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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.
| 41051. |
A particle with charge Q, moving with a momentum p, enters a uniform magnetic field normally. The magnetic field has magnitude B and is confined to a region of width d, where d lt P/(BQ). The particle is deflected by an angle thetain crossing the field. Then |
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Answer» <P>SIN `THETA=(BQd)/p` mv=p=BQr `r=p/(BQ)` sin `theta=(ED)/(CD)=d/r=(BQd)/p` 
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| 41052. |
A hypothetical train moving with aspeed of 0.6 c passes by the platform of a small station without being slowed down. The observes o the platform note that the length of the train is just equal to the length of the platform which is 200 m . (a) Find the rest length of the train(b) Find the length of the platform as measured by the obsevers in the train . |
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Answer» `L' = L (sqrt 1 - V^(2) / c^(2))` L (sqrt 1 - V^(2) / c^(2)) = 200 m / (sqrt 1- (0.6)^(2))` = 250 m. (b) The rest lenth of the platform is 200m. For the observes in the train, the platform is moving at a speed of 0.6c. The length as MEASURED by the observers in the train is , therefore , ` L' = 200m (sqrt 1- (0.6)^(2))=160m`. |
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| 41053. |
As the drawing shows, one microphone is located at the origin, and a second microphone is located on the +y axis. The microphones are separated by a distance of D = 1.50 m. A source of sound is located on the +x axis, its distances from microphones 1 and 2 being L_1 and L_2respectively. The speed of sound is 343 m/s. The sound reaches microphone 1 first, and then, 1.46 ms later, it reaches microphone 2. Find the distances L_1 and L_2 . |
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| 41054. |
Barrier potential of a p-njunction diode does not depend on |
| Answer» Solution :Barrier potential depends upon temperature, doping density and forward biasing. | |
| 41055. |
A motion is described by y = 3e^x.e^(-3t)where y,x arc in metrd and t is in seconds. |
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Answer» This represents EQUATION of progressive wave propagating along -X DIRECTION with `3ms^(-1)` |
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| 41056. |
The output of the following circuit is I when the input ABC is |
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Answer» 101 |
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| 41057. |
A rocket of mass 20kg has 180 kg of fuel. The exhaust velocity of fuel is 1.6 km/sec. Calculate the ultimate velocity of the rocket gained, when the rate of consumption of the fuel is 2kg/sec. (neglect gravity) |
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Answer» 3.7 km/sec |
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| 41058. |
What is the energyof the electron revolving in third orbit expressed in eV? |
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Answer» `1.51 eV` |
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| 41059. |
A multiconductor flat cable can have any number of wires in the range ____________ |
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Answer» `5-10` |
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| 41060. |
Dielectric constant of metals is |
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Answer» 1 |
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| 41061. |
State two characteristics of nuclear force. Why does the binding energy per nucleon decrease with increase in mass number for heavy nuclei like " "^(235)U ? |
| Answer» SOLUTION :The binding energy per nucleon decreases with increase in mass NUMBER for heavy nuclei like `" "_(92)^(235)U`. These nuclei contain a large number of protons which repel one another. As a result of this REPULSIVE force value of B.E. per nucleon decreases with increase in mass number and such heavier NUCLIDES are unstable. | |
| 41062. |
Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be: |
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Answer» 2 |
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| 41063. |
An electrical technician requires a capacitance of 2 uF in a circuit across a potential difference of 1 kV. A large number of 1 uF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors. |
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Answer» Solution :As potential difference required by the technician=1 kV = 1000 V and potential difference rating of each `1 MUF` capacitor `= 400 V` `:.` Minimum number of capacitros to be joined in SERIES `n= 1000/400 = 2.5` It means that atleast 3 capacitors should be joined in a series row, whose COMBINED capacitance will `be C/3 = (1muF)/3 = 1/3 muF` As technician requires a capacitance of `2 muF` , hence number of capacitor rows to be joined in parallel should be `m = (2muF)/(1/3muF)= 6` Minimum number of capacitors NEEDED by the technician `=nm = 3xx 6 =18`. The technician should arrange these 18 capacitors in 6 parallel rows, each row consisting of 3 capacitors in series. |
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| 41064. |
The temperature of a piece of metal is raised from 27^@C to 84^@C. The rate at which the material radiates energy increases radiates energy increases nearly |
| Answer» Answer :A | |
| 41065. |
Given circuit symbol , logical operation , truth table , and Boolean expression of AND , OR , NOT , NAND , NOR , and EX - OR gates |
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Answer» SOLUTION :• AND gate Circuit symbol Inputs Output A and B are inputs and Y is the output. It is a AND logic gate and hence A, B, and Y can have the (a) TWO input AND gate value of either 1 or 0.  BOOLEAN equation: Y=AB It performs logical multiplication and is different from arithmetic multiplication. Logic operation: The output of AND gate is high (1) only when all the inputs are high (1) The rest of the cases the output is low. Hence the output of AND gate is high (1) only when all the inputs are high • OR gate: Circuit Symbol A and B are inputs and Y is the output  Boolean equation: A+B=Y It performs logical addition and is different from arithmetic addition. Logic operation: The output of OR gate is high (logic 1 state) when either of the inputs or both are high • NOT gate: Circuit symbol A is the input and is the output. COD Boolean equation: `Y=overlineA` Logic operation: The output is the complement of the input. It is represented with an overbar. It is also called as inverter. The truth table infers that the output is when input A is and vice versa. The truth table of NOT. • NAND gate: A and B are inputs and Y is the output.  Boolean equation: `Y=overline(A.B)` Logic operation: The output Y equals the complement of AND operation. The circuit is an AND gate followed by a NOT gate. Therefore, it is summarized as NAND. The output is at logic zero only when all the inputs are high. The rest of the cases, the output is high (Logic 1 state). • NOR gate: Circuit symbol: A and B are inputs and Y is the output.  Boolean equation `Y=overline(A+B)` Logic operation Y equals the complement of OR operation (A OR B). The circuit is an OR gate followed by N este and is summarized as NOR. The output is high when all the inputs are low. The output is low for all other COMBINATIONS of inputs. Ex-OR gate: Circuit symbol A and B are inputs and Y is the output. The Ex-OR operation is denoted as`oplus` Boolean equation `Y=A.overlineB+overlinA.B` `Y=AoplusB` Logic operation The output is high only when either of the two inputs is high. In the case of an Ex-OR gate with more than two inputs, the output will be high when odd number of inputs are high.  
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| 41066. |
How many electrons, protons and neutrons are there in 12g of ._6C^12and in 14g of ._6C^14 ? (Take Avogadro number N=6xx10^23 ) |
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Answer» Solution :The number of atoms in 12g of `._6C^12` AVOGADRO number = `6xx10^23` The number of electrons in 12 g of `._6C^12=6xx6xx10^23` =`36xx10^23` The number of protons in 12 g of `._6C^12=36xx10^23` The number of neutrons in 12 g of `._6C^12` `=(A-Z)xx6xx10^23` `=6xx6xx10^23=36xx10^23` SIMILARLY number of electrons in 14G of `._6C^14=36xx10^23` Number of protons in 14g of `._6C^12=36xx10^23` Number of neutrons in 14 g of `._6C^14 =(A-Z)xx6xx10^23` `=(14-6)xx6xx10^23 = 48xx10^23` |
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| 41067. |
double-convex lensis tobemanufacturedfroma glassof refractive index1.55withbothfacesof the same radiusof curvatures . Calculatethe radiusofcurvaturerequiredif thefocallengthis tobe 20 cm ?Alsofind thefocallengthofthe lensif itis immersedin waterof refractiveindex 1.33 ? |
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Answer» Solution :given: `n_g =1.55 ,R = ? ,F=20cm, R_I = + R ,R_2 =- R ` `(1)/(f)=(n_g-1)[(1)/(R_1)-(1)/(R_2)]` `(1)/(20)= (1.55 -1) [1/R +1/R ]` `=0.55 xx2/R` `R=20xx 0.55 xx 2` ` R =22CM ` WHENTHE lensis IMMERSED in water `(1 )/(f_omega)=((n_g)/(n_omega)-1) [(1)/(R_1)-(1)/(R_2)]` `(1)/(f_(omega) )= ((1.55)/(1.33)-1)[(1)/(22)+(1)/(22)]` `(1)/(f_omega) =(1.1654 -1) 2/22` `(1)/(f_omega)= 0.1654 xx(2)/(22)` `(1)/(f_omega)=(0.3308)/(22)` `f_(omega)=(22)/(0.3308)` `f_(omega) = 66.5cm ` |
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| 41068. |
What are the differences in reflection by (i) a plane mirror, (ii) the wall of a building and (iii) a clean glass plate ? |
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Answer» Solution :(i) Regular reflection of light takes place in a plane mirror and a bright and distinct image is formed. (II) DIFFUSE reflection takes place from the wall of a building due to its rough SURFACE and no image is formed. The wall itself is seen with equal brightness. (iii) If light is incident on a clean glass plate, a small portion of the incident light is REFLECTED. Major portion of the light enters the glass plate and by refraction is transmitted to air through the other side, So, dim but distinct image is formed. |
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| 41069. |
An inhomegenous poorly conducting medium fills up the spacebetween plates 1 and 2 of parallel-platecapacitor. Itspermittivityand resistivityvary from values epsilon_(1), rho_(1) at the plate 1 to vlaues epsilon_(2), rho_(2) at plate 2. A dcvoltageis appliedto the capacitorthrouggh which a steadycurrent I flowsfrom plate1 to plate 2. Find the totalextraneous charge in the given medium. |
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Answer» Solution :By current conservation `(E (X))/(rho (x)) = (E (x) + d E(x))/(rho (x) + d rho (x)) = (d E (x))/(d rho(x))` This has the solution, `E_(x) = C rho (x) = (I rho (x))/(A)` Hence charge induced in the SLICE per unit area `d sigma = epsilon_(0) (I)/(A) [{epsilon (x) + d epsilon (x)} {rho (x) + d rho (x)} - epsilon(x)rho (x)] = epsilon_(0) (I)/(A) d [epsilon (x) rho (x)]` Thus, `dQ = epsilon_(0) Id [epsilon (x) rho (x)]` Hence total charge induced is by intergation, `Q = epsilon_(0) I (epsilon_(2) rho_(2) - epsilon_(1) rho_(1))` 
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| 41070. |
There are two identical square metallic plates kept on a rough horizontal floor at t=0, plates are given angular velocity omega and it is given that plate 1 is rotating about its centre A and plate 2 is rotating about one of its centre O. If t_(1) and t_(2) are the time taken by both the plates to come to rest then (t_(1))/(t_(2)) is (Both are independent cases) |
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Answer» Solution :`tau ALPHA mgd` `I.((DW)/(dt)) alpha mgd` and `I=K.md^(2)` THEREFORE `(dw)/(dt) alpha (1)/(d)`(Angular Acceleration of plate is independent of mass) Now apply superposition principal |
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| 41071. |
A short bar magnet placed with its axis at 30^@ to a uniform magnetic field of 0.02 T experiences a torque of 0.060 Nm. (i) Calculate magnetic moment of the magnet, and (ii) find out what orientation of the magnet corresponds to its stable equilibrium in the magnetic field. |
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Answer» Solution :Here B= 0.02 T, `THETA= 30^@`and `tau = 0.060 Nm` (i) `THEREFORE ` Magnetic moment of the magnet `m =(tau)/(B sin theta) = (0.060)/(0.02 xx 0.5000) = 6 A m^2` (II) For stable equilibrium of the magnet in the magnetic field `theta = 0^@` |
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| 41072. |
The equation of a stationary wave is y = 2 sin ((pi x)/(15)) cos( 48pi t). The distance between a node and its next antinode is. |
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Answer» 22.5 UNITS |
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| 41073. |
A physical quantity is represented by X = M^(@)L^(b)T^(-c) . If the percentage error in the measurement of M,L and T are 2alpha%,Beta%,3gamma%, respectively then maximum percentage error in X is |
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Answer» `(aalpha+B BETA-c GAMMA)%` |
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| 41074. |
A point object at a distance of 36 cm from a convex lens of focal length 10 cm, is moved by 10 cm in 2 sec along principal axis towards the lens. Then image will aslo change its position. a. Write the law which relates object and image distance from the lens. b. Find the inwal and final position of the image and calculate average speed of image . c. A man argues that the image will move uniformly at the same speed as that of object What is your opinion ? Justify. |
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Answer» Solution :a.Lens maker.s formula `(1)/(f) = (1)/(v) - (1)/(U)` b. `(1)/(f) = (1)/(v) -(1)/(u) ` `v_(i) = (uf)/(u + f) = (- 36 xx 10)/(-36 + 10) = (-360)/(-26) = 13.84` cm `Y_(f) = (uf)/(u + f) = (- 26 xx 10)/(- 26 + 10) = (-260)/(-16) = 16.25`cm speed = `("distance " )/("time") = (16.25 - 13.84)/(2) = 1.205 MS^(-1) ` c. No. speed of the image is not unifomr eventhough speed of the OBJECT changes UNIFORMLY. |
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| 41075. |
Explain reasons why the sun looks reddish at sunrise and sun set? |
| Answer» Solution :During sunrise or sunset the sun is NEARER to the horizon. Sunlight travels maximum thickness of the atmosphere. So, sorter waves of BLUE REGION are scattered away by the atmosphere. Red waves of LONGER wave-length are LEAST scattered and reach the observer. So, the sun appears red. | |
| 41076. |
A conducting gas is in the form of a long cylinder. Currents flows through the gas along the length of the cylinder. The current is distributed uniformly across the cross-section of the gas. Disregard thermal and electrostatic forces among the gas molecules. Due to the magnetic fields set up inside the gas and the forces which they exert on the moving ions, the gas will tend to |
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Answer» expand |
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| 41077. |
V_C is the ideal voltmeter in the figure. Resistance of the resistor shown is R. Initially the switch is in position 2 when charging of the capacitor starts. Initially the capacitor was uncharged. The switch in circuit shifts automatically from 2 to 1 when V_C gt 2V//3 and goes back to 2 from 1 when V_C lt V//3. The ideal voltmeter reads voltage across capacitor as plotted. What is the period T of the wave from in terms of R and C ? , |
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Answer» `3RC 1n 3 ` For this , `V/3 = (2V)/2 E^(-t_1//tau)` or `t_1 = tau1n 2 = RC 1n2` `t_2` is time for CHARGING of capacitor from `V//3 "to" 2V//3`. So `int_(VC//3)^(2VC//3) (dq)/(VC-q) = int_(0)^(t_2) (dt)/(RC)` or `t_2= RC1n 2` `T = t_1 + t_2 = 2RC 1n 2 `  .
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| 41078. |
Modulation is a process employed to superimpose a carrier wave with a signal wave. In amplitude modulation, what will change? |
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Answer» Frequency |
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| 41079. |
Draw phasor diagram for a RLC series circuit connected to ac voltage source. |
Answer» SOLUTION :
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| 41080. |
A ball is thrown upwards from the ground. It is at a height 100 m in upward and downward journeys at time t_1 andt_2 respectively. If g = 10 m/s^(2), then the product of t_1 . t_2 is: |
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Answer» 10 or `5t^2 -ut+100=0` There are two roots for this equation .Let `t_1` and `t_2` be the value. The PRODUCT of the roots `t_1.t_2=(100)/(5)=20` |
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| 41081. |
What is the phase relation between current and emf in an AC circuit containing a cpacitor only ? Sketch a graph showing the variation the reactance of a capacitor with frequency. |
Answer» SOLUTION :CURRENT LAGS behind the applied VOLTAGE by `(pi)/(2)` in an inductive circuit. 
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| 41082. |
The density of a liquid of coefficient of cubical expansion gamma is d at 0^(@)C. When the liquid is heated to a temperature T, the change in density will be : |
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Answer» `(-gamma T d_(0))/((1+gamma T))` `d_(t) =(d_(0))/(1+gamma T)` `therefore` Change is density `=d_(e)-d_(0)=(d_(0))/(1+gamma T)-d_(0)` `=(d_(0)-d_(0)-gamma Td_(0))/(1+gamma T)=(-gamma Td_(0))/(1+gamma T)` Thus, correct choice is (a). |
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| 41083. |
A solid cube floats in a liquid of density rho with half of its volume submerged as shown. The side length of the cube is a, which is very small compared to the size of the container. Now a liquid of density (rho)/(3) (immiscible with the liquid already in the container) is poured slowly into the container. The column of the liquid of density (rho)/(3) that must be poured so that the cube is fully submerged (with its top surface coincident with the surface of the poured liquid) is : |
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Answer» `(a)/(3)` Then, because cube is in equilibrium `rho_(o)a^(3)G=rho((a^(3))/(2))g` `impliesrho_(o)=(rho)/(2)` Let the final equilibrium POSITION of the cube be as SHOWN Then, `rho_(o)a^(3)g=((rho)/(3))(a^(2)x)g+rho(a-x)(a^(2))g` `impliesx=(3)/(4)a` 
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| 41084. |
The density of a rod of 1 m and of non-uniform structure is given by rho(x)=a(1+bx^(2)), where a and b are constant such that alexle1. The centre of mass of the rod will be at : |
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Answer» `(3[2+b])/(4[3+b])` |
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| 41085. |
Rest mass energy of an electron is 0.51 MeV. If this electron is moving with a velocity 0.8 c (where c is velocity of light in vacuum), then kinetic energy of the electron should be. |
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Answer» 0.28 MeV |
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| 41086. |
Two long, parallel copper wires of diameter 4.0 mm carry currents of 7.0 A in opposite directions. (a) Assuming that their central axes are 20 mm apart, calculate the magnetic flux per meter of wire that exists in the space between those axes. (b) What percentage of this flux lies inside the wires? (c) Repeat part (a) for parallel currents. |
| Answer» SOLUTION :(a)7.8 μWb/m, (B) 22%, (C) 0 | |
| 41087. |
नदियों का आर्थिक महत्व क्या है? |
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Answer» जल विद्युत का उत्पादन |
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| 41088. |
An electron (mass = 9xx10^(-31)kg, charge = 1.6xx10^(-19)C) whose kinetic energy is 7.2xx10^(-18)joule is moving in a circular orbit in a magnetic field of 9xx10^(-5)Wb//m^(2). The radius of the orbit is _____ . |
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Answer» 1.25 cm `q=1.6xx10^(-19)C` `m=9xx10^(-31)kg` `B=9xx10^(-5)Wb//m^(2)` Radius of CIRCULAR path `r=sqrt(2mE)/(qB)` `thereforer=sqrt((2xx9xx10^(-31)xx7.2xx10^(-18))/(1.6xx10^(-19)xx9xx10^(-15)))` = 0.25 m = 25 cm |
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| 41089. |
In Fig. , a French submarine and a U.S. submarine move toward each other during maneuvers in motionless water in the North Atlantic. The French sub moves at the speed v_v = 48.00 km/h, and the U.S. sub at v_(US)= 72.00 km/h. The French sub sends out a sonar signal (sound wave in water) at 1.560 xx 10^3Hz. Sonar waves travel at 5470 km/h. (a) What is the signal's frequency as detected by the U.S. sub? (b) What frequency is detected by the French sub in the signal reflected back to it by the U.S. sub? |
| Answer» SOLUTION :`(a) 1.595 XX 10^3 HZ, (B ) 1.630 xx 10^3 Hz` | |
| 41090. |
If you were driving a car, what kind of mirror would you prefer to use for observing traffic at your back ? |
| Answer» Solution :CONVEX MIRROR, because it has wider field of view, it.s IMAGE is ERECT and virtual. | |
| 41091. |
A prism of angle A has one surface silvered. Light rays falling at an angle of incidence 2A on first surface and return through the same path after suffering refraction at the second silvered surface. Refractive index of prism material is: |
| Answer» Answer :D | |
| 41092. |
In a reflective astronomical telescope we use a ________ of larger aperture and large focal length as the _________ |
| Answer» SOLUTION :CONCAVE MIRROR, OBJECTIVE of TELESCOPE | |
| 41093. |
Match the cprresponding entries of Coolumn 1 with Column 2. [Where m is the mangnification produced by the mirroe {:(Column1,Column2),(A. m=-2,a."Convex mirror"),(B.m=-1/(2),b."Convave mirror"),(C.m=+2,c."Real image"),(D. m=+1/(2),d."Virtual image"):} |
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Answer» `A rarr a` and C , `B rarr a` and `d, Crarr a` and `b, D rarr c` |
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| 41094. |
There is enough for everybody's need and not for any body's greed," who among the following has given the above statement? |
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Answer» VINOBA Bhave |
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| 41095. |
Demostrate that is the case of a steady flow of an ideal fluid turns into Bernoulli equation. |
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Answer» Solution :The Euler's equation is `rho(dvecv)/(dt)=vecf-vecnablap=-vecnabla(p+rhogz)`, where z is vertically upwards. Now `(dvecv)/(dt)=(delvecv)/(delt)+(vecv*vecnabla)vecv`. But `(vecv*vecnabla)vecv=vecnabla(1/2v^2)-vecvxxCurlvecv` (2) we consider the steady (i.e. `delvecv//delt=0`) flow of an incompressible fluid then `rho`= CONSTANT. and as the MOTION is irrotational CURL `vecv=0` So from (1) and (2) `rhovecnabla(1/2v^2)=-vecnabla(p+rhogz)` or, `vecnabla(p+1/2rhov^2+rhogz)=0` Hence `p+1/2rhov^2+rhogz=const ant`. |
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| 41096. |
A man holds a 2 kg book between his palms. so that each hand exerts the same horizontal force on the book. The coefficient of static friction between the palms and the book is 0.4 and g = 10ms^(-2). If the book is prevented from falling, the least force exerted by each hand on the book is |
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Answer» 50 N |
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| 41097. |
A simple pendulum is suspended vertically to the ceiling of compartment in a stationary train. If the 'train' is constantly accelerated by acceleration 'a' the angle theta which string makes with vertical is : |
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Answer» zero  `T sin theta=ma` `T cos theta=mg` `:.""tan theta=a//gimpliestheta=tan^(-1)(a//g)` THUS correct choice is (B). |
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| 41098. |
In case of hydrogen and oxygen at N.T.P., which of the following quantities is/are the same? |
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Answer» average momentum PER molecule Clearly both `K_(1)andK_(2)` are equal for hydrogen and OXYGEN. |
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| 41099. |
A particle of mass mis located in a spherically symmetrical potential well U(r )=0 for r lt r_(0) and U(r )=U_(0) for r gt r_(0). (a) By means of the substitution Psi(r )= xi(r )//r find the equation defining the proper values of energy E of the particle for E lt U_(0), when its motion is described by a wave function Psi(r ) depending only on r. Reduce that equation to the form sinkr_(0)+-kr_(0)sqrt( ħ^(2)//2mr_(0)^(2)U_(0)), where k=sqrt(2mE//) ħ (b) Calculate the value of the quantity r_(0)^(2)U_(0) at which the first level appears. |
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Answer» Solution :If we put `Psi=(CHI(r ))/(r )` the equation for `chi(r )` has the from `chi''k^(2)chi=0,0 LE r lt r_(0)` and `chi''-alpha^(2)chi=0r_(0) lt r lt oo` where `k^(2)=(2mE)/( ħ^(2)),alpha^(2)=(2M(U_(0)-E))/( ħ^(2))` The boundary condition is `{:(chi(0),,,=,0),(and chi,chi',are conti n uous at,r=,r_(0)):}}` These are exactly same as in the one dimensional problem in problem (6.85) We therefore omit further details. |
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| 41100. |
A 20 V battery of internal resistance 1 Omega is connected to three coils of 12 Omega, 6 Omega and 4 Omega in parallel, a resistor of 5Omega and a reversed battery (emf=8 V and internal resistance =2 Omega), as shown in Fig. 4.69. Calculate the current in each resistor and the terminal potential difference across each battery. |
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Answer» |
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