+I_(b))omega _(i)^(2)`B. `(1)/(2) (I_(t)^(2))/((I_(t)+I_(b))omega _(i)^(2)`C. `(I_(b)-I_(t))/((I_(t)+L_(b)))omega _(i)^(2)`D. `(1)/(2)(I_(b)I_(t))/((I_(t)+I_(b)))omega_(i)^(2)` |
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Answer» Correct Answer - D By Conservation of angular momemtum `I_(t)omega _(i)(I_(t)+I_(b))omega_(f) implies omega_(f)=((I_(t))/(I_(t)+I_(b)))omega_(i)` Loss in kinetic energy `=(1)/(2) I_(t)omega_(i)^(2)-(1)/(2)(I_(t)+I_(B))(omega_(f)^(2))` `=(1)/(2)((I_(b)I_(t))/(I_(b)+I_(t)))omega_(i)^(2)` |
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| 1707. |
`C` and `Si` both have same lattice structure, having `4` bonding electrons in each. However, `C` is insulator whereas `Si` is intrinsic semiconductor. This is becauseA. The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the thirdB. The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit.C. In case of C the valence band is not completely filled at absolute zero temperatureD. In case of C the conduction bans is partly filled even at absolute zero temperature |
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Answer» Correct Answer - A The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third. |
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| 1708. |
Which one of the following condition will favour maximum formation of the product in the reaction. `A_(2)(g) + B_(2)(g)hArrX_(2)(g) Delta_(r)H = - X` kJ ?A. High temperature and high pressureB. Low temperature and low pressureC. Low temperature and high pressureD. High temperature and low pressure |
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Answer» Correct Answer - C `A_(2)(g) + B_(2)(g) hArr X_(2)(g), DeltaH=-xKJ` On increasing pressure equilibrium shifts in a direction where pressure decreases i.e. forward direction. On decreasing temperature , equilibrium shifts in exothermic direction i.e., forward direction. So high pressure and low temperature favours maximum formation of product. |
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| 1709. |
Which of the following ions exhibits d-d transitions and paramagnetism as well?A. `MnO_(4)^(-)`B. `Cr_(2)O_(7)^(2-)`C. `CrO_(4)^(2-)`D. `MnO_(4)^(2-)` |
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Answer» Correct Answer - D `CrO_(4)^(2-) implies Cr^(6+)=[Ar]` Unpaired electron (n)=0, Diamagnetic `Cr_(2)O_(7)^(2-) implies Cr^(6+)=[Ar]` Unpaired electron (n)=0, Diamagnetic `MnO_(4)^(2-)=Mn^(6+)=[Ar]3d^(1)` Unpaired electron (n)=1, Paramagnetic `MnO_(4)^(-)=Mn^(7+)=[Ar]` Unpaired electron (n)=0, Diamagnetic |
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| 1710. |
Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified `KMnO_(4)` for complete oxidation ?A. `FeC_(2)O_(4)`B. `Fe(NO_(2))_(2)`C. `FeSO_(4)`D. `FeSO_(3)` |
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Answer» Correct Answer - C ` underset(v.f=5)(KMnO_(4))+underset(v.f.=3)(FeC_(2)O_(4))toMn^(+2)+Fe^(+3)+CO_(2)` `underset(v.f=5)(KMnO_(4))+underset(v.f.=5)(Fe(NO_(2))_(2))toMn^(+2)+Fe^(+3)+NO_(3)^(-)` Similarity of moles of `Fe(NO_(2))_(2)` is x `n_(KMnO_(4))=(5x)/(5)=x` `underset(v.f=5)(KMnO_(4))+underset(v.f=1)(FeSO_(4))toMn^(+2)+Fe^(+3)` and also the moles of `FeSO_(4)` is x `n_(KMnO_(4))=(x)/(5)` `underset(v.f=5)(KMnO_(4))+underset(v.f=3)(FeSO_(3))toMn^(+2)+SO_(4)^(2-)+Fe^(+3)` and also the moles of `FeSO_4` is x `n_(KMnO_(4))=(3x)/(5)` |
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| 1711. |
One mole of an ideal monatomic gas undergoes a process described by the equation `PV^(3)`= constant. The heat capacity of the gas during this process isA. `(3)/(2)R`B. `(5)/(2)R`C. `2R`D. `R` |
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Answer» Correct Answer - D `PV^(3)=C(n=3)` Specific heat in polytropic process `PV^(n)=K` `C=C_(V)+(R)/(1-n)` `=(3)/(2)R+(R)/(1-3)` `=(3)/(2)R-(R)/(2)=R`. |
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| 1712. |
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, isA. `6.25%`B. `20%`C. `26.8%`D. `12.5%` |
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Answer» Correct Answer - C Efficiency of ideal heat engine, `eta=(1-(T_(2))/(T_(1)))` `T_(2)` : Sink temperature `T_(1)`: Source temperature `% eta=(1-(T_(2))/(T_(1)))xx100` `=(1-(273)/(373))xx100` `=((100)/(373))xx100=26.8%` |
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| 1713. |
The work function of a surface of a photosensitive material is `6.2 eV`. The wavelength of the incident radiation for which the stopping potential is `5 V` lies in theA. Infrared regionB. X-ray regionC. Ultraviolet regiD. Visible region |
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Answer» Correct Answer - C `E_("incident")=W+K_("max"),K_("max")=eV_(0)` `hv=hv_(0)+eV_(0)," stopping potential "=V_(0)` `(hc)/(lambda)=6.2e+5e` `lambda=(hxxc)/(11.2xx1.6xx10^(-19))=(6.63xx10^(-34)xx3xx10^(8))/(11.2xx1.6xx10^(-19))` `=1.1xx10^(-7)m` |
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| 1714. |
Two radioactive materials `X_(1)` and `X_(2)` have decay constants `5 lambda` and `lambda` respectively. If initially they have the same number of nuclei, then the ratio of the number of muclei of `X_(1)` to that of `X_(2)` will be `1/e` after a timeA. `(1)/(4lambda)`B. `(e)/(lambda)`C. `lambda`D. `(1)/(2)lambda` |
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Answer» Correct Answer - A `x_(1)=N_(0)e^(-5lambdat)` `x_(2)=N_(0)e^(-lambdar)` `(x_(1))/(x_(2))=e^(-5lambdat+lambdat)` `(1)/(e)=e^(-4lambdat)` `e^(-1)=e^(-4lambdat) rArr t=(1)/(4lambda)` |
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| 1715. |
Which will make basic buffer?A. 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOHB. 50 mL of 0.1 M NaOH + 25 mL of 0.1 M `CH_(3)COOH`C. 100 mL of 0.1 M `CH_(3)COOH` + 100 mL of 0.1 M NaOHD. 100 mL of 0.1 M HCl + 200 mL of 0.1 M `NH_(4)OH` |
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Answer» Correct Answer - D Acid-base titration `underset("10 m mole")(HCl)+underset(20-10)underset("20 m mole")(Na_(4)OH)tounderset("10 m mole")(Na_(4)Cl)` HCl is limiting reagent Solution contain `underset((WB))(NH_(4)OH)&underset((SAWB))(NH_(4)Cl)` The basic buffer will form |
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| 1716. |
For a cell involving one electron `E_(cell)^(0)=0.59V` and 298K, the equilibrium constant for the cell reaction is: [Given that `(2.303RT)/(F)=0.059V` at `T=298K`]A. `1.0xx10^(30)`B. `1.0xx10^(2)`C. `1.0xx10^(5)`D. `1.0xx10^(10)` |
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Answer» Correct Answer - D Nearest equation `E_(cell)^(ɵ)=E_(cell)^(ɵ)(-0.059)/(n)logQ_(C)` At equilibrium Ecell `=0,Q_(C)=K_(C)` `E_(cell)^(ɵ)=(-0.059)/(n)logK_(C)" Value of "E_(cell)^(ɵ)=0.59V` `0.59=(0.059)/(1)logK_(C)" value of "n=1` `K_(C)="antilog "10` `K_(C)=1xx10^(10)` |
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| 1717. |
`[Cu(NH_3)_4]^(+2)` reacts with `HNO_3` in excess of water gives :A. `Cu(OH)_2`B. `Cu(NO_3)_2`C. `Cu(H_2O)^(–2)`D. None of the above |
| Answer» Correct Answer - B | |
| 1718. |
Which of the following compound contain zero oxidation state of Fe :A. `[Fe(CN)_6]^-4`B. `[Fe(CN)_6]^-3`C. `Fe(CO)_5`D. All of the above |
| Answer» Correct Answer - C | |
| 1719. |
A compound contains `C=40%,O=53.5%`, and `H=6.5%` the empirical formula formula of the compound is:A. `CH_2O`B. `CH_4O`C. `CH_4O_2`D. CHO |
| Answer» Correct Answer - A | |
| 1720. |
Character which is closely related to human evolution:A. Disappearance of tailB. Reduction in size of jawsC. Binocular visionD. Flat nails |
| Answer» Correct Answer - B | |
| 1721. |
Which of the following is initiation codon:A. UAGB. AUCC. AUGD. CCU |
| Answer» Correct Answer - C | |
| 1722. |
First cloned animal:A. Dolly sheepB. Polly sheepC. Molly sheepD. Dog |
| Answer» Correct Answer - A | |
| 1723. |
Which one of the following is a matching set of a phylum and its three examples?A. Cnidaria – Bonellia, Physalia, AureliaB. Platyhelminthes-Planaria, Schistosoma, EnterobiusC. Mollusca-Loligo, Teredo, OctopusD. Porifera-Spongilla, Euplectella, Pennatula |
| Answer» Correct Answer - A | |
| 1724. |
A major breakthrough in the study of cells came with the development of electron microscope. This is becauseA. The resolution power of the electron microscope is much higher than that of the light microscopeB. The resolving power of the electron microscope is 200-350 nm as compared to 0.1-0.2 nm for the light microscopeC. Electron beam can pass through thick materials, whereas light microscopy requires thin sectionsD. The electron microscope is more powerful than the light microscope as it uses a beam of electrons which has wavelength much longer than that of photons |
| Answer» Correct Answer - C | |
| 1725. |
An important evidence in favour of organic evolution is the occurrence of `:`A. Homologous and vestigial organsB. Analogous and vestigial organsC. Homologous organs onlyD. Homologous and analogous organs |
| Answer» Correct Answer - A | |
| 1726. |
Exponential growth is shown byA. Unicellular formsB. A cell in tissue cultureC. EmbryoD. Multicellular plants |
| Answer» Correct Answer - B | |
| 1727. |
Homologous organs areA. Wings of cockroach and wings of batsB. Wings of insects and wings of birdsC. Air bladder of fishes and lungs of frogD. Pectoral fins of fishes and forelimbs of horse |
| Answer» Correct Answer - D | |
| 1728. |
Which of the following is a secondary air pollutant?A. PANB. COC. `NO_2`D. `SO_2` |
| Answer» Correct Answer - A | |
| 1729. |
Insulin differs from growth hormone in the fact that itA. Increases activity of m-RNA and RibosomesB. Increase the permeability of cell membraneC. Affects metabolism of fats by inducing lipogenesisD. Increasing protein synthesis |
| Answer» Correct Answer - C | |
| 1730. |
Forest cover of India according to "State of Forest Report 1997" of Ministry of Environment and Forest is :A. 0.11B. 0.195C. 0.17D. 0.187 |
| Answer» Correct Answer - B | |
| 1731. |
Nitrogen fixing bacteria convert:-A. `N_2toNH_3`B. `NH_4^(+) to NitratesC. `NO_2 to NO_3`D. `NO_3 to N_2` |
| Answer» Correct Answer - A | |
| 1732. |
During injury mast cells secreteA. HistamineB. HeparinC. ProthrombinD. Antibodies |
| Answer» Correct Answer - A | |
| 1733. |
A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by `1 kg`, the time period of oscillations becomes 5s. The value of m in kg isA. `(3)/(4)`B. `(4)/(3)`C. `(16)/(9)`D. `(9)/(16)` |
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Answer» Correct Answer - D `3=2pisqrt((m)/(k))` `5=2pisqrt((m+1)/(k))` `(3)/(5)=sqrt((m)/(m+1))` `(9)/(25)=(m)/(m+1)` `9m+9=25m` `16m=9` `m=(9)/(16)` |
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| 1734. |
Energy per unit volume for a capacitor having area A and separation d kept at potential diffeence V is given by : -A. `1/2epsilon_0 V^2/d^2`B. `1/(2epsilon_0) V^2/d^2`C. `1/2CV^2`D. `Q^2/"2C"` |
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Answer» Correct Answer - A Energy density =`1/2in_0v^2/d^2` |
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| 1735. |
If `|vecA+vecB|=|vecA|+|vecB|`, then angle between `vecA` and `vecB` will beA. `90^@`B. `120^@`C. `0^@`D. `60^@` |
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Answer» Correct Answer - B `|vecA+vecB|^2=A^2+B^2+2AB cos theta` `rArr A^2=A^2+A^2+2A^2 cos theta` `rArr cos theta =-1/2 rArr theta=120^@` |
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| 1736. |
A cylindrical rod having temperature `T_(1)` and `T_(2)` at its ends. The rate of flow of heat is `Q_(1) cal//sec`. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat `Q_(2)` will beA. `4Q_1`B. `2Q_2`C. `Q_1/4`D. `Q_1/2` |
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Answer» Correct Answer - B Heat flow rate =`(KA(T_1-T_2))/L=Q` when linear dimensions are doubled `A_1 prop r_1^2 " " L_1=L` `A_2 prop 4r_1^2 " " L_1=2L_1` so `Q_2=2Q_1` |
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| 1737. |
Function of telomeres in nucleus:A. Pole ward movementB. To initiate the RNA synthesisC. To seal the end of ChromosomeD. To recognize the homologous chromosome |
| Answer» Correct Answer - C | |
| 1738. |
Barrier potential of a `p-n` junction diode does not depend onA. diode designB. temperatureC. forward biasD. doping density |
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Answer» Correct Answer - A Barrier potential of a p-n junction diode does not depend on diode design |
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| 1739. |
The mass of proton is `1.0073 u` and that of neutron is `1.0087 u` (`u=` atomic mass unit). The binding energy of `._(2)He^(4)` is (mass of helium nucleus `=4.0015 u`)A. 0.0305 JB. 0.0305 ergC. 28.4 MeVD. 0.061 u |
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Answer» Correct Answer - C BE=`Delta`m x 931 =[2(1.0087 + 1.0073)-4.0015] x 931 =28.4 MeV |
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| 1740. |
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60^(@) is W. Now the torrue required to keep the magnet in this new position isA. `(W)/(sqrt(3))`B. `sqrt(3)W`C. `(sqrt(3)W)/(2)`D. `(2W)/(sqrt(3))` |
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Answer» Correct Answer - B `W=MB(cos^(@)-cos60^(@))` `W=MB(1-(1)/(2))=(MB)/(2)` required torque for this position `tau=Mbsintheta` `=MBsin60^@` `=(sqrt(3))/(2)MB=sqrt(3)W` |
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| 1741. |
If potential (in volts) in a region is expressed as `V (x, y, z) = 6xy - y + 2yz`, the electric field (in `N//C)` at point `(1, 1, 0)` isA. `-(6hati+9hatj+hatk)`B. `-(3hati+5hatj+3hatk)`C. `-(6hati+5hatj+2hatk)`D. `-(2hati+3hatj+hatk)` |
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Answer» Correct Answer - C `V=6xy-y+2yz` `E=-(delV)/(delx)hati-(delV)/(dely)hatj-(delV)/(delz)hatk` `E=-6yhati-(6x-1)hatj-2yhatk` AT (1,1,0) `E=-6hati-5hati-2hatk` `E=-(6hati+5hatj+2hatk)` |
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| 1742. |
A ball is thrown vertically downwards from a height of 20m with an intial velocity `v_(0)`. It collides with the ground, loses 50% of its energy in collision and rebounds to the same height. The intial velocity `v_(0)` is (Take, g =10 `ms^(-2)`)A. 10 `ms^(-1)`B. 14 `ms^(-1)`C. 20 `ms^(-1)`D. 28 `ms^(-1)` |
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Answer» Correct Answer - C `0.5(mgh+(1)/(2)mv^(2))=mgh` `0.25mv^(2)=0.5mgh` `v=sqrt(2gh)` `=sqrt(2xx10xx20)=sqrt(400)=20m//s` |
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| 1743. |
A solid sphere is in rolling motion. In rolling motion a body prosseses translational kinetic energy `(K_(t))` as well as rotational kinetic energy `(K_(r))` simutaneously. The ratio `K_(t) : (K_(t) + K_(r))` for the sphere isA. `10:7`B. `5:7`C. `7:10`D. `2:5` |
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Answer» Correct Answer - B `K_(t)=(1)/(2)mv^(2)` `K_(t)+K_(r)=(1)/(2)mv^(2)+(1)/(2)Iomega^(2) =(1)/(2) mv^(2)+(1)/(2)((2)/(5)mr^(2))((v)/(r))^(2)` `=(7)/(10)mv^(2)` So, `(K_(t))/(K_(t)+K_(r))=(5)/(7)` |
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| 1744. |
A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnetic is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains horizontal potential energy. the work required to do this comes fromA. The lattice structure of the material of the rodB. The magnetic fieldC. The current sourceD. The induced electric field due to the changing magnetic field |
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Answer» Correct Answer - C Energy of current source will be converted into potential energy of the rod. |
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| 1745. |
An inductor `20 mH`, a capacitor `100 muF` and a resistor `50 Omega` are connected in series across a source of emf, `V=10 sin 314 t`. The power loss in the circuit isA. 2.74 WB. 0.43 WC. 0.79 WD. 1.13 W |
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Answer» Correct Answer - C `P_(av)=((V_(RMS))/(Z))^(2)R` `Z=sqrt(R^(2)+(omegaL-(1)/(omegaC))^(2))=56 Omega` `therefore P_(av)=((10)/((sqrt(2))56))^(2)xx50=0.79 W` |
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| 1746. |
There object, `A` : (a solid sphere), `B` : (a thin circular disk) and `C` : (a circular ring), each have the same mass `M` and radius `R`. They all spin with the same angular speed `omega` about their own symmetry axes. The amount of work `(W)` required ot bring them to rest, would satisfy the relationA. `W_(B) gt W_(A) gt W_(C)`B. `W_(A) gt W_(B) gt W_(C)`C. `W_(C) gt W_(B) gt W_(A)`D. `W_(A) gt W_(C) gt W_(B)` |
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Answer» Correct Answer - C Work done required to bring them rest `Delta W=Delta KE` `DeltaW=(1)/(2)Iomega^(2)` `DeltaW prop I` for same `omega` `W_(A) : W_(B) : W_(C) =(2)/(5)MR^(2) : (1)/(2)MR^(2) : MR^(2)` `=(2)/(5): (1)/(2):1` `=4:5:10` `implies W_(C) gt W_(B) gt W_(A)` |
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| 1747. |
In the AOB system of blood,group,if both antigens are present but no antibody,the blood group of the individual would be :A. OB. ABC. AD. B |
| Answer» Correct Answer - B | |
| 1748. |
Bicarpellary gynoecium and oblique ovary occurs inA. MustardB. BananaC. PisumD. Brinjal |
| Answer» Correct Answer - D | |
| 1749. |
The kind of tissue that forms the supportive structure in our pinna (external ears) is also found in -A. tip of the noseB. vertebraeC. nailsD. ear ossicles |
| Answer» Correct Answer - 1 | |
| 1750. |
What happens if bone of frog is kept is dilute hydrochloric acid:A. Will become flexibleB. Will turn blackC. Will Break in piecesD. Will shrinke |
| Answer» Correct Answer - A | |