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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.
| 11651. |
1)313.In the given circuit, the effective resistance betweenpoints A and E will beBRwR32R2R ZZ.wwFW2R(2) 3(1) R(4) 2RNIDin field intensity produced due toum at |
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Answer» R will be the correct answer |
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| 11652. |
2. Why does not sound travel in vacuum?3. Why is vo |
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Answer» Avacuumis an area without any air, likespace. Sosoundcannottravelthroughspacebecause there is no matter for the vibrations to work in. thank you |
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| 11653. |
. Crater wear leads to(a) increase in cutting temperature(c) friction and cutting forces(b) weakening of tool(d) all of these |
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Answer» Crater wear is localized to the rake side of the insert. It is due to a chemical reaction between the workpiece material and the cutting tool and is amplified by cutting speed. Excessive crater wear weakens the cutting edge and may lead to fracture. The answer is option D Like my answer if you find it useful! |
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| 11654. |
(1) The speed oflight in a transparent mediumis 2.4 x 10^8 m/s. Calculate the absolute refractiveindex of the medium. |
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Answer» Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium = 3 x 10^8/2.4×10^8= 3 x 10^8/2.4×10^8= 1.25 |
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| 11655. |
The speed of light in a transparent medium is 0.6 times that of its speed in vaccumm.What is the refractive index of the medium? |
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| 11656. |
ea's OT CTs Sce53. How is the refractive index ot anopticel medium related tothe speed of light-in that medium? |
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| 11657. |
What is the use of fence in SI system |
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Answer» The SI unit of force is thenewton, symbol N. The base units relevant to force are: the metre, unit of length, symbol m, thekilogram, unit of mass, symbolkg, thesecond, unit of time, symbol s. |
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| 11658. |
3) Find the percent increase in the magnetic field Bwhen the space within a current-carrying toroidis filled with aluminum. The susceptibility ofH i100perma-5aluminum is 2.1 x 10(Ans. : 2.1 × 10-3) |
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| 11659. |
A tree 12m high, isbroken by wind in such a way that is top touches the ground and makes 60° withground. At what height from the bottom the tree is broken by the wind? |
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Answer» if want more clear explanation tell me |
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| 11660. |
42) Two smal balls, having equal positive charge q coulomb are suspended by two insulatingstrings of equal length /metre from a hook fixed to a stand. The whole set up is taken in asatellite into space where there is no gravity. What is the angle between the two strings and thetension in each string / |
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Answer» The angle between the two strings is 180 degree.As there would be no gravity the mg force would be zero and hence all the objects will be start floating w.r.tto each other. |
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| 11661. |
The density of a material in c.g.s system ispc In a system of units n which units oflength is 2cm, and unit of mass is 4g, What isthe numerical value of density of the material is1) 42)23) 14) 0.5 |
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Answer» Option (1) is correct as:- |
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| 11662. |
9.2 Which would require agreater force-accelerating a 2 kg massat 5 m s2 or a 4 kg mass at 2 m s2? |
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Answer» Force = mass × accelerationSo , F = ma So when m = 2kg and a = 5m/s²F = ma = 2× 5 N = 10 N When m = 4 Kg and a = 2m/s²F = ma = 4 × 2 = 8 N First when m = 2kg and a = 5m/s² have greater force |
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| 11663. |
Example 9.2 Which would require agreater force-accelerating a 2 kg massat 5 m s2 or a 4 kg mass at 2 m s2? |
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Answer» Force = mass × accelerationfor First bodyF(1) = 2× 5 = 10 NFor second bodyF(2) = 4 × 2 = 8 Nso first body needs more force |
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| 11664. |
Q 13)If x - a bt +ct? ,where x is in metres and t is in seconds. What are the units of a , b ,and c? |
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Answer» The principle of homogeneity of units,, ie. if the left side of an equation is a distance, then each term on the right hand side of that equation, also need to be a distance and seperately. So units of ‘a’ need to be those of distance, unit of b×t are also a distance. Therefore the unit of b = distance/time. Similarly, the units of c t ² also need to be a distance. So units of c = distance /t². Thus units of a = metre, unit of b = metre per second (a velocity or speed), and the unit of c = metre per second per second ( unit of acceleration). This equation reminds us of the equation of motion of a particle in a straight line, with a velocity b metre per second, and an acceleration c metre per second per second. In the equation x gives the displacement of the particle at time t, when the initial displacement of the particle at time t= 0 is ‘a’ metre. tq |
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| 11665. |
161Two small balls, having equal positive charge qcoulomb are suspended by two insulating strings ofequal length I metre from a hook fixed to a standThe whole set up is taken in a satellite into spacewhere there is no gravity. What is the anglebetween the two strings and the tension in eachstring ?[IIT 86] |
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| 11666. |
hysics5. Two particles A and B having charges of+2.00×10-6 C and of-400x10-6C respectively areheld fixed at a separation of 200 cm. Locate the point(s)on the line AB where (a) the electric field is zero (b) the49electric potential is zero.nroduçes an electric field of magnitudeWhat, is the |
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| 11667. |
0.1.2. The electrostatic force on a small sphere of charge 0.4 pC due to another small srge-0.8 pC in air is 0.2 N.(a) What is the distance between the two spheres ? |
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| 11668. |
Two small spheres each carrying a charge q are placedmetre apart. If one of the sphere is taken around the otherone in a circular path of radius r, the work done will beequal to[MP PET 1995: MNR 1998; Pb. PMT 2000; |
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Answer» work done is zero, as a circular path around a small positively charged sphere is its equipotential surface and work done on carrying a charge through a equipotential surface is zero |
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| 11669. |
el 3U IVTwo fixed charges A and B of 5C each are separated bydistance of 6 m C is the mid point of the line joining A andB. A charge 'Q' of 5C is shot perpendicular to the lineJoining A and B through C with a kinetic enery of 0.06 J.The charge 'Q' comes to rest at a point D. The distance CDIKCET 2012)(a) Sm(b) 3 m(c) 3V3 mSell amI nder the influence of the Cou |
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Answer» Dear student I suppose the charge q = - 5 μC starts with a KE of 0.06 J from point C and stops at point d. Initial total energy of the charge q = KE + PE = 0.06 + 9*10⁹ * 5 *10⁻⁶ * (-5 *10⁻⁶)/3 + 9*10⁹ * 5*10⁻⁶*(-5*10⁻⁶)/3 J = 0.06 - 0.075 - 0.075 J = - 0.09 J Let ad = bd = x m Final energy = PE (as KE = 0) = - 2 * 9*10⁹ * 5*10⁻⁶ * 5 * 10⁻⁶ /x J = - 0.450/x J Hence, using Energy conservation we get: 0.450/x = 0.09 x = 5 mcd = √(x² - 3²) = 4 m the answer could be option d |
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| 11670. |
Three mass point m1, m2 and m3 are located at the vertices of an equilateral triangle of length a. What is the momentof inertia of the system about an axis along the altitude of the triangle passing through m,?10·[a /4(m,+ my] |
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| 11671. |
Find the mass M of the hanging block which will prevent the smaller block fromi sipping over the triangularbłock. All the surfaces are frictionless and the string and pulley are light. (use the concept of pseudoforce andresolve the forces into components)that the pulley and |
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| 11672. |
47) The acceleration of 2 kgmass in the given figure is (all surfaces are smooth and pulleyare massless and frictionless)(Dg/2Ans:4(2)g/33)g/4(4)g/6 |
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| 11673. |
44. Two masses M, and M2 are attached to the ends ofa light string which passes over a massless pulleyattached to the top of a double inclined smooth planeof angles of inclination α and β. The tension in thestring is:M1 fixedM2(sin β) g(A) M+ M2M i (sin α ) gM , M2 (sin β + sin α)gM1 + M2(D) Zero |
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Answer» option C is correct. c is the best answer for ur questions |
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| 11674. |
2. Does an electric charge experience a force due to the field produced by itself? (yes/No) |
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Answer» give one reason why a gas completely fills the vessel in which it is kept No the charge does not feel the force of the field produced by it.This electric field rather exert force on other charges in it's range of influence. |
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| 11675. |
under this repulsion12. Two insulating small spheres are rubbed against eachother and placed 1 cm apart. If they attract each otherwith a force of 01N, how many electrons wereone sphere to the other duringtransferred fromrubbing?un to the electric force between |
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| 11676. |
two fixed point charges +4e and+e units are seprated by a distance a where should the third charge be placed for it to be in equillibrium |
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Answer» I hope ,this solution help you to understand |
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| 11677. |
Ablock ofmass mi-2kg。n a smooth inchnedplane at angle 30° is connected to a second blockof mass m2 -3 kg by a cord passing over africtionless pulley as shown in fig. The accelerationof each block is- (assume g 10 m/sec |
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| 11678. |
A bomb at rest explodes into two parts ofmasses m1 and m2 . If the momentums ofthe two parts be p1 and p2, then their kineticenergies will be in the ratio of-(A) my / m2(C) P1 / P(B) m2 / m1(D) P2 P1um is enuivalent |
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Answer» 1 aahe karan magation due to gravity formula is M1/m2 |
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| 11679. |
three blocks of masses m1, m2, and m3 are placed in contact with each other in a frictionless table. a force is applied on the heaviest mass m1; find the acceleration of m3 |
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Answer» PLEASE provide the diagram |
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| 11680. |
A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is attachedTo the other end a mass of 10 kg is attached. The tension is the thread is:(A) 24.5 N(C) 79 N(B) 2.45 N(D) 73.5 N |
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Answer» Let m₁ = 6kg and m₂ = 10kg 10 kg mass is heavier than 6 kg mass . So, 6 kg moves upward and 10 kg moves downward. Let acceleration generated due to motion of bodies is a Now, apply Newton's 2nd law, For 1st body , T - m₁g = m₁a ----(1) For 2nd body,m₂g - T = m₂a -----(2) Solving both equations we get , T = 2m₁m₂g/(m₁ + m₂) Now, put the values of m₁, m₂ and ge.g., T = 2 × 6 × 10 × 9.8/(6 + 10) T = 12 × 98/16 T = 294/4 N = 73.5 N Hence, tension in string = 73.5 N option (d) is correct |
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| 11681. |
(d) 10 N16. Two masses m, and m2 are attached to a string whichpasses over a frictionless smooth pulley. Whenm 10kg. m26kg. the acceleration of masses is(a) 20 m/s2(b) 5m/s(c) 2.5 m/s(d) 10m/s |
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Answer» Why the value 4g change in 40 |
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| 11682. |
Two blocks of masses mand m2 are connected by ag of spring constant k (figure 9-E15). The block ofmass ma is given a sharp impulse so that it acquires avelocity v towards right. Find (a) the velocity of thecentre of mass, (b) the maximum elongation that thespring will suffer.m12VoFigure 9-E15 |
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| 11683. |
oueA charge 9 is divided into 2 parts such that they experience maximum electric repulsionwhen placed at a certain distance apart? Find the ratio of two charge in which chargedis divided? |
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| 11684. |
5.A charge Q is divided in two parts q and Q-9. What is value of q for maximumforce between them?(b)(C)Q (d) |
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Answer» (d) is correct option |
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| 11685. |
Two fixed charges A and B of 5HC each are separated by adistance of 6 m. C is the mid point of the line joining A andB. A charge Q of -5 AC is shot perpendicular to the linejoining A and B through C with a kinetic energy of 0.06 JThe charge 'Q comesis3.to rest at a point D. The distance ClD[KCET 2012](a) 3m(b) 3 m(c) 3/3m(d) 4mold of charge +Q. a |
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Answer» Answer:d)4mExplanation |
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| 11686. |
d)6m,1:6decreases by 36 % of its value on the surface of the earth is:(Assume radius of the earth is R)(BHU 2000)(d) 4 R4 |
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Answer» so Option a is correct |
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| 11687. |
A 3000 N block is placed on an inclined plane as shown in fig. Find the maximum value ofW for equilibrium if tipping does not occur. Assume coefficient of friction as 0.2.FrictionlessPulley3000 N30 |
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| 11688. |
Figure 5-E926 A constant force F-m2g/2 is applied on the block ofmass m, as shown in figure (5-E10). The string and thepulley are light and the surface of the table is smooth.Find the acceleration of m,m1m2Figure 5-E10 |
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| 11689. |
Draw the magnetic field lines due to a currentrrying loop |
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| 11690. |
Two point charges repel each other with a force of 100 N. One of the charges isincreased by 10 % and other is decreased by 10%. The new force of repulsion at thesame distance will be55 N77 N99 N100 N |
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| 11691. |
(3) 30f pulleyand all strings as shown are massless100 N. Calculate weight of theblock (g = 10 m/s)60%30%T.2(1) 50 N(3) 50/3 N(2) 100 N(4) 100-3 N |
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| 11692. |
placedTwo positive charges q, and 2 (q, > q2) area certain distance apart. Depict the electric field linesdue to the system of charges. |
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| 11693. |
1. Two positive charges q, and q, (q, >q2) are placeda certain distance apart. Depict the electric field linesdue to the system of charges. |
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| 11694. |
Two charges 5 x 10- C and -3 x 10-8 C are located 16 cm apart. Atwhat point(s) on the line joining the two charges is the electricpotential zero? Take the potential at infinity to be zeto. |
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| 11695. |
Two point charges repel each other with a force of 100 N. One of the charges isincreased by 10% and other is decreased by 10%. The new force of repulsion at thesame distance will be55 N77 N99 N100 N |
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| 11696. |
のマ市(5V3 RT 4 m/s) |
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Answer» thanks |
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| 11697. |
Two electric charges + g and + 4 q are placed at adistanceposition of the point on the line joining the twocharges where the electric field is zero.7.6a apart on a horizontal plane. Find the |
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Answer» For electric field to be zero, elctric field due to +q should be equalto electric field due to +4q. Let us take that point to be at x distance from +q and 6a-x from +4q +q/x²= +4q/ (6a-x)², here 1/4piE0will be cancelled on both sides Taking square roots on both sides we get, 1/x =2/ 6a-x on solving we get, 6a-x = 2x , 6a=3x , x=2a and 6a-x= 4a So, x is at a distance of 2a from +q and 4a from+4q. Hope it helps. thank you for help me |
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| 11698. |
9. Two point charges +49 and +q areseparated by a distance 'a' apart whereshould a third charge (e) be placed that thesystem be in equilibrium? What will be themagnitude and sign of (e)? |
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| 11699. |
andfan(a).betA05]1. Three point charges are placed at the followingpoints on the X-axis : 2C at x 0, -3C atx 40 cm and -5C at x = 120 cm. Calculate theforce on the - 3C charge.(Ans. 0.5485 N, towards left)+20CM -34C-54Co40 cm120 cmF |
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Answer» Fe= (-9.0 * 10^9) * (2.0 * 10^-6)(3.0 * 10^-6) / 0.40^2= -54 * 10^-3 / 1.6 * 10^-1= -0.3375 N Fe= (-9.0 * 10^9) * (-5.0 * 10^-6)(3.0 * 10^-6) / 0.80^2= 135 * 10^-3 / 6.4 * 10^-1= -0.2109 N a = -0.55N |
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| 11700. |
A bar-magnet placed in a uniform magnetic field of0.3 T, with its axis at 30° to the field, experiences atorque of 0.06 N m. Find the magnetic moment of thebar-magnet.0.4 A m |
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