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.
| 9501. |
7.A particle moves from point A to point B as shown in the figure. Find the displacement of particle.(OA = 3 cm, OB = 4 cm) |
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| 9502. |
A faulty thermometer reads 5°C is melting iceand 99°C in dry system at normal pressure.Find the correct temperature in fahrenheitwhen termometer reads 52°C. |
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| 9503. |
what is the normal BP of human |
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Answer» 120/80mmhg human ka bp hota h Normal BP of human is 120/80 |
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| 9504. |
PAGE NoDATEConver sien factor behweos sr and CGSu nit 따-pres.care |
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Answer» by different method |
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| 9505. |
Aone litre flask contains some mercury. It isfound that at different temperatures the volumeof air inside the flask remain same. The volumeof mercury taken in the flask is (coefficient oflinear expansion of glass is 9x10-o C andcoefficient of volume expansion of Hg is-6/01.8x10/°C). EAMCET 2008M]1) 150ml 2) 750ml 3)1000ml 4)700ml |
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| 9506. |
4. The graph shows the change x in the lengtha wire by the application of a force Fatdifferent temperatures T and T, respectiveThen1)T, >1235 <T3) T, -T,4) T, is not equal to T, |
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Answer» 2 is correct answer. 2) is the right answer of the following |
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| 9507. |
4 While deciding the unit for heat,which temperatures interval ischosen? Why? |
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Answer» Unit of heat is independent of choice of temperature interval. Unit of heat is say Joule in S.I and Calories in C.G.S, is independent of temperature, temperature interval, temperature gradient etc. |
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| 9508. |
LOliverla speeu Ul 34RITET INTU 11179A particle is moving in a circular path of radius r. What will be the displacement after a half circle? |
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| 9509. |
नस्रप्कड 0y SOl उसने= ey O S ngj\ 2 नि रस ० Oपृ निगल जप, o mw % XoAQGD 0 9 m_g 5oo T PO PWNEsg w ey o\L S नहर्ठ-छ9 ¢Gy 5| s TOETR TR O :| |
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Answer» U= -15cmv= 5cmf= ?and to find a radius of curvature formula isR=2fso, using mirror formula 1/f=1/u+1/vputting values 1/f=1/-15+1/51/f=5+(-15)/751/f= -10/75cross multiple-10f=75f=75/-10f= -7.5So, R=2f2×-7.5R= -15 |
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| 9510. |
calulaleoteand(S/㏠|
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| 9511. |
A Q. 2. A compound microscope consists of an objectiveof focal length 1 cm and eyepiece of focal length5 cm separated by 12.2 cm. (i) At what distancefrom the objective should an object be placed sothat the final image is formed at least distance |
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| 9512. |
.6. (a) Define "power of accommodation of eye"(b) A person uses a lens of power -2.0D for correcting his distant vision. Porcorrecting his near vision he uses a lens of power +1.5D. What is the focal length ofthe lenses required to correct this disorder?(3M) |
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Answer» (a) The process by which the ciliary muscles change the focal length of an eye lens to focus distant or near objects clearly on the retina is called the power of accommodation of the eye. (b) For the first one P = -2.0D And we know that,P= 1/ f Therefore, f = 1/ P = 1/ -2.0 = -0.5m = -0.5 ×100 = -50 cmThis is a concave lens. Now for the second oneP = 1.5DP= 1/ f f = 1/ P = 1/ 1.5 = 2/3 × 100 = 66.6 cmThis is a convex lens. |
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| 9513. |
.6. (a) Define "power of accommodation of eye".(b) A person uses a lens of power -2.0D for correcting his distant vision. Forcorrecting his near vision he uses a lens of power +1.5D. What is the focal length ofthe lenses required to correct this disorder?(3M) |
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Answer» a. Power of accommodationis the ability of theeyelens to focus near and far objects clearly on the retina by adjusting its focal length.Power of accommodationof theeyeis limited. It implies the focal length of theeyelens cannot be reduced beyond certain minimum limit. P=1/f-2.0=1/ff=1/-2.0f=-0.5cm P=+1.5DP=1/f+1.5 D=1/ff=1/1.5f=0.66cm |
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| 9514. |
falls through a distagnitude 2.0 x 10' Nversed keeping its mthe same distance [Fse. Contrast the situa |
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| 9515. |
falls through a distagnitude 2.0 x 10' Nversed keeping its nthe same distance [Fse. Contrast the situa |
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| 9516. |
4 (a) Define power. What is its unit?ninl unit of energy? Est |
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Answer» Power is work done per unit time. SI unit of power is watts. |
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| 9517. |
A sound wave hos frequency 2 ikhz and wweiength 40 cm ina sven medium.Calculate the tme taken by 1t to travel 1.6 kmin a given medium. |
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Answer» Given: Frequency = n=2kHz= 2x1000 Hz Wavelength=lambda= 40cm=40/100=0.4m Speed of sound: V= frequency * wavelength =2×1000×0.4 =800m/s Speed = Distance/ time Distance= 1.6 km= 1.6x1000 m Time= distance/ speed = 1.6x1000/ 800 =16×100/800 =2sec Therefore time required to travel 1.6km is 2 sec. |
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| 9518. |
Datserly companionthe dist ance tHaulled by a body in the nh second is pusn byen inueloity and ocab -o the body |
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Answer» what type of answer is this |
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| 9519. |
(i) Phosphorus is very....non-metal. |
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Answer» Ans :- Phosphorus is very reactive non metal |
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| 9520. |
MODEL TEST PAPER-2Speed of a car is 20 m/s. How long will it take to cover the distance ofkm?19.36 |
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Answer» Speed = Distance / Time Time = Distance / Speed Speed = 20m/s Distance = 36 km = 36000 m Time = 36000/20 = 1800 s = 30 min If you find this answer helpful then like it. |
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| 9521. |
A coil of area 0.15 m13. 200 V.and 50 turns is normalto a magnetic field which changes from 5 x 10° to2 x 10 Wb m in 30 ms. Compute the emf inducedAns. 0.75 V.in the coil. |
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| 9522. |
21/25)4ΊWith the help of a cireuit diagram prove thatwhen a number of resistors are connected inparallel, the reciprocal of equivalent resistanceof the combination is equal to the sum of thereciprocals of the individual resistances of theresistorsFind the resistance between A and B in thefollowing network |
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Answer» those 2 resistance are in series so those will be added up 2+2= 4 ohmafter that there will be 2 resistanceswhose value is 4 ohm , 4 ohm each so these will be in parallel and the resultant will be 4*4/8= 2 ohm |
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| 9523. |
Two resistors 3 \Omega \text { and } 6 \Omega are combined in parallel. If the combination is connected to a battery ofemf 10 V and negligible internal resistance, determine the current drawn from the battery. |
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Answer» Resistors are in parallelso 1/R=1/3+1/6=(2+1)/6=3/6=1/2so R=2V=IRso current=10/2=5A |
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| 9524. |
34 There are n exactly identical resistorseach having resistance R. If the resultantresistance when joined in parallel is λ, thenon connecting them in series the resistancewill come out be(a) y(b) n2λ(d) n3λtoe hetween 4 and B isn3 |
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| 9525. |
40. A parallel combination of two resistors of 1 Ω each, isconnected in series with a 1.5 2 resistor. The totalcombination is connected across a 10 V battery. The currentDCE 2004]flowing in the circuit is |
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| 9526. |
A parallel combination of two resistors of 1 2 each, isconnected in series with a 1.5 2 resistor. The totalcombination is connected across a 10 V battery. The currentflowing in the circuit is[DCE 2004) |
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| 9527. |
3. n equal resistors are first connected in seriesand then connected in parallel. What is the ratioof the maximum to the minimum resistance?(a) n(b) 1/n^2c) n ^2d) 1 / n |
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| 9528. |
1.Figure shows a small block A of mass m kept at the left end of a plank B of massM 2m and length l. The system can slide on a horizontal road. The system isstarted towards right with the initial velocity v. The friction coefficients between theroad and the plank is 1/2 and that between the plank and the block is 1/4. Find (a) thetime elapsed before the block separates from the plank. (b) displacement of blockand plank with respect to ground till that moment ?the horizontal surface below the bigger blo |
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| 9529. |
2. Obtain an expression for the effective resistance when a number ofresistors are connected in parallel.3 a Anabi |
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Answer» 1/R=1/R1 +1/R2 +..... |
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| 9530. |
1.961284527712 cells of each emf 2V are connected in series among them, if 3 cells are connectedwrongly. Then the effective emf. Or the combination is1.18V2. 12V3.24V4. 6V |
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Answer» Option B is the correct answer. tq |
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| 9531. |
SECTION-BPulley Block System28. Figure shows a uniform rod of mass 3 kg and oflength 30 cm. The strings shown in figure arepulled by constant forces of 20 N and 32 N Theacceleration of the rod is10cm 20cm20N32N |
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| 9532. |
Q.17. Derive an expression for equivalent resistance of 3 resistors connected in parallel. ( |
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Answer» Parallel combinationHere three resistances are connected in parallel.In this case the potential difference across the ends of all the resistances will be same and it will be equal to the voltage of the battery used. |
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| 9533. |
9How will you conclude that the same potentialdifference (voltage) exists across three resistorsconnected in a parallel arrangement to a battery? |
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Answer» For parallel connectingthe resistors are all connected across the battery. So the same potential difference acts across all the three resistors and the potential difference is equal to the emf of the battery. by connecting in parallel combination is the correct answer of the given question |
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| 9534. |
A cell supplies a current of 1.2 A through tworesistors each of 2 2 connected in parallel. Whenthe resistors are connected in series, it supplies acurrent of 0.4 A. Calculate : (i) the internalresistance, and (ii) e.m.f. of the cell.Ans. (i) 0-5 2. (i) 1-8 V |
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| 9535. |
Thrae qual resistors Connected in Sariesacxass a batter dissipate 10w PawerWhat will be-the-Pawer dissipated.tthoSame resis tor ae connecteal in parll-across the . Same battary |
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Answer» When the three resistors are in series, effective resistance in the circuit = 3R.Power dissipated in the series circuit = Pseries= V^2/(3R) = 10 W ⇒V^2/(R) = 30 WWhen the three resistors are in parallel, effective resistance in the circuit = R/3.Power dissipated in the parallel circuit = Pparallel= V^2/(R/3) = 3V^2/(R) = 3 x 30 W = 90 W. |
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| 9536. |
a 0.5mL27.From given groups of animals an open circulatory system is present in which of the followine?a) Molluscab) Arthropodac) Coelenteratad) Annelida(ii) a and c(i) a and b(iii) b and o(iv) b andd |
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Answer» (i) a and b. This is the correct option because only Mollusca and Anthropoda have open circulatory system. Please hit the like button if this helped you Thank you very much |
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| 9537. |
Scombinayenon tuos a resiswrould conneet three reststors, each of resistanece 6 12, so that thetanee of u 9 2.( 42. |
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| 9538. |
24. Mass of an object is 30 kg. What is its weight on theearth?(A) 29.4 N(C) 5 N(B) 294 N(D) 30 N |
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Answer» On Earth,g=9.8m/s2. So, the object's weight on Earth will be: W=30kg⋅9.8m/s2 =294 N |
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| 9539. |
53. Determine the sine of the angle between thevectors (3i + 2j +4k) and (2i -2j- 4k). |
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Answer» Dot product of two vectors is (3i+2j+4k)•(2i-2j-4k) = (6-4-16) = -14 now the magnitude of these vectors are √(3)²+(2)²+(4)² = √9+4+16 = √29 and √(2)²+(-2)²+(-4)² = √24 = 2√6 now angle between them is cos∅ = dot product/(magnitude1*magnitude2) = -14/(2√6*√29) = -7/√174 ∅ = cos-¹(-7/√174) |
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| 9540. |
7. A partide moves from position T -(3i+2j-6k) m to position i -(14j+13+9km under the action of forea(4i+ j+3k) N. Find the work done. |
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Answer» 100 J is the answer |
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| 9541. |
If A = i-2j +3k.B=41+4j-4k andC-5-8jt9k then A+ B+0 is |
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Answer» A+B+C = i(1+4+5) +j(-2+4-8) +(3-4+9)k = 10i-6j+8k|A+B+C|= √(10)²+(-6)²+(8)² = √200 = 10√2 |
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| 9542. |
(OQ The accelerations of a particle as seen from two framess of a particle as seen fromS, and S, have equal magnitude) 4 m/s(a) The frames must be at rest with respect to eachother.(b) The frames may be moving with respect to each otherbut neither should be accelerated with respect to theother.(c) The acceleration of S, with respect to S, may eitherbe zero or 8 m/s.(d) The acceleration of S2 with respect to S, may beanything between zero and 8 m/s 2 |
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Answer» suppose the acceleration of particle is a3and for frame s1 , it is a1. and for frame s2 , it is a2 now from frame 1.the magnitude of acceleration of particle is 4m/s²so, |a3-a1| = 4 from frame 2.the magnitude of acceleration of particle is 4m/s²so, |a3-a2| = 4 on adding we can get max acceleration of 8 and minimum acceleration of 0.since acceleration is added and subtracted vectorily so the relative acceleration between s1 and s2 can be in between 0 to 8. option D |
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| 9543. |
10. The velocity (in m/s) of an object changes from Vi-10i +4j+2k to v2 -4i 2j 3k in5 second. Find the magnitude of average acceleration |
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| 9544. |
I rod AB of mass m and length initially standsvertically with the end B above end A. The end A of therod is pivoted. The rod is free to rotate is a vertical planeabout a horizontal axis passing through the end A. X andy directions are as shown. Y-axis is parallel to the rod.The rod is then given a negligible push to a side and the rodrotates. When the rod has turned through an angle of 90°a) the angular speed of the rod isb) the I - component of the pivot reaction isthe Y-component of the pivote reaction is 4mgd) the X-component of the pivote reaction is in thenegative I - directionROUGH WORK |
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| 9545. |
Give three examples to show the diffĂŠerence between changes that can be reversed andchanges that cannot be reversed. |
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| 9546. |
Weigon the surface of the Earht of an object changes from place to placeth |
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Answer» Weight of an object varies from place to place on earth due to variation in gravity at different places. There are factors due to which acceleration due to gravity may vary. Like altitude, latitudes and even the spinning of our planet may vary acceleration due to gravity at different places on earth. |
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| 9547. |
projection is v(2g(b+a/16 b).3.) A 50 g ball hits a wall at an angle 45° with speed 20 m/s and rebounds with the same speedperpendicular to the direction of incidence. Find the impulse.i) A car is moving with a speed x m/s can be stopped by a force in distance y. If mass of the car isincreased by 50% what would be new stopping distance in terms of force remaining sameOrii) Find a vector whose magnitude is 10 and which is perpendicular to the vectors A:2H3-4k andboth.RSi Ru in |
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Answer» Impulse I = change in momentum I= mv’-mv we will take vertival and horizontal component angle theta = 45’ I= (mvcos45°-mvsin45°)-(mvcos45°-mvsin45°) vertical resolution cancelout & horizontal adds up I=2mvsin45 I=2*0.05*20*1\root2 = 1\root2 N-s please specify the question that you want solution for |
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| 9548. |
dydx22. IfyFX2 loge, then find the value of |
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| 9549. |
two co facial circular loop L1 and L2 of radii 3cm and 4cm .what should be the magnitude and direction of current in the loop 2 so that the net magnetic field at the point o is zero |
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| 9550. |
111. A sphere of mass m moving with a constant velocityhits another stationary sphere of the same mass. If‘e, is the coefficient of·restitution. Then the ratio ofthe velocities of the two spheres after collision willbe1+ eb)c)d)1+e |
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