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.
| 7951. |
A cylindrical bar magnet is kept along the axis of a circular coil and near it as shown in figure.The magnet is rotated in case (a) about its own axis and in case (b) about axis perpendicular tothe length of magnet. In which case will there be an induced e.m.f. at the terminals of the coil?29.CoilCoilNScase (a)case (b)(1) case (a)(3) both case (a) and case (b)(2) case (b)(4) Neither case (a) nor case (b) |
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Answer» We know magnet is placed along the axis of the circular coil, hence axis of the magnet is parallel to the axis of the circular coil.1) When magnet is rotated about its own axis then there will be very negligiblechange in the flux linked with the coil. Hence EMF induced is less. 2) IN the second case as magnet is rotated in perpendicular direction then flux linked with the coil decreases then become zero and then again increases and reached to the maximum value.Hence emf induced is more in this case.option 3 |
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| 7952. |
intibiotics., affecS aoSincase of bactehiothe case |
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Answer» Bacteria have cell wall. Antibiotics block the formation of cell wall in bacteria which make them weak. therefore, bacteria die. Virus don't have cell walls. That's why antibiotics are not effective against them. |
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| 7953. |
ーーwhat is Trop dene matchin-in-transformedTmedene matching intonoeり,'' |
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Answer» Inelectronics,impedance matchingis the practice of designing theinput impedanceof anelectrical loador theoutput impedanceof its corresponding signal source tomaximize the power transferor minimizesignal reflectionfrom the load. In the case of a complex source impedanceZSand load impedanceZL, maximum power transfer is obtained whenzs= zl where the asterisk indicates thecomplex conjugateof the variable. WhereZSrepresents thecharacteristic impedanceof atransmission line, minimum reflection is obtained when |
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| 7954. |
Which of the following is a poisonous fish?hammerhead sharkgoldfishlion fishkiller whale |
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Answer» lion fish |
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| 7955. |
A circular disc of radius R and thickness has moment of inertia /about an axis passing through its centre andperpendicular to its plane. It is melted and recasted into a soid sphere. The moment of inertia of the sphere about itsdiameter as axis of rotation is21(D) 10 |
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| 7956. |
32.A circular disc of radius R and thicknessand perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of thesphere about its diameter as axis of rotation isAlhas moment of inertia r about an axis passing through its CentreB)c)D)1o |
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| 7957. |
length ofand A-B where A-+3 k |
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Answer» Given A = 3i+2j and B = i-2j+3k so, A+B = 4i+3k and A-B = 2i+4j-3k |
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| 7958. |
12. Calculate the radius of gyration of a cylindrical rodof mass M and length L about an axis of rotationperpendicular to its length and passing through its[MNREC 96](Ans. K = LN12)centre. |
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Answer» √12 = 2√3 |
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| 7959. |
LOS44. Calculate the equivalent capacitance between thepoints A and B in the given circuit. If 100 V PD isestablished between A and B then what will be thecharge on the capacitor closest to?3 ALF3 jLF3 LF3 uF3uF |
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| 7960. |
lan6 VI)The resultant resistanceii) The total current) The voltage across 7 Ω resistanceii |
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Answer» i) Resultant R = 7+(10*5)/(10+5) = 7+50/15 = 7+10/3 = 31/3 Ω ii) Total current = 6v/(31/3) = 18/31Ω iii) V across 7Ω = i*r = 7*18/31 = 126/31 V |
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| 7961. |
What is the value of unknown resistor R in Fig. if the voltage drop across the500 Ί resistor is 2.5 V? All resistances are in ohms.A5505012 V500 |
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Answer» drop on 50 ohm resistance=2.5*50/500=0.25 vdrop across cd=2.5+0.25=2.75vdrop across 550 ohm=12v-2.75v=9.25 vi=9.25/550=0.0168 Ai2=2.5/500=0.005Ai1=0.0168-0.005=0.0118AR=2.75/0.0118=233 ohm |
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| 7962. |
The voltage across a wire is (100 ± 5)V andthe current passing through it is (10±0.2)A. Find the resistance of the wire.Solution |
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| 7963. |
lan.r Vel) The resultant resistanceii) The total currentiii) The voltage across 7 Ω resistance |
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Answer» i) Resultant R = 7+(10*5)/(10+5) = 7+50/15 = 7+10/3 = 31/3 Ω ii) Total current = 6v/(31/3) = 18/31Ω iii) V across 7Ω = i*r = 7*18/31 = 126/31 V |
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| 7964. |
1. What is the charge stored when the voltage acrossa 50 uF capacitor is 9 V? |
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| 7965. |
ProblemFind the value of voltage across a 10 ohms resistor in thecircuit shown (Fig 4) when the current of 2 amps flowsthrough the 10 ohms resistor.SolutionVoltage across 10 ohmsV = IXR= 2 x 10 = 20 volts. |
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Answer» 20 is answer to question 20 volts is the correct answer of the given question |
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| 7966. |
Taking frequency f, velocity v and density p to be thefundamental quantities, then the dimensional formula formomentum will be(a) [pvtf ](c) [ pvf ](b) [ pv/(d) [pv2f]3 f-1 |
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| 7967. |
Calculate the energy transferred by a 5 A current flowing through a resistor of2 ohms for 30 minutesSolution. Wewill first calculate the power by using the given values of current and resistance. This can |
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Answer» Cᴜʀʀᴇɴᴛ => 5 ᴀᴍᴘᴇʀᴇꜱRᴇꜱɪꜱᴛᴀɴᴄᴇ => 2 ᴏʜᴍꜱ Pᴏᴡᴇʀ => (5)² × 2 =25 × 2=30 ᴡᴀᴛᴛꜱ=50 / 1000 ᴋW= 0.05 ᴋᴡTɪᴍᴇ=> 30 ᴍɪɴ=30 / 60=½ ʜʀꜱ= 0.5 ʜʀꜱEɴᴇʀɢʏ = P × ᴛ= 0.05 × 0.5=0.025 Kᴡʜ |
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| 7968. |
What are the fundamental units in S.I. system ? Namethem along with their symbols. |
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Answer» The SI base units and their physical quantities are themetreformeasurementoflength, thekilogramformass, thesecondfor time, theamperefor electric current, thekelvinfortemperature, thecandelaforluminous intensity, and themolefor amount of substance. |
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| 7969. |
If force (F), velocity (V) and time (T) are taken as fundamental units, then find the dimensions of mass |
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Answer» We know that force = mass*acceleration = mass*V/t SO, F = m*(V)/t => m = F-t/V = Ft/V |
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| 7970. |
2. Differentiate fundamental units and derived unitsby giving examples. Write the popular units. |
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Answer» The fundamental units defined by the International System of Units. They are not dependent upon any other units, and all other units are derived from them. Example: meter, kilogram A derived unit is anSIunit of measurement made of a combination of theseven base units. Example: Newton which is m×kg/sec². can u explain more about this fundamental units are something that is taken as convection. But derived units are a combination of fundamental units. see above-given examples, how newton is derived. |
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| 7971. |
τη:Two identical resistors of 20 Ω each are connected in parallel. This combination, in turn,is connected to a 10 Ω resistor. The equivalent resistance of the combination will be: |
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Answer» Series nR = 2×20 =40ohm parallel R/2 = 20/2 = 10ohm Ratio = 40/10 = 4:1 wrong |
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| 7972. |
The three blocks shown move with constant velocities. Find the velocity of block A and B2应直 |
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Answer» (Va + Vp2)/2 = Vp1 ( that is 0) => Va = -Vp2 = -10m/s ↑ now (Vb + Vc)/2 = Vp2=> Vb = 2(Vp2) - Vc = 2*(10) -2 = 18 m/s ↑ |
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| 7973. |
215 You have three identical resistance than usingall of them how many maximum combination youcan madee(1) 2 (2) 3 4 (4) 5 |
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Answer» arrangement 1: when all connected in parallel.. arrangement 2: when all connected in series... arrangement 3:: when 2 connected in series...in parallel with 3rdone... arrangement :4 when 2 connected in parallel..in series with 3rdone. thanks |
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| 7974. |
जनटर %Xy 3106 (XL—I-33) 5 |
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Answer» (x + 3)/(x-3) + 6(x-3)/(x+3) = 5 Let (x+3)/(x-3) = y y + 6/y = 5 y^2 + 6 = 5y y^2 - 5y + 6 = 0 y^2 - 3y - 2y + 6 = 0 y(y - 3) - 2(y - 3) = 0 (y - 2)(y - 3) = 0 y = 2, 3 For y = 2x + 3/x - 3 = 2x + 3 = 2x - 6x = 9 For y = 3x + 3/x - 3 = 3x + 3 = 3x - 92x = 12x = 12/2 = 6 |
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| 7975. |
“ 33 5: shern P ey v 0070 vegu i |
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Answer» work done =change in kinetic energy This is the work energy theorm. The relationship between Final KE abetween work done on a body and its kinetic energy is very nicely captured byWork-Kinetic Energy Theorem. This theorem essentially says that if a force is exerted on a body, the force does work on the body (often described as a dot product of Force and displacement). Then this work done is equal to the difference and initial KE. Or Work Done by a Force, W = KE (final) - KE (initial). |
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| 7976. |
16. Three blocks A, B and C are suspended as shown in the" figure. Mass of cach blocks A and C is m. If system isin cquilibriurn and mass of B is M, then:(A) M 2 m(C) M> 2 m(B) M<2 m(D) M mtion is shown in |
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| 7977. |
m-2kgThree identical blocks of masses m-2 kg are dra force F 10-2 N with an acceleration of 0-6 m 21byfrictionless surface [Fig 1.66]. What is the tension (iin the string between the blocks B and C? Ans.78NaredrawnPA |
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Answer» thanku so much |
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| 7978. |
A circular platform is mounted on a frictionlessvertical axle. Its radius R 2 m and its moment ofinertia about the axle is 200 kg m2. It is initially atrest. A 50 kg man stands on the edge of theplatform and begins to walk along the edge at thespeed of 1 ms-1 relative to the ground. Time takenby the man to complete one revolution is[AIPMT (Mains)-2012]3π(1) t S(2) S2(3) 2Tt S(4) SA 2 |
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Answer» according to law of conservation of momentum |
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| 7979. |
The average magnitude of force necessary to stopa hammer moving with 30 Ns momentum in 0.5 sis(2) 10 N(3) 30 N(4) Infiniteau |
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Answer» Force = momentum /timeMomentum = 30 Ns, time =.5 s Force = 30/.5 = 300/5 = 60 N Correct option is (1) |
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| 7980. |
Distance of the centre of mass of a solid uniform conefrom its vertex is z. If the radius of its base is R andits height is h then z, is equal to(a)5h/8(b)3h^2/8R(c) h^2/4Rd) 3h/4 |
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Answer» The answer is Option 2. |
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| 7981. |
Example 5. A particle is projected with velocity2 gh, so that itjust clears two walls of equal height h, which are at a distanceof 2h from each other. Show that the time interval of passingbetween the two walls is 2 |
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| 7982. |
If the angle of incidence and angle ofemergence of a light ray falling on a glass slabare i and e respectively, prove that, i- e.Ans. |
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| 7983. |
Prove the following statements.a. If the angle of incidence and angle ofemergence of a light ray falling on aglass slab are i and e respectively, provethat, i e. |
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Answer» Assume ABCD to be the glass plate with sides AB parallel to CD. Let the refractive index of air and glass be n1and n2. Consider a ray of light EF incident on surface AB of the glass slab with angle of incidence,∠i.The light ray is refracted into the glass slab along the path FG and towards the normal, NN’ .∠r1is the angle of refraction.The light ray is refracted out of the glass slab at the surface CD along the path GH. The angle of incidence at this surface is∠r2. The light ray emerges out of the slab forming∠e as angle of emergence. The surfaces AB and CD are parallel and FG is the transversal,Therefore,∠r1=∠r1 Applying Snell’s Law at the surface AB,sinisinr1=n2n1.......(1)Similarly, at surface CD, according to Snell’s Lawsinr2sine=n1n2.......(2)Multiplying equation 1 and 2, sinisinr1×sinr2sine=n2n1×n1n2sinisinr1×sinr2sine=1 As AB is parallel to CD, the perpendiculars to the AB and CD, NN’ and MM’ are also parallelTherefore, alternate interior angles, r1and r2are congruent i.e.r1=r2Hence,sinisinr1×sinr1sine=1sinisine=1Therefore,sini=sineor,i=e Hence, angle of incidence equals angle of emergence. |
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| 7984. |
, Prove the following statements.If the angle of incidence and angle ofa.emergence of a light ray falling on aglass slab are i and e respectively, provethat, i e. |
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Answer» Let ABCD is the glass slabAssume that the air Index and glass index = n₁ and n₂From the diagram ,Consider a light ray WX makes the∠i on the surface AB.Reflected light XY to the Normal NN' makes the angle ∠r₁Refracted light YZ outside the slab CD makes the angle∠₂Here we observe that AB and CD are parallel and XY is transversel. Applying snell law on first surface (AB) & CD we get simultaneously -----→(a) ----→(b) On multiplying eq. (a) & (b) , we get orHence,sin i = sin e ∴ |
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| 7985. |
A high-jumper successfully clears the bar. Is it possible16.that his centre of mass crossed the bar from below it?Try it with appropriate figures. |
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| 7986. |
A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line betweenangular limits-o and ф For an angular dsplacement®' [I0! < 0] the tension in the string and velocity of thebob are T and v respectvely. The following relations hold good under the above conditions:6.UIT 19861Mv2(A) T cos θ=Mg- Mg cos0yfej Tangential accleration g sin 0(D) T Mg cose |
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Answer» there are to option correct and unable to understand |
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| 7987. |
34.તો આ અતર S નું મૂર્વ રીપી. 9 TUITIA body falling for 2 seconds covers a distance S equal tothat covered in next second. Taking g=10m/sS-[EAMCET (Engg.) 1995)(a) 30 m(b) 10 m(c) 60 m(d) 20 m |
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Answer» distance travelled by the body in 2 seconds is,s=ut+12gt2 =u×2+12×g×2×2 =2u+2g (i)Distance travelled by the body in next second is,s=u+g22n-1 =u+g22×3-1 =u+5g2 (ii)Since, the body covers equal distance, so from equation (i) and (ii), 2u+2g=u+5g2 or, u=g2 Now putting the value of u in equation (i), we get, s=2u+2g =2×g2+2g =3g=3×10=30 m . option a is correct answer --- I hope it helps u |
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| 7988. |
16.A concave mirror of focal length 'f, is placed at adistance of 'd' from a convex lens of focal length'f'. A beam of light coming from infinity and fallingon this convex lens-concave mirror combinationreturns to infinity. The distance 'd' must equal :(2) -24, +40 dit(1) 2f1 + f2(3) f1 + f2(4) -fi +12 |
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Answer» 1) is the correct answer (1) 2f1+f2 is the answer |
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| 7989. |
16.The current through a wire depends on time as i=1,+ at, where is = 10 A and a = 4 A/s.Find the charge crossed through a section of the wire in 10 seconds, and average currentfor that interval(A) 30 C, 300 A (B) 60 C, 30A (C) 300 C, 30A (D) of these |
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Answer» D) is the correct answer a) is the correct answer |
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| 7990. |
65. In the given figure the current through the 20 V batteruit16v [ 16vII20v223342(a) 11 A(b) 12 A(c) 7 A(d) 14 A66 In the nine. |
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Answer» Option A 11 amp |
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| 7991. |
SECTION -C400 Nm acting on abody of massupkg, produces anCalculate the moment of inertia & radirus of gyrationueofacceleration of 20 rad/s.of body. |
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| 7992. |
SECTION C16 A torque of 400 Nm acting on a body of mass kg. produces an angularacceleration of 20 rad/s. Calculate the moment of inertia & radius of gyrationof body.ion Derive the |
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| 7993. |
Calculate R.P of microscope with cone angle of 60 falling light of400 nm in Air. |
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Answer» RP=2sinA/1.22xwavelengthRP=2*√3/2/1.22*400*10^-10RP=√3/488*10^-10RP=√3*10^10/488RP=1.732*10^10/488RP=1732*10^7/488RP=3.54*10^7 |
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| 7994. |
. When we press the bulb of a dropper with its nozzle kept in water, air inthe dropper is seen to escape in the form of bubbles. Once we release thepressure on the bulb, water gets filled in the dropper. The rise of water inthe dropper is due to(a) pressure of water.b) gravity of the earth(c) shape of rubber bulb.(d) atmospheric pressure.URE |
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| 7995. |
A train has to negotiate a curve of radius 400 m. By how much height should the outer rail be raisedwith respect to inner rail for a speed of 48 km/hr? The distance between the rails is 1 mF4. |
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Answer» wrong answer |
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| 7996. |
9. (a) Water does not run out of a dropper unless its rubberbulb is pressed. |
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Answer» Water doesn't come out because the water cannot be replaced by air when the water is squeezed out. This is because the water would spill out, but air pressure is in the way If the hole at the end of the dropper was bigger, then the water could spill out. That is why the water spills out when you pour a drink for yourself. The water spills out , and at the same time air is spilling into the bottle. no wrong....the actual answer is because of same atmospheric pressure and water pressure inside the dropper |
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| 7997. |
Q.16 Let nr and n be respectively the number ofphotons emitted by a red bulb and a bluebulb of equal power in a given time(A) nr nb(B) nr x nb(C) nr nb(D) the information is insufficient to get arelation between n, and np |
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| 7998. |
48. A car is moving at a speed of 40 m/s on a circulartrack of radius 400 m. This speed is increasing atthe rate of 3 m/s2. The acceleration of car is |
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Answer» There are 2 components of acceleration. Towards center: V2/R = 40 x 40 /400 = 4 m/s2 Tangent( orthogonal to center) = 3 m/s2 Resultant of 1 and 2 above = 5 m/s2 can u explain clearly??i don't understand As the car is moving along a circular path, it will experience two types of accelerations:1) Tangential accelerationThis is the acceleration that occurs in a straight line, whose value is given to us as 3 m/s22) Centripetal accelerationThis is what causes objects moving in a circular path to be flung outwards. Its value can be calculated from the formula A= v2/r, where v is linear velocity and r is the radius of the path. Putting all values into the formula, we get the value of centripetal acceleration as 4m/s2.Always, tangential or linear acceleration is along a tangent to the path, while centripetal acceleration along the radius vector of the path. So they always lie perpendicular to each other. The net acceleration is given by sqrt(A2+a2), where A and a are linear and centripetal acceleration respectively.Hence we get the value of acceleration as 5 m/s2. |
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| 7999. |
5. A car is moving with a speed of 30 m/s on a circulartrack of radius 500 m. Its speed is increasing at arate of 2.0 m/s2. Determine the magnitude of itsAns. 2.7 m s 2.9.acceleration |
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Answer» The centripetal acceleration is = v²/r = 30*30/500 = 9/5 m/s² = 1.8 m/s² and the tangential acceleration is = 2m/s² so net acceleration = √(2)²+(1.8)² =√7.24 = 2.7 m/s² thanks |
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| 8000. |
हाफ I3 B) |
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Answer» Anatural resourceis anything that people can use which comes from thenaturalenvironment.Examplesofnatural resourcesare air, water, wood, oil, wind energy, iron,andcoal. The resources which we obtain from nature are called natural resource. Ex. Water, sun light, wind, etc |
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