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.
| 801. |
og. Find the accelerations a) a2, a3 of the three blocksre (6-E8) if a horizontal force of 10 N is23.shown in figuapplied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block.Take g = 10 m/s2μ,-0.22 kg2μ3= 0.07 kg |
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Answer» where is the fbd |
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| 802. |
As shown in the figure, two equal masses each of 2 kg are suspendedfrom a spring balance. The reading of the spring balance will be :-2kg2kg(1)(2)(3)Zero2 kg4 kgBetween zero and 2 kg |
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Answer» It will be 0 as its in balanceOption-A (1) zero -2-0-2- right answer the right answer is= 0 |
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| 803. |
Section -II : Single Choice Questions31The small marble is projected with a velocity of 10 ms-1 in a direction 450 from the horizontal y-directiosmooth inclined plane. Calculate the magnitude v of its velocity after 2s7.ㄨㄡ10 ms4510(A)-1(C) 10ms1(D) 5v2s(B) 5 Tns |
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| 804. |
A wheel has moment of inertia 5 × 10-3 kg m2and is making 20 rev/sec. The torque needed tostop it in 10 sec is × 1()"LN-m :-(1) 2π(2) 2.5n(3) 4π(4) 4.5m |
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Answer» 2pi/100 Nm |
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| 805. |
4.5What is the magnitude of magnetic force per unit length on a wirecarrying a current of 8 A and making an angle of 30° with thedirection of a uniform magnetic field of 0.15 T? |
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| 806. |
A 40 kg flywheel in the form of a uniform circular disc 1 m radius is making 120 rpm. Calculate the angularmomentum[251.2 Kgm%) |
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Answer» Mass, m = 40kg Radius, r = 1m Angular velocity is, w = 120 rev/min = (120)(2π)/60 rad/s= 4π rad/s MI of the ring is, I = mr²/2= (40)(1)2kg m2= 40 kg m2 Now, Angular momentum, L = Iw = (40)(4π) = 251.2 kg m2s-1 can you plss explain 2nd step where I=mr2/2 moment of inertia of a disc=mr²/2 then it will be 40×(1)2/2=20 |
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| 807. |
EXERCI. A wheel is making revolutions about its axis withuniform angular acceleration. Starting from rest, itreaches 100 rev/sec in 4 seconds. Find the angularacceleration. Find the angle rotated during these fourseconds. |
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| 808. |
is 13. What can be said about the centre of mass of a uniformhemisphere without making any calculation? Will itadistance from the centre be more than r/2 or less thanr/2?le8 |
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| 809. |
in the circuit shown, value of R in ohm whichresult in no current through the 30 V battery is130 V| 50 v LER3200102 |
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Answer» r-30 par hoga of the year and |
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| 810. |
Q.46 A force of (2î -4j +2k) Newton acts ata point (3i + 2j 4K) metre from theorigin. The magnitude of torque is(A) zero(C) 0.244 N-m(B) 24.4 N-rm(D) 2.444 N-m |
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Answer» thanku |
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| 811. |
Gravitational potential at a point due to a uniform solid sphere |
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| 812. |
Example 9.4 A force F = (2 + x) acts on a particle in x-direction where F is innewton and x in metre. Find the work done by this force during a displacementfrom x = 1.0 m to x = 2.0 mL |
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| 813. |
Example 9.4 A force F = (2 + x) acts on a particle in x-direction where F is inewton and x in metre. Find the work done by this force during a displacementfrom x = 1.0 m, to x = 2.0 m. |
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| 814. |
3. A force F (3xi +4j) Newton (where x is in metres) acts on a particle which moves from a position3-3(2m, 3m) to (3m, Om). Then the work done is(A) 7.5J(B)-12J(C) -4.5 J(D) +4.5 J |
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| 815. |
The vector sum of the forces of 10 newton and 6 newton can be:A ) 2NB) 8NC ) 18 N D ) 20 N |
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Answer» 8N , because sum will always be in between 10-6 = 4N to 10+6 = 16N |
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| 816. |
The mass of an oxygen atom is 16.00 u. Find itsmass in kg.5 |
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| 817. |
esFour particles each of mass M, are located at the vertices of a square with side L. Thegravitational potential due to this at the centre of the square isn, menu- |
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Answer» Net force acting on any one particle M=GM²/(2R)²+GM²/(R√2)²Cos45°+GM²/(R√2)²Cos45° M=GM²/R² [1/4+1/√2] This force will equal to centripetal force So,. Mv²/R=GM²/R²[1/4+1/√2] V=√[GM²/4R(1+2√2)] V=1/2√[GM/R(2√2+1)] |
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| 818. |
8. Suppose the gravitational potential due to a smallsystem is k/r at a distance r from it. What will be thegravitational field? Can you think of any such system?What happens if there were negative masses? |
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| 819. |
Gravitational Potential energy |
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Answer» In classical mechanics, the gravitational potential at a location is equal to the work per unit mass that would be needed to move the object from a fixed reference location to the location of the object. It is analogous to the electric potential with mass playing the role of charge |
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| 820. |
differentkindsofcelestialbodiesotherthantheplanetsandtheirmoons,fodfldfi11. Name threesolar system. |
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Answer» asteroids,comets and meteors the smallest planet is |
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| 821. |
what is celestial bodies |
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Answer» Anastronomical objectorcelestial objectis a naturally occurringphysical entity, association, or structures that exists in theobservable universe.Inastronomy, the termsobjectandbodyare often used interchangeably. However, anastronomical bodyorcelestial bodyis a single, tightly bound, contiguous entity, while an astronomical or celestialobjectis a complex, less cohesively bound structure, which may consist of multiple bodies or even other objects with substructures. |
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| 822. |
bodies 11, 112 and my are hanging onm are hanging on a stringhown below. If my mim, 1strings connectingan17. Three bodies m, mover a fixed pulley as shown bele2: 3 find the ratio of tensions ibodies m, to me and m, to mymaLaval |
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| 823. |
2. Which of the following is not a fossil fuel ?(a) coal(b) petroleum(c) biogas(d) wood |
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Answer» [b] petroleum because it is limited and it take 300 years to 350 years to make fossil fuel |
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| 824. |
An electric fan has blades of length 30 cm mea-sured from the axis of rotation. If the fan is rotat-ing at 120 rpm, the acceleration of a point on thtip of the blade is(1) 1600 ms-2(3) 23.7 ms-2(2) 47.4 ms(4) 50.55 ms-2 |
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Answer» since the rotation is 120 rpm jntead of 1200 in the solution, fhe correct option will be (b) 47.4m/s^2 |
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| 825. |
in the circuit shown, value of R in ohm whidresult in no current through the 30 V battery is50 V30 V2003102 |
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| 826. |
the circuit shown, value of R in ohm whicsult in no current through the 30 V battery is50 V30 Vwwww{202W102 |
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| 827. |
the circuit shown, value of R in ohm whicesult in no current through the 30 V battery is| 50 V30 VZR3202 3102 |
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| 828. |
In the circuit shown, value of R in ohm whicresult in no current through the 30 V battery is| 50 vL30 V3202102 |
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Answer» R=30 ohms Stups are given 👆 yes the above solved is right |
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| 829. |
In the circuit shown, value of R in ohm which willresult in no current through the 30 V battery is50 V30 Vw200 }102 |
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| 830. |
33 Three equal weights A,LILLLLLB, C of mass 2 kg eachare hanging on a stringpassingovera fixed frictionless pulleyas shown in the fig. Thetension in the string connectingweights B and Cis-(A) zero(B) 13 Newton(C) 3.3 Newton(D) 19.6 Newton |
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| 831. |
(a) doubled (6 e0ulles Tou13. who among the following gave first the experimental value of G.uced to haltt The ima(a) Cavendish(b) Newton(c) Copernicusmaged Taylor(a) bet |
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Answer» Cavendish |
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| 832. |
A body of mass 300 gm is at rest. What force inNewton will you have to apply to move it through200 cm in 10 sec.? |
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Answer» t= 10 seconds S distance= 200 cm=2 metres mass = 300 grams= 0.3 kilograms Velocity=S/t=2/10=0.2 metre/ sec initial velocity u= 0 Acceleration = (v-u)/t ⇒a=(0.2-0)/10 ⇒a=0.02 metre/sec² Now mass = 0.3 kg So force=m×a ⇒F=0.3×0.02 ⇒Force=0.006 Newtons. |
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| 833. |
Q. 29. What is meant by gravitational potentialenergy? |
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Answer» Gravitational potential energychanges into kineticenergy. The equation forgravitational potential energyis GPE = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 on Earth), and h is the height above the ground in meters. |
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| 834. |
an artificial satellite of mass 1000kg revolves around the earth in circular orbit of radius 6500km calculate (a) orbital velocity (b) orbital kinetic energy (c) gravitational potential energy |
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| 835. |
11.The unit vector perpendicular to the vectors 6i +2j+ 3kand Зі-6-2k is-2i-3j+6k2i +3j-6k2i-3j-6k2i+3j+6 |
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Answer» The unit vector is given by cross product of these vectors , divided by the magnitude of that vector. the cross product is 14i+21j-42j so, the magnitude is √14²+21²+42² = 49 so the unit vector is.(14i+21j-42j)/49 = (2i+3j-6k)/7 option C. |
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| 836. |
The component of vector A=2i+3j, along the vector i+] is |
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Answer» The component of A along i+j (magnitude = √1²+1²=2)= (2i+3j)•(i+j)/√2= 5/√2 |
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| 837. |
Enample a5 A teain starting fromattains a velocity of 72 km r in5 minutes. Assuming that tiapoeleration is uniform. find (l tirasceleration and (i) the distanevaveled by the train for attaining thvery |
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Answer» Velocity=72 kmh⁻¹=72×5/18=20 ms⁻¹Time taken=5 minutes=5×60=300 si) acceleration=(20/300) ms⁻²=0.067 ms⁻²ii)Distance=20×300=6000 m It is wrong |
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| 838. |
5.What is TIR? Explain optical fiber and its use in communication. |
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Answer» Total internal reflection is the phenomenon which occurs when a propagated wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface. Fiber-optic communication is a method of transmitting information from one place to another by sending pulses of light through an optical fiber. ... Optical fiber is used by many telecommunications companies to transmit telephone signals, Internet communication, and cable television signals. |
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| 839. |
L muกา1Tre ale of cc app hedC Gan |
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Answer» Kindly post the complete question. |
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| 840. |
the circuit shown, value of R in ohm whichsult in no current through the 30 V battery is50 V130 VEwwww{2002102w |
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| 841. |
In the circuit shown, value of R in ohm whilresult in no current through the 30 V battery is130 V| 50 v ,ZR3202 }102 |
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| 842. |
a block of mass 1 kg connected with a spring of force constant 100 newton per metre is suspended to the ceiling of lift moving upward with constant velocity 2 metre per second calculate the extension produced in spring |
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| 843. |
of R in ohm which willIn the circuit shown, value of R in ohm wresult in no current through the 30 V battery is50 V30 V32022102 |
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| 844. |
7. Two small identical electrical dipoles ABand CD, each of dipole moment 'p' are kept!at an angle of 120° as shown in the figure.What is the resultant dipole moment of thiscombination ? If this system is subjected toelectric field (E) directed along + X direction.what will be the magnitude and direction ofthe torque acuing on this? [Delhi 2011]+994120x-9.B+90C |
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| 845. |
Two particles A and B, having opposite charges2.0×10-6 C and -20×10-6C,separation of 1.0 em. (a) Write down the electric dipolemoment of this pair. (b) Calculate the electric field at aare placed at a |
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| 846. |
in the circuit shown, value of R in ohm whichresult in no current through the 30 V battery is50 V130 V3202102 |
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| 847. |
Electric Dipole Moment |
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| 848. |
Q.11. What is electric dipole? Define electric dipole moment and state SI unit.OR |
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Answer» An electric dipole is a separation of positive and negative charges. The simplest example of this is a pair of electric charges of equal magnitude but opposite sign, separated by some It's SI unit is coulomb-metre. ideal electric. dipole |
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| 849. |
08)What is the formula of electric dipole moment. |
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Answer» The electric dipole moment is a measure of the separation of positive and negative electrical charges within a system The unit is coulomb-meter;p = qdthe magnitude of the charges multiplied by the distance between the two. |
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| 850. |
11) Define electric dipole moment and state its SI unit. |
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Answer» TheSI unitsforelectric dipole momentare coulomb-meter (C.m); however, the most commonunitis the debye (D). Theoretically, anelectric dipoleisdefinedby the first-order term of the multipole expansion; it consists of two equal and opposite charges that are infinitely close together |
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