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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.
| 651. |
Which one of the following has least similar characters ? |
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Answer» family |
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| 652. |
The photoelectric work function for a metal surface is 4.125 eV. The cut off wavelength for this surface is |
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Answer» 3000 `Å` |
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| 653. |
A capacitor, a resistor and 4 henry inductor are connected in series to an a.c. source of 50 Hz. Calculate capacitance of capacitor if the current is in phase with voltage. |
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Answer» Solution :When V and I in PHASE `X_(L) = X_(C ) v = (1)/(2PI) (1)/(sqrt(LC))` `C = (1) /(4 pi^2 v^2 L)= (1)/(4 pi^2 XX 50 xx 50 xx (4)/(pi^2))` `= 2.5 xx 10^(-5) = 25 muF.` |
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| 654. |
A radio - active sample of half- life 10 days contains 1000 x nuclei . Number of original nuclie present after 5 days is |
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Answer» 707x Here `n=(1)/(2)` and `N_(0)=1000x` `THEREFORE N=(1000x)/(2^((1)/(2)))=(1000x)/(SQRT(2))=0.707xx1000x` `N=707x` |
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| 655. |
The value of 'h' for particular capillary tube and particular liquid comes out be 4 cm from the formula, h = 2Tcostheta//rhorg . But if the length of the tube is 3 cm only. Then |
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Answer» The liquid will OVERFLOW from TUBE |
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| 656. |
Impendence of a circuit may also be calculated using impendence triangle. Explain. |
Answer» Solution : From the figure ` TAN phi = ( (omega L - (1)/(omega C) )/(R )) = ( (X_L - X_C)/( R ) ) and Z = sqrt(R^2 + (X_L - X_C)^2)` |
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| 657. |
A transformer has 450 turns in its primary coil and 30 turns in its secondary coil. Which one of the following statements concerning this transformer is true? |
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Answer» <P>The turns RATIO is 15 for this transformer. |
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| 658. |
The speed corresponding to the peak of Maxwell-Boltzmann molecular distribution curve is |
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Answer» the root-mean-square SPEED of the gas molecules |
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| 659. |
Approximate thickness of oil film to observe interference of light (due to which it looks coloured) is |
| Answer» Answer :B | |
| 660. |
What is active region for transistor? |
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Answer» |
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| 661. |
There are three sources of sound of equal intensities with frequencies 400, 401 and 402 Hz. The no. of beats per second is : |
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Answer» 0 |
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| 662. |
In satellite communication 1) the frequency lised lies between 5MHz and 10MHz 2)the uplink and downlink frequencies are different 3)the orbit of geostationary satellite lies in the equatorial plane at an inclination of 0^(@) In the above statements |
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Answer» Only 2 and 3 TRUE |
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| 663. |
The minimum refractive index of a right-angled prism to turn a beam of light falling normally on its one face is |
| Answer» Answer :C | |
| 664. |
The valency of the impurity atom added to make germanium crystal into n-type semi-conductor have …. |
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Answer» 6 |
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| 665. |
An uniform electric field of strength E exists in a region. An electron of mass m enters a point A perpendicular to x - axis with velocity V. It moves through the electric field and exists at point B. The components of velocity at B are shown in (Fig. 3.156). At B the y - component of velocity remain uncharged. . Find electric field. |
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Answer» `(2 m aV^2)/(e d^2)` ACCLERATION of ELECTRON along x - axis is `a_x = eE//m`. Time taken to go from A to B is `t = d//V`. Along x - axis `a = (1)/(2) (a_x) t^2 = (1)/(2) (e E)/(m) (d/V)^(2)` or `E = (2 m a V^2)/(e d^2)` `v_1 = a_x t = (e E)/(m) d/V = (e)/(m) d/V (2 m a V^2)/(e d^2) = (2 a)/d V` Velocity at `B is sqrt(V^2 + v_1^2) = Vsqrt (1 + ((2 a)/d)^2)` RATE of work done by field at `B` is `F v_1 = e E (2 a)/d V = (2 a e V)/d (2 m a V^2)/(e d^2) = (4 m a^2 V^3)/(d^3)`.  .
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| 666. |
Figure (i) and (i) show the displacement-time graphs of two particles moving along the x-axis. We can say that |
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Answer» both the particles are having a uniformly retardedmotion  When body is moving with constant RETARDATION, the time displacement graph is a curve with bend downwards. 
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| 667. |
An alpha particle is equivalent to.............or.............. . |
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Answer» |
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| 668. |
In Young's double slit experiment, the two slits act as coherent sources of equal amplitude 'a' and of wavelength lambda. In another experiment with the same setup the two slits are sources of equal amplitude 'a' and wavelength lambda but are incoherent. The ratio of intensities of light at the mid point of the screen in the first case to that in the second case is .... |
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Answer» `2:1` `I_(max) = (a + a)^(2) = 4a^(2)` Now, when sources are INCOHERENT, no interference occurs, intensity at mid point is `I=I_(1)+I_(2) RARR a^(2)+a^(2)=2A^(2)` `:. (I_(max))/(I)=(4a^(2))/(2a^(2))=2:1` |
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| 669. |
Hydrogen atom does not emit X-rays because : |
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Answer» it has SINGLE electron |
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| 670. |
The interference pattern of two identical slits separated by a distance d=0.25 mm is observed on a screen at a distance of 1 m from the plane of the slits. The slits are illuminated by monochromatic light of wavelength 589.3 nm (sodium D) travelling perpendicular to the plane of the slits. Bright bands are observed on each side of the central maxima. Calculate the separation between adjacent bright bands ? |
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Answer» Solution :FRINGE WIDTH `(beta)=(Dlambda)/(d)=((1m)(589.3xx10^(-9)m))/((0.25xx10^(-3))m)` `=2XX10^(-3)m=2mm`. |
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| 671. |
A 100m long antenna is mounted on a 500m tall building. The complex can become a transmission tower for waves with a lambda |
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Answer» `-400` m |
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| 672. |
A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 35 cm from the wire ? |
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Answer» SOLUTION :For an EXTREMELY long STRAIGHT thin current carrying wire, magnetic field PRODUCED at perpendicular distance r is, `B=(mu_(0)I)/(2pir)` `thereforeB=((4pixx10^-7)(35))/((2pi)(0.2))` `thereforeB=3.5xx10^(-5)T` 
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| 673. |
A particle moves in x-y plane according to the equation bar(r ) = (hat(i) + 2hat(j)) A cos omega t. The motion of the particle is |
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Answer» on a STRAIGHT line |
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| 674. |
Statement-1 : Water kept in an open vessel will quickly evaporate on the surface of the moon. Statement-2 : The temperature at the surface of the moon is much higher than boiling point of water. |
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Answer» Statement-1 is true, Statement-2 is false. |
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| 675. |
In an electromagnetic wave in free space the rms value of the electric field is 3Vm^(-1). The peak value of the magnetic field is |
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Answer» `1.414xx10^(-8)T` |
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| 676. |
A square current carrying loop made of thin wire and having a mass m =10g can rotate without friction with respect to the vertical axis O O_1, passing through the centre of the loop at right angles to two opposite sides of the loop. The loop is placed in a homogeneous magnetic field with an inductionB = 10^(-1)T directed at right angles to the plane of the drawing. A current I = 2A is flowing in the loop. Find the period of small oscillations that the loop performs about its position of stable equilibrium. |
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Answer» Solution :`T_0 = 2pi sqrt(m)/(6IB) = 0.57 s` [Hint : Restoring torque equation : `I AB SIN THETA = - I_0 theta,` where `A = a^2` and `I_0 = m/4 [(a^2)/(12) xx 2 + (a/2)^2 xx 2 ]` `= m /4 xx 2/3 a^2` `rarr (m/6 a^2)overset(..)theta = -a^2 B theta rArr overset(..) = - ((6IB)/(M)) theta ]`. |
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| 677. |
A block mass 'm' is kept on the ground as shown in figure. If answer is no, draw action reaction pair. |
Answer» SOLUTION :Pair of 'mg' of BLOCK acts on EARTH in OPPOSITE DIRECTION 
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| 678. |
A balance point in a meter bridge experiment is obtained at 30 cm from the left. If right gap contains 3.5Omega, what is the resistance in the left gap? |
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Answer» Solution :The balancing length from left `L=20cm` RESISTANCE in right gap `R=3.5Omega` `R=Sxx(l)/((100-l)),R=3.5xx(30)/(70)rArr R=1.5Omega` |
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| 679. |
The distance of two points on the axis of a magnet from its centreis 10 cmand 20 cm respectively the ratio of magnetic intensity at these points is 12.5 :1 the length of the magnet will be |
| Answer» ANSWER :D | |
| 681. |
To move an airplane through the air, thrust is generated by some kind of propulsion system. Beginning with the Wright brothers' first flight, many airplanes have used internal combustion engines to turn propellers to generate thrust. Today, most general aviation or private airplanes are powered by internal combustion (IC) engines, much like the engine in your family automobile. When discussing engines, we must consider both the mechanical- operation of the machine and the thermodynamic processes that enable the machine to produce useful work. We consider the thermodynamics of a four-stroke IC engine. Figure shown represents such a cycle which is used in all Internal combustion engines. The figure shows a p-V diagram of the cycle. Using the engine stage numbering system, we begin at the lower left with Stage 1 being the beginning of the intake stroke of the engine. The pressure is near atmospheric pressure and the gas volume is at a minimum. Between Stage 1 and Stage 2 the piston is pulled out of the cylinder with the intake valve open. The pressure remains constant, and the gas volume increases as fuel/air mixture is drawn into the cylinder through the-intake valve. Stage 2 begins the compression stroke of the engine with the closing of the intake valve. Between Stage 2 and Stage 3, the piston moves back into the cylinder, the gas volume.decreases, and the pressure increases because work is done on the gas by the piston. Stage 3 is the beginning of the combustion of the fuel/air mixture. The combustion occurs very quickly and the volume remains constant. Heat is released during combustion which increases both the temperature and the pressure, according to the equation of slate. Slage 4 begins the power stroke of the engine. Between Stage 4 and Stage 5, the piston is driiven towards the crankshaft, the volume in increased, and the pressure falls as work is done by the gas on the píston. At Stage 5 the exhaust valve is opened and the residual heat in the gas is exchanged with the surroundings. The volume remains constant and the pressure adjusts back to atmospheric conditions. Stage 6 begins the exhaust stroke of the engine during which the piston. moves back into the cylinder, the volume decreases and the pressure remains constant. At the end of the exhaust stroke, conditions have returned to Stage 1 and the process repeats itself. During the cycle, work is done on the gas by the plston between stages 2 and 3. Work is done by the gas on the piston between stages 4 and 5. The difference between the work done by the gas and the work done on the gas is the area enclosed by the cycle curve and is the work produced by the cycle. The work times the rate of the cycle (cycles per second) is equal to the power produced by the engine. Choose.the correct statement. |
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Answer» During the compression stroke, the GAS does positive WORK. |
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| 682. |
To move an airplane through the air, thrust is generated by some kind of propulsion system. Beginning with the Wright brothers' first flight, many airplanes have used internal combustion engines to turn propellers to generate thrust. Today, most general aviation or private airplanes are powered by internal combustion (IC) engines, much like the engine in your family automobile. When discussing engines, we must consider both the mechanical- operation of the machine and the thermodynamic processes that enable the machine to produce useful work. We consider the thermodynamics of a four-stroke IC engine. Figure shown represents such a cycle which is used in all Internal combustion engines. The figure shows a p-V diagram of the cycle. Using the engine stage numbering system, we begin at the lower left with Stage 1 being the beginning of the intake stroke of the engine. The pressure is near atmospheric pressure and the gas volume is at a minimum. Between Stage 1 and Stage 2 the piston is pulled out of the cylinder with the intake valve open. The pressure remains constant, and the gas volume increases as fuel/air mixture is drawn into the cylinder through the-intake valve. Stage 2 begins the compression stroke of the engine with the closing of the intake valve. Between Stage 2 and Stage 3, the piston moves back into the cylinder, the gas volume.decreases, and the pressure increases because work is done on the gas by the piston. Stage 3 is the beginning of the combustion of the fuel/air mixture. The combustion occurs very quickly and the volume remains constant. Heat is released during combustion which increases both the temperature and the pressure, according to the equation of slate. Slage 4 begins the power stroke of the engine. Between Stage 4 and Stage 5, the piston is driiven towards the crankshaft, the volume in increased, and the pressure falls as work is done by the gas on the píston. At Stage 5 the exhaust valve is opened and the residual heat in the gas is exchanged with the surroundings. The volume remains constant and the pressure adjusts back to atmospheric conditions. Stage 6 begins the exhaust stroke of the engine during which the piston. moves back into the cylinder, the volume decreases and the pressure remains constant. At the end of the exhaust stroke, conditions have returned to Stage 1 and the process repeats itself. During the cycle, work is done on the gas by the plston between stages 2 and 3. Work is done by the gas on the piston between stages 4 and 5. The difference between the work done by the gas and the work done on the gas is the area enclosed by the cycle curve and is the work produced by the cycle. The work times the rate of the cycle (cycles per second) is equal to the power produced by the engine. Consider that the above diagram represents an ideal such cycle then choose the incorrect statement fora corresponding real cycle. |
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Answer» The area of the p-V diagram for the REAL cycle will be lesser than that for ideal cycle. |
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| 683. |
Particles obtained from natural radioactive substances are called ? |
| Answer» SOLUTION :`alpha-paricles` | |
| 684. |
When will be distilled books like distilled waters? |
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Answer» When the BOOKS aren’t great |
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| 685. |
de-Broglie wavelength ofamoving particle of kinetic energy K is given by the expression lamda=(h)/(sqrt(mK)). |
| Answer» SOLUTION :FALSE: `LAMDA=(H)/(SQRT(2mK))`. | |
| 686. |
ABCD is a square of side 2m. Charges of 5nC, + 10nC and -5nC are placed at corners A,B and C respectively .What is the workdone in transferring a charge of 5nC from 'D' to the point of intersection of the diagonals ? |
Answer» SOLUTION : In the Fig. .O. represents the point of intersection of the diagonals. AO=BO=CO=DO=`sqrt2` Total electric potential at .O. is `V_0=1/(4piepsilon_0)[q_1/(AO)+q_2/(BO)+q_3/(CO)]` `=9xx10^9[(5xx10^(-9))/sqrt2+(10xx10^(-9))/sqrt2-(5xx10^(-9))/sqrt2]` `=9xx10^9xx10^(-9)[5/sqrt2+10/sqrt2-5/sqrt2]` =9[3.536+7.072-3.536] `V_0`=63.648 volt Total electric potential at D is `V_D=1/(4pi in_0)[q_1/(AD)+q_2/(BD)+q_3/(CD)]` `=9xx10^9 [(5xx10^(-9))/2+ (10xx10^(-9))/(2SQRT2)-(5xx10^(-9))/2]` `=9xx10^9xx10^(-9)[5/2+5/sqrt2-5/2]` =9[25+3.536-2.5] = 31.824 volt `therefore` Potential difference between O and D is `V=V_0-V_D`=63.648-31.824=31.824 volt `therefore` Work required to transfer a charge of `5xx10^(-9)` C FROMD to O is `w=Vq=31.824xx5xx10^(-9)` `w=159.12xx10^(-9)` Joule . |
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| 687. |
25^(@) C is equal to...... |
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Answer» 298K |
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| 688. |
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall without falling when the floor is suddenly removed ? |
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Answer» Solution :The frictional force f (vertically upwards) opposes the weight MG. The man remains stuck to the wall after the floor is removed if `mg le f_(L) i.e., mg LT MU m R omega^(2)`. The minimum ANGULAR speed of rotation of the cylinder is `omega_(MIN)=sqrt((g)/(mu R))=sqrt((9.8)/(0.15xx3))=4.67 s^(-1)`. |
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| 689. |
A TV transmitting antenna is 125 m tall. How much service area can this transmitting antenna cover, if the receiving antenna is at the ground level? Radius of earth-6400 km |
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Answer» SOLUTION :`h_(T)=125M` and R=`6400xx10^(3)m` `d=SQRT(2h_(1)R)=sqrt(2xx64xx10^(5)xx125)=40xx10^(3)m`=40km Area covered A=`pid^(2)=3.14xx(40)^(2)=5024km^(2)` |
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| 690. |
Find the ratio of the number of the atoms of gaseous sodium in the state 3P to that in the ground state 3S at a temperature T==2400K. The spectral line corresponding to the transition 3P rarr 3S is known to have the wavelength lambda=589nm. |
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Answer» Solution :We have `(N)/(N_(0))=(g)/(g_(0))E^(-h omega//kT)=(g)/(g_(0))e^(-2pi h c//lambdakT)` Here `g=` degeneracy of the `3P` state`=6, g_(0)=` degeneracy of the `3S` state `=2` and `lambda=` wavelength of the `3P rarr 3S` line `((2pi ħc)/(lambda))=` energy DIFFERENCE between `3P & 3S` levels. Substitution gives `(N)/(N_(0))= 1.13xx10^(-4)` |
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| 691. |
A pulse is propagating on a long stretched string along its length taken as positive x-axis. Shape of the string at t = 0 is given by y=sqrt(a^(2) -x^(2)) when|x| le a=0when |x| > a What is the general equation of pulse after some time.t. , if it is travelling along positive x-direction with speed V ? |
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Answer» `y(x,t) = SQRT(a^(2) -(x+VT)^(2))`, when `|x|Vt| le a = x+ Vt` when `|x + Vt| ge a` |
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| 692. |
State Biot - Savart's law. |
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Answer» Solution :The magnitude of magnetic field `vec(DB)` at a POINT P at a distance r from the small elemental length taken on a conductor carrying CURRENT varies (i) directly as the strength of the current I (ii) directly as the magnitude of the length element `vec(dl)` (III) directly as the sine of the angle ( say , `theta`) between `vec(dl)" and " hat r` . (iv) inversely as the square of the distance between the point P and length element `vec(dl)`. This is expressed as `dB propto (Idl)/r^(2) sin theta` `dB = K (Idl)/r^(2) sin theta` where `K = mu_(0)/(4pi) `in SIunits and K = 1 in CGSunits. Is vector notation, `d vecB = ( mu_(0))/(4pi) (I vec(dl) xx hatr)/r^(2) ` Here vector `vec(dB)` is perpendicular to both I `vec(dl)` ( pointing the direction of currentflow) and the unit vector `hatr` directed from `vec(dl)` toward point P . |
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| 693. |
The following configuration of gates is equivalent to |
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Answer» `HAND` Output of `G_(1)=BAR(A.B)` Output of `G_(3)` `=(A+B).bar(A.B)=(A+B).(bar(A)+bar(B))` If `A=1` and `B=1`, then `A+B=1` `A+B=0` So, output is zero, If `A=0` and `B=0`, then `A+B=0` `bar(A)+bar(B)=1+1=1` So, output is zero. Clearly, the given combination is `XOR` gate. |
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| 694. |
In a transistor circuit, when the base current is increased by 50mu A keeping the collector voltage fixed at 2 V. the collector current increases by 1 mA The current gain of the transistor is |
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Answer» 20 |
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| 695. |
A magician during a show makes a glass lens with n = 1.47 disppear in a trough of liquid:-Could the liquid be water? |
| Answer» SOLUTION :No, the REFLECTIVE INDEX of WATER is 1.33 | |
| 696. |
The work done to move a charge on an equipotenital surface is |
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Answer» Infinity |
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| 697. |
Assertion: The loss of strength of a signal while propagating through a medium is known as attenuation. Reason: Transmitter helps to avoid attenuation. |
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Answer» |
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| 698. |
I-V characterstic of four devices are shown in Fig. 15.1 Identify devices that can be used for modulation: |
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Answer» `i` and `iii` Square law modulator is used for modulation purpose. Charactersitics shown by (`i`) and (`iii`) correspond to linear devices. And by (`ii`) and same part of (`iv`) corresponds to square law device which SHOWS non-linear relations. Hence (`ii`) and (`iv`) can be used for modulation. |
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| 699. |
Using the betatron condition, demonstaratethat thestrengthof the eddy-currentfield has the extremum magnitude on anequiolibrium orbit. |
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Answer» SOLUTION :The INDUCED electric field (or eddy CURRENT field) is given by, `E(r) = (1)/(2pi r) (d)/(dt) int_(0)^(r) 2pi r' (r') B (r') dr'` Hence, `(dE)/(dr) = -(1)/(2pi r^(2)) (d)/(dt) int_(0)^(r) 2pi r' B (r') dr' + (dB (r))/(dt)` Thisvansihesfor `r = r_(0)` by thebetatron condiction, where `r_(0)` is the radiusof theeuqlibrium orbit. |
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| 700. |
A planetof massm movesin a ellipticalorbitaroundan unknownstarof massM suchthatitsmaximuman minimumdistances fromthe starare equaltor_1 and R_2respectively. Theangularmomentumof theplanetrelativetothe centreof thestar is |
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Answer» `m sqrt((2GM r_1 r_2)/(r_1 +r_(2)))` ACCORDING to theof conservationof ANGULARMOMENTUM ` mv _1r_1 = mv _2 r_2 ` ` impliesv_2 = (v_1r_1)/(r_2)` FROMTHE lawof conservationof totalmechanicalenergy . `(-GMM)/(r_1)+(1)/(2)mv _1^(2) =-(GMm)/(r_2)+(1)/(2) mv_(2)^(2)` .... (ii ) FromEqs.(i )and (ii )we get `V_1 =sqrt((2Gmr_2)/((r_1 +r_2)r_1)` Angularmomentum, ` L = mv _1 r_1 = m (sqrt((2GMr_2)/((r_1+r_2)r_1)))xxr_1` ` L= m sqrt((2GMr_1 r_2)/(r_1 +r_2))` |
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