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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.
| 38001. |
In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500 Å, the first minimum is observed, at angle 30^(@). The first secondary maximum is observed at an angle theta = |
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Answer» `SIN ^(-1)(1)/(SQRT(2))` |
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| 38002. |
The effective current in a 50 c.p.s. A.C. circuit is 20A. What is the peak value of the current and current after 1/600 secafter it was zero? |
| Answer» SOLUTION :28.28 A, 14.14 A | |
| 38003. |
In a transistor beta=45, the change in the voltage across 5 kOmega resistor which is connected in collector circuit is 5V. Find the change in base current. |
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Answer» Solution :Change in the COLLECTOR current `DeltaI_(C)=DeltaV//R_(L)` `DeltaI_(C)=(5)/(5xx10^(3))=10^(-3)A=1mA` But `BETA=(DeltaI_(C))/(DeltaI_(B))` `DeltaI_(B)=(DeltaI_(C))/(beta)=(10^(-3))/(45)=0.022mA`. Change in BASE current `(DeltaI_(B))=0.022mA` |
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| 38004. |
As the mass number "A" increases, the binding energy per nucleon in a nucleus |
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Answer» increases |
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| 38005. |
A metal block with a heater in it is placed in a room at temperature 293 K. When the heater is switched on it is observed that the temperature of the block rises at the rate of 2^(@)C//sand when its temperature rises to 30^(@)C, it is switched off. Just after when heater is switched off, it is observed that the block cools at 0.2^(@)C//s.If Newton's law of cooling is assumed to be valid, find the power of the heater. Also find the thermal power radiated by the block when it was at 30^(@)C and at 25^(@)C.Given that the heat capacityof the block is 80 J//^(@)C. |
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| 38006. |
An initially stationary device lying on a frictionless floor explodes into two pieces and slides acros the floor. One piece is moving in |
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Answer» positive y-direction `p_(i)=p_(F)` and initial momentum, `p_(i)=m u=m(0)=0` `:. p_(f)` should also be zero. Hence, other piece will move in negative x-direction. |
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| 38007. |
A coil having n turns and resistance R Omega is connected with a galvanometer of resistance 4 R Omega. This combination is moved in time t sec from a magnetic field W_(1) weber to W_(2) weber. The induced current in the circuit is |
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Answer» `-((W_(2)-W_(1)))/(5Rnt)` `rArr IR=-n.(d phi)/(dt) rArr I=-(n)/(R ).(d phi)/(dt)` `rArr I=-(n)/(R_(1))((W_(2)-W_(1)))/((t_(2)-t_(1)))`.USE this FORMULA for I `I=(-1)/((R+4R)).(n(W_(2)-W_(1)))/(t)rArr I=-(n(W_(2)-W_(1)))/(5Rt)`. |
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| 38008. |
If radius of O_(2) molecule =40A. T=27^(@)C annd P=1atm. Find the time of relaxation. |
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Answer» `10^(-10)sec` Now `n=(N)/(V)=(muN_(a))/(V)` `PV=muRT` `(mu)/(V)=(P)/(RT)` so `n=(P)/(RT)xxN_(a)` `tau=(sqrt(m_(o))RT)/(sqrt(2)mu.PN_(a)d^(2)sqrt(3RT))` `tau=(sqrt(m_(o)RT))/(sqrt(6)xx3.14xx10^(5)xx6.02xx10^(23)xx16xx10^(-18))` `=(sqrt(32xx3xx8.3xx10^(-1)))/(sqrt(6)xx3.14xx10^(5)xx6.02xx10^(23)xx16xx10^(-18))` `=(sqrt(96xx0.83))/(sqrt(6)xx3.14xx6.02xx16xx10^(10))` `=(4xxsqrt(0.83))/(3.14xx6.02xx16xx10^(10))` `=(0.9)/(3.14xx6xx4xx10^(10))` `=0.01xx10^(-10)` `=10^(-12)` sec |
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| 38009. |
Two watch glasses of radii of curvature 10cm and 30cm are cemented at the edges to form an air convex lens. (a) What is its focal length in air and water? (b) Is it convergent or divergent in water ? |
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| 38010. |
The speed of light in an isotropic medium depends on, |
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Answer» its intensity |
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| 38011. |
The current through an inductive circuit of inductance 4mH is i = 12 cos 300t ampere. Calculate : (i) Reactance of the circuit. |
| Answer» SOLUTION :(i) REACTANCE `X_(L) = OMEGA L= 300 xx 4xx 10^(-3) = 1.2 Omega`. | |
| 38012. |
The electromagnetic waves that propagate in space is : |
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Answer» Longitudinal |
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| 38013. |
How does wavefront division provide coherent source? |
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Answer» SOLUTION :(i) This is the most commonly used method for producing two coherent SOURCES. A point source produces spherical WAVEFRONTS. (ii) All the points on the wavefront are at the same phase. If two points are CHOSEN on the wavefront by using a DOUBLE slit, the two points will act as coherent sources. |
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| 38014. |
The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00 cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32 cm below the top surface of the slab. Calculate the dispersive power of the material. |
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Answer» Again` mu_r=(Real depth)/(APPARENT depth)` `=2.00/1.32=1.515` so, DISPERSIVE POWER `(mu_v-mu_r)/(mu_y-1)` `=0.014/(1.515-1)` `=0.014/0.515=0.027` |
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| 38015. |
In the circuit given below, what will be the reading of the voltmeter? |
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Answer» 300 V |
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| 38016. |
Draw block diagram of a reciever |
Answer» SOLUTION :
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| 38017. |
Two parallel metal plates having charges +Q and -Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will |
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Answer» BECOME ZERO |
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| 38018. |
The phase difference between the alternating current and emf is pi/2. Which of the following cannot be the constituent of the circuit ? |
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Answer» C alone |
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| 38019. |
In a ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.50 s later, at distance d = 25.0 m from the building and at angle theta=60.0^(@) with the horizontal. (a) Find h. (Hint: One way is to reverse the motion, as if on video.) What are the (b) magnitude and ( c) angle relative to the horizontal of the velocity at which the ball is thrown? (d) Is the angle above or below the horizontal? |
| Answer» SOLUTION :(a) 32.3 m, (B) 21.9 m/s, ( C) `40.4^(@)`, (d) below | |
| 38020. |
What is focal length of a plane mirror ? |
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Answer» SOLUTION :For a plane mirror, the distance of the object from the plane mirror is EQUAL to the distance of the image from the mirror. `1/f=1/u+1/u= (-1)/u+1/u` `therefore u-u, 1/theta` and `f=prop` |
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| 38021. |
A train is moving in the north-south direction with a speed of 108 kmh^(-1). Find the e.m.f. generated between two wheels, if the length of the axle is 2m. Assume that the vertical component of earth's field is 8.0 xx 10^(-5) Wbm^(-2) |
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| 38022. |
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximumdistance at which these dots can be resolved by the eye is, (take wavelength of light, lamda = 500 nm) |
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Answer» 1 m `D=(Yd)/(1.22lambda)=(1xx10^(-3)xx3xx10^(-3))/(1.22xx500xx10^(-9))=5M` |
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| 38023. |
When current is increasing in the coil, the direction of induced emf is ….. the direction of current. |
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Answer» same as  Arrow of right SIDE SHOWS increment and arrow of left side shows DECREMENT. |
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| 38024. |
When Igm of U^(235)is completely annihilated energy liberated is E_1and when l gm of U_(235)completely undergoes fission the energy liberated is E_(2) ,then |
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Answer» `E_(1) GT E_2` |
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| 38025. |
A person in front of a mountain is beating a drum at the rate of 40 per minute and hears no distinct echo. If the person moves 90 m closer to the mountain, he has to beat the drum at 60 per minute to not hear any distinct echo. The speed of sound is |
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Answer» `320ms^(-1)` |
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| 38026. |
Match List-I with List II |
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Answer» a-h, b-g, c-e, d-f |
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| 38027. |
(A) :Excited nucleus can emit gammaradiation, but excited electron can not emit gammaradiation (R): Energy levels of excited nucleus is in Mev, excited electron is in ev and gammaradiation has energy in Mev 4. |
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Answer» Both .A. and .R. are TRUE and .R. is the correct EXPLANATION of .A. |
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| 38028. |
A step up transformer operates on a 230 V line and supplies a current of 2A. The ratio of primary and secondary windings is 1:25.The primary current is |
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Answer» 12.5A |
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| 38029. |
SI unit of magnetic permeability mu_0 (" or " mu) is |
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Answer» `A m^(-1)` |
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| 38030. |
(a) A mobile phone lies along the principal axis of a concave mirror. Show, with the help of a suitable diagram, the formation of its image. Explain why magnification is not uniform. (b) Suppose the lower half of the concave mirror's reflecting surface is covered with an opaque material. What effect this will have on the image of the object? Explain. |
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Answer» SOLUTION :(B) As the laws of reflection are true for all points of the MIRROR, the height of the whose image will be PRODUCED. However, as the AREA of the reflecting surface has been reduced, the image intensity will be reduced. In other words, the image produced will be less bright. |
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| 38031. |
Action of heat on a mixture of sodium propionate and sodalime produces : |
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Answer» methane |
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| 38032. |
White light is used to illuminate the two slits in Young's double slit experiment. The separation between the slits is d and the distance between the screen and the slit is D (gt gt d). At a point on the screen directly in front of one of the slits, certain wavelengths are missing. The missing wavelengths are (here m= 0, 1, 2, ........ is an integer): |
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Answer» `LAMBDA=(d^(2))/((2m+1)D)` |
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| 38033. |
Assume that light of wavelength 6000A is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch? |
| Answer» SOLUTION :A 100 inch telescope implies that 2a=100 inch=254cm.Thus if `lamda=6000Å=6xx10^(-5)CM`then`DELTA theta~~=(0.61xx6xx10^(-5))/127~~2.9xx10^(-7)` RADIANS | |
| 38034. |
Resistance of ammeter made by combination of 20Omega galvanometer and 2Omega shunt is ______ |
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Answer» `0.18Omega` `thereforeR=(GS)/(G+S)=(20xx2)/(20+2)=40/22~~1.8Omega` |
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| 38035. |
95 तथा 152 का महत्तम समापवर्तक (HCF ) है : |
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Answer» 57 |
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| 38036. |
A moving coil galvanometer give full scale deflection when a current of 0.005 A is passing through its coil. It is converted into voltmeter reading upto 5 V using an external of 975Omega.What is the resistance of the galvanometer coil? |
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Answer» `25Omega` |
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| 38037. |
In C.G.S. system the magnitude of the force is 100 dyne. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is(new units) |
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Answer» `0.036 ` |
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| 38038. |
Newton's laws of motion are applicable in all inertial reference frames. Some physical quantities, when measured by observers in different reference frames, have exactly the same value. Such physical quantities are called invariant. In Newtonian mechanicsmass, time and force are invariant quantities . On the other hand, some physical quantities, when measuredby observer in different referenceframes, do not have the same value. Sigmae physical quantities are called not invariant . In Newtonian mechanicsdisplacement, velocity and work ( which is the dot product of force and displacement) are not invariant. Also kinetic energy(=1/2mv^(2)) is not invariant. Physicists believe that all laws of physics are invariant in all inertialframes, i.e. the work-energy principle states that the change in the kinetic energy of a particle is equal to the work done on it by the force. Although, work and kinetic energy are not invariant in all reference frames, the work-energy principle remains invariant. Thus even though differentobservers measuring the motion of the same particle find different valuesof work and change in kinetic energy, they all find the work energy principle holds in their respective frames. Which of the following quantities is invariant? |
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Answer» Work |
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| 38039. |
Newton's laws of motion are applicable in all inertial reference frames. Some physical quantities, when measured by observers in different reference frames, have exactly the same value. Such physical quantities are called invariant. In Newtonian mechanicsmass, time and force are invariant quantities . On the other hand, some physical quantities, when measuredby observer in different referenceframes, do not have the same value. Sigmae physical quantities are called not invariant . In Newtonian mechanicsdisplacement, velocity and work ( which is the dot product of force and displacement) are not invariant. Also kinetic energy(=1/2mv^(2)) is not invariant. Physicists believe that all laws of physics are invariant in all inertialframes, i.e. the work-energy principle states that the change in the kinetic energy of a particle is equal to the work done on it by the force. Although, work and kinetic energy are not invariant in all reference frames, the work-energy principle remains invariant. Thus even though differentobservers measuring the motion of the same particle find different valuesof work and change in kinetic energy, they all find the work energy principle holds in their respective frames. Choose the invariant quantities from the following |
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Answer» mass |
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| 38040. |
Consider the charges q, q and -q placed atvertices of an equilateral triangle as shown it figure. What is the force on each charge ? |
Answer» Solution : Situation given in the statement is depicted in the following figure. Here `|q_(1)| = |q_(2)| = |q_(3)|=Q`  Here since AABC is an equilateral TRIANGLE, we have AB = BC = CA = I Magnitudes of mutual Coulombian forces are equal to F. `F_(12) = F_(21) = F_(23) = F_(13) = F_(31) = F = (kq^(2))/l^(2)`....(1) Resultant Coulombian force on `q_(1)` is: `vecF_(1) = vecF_(12) + vecF_(13)` `therefore F_(1) = sqrt(F^(2) + F^(2) + 2F F cos (120^(@))) (therefore F_(12) = F_(13) =F)` `=sqrt(2F^(2) + 2F^(2) (-1/2))` (Here, `theta = 120^(@)`) `=sqrt(F^(2))` `therefore F_(1) = F` (in the direction parallel to `bar(BC)`).....(2) Resultant Coulombian force on `q_(3)`is, `vecF_(3) = vecF_(31) + vecF_(32)` `therefore F_(3) = sqrt(F^(2) + F^(2) + 2F F cos 60^(@)) (therefore F_(31) = F_(32) = F)` `therefore F_(3) = sqrt(2F^(2) + 2F^(2)(1/2))` (Here `theta = 60^(@)`) `=sqrt(3F^(2))` `therefore F_(3) = sqrt(3F)` (in the direction of `vec(CD)`) where D is the midpoint of `bar(AB)` ) (or along the direction bisecting `angle(ACB)`) `therefore vecF_(B) = {(F cos 60^(@))(-hati) + (F sin 60^(@)hatj)}+ Fcos 60^(@) hati + F sin 60^(@) hatj + sqrt(3)F(-hatj)` `=-F/2hati + sqrt(3)/2 Fhatj + F/2hati + sqrt(3)/2 Fhatj - sqrt(3)Fhatj` `=sqrt(3)Fhatj - sqrt(3)Fhatj` `therefore vecF_(B) = vec0` `therefore vecF_(B) =0` Here, if given point charges have equal mass then resultant gravitational force on this isolated system will also be zero. Another alternative Method (Aliter) : Here, resultant Coulombian force acting on whole system is, `vecF_(B) = vecF_(1) + vecF_(2) + vecF_(3)` `therefore =(vecF_(12) + vecF_(13) + vecF_(21) + vecF_(23) + vecF_(31) + vecF_(32))` `=(vecF_(12) + vecF_(21)) + (vecF_(13) + vecF_(31)) + (vecF_(23) + vecF_(32))` `=vec0 + vec0 + vec0` `=vec0` (`therefore` According to Newton.s third law) `vecF_(12) =-vecF_(21).vecF_(13) =-vecF_(31)` and `vecF_(23) =-vecF_(32)` |
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| 38041. |
The energy gap between conduction band and valence band is of the order of 0.07 V It is a/an |
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Answer»
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| 38042. |
The unit of energy in atomic physics are |
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Answer» erg |
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| 38043. |
Figure shows a block A on a smooth surface attached with a spring of force constant k to the ceiling. In this state spring is in its natural length l . The block A is connected with a massless and frictionless string to another identical mass B hanging over a light and smooth pulley. Find the distance moved by A before if leaves contact with the ground. |
Answer» Solution : Due to weight of BLOCK B, it moves down and PULLS the block A now as block A and B move, spring gets stretched and becomes inclined as its lower end is attached to the block A. It will break off from the GROUND below it when the VERTICAL component of the spring force on block A will balance its weight mg Let it happens when A moves by a distances as shown in figure. At this instant let the spring be inclined at an angle with the vertical If the stretch in the spring at this instant is x, then it is given as `x = l sec theta - l ` or ` x = l ( sec theta - 1)` ........... (1) If mass A breaks off from ground below it , we have ` kx cos theta = mg ` From equation (1) and (2) substituting the value ofx, we get `kl( sec theta - 1) cos theta = mg or kl(1- cos theta ) = mg or cos theta =1 -(mg)/(kl) or tan theta =([k^(2)l^(2) (kl - mg)^(2) ]^(1//2))/(kl-mg)` At this instant the distance travelled by mass A and B is given by ` s=l tan theta (or) s=l([k^(2)l^(2)(kl-mg)^(2)]^(1//2))/(kl-mg)` |
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| 38044. |
The viscosity of an ideal liquid is |
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Answer» a)1 |
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| 38045. |
In an optical fibre, the light signal is transmitted by |
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Answer» Refraction at the CORE-CLADDING interface |
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| 38046. |
__________are used in speed guns to time fast moving cricket balls and automobiles. |
| Answer» SOLUTION :MICROWAVES | |
| 38047. |
What is the probability of alpha-particles with kinetic energy? 5j MeV tunnelling through the potential barrier of a polonium nucleus? |
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Answer» `w=e^(-a),"where "a=(2R)/hsqrt(2m(U_(0)-K))` Here `U_(0)` is the height of the Coulomb barrier, and K is the kinetic energy of the alpha-particle. For data on the radius of the polonium NUCLEUS and on the barrier height. |
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| 38048. |
{:((i)" Current",(a) V_(d)=atau),((ii)" Drift velocity",(b) I=(Q)/(t)),((iii)" Current density",(c )V=IR),((iv)" Ohm's Law",(d) J=(I)/(A)):} |
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Answer» |
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| 38049. |
A smooth inclined plane of angle of inclination 30^(@) is placed on the floor of a compartment of a train moving with a constant acceleration a. When a block is placed on the inclined plane, it does not slide down or up the plane. The acceleration a must be |
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Answer» g Then resolving `VECG` and `veca` into components The block will stay at rest when a `COS theta = g sintheta` `a=g tan theta` `since theta = 30^(@) :. a=(g)/(sqrt3)` Hence correct choice is (d) 
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| 38050. |
Calculate energy of photon having 6840 Å wavelength (hc=12400 eVÅ) |
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Answer» 1.81 EV `THEREFORE f=(c )/(lambda) f=( c)/(lambda)implieshf=(hc)/(lambda),hc=12400 eVÅ` ENERGY of PHOTON =`(12400 eVÅ)/(6840Å)=1.81 eV` |
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