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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.
| 2401. |
telephones, TV, radio ransmission, satelite, functioning are examples of? |
| Answer» SOLUTION :COMMUNICATION | |
| 2402. |
The first & second stage of two stage rocket separately weigh 100 kg and 10 kg and contain 800kg and 90kg fuel respectively. If the exhaust velocity of gases is 2 km/sec then velocity of rocket after second stage is (nearly) (log_(10)5 = 0.6990) (neglect gravity) |
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Answer» `7.8 XX 10^(3)` m/s |
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| 2403. |
Maximum frequency of emission is obtained for the transition |
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Answer» n = 2 to n = 1 |
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| 2404. |
The current I through a rod of a certain metallic oxide is given by I = 0.2 V^(1//2) , where V is the potential difference cross it. The rod is connected in series with a resistance to a 6 volt battery of negligible internal resistance. What value should the series resistance have so that : A) the current in the circuit is 0.4 A B) the power dissipated in the rod is twice that dissipated in the resistance. |
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Answer» `10 OMEGA, 10 Omega` |
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| 2405. |
If mu_(0) be the magnetic permeability are epsilon_(0) the electric permittivity of free space then speed of em waves in freee space is given as : c = ______________. |
| Answer» SOLUTION :`(1)/(SQRT(mu_(0)epsilon_(0)))` | |
| 2406. |
A pn junction diode in which current carriers are generated by |
| Answer» Answer :A | |
| 2407. |
For a transistor , beta = 100 . The value of alpha is |
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Answer» 0.01 `ALPHA = BETA/(1+beta)=100/01=0.99` |
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| 2408. |
barAand barBare twovectorswhichactatapoint simultanouslyso thattheyformtheadjacentsidesof aparallelrgrams.ifbarCandbarDare thediagonalsof theparallelogram findthearea ofparallelogram . |
| Answer» SOLUTION :`| vecAxx VECB |, 1/2 | vecCxx VECD|` | |
| 2410. |
If a size of particle is a and wavelength of light islambda , for a lt lt lambda scattering is directly proportional to ....... |
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Answer» `(1)/(lambda^4)` |
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| 2411. |
The Balmer series for the H-atom can be observed ... |
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Answer» if we measure the FREQUENCIES of light emitted when an excited atom falls to the ground state. `rArr` THUS, options (B) and (D) are correct. |
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| 2412. |
In a communication system noise does not affect the signal |
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Answer» at the transmitter |
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| 2413. |
A particle of charge q, mass m starts moving from origin under the action of an electric field vec(E)=E_(0)hat(i) magnetic field vec(B)=B_(0)hat(k).Its velocity at (x,0,0) is 6hat(i)+8hat(j). The value of x is |
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Answer» `(25M)/(qE_(0))` |
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| 2414. |
निम्नलिखित कोण किस चतुर्थांश में हैं? ज्ञात कीजिए । 320^(@) |
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Answer» । चतुर्थांश |
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| 2415. |
A coil of 100 turns carries a current of 5 A and creates a magnitude flux of 10^(-5) T m^(2) per turn. The value of its inductance is |
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Answer» 0.05 mH Total magnetic flux `Nphi_(B)=LI` `RARR L=(N phi_(B))/(I)=(100 xx 10^(-5))/(5)=2 xx 10^(-4) H=0.20mH` |
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| 2416. |
An object is placed at a distance of 21 cm from a concave mirror of radius of curvature 10 cm. Now a glass slab of thickness 3 cm and refractive index 1.5 is placed between the object and the mirror. Determine the position of the final image. The distance of the nearer face of the slab from the mirror is 1 cm. |
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| 2417. |
The logic gates giving output '1' for the inputs of '1' and '0' are |
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Answer» AND and OR |
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| 2418. |
Eddy currents were invented by ...... scientist. |
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Answer» Lenz |
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| 2419. |
A diver under water, looks obliquely at a fisherman standing on the bank f a lake. Would the fisherman look taller or shorter to the diver than what he actually is ? |
| Answer» SOLUTION : TALLER | |
| 2420. |
The production of X-rays is a phenomenon for the conservation of: |
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Answer» MASS into POTENTIAL ENERGY |
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| 2421. |
Which of the following quantities remain constant in a step down transformer ? |
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Answer» current |
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| 2422. |
An object is placed at 20 cm in front of a concave mirror produces three times magnified real image. What is focal length of the concave mirror ? |
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Answer» 15 cm `-3f + 60 = f` `f = -15 cm` SINCE MIRROR is concave f = 15 cm |
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| 2423. |
The concentration of electrons in a semiconductor is 3 xx 10^(13)//cm^3and hole concentration is 5 xx 10^(14) //cm^3. The semiconductor is |
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Answer» n-type |
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| 2424. |
Find the equivaent capacitance of the circuit between point A and B. |
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Answer» 2C |
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| 2425. |
Two wires of equal lengths are bent in the form of two loops. One of the loops is square shaped whereas the other loop is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reason. |
| Answer» Solution :For a WIRE of given LENGTH, the circular loop has GREATER area than the square loop. So the circular loop will EXPERIENCE greater TORQUE in the magnetic field, because torque a area of the loop. | |
| 2426. |
The number of turns in a coil of Galvanometer is tripled, then |
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Answer» Both VOLTAGE and CURRENT SENSITIVITY REMAINS constant |
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| 2427. |
What do you understand by "hysterisis"? How does this property infulence the choice of materials used in different appliances where electromagnets are used ? |
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Answer» Solution :Cycle of magnetisation : When a ferromagnetic SPECIMEN is slowly magnetised, the intensity of magnetisation varies with magnetic field through a cycle is called cycle of magnetisation. (2) Hysterisis : The LAGGING of intensity of magnetisation (I) and magnetic induction (B) behind magnetic field intensity (H) when a magnetic specimen is subjected to a cycle of magnetisation is called hysterisis. (3) Retentivity : The value of 1 for which H = 0 is called retentivity or residual magnetism. (4) Coercivity : The valueof magnetising force required to reduce I is zero in reverse direction of H is called coercive force or coercivity. (5) Hysterisis curve : The curve represents the relation between B or I of a ferromagnetic material with magnetising force or magnetic intensity H is known as Hysterisis curve. (6) Explanation of hysterisis loop or curve : (a) In fig, a closed curve ABCDEFA in H - I plane, called hysteris loop is shown in fig. (b) When ferromagnetic specimen is slowly magnetised, I increases with H. (c) Part OA of the curve shows that I increases with H. (d) At point A, the value I becomes constant is called saturation value. (e) At B, I has some value while H is zero. (f) In fig. BO represents retentivity and OC represents coercivity. (7) Uses : The properties of hysterisis curve, i.e., saturation, retentivity, coercivity and hysterisis loss help us to choose the material for specific purpose. (i) Permanent magnets : A permanent magnet should have both large retentivity and large coercivity. Permanent magnets are used in galvanometers, voltmeres, ammeters, etc. (ii) An electromagnet core : The electromagnet core material should have maximum induction field B even small fields H, low hysterisis loss and high initial permeability. (iii) TRANSFORMER cores, Dynamocore, Chokes, Telephone DIAPHRAGMS : The core material should have high initial permeability, low hysterisis loss and high specific resistance to reduce eddy currents. Soft IRON is the best suited material. 
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| 2428. |
Which of the following represent gamma -decay ? |
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Answer» `""^(A)X_(Z)+GAMMA=""^(A)X_(Z-1)+a+b` |
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| 2429. |
Which of the following is not true about electric charge ? |
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Answer» CHARGE is a scalar quantity |
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| 2430. |
The author was away from his grandmother for how many years ? |
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Answer» 10 years |
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| 2431. |
An AC source producing emf V=V_(0) "["sin omega t+sin 2omegat"]" is connected in series with a capacitor and a resistor. The current found in the circuit is |
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Answer» `i_(1)-i_(2)` |
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| 2432. |
The angle of dip at place is delta. If dip is measured in a plane making an angle theta with the magnetic meridian, the apparent angle of dip delta_(1) will be equal to |
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Answer» `tan_(-1) (tan delta SEC THETA)` Again, `tandelta_(1) = V/(H. cos theta) = (tandelta)/(cos that)=tan delta sec theta` `implies delta_(1) = tan^(-1) (tantheta sec theta)` |
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| 2433. |
An inductor of reactance 1 Omegaand a resistor of 2Omega are connected in series to the terminals of a 6 V (rms) a.c. source. The power dissipated in the circuit is |
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Answer» 8 W `lt P gt =V_"rms" I_"rms" cos phi` …(1) but `I_(rms)=V_(rms)/Z` but `Z=sqrt(R^2+X_L^2)=sqrt((2)^2 + (1)^2)=sqrt5Omega` `therefore I_(rms)=V_(rms)/Z=6/sqrt5`A ….(2) cos `phi=R/Z=2/sqrt5`...(3) Putting the values of equation (2) and (3) in equation (1), `lt P gt =6xx6/sqrt5xx2/sqrt5` `=72/5` `therefore lt P gt` =14.4 W |
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| 2434. |
Figure shows a wire ring of radius a that is perpendicular to the general direction of a radially sysmmetric, diverging magnetic field. The magnetic field at the ring is everywhere of the same magnitude B, and its direction at the ring everywhere makes an angle thetha with a normal to the plane of the ring. thge twisted lead wires have no effect on the problem. Find the magnitude of the force the field exerts on the ring if the ring carries a current i. |
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| 2435. |
Two long straight parallel wires A and B are placed 50 cm apart and carry currents 20 amp and 15 amp respectively. A point 'P' is 40 cm from wire A and 30cm from wire B. Find the magnitude of the resultant magnetic field at 'P'. |
| Answer» SOLUTION :`SQRT(2) XX 10^(-5) T` | |
| 2436. |
A bullet (m_(1) = 25 g) is fired with a velocity 400 ms^(-1) get embedded into a bag of sand (m_(2) = 4.9 kg) suspended by a rope. The velocity of the bag is nearly : |
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Answer» `0.2 ms^(-1)` `(25)/(100)xx400 = (4.9 +0.025) upsilon` `upsilon=(10)/(4.925) = 2.030 ~~ 2 mg^(-1)` (d) is the choice. |
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| 2437. |
यदि X=sqrt2-1 हो तो 1/x का मान होगा |
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Answer» `sqrt2-1` |
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| 2438. |
In a transformer, the number of turns in the primary and secondary coils are 1000 and 3000 respectively. If the primary is connected across 80V AC, the potential difference across each turn of the secondary will be |
| Answer» Answer :D | |
| 2439. |
Which is the instrument to detect the charge on the body ? Explain it with diagram. |
Answer» Solution :A SIMPLE apparatus to detect charge on a body is the simple gold-leaf electroscope as shown in FIGURE.  It consists of a vertical metal rod housed in a box, with two thin gold leaves attached to its bottom end. When a charged OBJECT touches the metal knob at the top of the rod, charge flows on to the leaves and they diverge. The degree of divergence is an indicator of the AMOUNT of charge. |
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| 2440. |
Draw a labelled ray diagram showing the formation of a final image by a compound microscope at the least distance of distinct vision. |
Answer» Solution :RAY diagram of COMPOUND MICROSCOPE for FORMATION of image at the near point of the eye. 
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| 2441. |
A solid spherical ball and a hollow spherical ball of two different materials of densities p_(1) and p_(2) respectively have same outer radii and same mass. What will be the ratio the moment of inertia (about an axis passing through the centre) of the hollow sphere to that of the solid sphere? |
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Answer» `(p_(2))/(p_(1))(1-(p_(2))/(p_(1)))^(5/3)` ALSO, Rsolid = Rhollow = R (SAY) and `M_("solid") = M_("hollow")` `M=4/3pip_(1)R^(3)` `=4/3pip_(2)[R^(3)-R_("INNER")^(3)]rArrR^(3)-R_("inner")^(3)=(p_(1))/(p_(2))R^(3)` `rArr_("inner")^(3)=R^(3)[1-(p_(1))/(p_(2))]` `rArrR_("inner")=R[1-(p_(1))/(p_(2))]^(1//3)` ..(i) Now, ratio of their moment of inertia about the central axis will be `(I_("hollow"))/(I_("solid"))=(4/3piR^(3)p_(2)xx2/5R^(2)-4/3piR_("inner")^(3)p_(2)xx2/5R_("inner")^(2))/(2/5xx4/3piR^(3)p_(1)xxR^(2))` `=(p_(2))/(p_(1))[(R^(5)-R_("inner")^(5))/(R^(5))]=(p_(2))/(p_(1))[1-((R_("inner"))/R)^(5)]` `(I_("hollow"))/(I_("solid"))=(p_(2))/(p_(1))[1-[1-(p_(1))/(p_(2))]^(5//3)]` [From equation (i)] 
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| 2442. |
One end of a heavy uniform rod AB can slide along rough horizontal guiding surface CD with the help of massless ring as shown in the figure. The dimension of ring is negligible. BE is the ideal string if angleEBA is right angle and alpha is the angle between rod AB and horizontal when the rod is on the verge of sliding. Find the coefficient of friction between ring and horizontal guiding surface CD. |
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Answer» `(tanalpha)/(2+tan^(2)ALPHA)`  SINCE BODY is in equilibrium under the influence of three forces only so they must e CONCURRENT. USING Lami's theorem we can write `tanalpha=(L)/(2x)` and `tan(alpha+lamda)=(L)/(x)` `tan(alpha+lamda)=2tanalphaimplies(tanalpha+tanlamda)/(1-tanalphatanlamda)=2tanalpha` `impliestanalpha+tanlamda=2tanalpha-2tan^(2)alphatanlamda` `impliestanlamda=mu=(tanalpha)/(1+2tan^(2)alpha)` |
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| 2443. |
The wave nature of electron was verified by |
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Answer» PHOTOELECTRIC EFFECT |
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| 2444. |
Consider a sphere of radius R having charge q uniformly distributed insider it. At what minimum distance from surface the electric potential is half of the electric potential at its centre? |
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Answer» R |
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| 2445. |
A closely wound solenoid of 800 turns and area of cross-section 2.5 xx 10^(-4) m^2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? |
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Answer» Solution :Here N= 800 , A = `2.8xx10^(-4) m^2 ` and I = 3.0 A `therefore ` MAGNETIC MOMENT of solenoid m = NIA =` 800xx3.0xx2.5xx10^(-4)=0.60 A m^2` The axis of the solenoid acts like the axis of a bar magnet . One end of solenoid BEHAVES as N-pole and the other end as S-pole . Exact direction of FIELD can be determined by applying the right hand rule. |
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| 2446. |
Only circular part of the wire shown in the figure has resistance sigam per unit length. The radii of the circles are r and R and /_POQ=alpha (in radian). If a potential difference V_(P)-V_(Q)=V is applied across the point P and Q, the magnetic field at point O is |
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Answer» `(mu_(0))/(4pi)(V)/(SIGMA)[(2pi-ALPHA)/(r^(2))-(alpha)/(R^(2))]` |
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| 2447. |
lim_(Nrarroo) sum_(r=1)^n (2r-1)/2^ris equal to :- |
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Answer» Solution :`S=1/2+3/2^2 + 5/2^3 + ….OO` `S/2=1/2^2+3/2^3 + …… oo` `S/2=1/2+2(1/2^2+1/2^3+….oo)` `=1/2+1/2 2 RARR S=3` |
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| 2448. |
Three projecties A , B and C are projected at an angle of 30^(@) , 60^(@) respectively , If R_(A), R_(B) and R_(C)are ranges of A, B and C respectively then (velocity of projection is same for A, B and C |
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Answer» `R_(A)=R_(B)=R_(C)` Range is same for both `30^(@)AND60^(@)` as they are complementary angles. `therefore R_(A)=R_(C )ltR_(B)` |
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| 2449. |
Explain free electron model of conduction. |
| Answer» Solution :We know that electrons from OUTERMOST orbit which participate in bonding phenomenon during the formation of a crystal. Electrons from outermost orbits can be TWO types, namely valence electrons and conduction electron. The energy level of valence electron is lower and it remains BOUND to the nucleus. valence electrons are responsible for bonding. The energy level of conduction electrons is high. Conduction electrons cannot reamain bound to a particular nucleus. Conduction electrons can move freely inside the volume of material . energy of free electron is higher but it is not sufficient to escape from the system because when one conduction electron escapes, then the remaining system attains positive CHARGE and attracts back the electron into the system. These conduction electrons are know as free electrons, metallic conduction conduct electricity through them. Insulating MATERIALS do not have sufficient free electrons to cannot electricity through them . | |
| 2450. |
Oil flows in a pipeline at a speed of 0.8 m/s. The oil flow rate is 2 xx 10^3 tons/h. Find the diameter of the pipeline. 19.2. The internal diameter of a nozzle is 2 cm. It emits |
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