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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.
| 13251. |
An athlete completes one round of a circular track of diameter200 m in 40 s. What will be the distance covered and thedisplacement at the end of 2 minutes 20 s? |
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| 13252. |
1. An athlete completes one round of a circular track of diamet290-mm an,' What will be the distance covered and thedisplacement at the end of 2 minutes 20 s? |
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| 13253. |
26. Assuming that the moon completes one revo-lution in a circular orbit around the Earth in 27.3days. Calculate the acceleration of the moon ! |
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Answer» thanks |
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| 13254. |
12) Siate and explain triangle law of vector addition) Prove the relationV= ηλ where, V is velocity of waven is frequency and λ is wavelength |
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Answer» Triangle law of vector additionstates that when twovectorsare represented by two sides of atrianglein magnitude and direction taken in same order then third side of thattrianglerepresents in magnitude and direction the resultant of thevectors. This simply means that, if you have a two vectors that represents the two sides of the triangle then the third side of that triangle will represent their resultant. Example In a time of one period, the wave has moved a distance of one wavelength. Combining this information with the equation for speed (speed = distance/time), it can be said that the speed of a wave is also the wavelength/periodsSince the period is the reciprocal of the frequency, the expression 1/f can be substituted into the above equation for period. Rearranging the equation yields a new equation of the form: Speed =Wavelength×Frequency The above equation is known as the wave equation. It states the mathematical relationship between the speed (v) of a wave and its wavelength (λ) and frequency (f). Using the symbolsv, λ, andf, theequation can be written as v=f×λ |
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| 13255. |
Q.22. What is the triangle law vector addition? |
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Answer» Triangle law of vector additionstates that when twovectorsare represented by two sides of atrianglein magnitude and direction taken in same order then third side of thattrianglein opposite represents in magnitude and direction the resultant of thevectors. |
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| 13256. |
... . . A l/ l R 1-5 and l A × B-8 Angle between A and B isacute, then A B is(a) 6(b) 3(c) 4(d) 7 |
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| 13257. |
BsinA+BcoseLet direction of R makes angle a with A thaprove this formula for direction ofresultant vector of parallelogram vectoraddition law. |
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| 13258. |
of R in ohm which willin the circuit shown, value of R in ohm wresult in no current through the 30 V battery is| 50 v,130 VSR3202 }102 |
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| 13259. |
A car travels 4 kilometre in 3 minutes Find the speed of Car in centimetre per second |
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Answer» distance=4 km=4000×100 cm=400000cm time=3 mins=3×60 secs=180 secs so,speed=distance/time =400000÷180=2222.222.... |
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| 13260. |
49. The practical unit of electric dipole moment is(a) coulomb-metre(b) newton/coulomb(c) Joule/coulomb(d) debye |
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Answer» The SIunitsforelectric dipole momentare coulomb-meter (C.m).But common unit is debye (d) |
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| 13261. |
in the circuit shown, value of R in ohm whisresult in no current through the 30 V battery is| 50 vL130 V{202102 |
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| 13262. |
Your friend has ruptured his muscle after slipping on the road. What four steps 2of first aid would you adopt ? |
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Answer» Rest the muscle.For either a first or second degree injury, medical care is not typically needed. You can treat these injuries with RICE, a mnemonic for the treatment. "R" stands for "rest" the injured area. Ice the area.Apply ice, either in the form of a bag of frozen peas or crushed/cubed ice in a protective plastic wrapping. Wrap the ice pack in a cloth or thin towel before you use it. Apply the ice to the affected area for 15 to 20 minutes every 2 hours for the first 2 days after the injury. Compress the muscle.You can also wrap the injured area with an ace bandage to protect the area for the first 48 to 72 hours. Be sure to wrap it snugly, but not too tight. Elevate the injured limb.You can also elevate the injured limb above your heart to help reduce the swelling. Place the limb up on some pillows and lay down. Make sure that you are in a comfortable position. |
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| 13263. |
3. Find the expression for the gravitational potentialenergy of a body of mass m at a height h |
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| 13264. |
19. A bucket full of water is on the first flooryour house and another identical bucketwith same quantity of water is kept on the secondfloor. Which of the two has greater potentialenergy?20. Write the expression for the gravitationalpotential energy explaining the meaning of thesymbols used.21. A body of mass m is moved from ground to aheight h. If force of gravity on mass of 1 kg is gnewton, find : (a) the force needed to lift the body.(b) the work done in lifting the body and (c) thepotential energy stored in the body. |
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| 13265. |
What average force is necessary to stop a bulletof mass 20 gm and speed 250 m/sec as it pen-etrates wood to a distance of 12 cm:(a) 3.4 x103 newton(c) 4.0 x10° newton(b) 5.2 x103 newton(d) 3.6 x 103 newton |
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Answer» Using kinematics eq.v^2 =u^2 - 2as0 = (250 )^2 - 2*0.12*a0.24a=62500a= 62500/0.24nowF = ma= 20/1000 * 62500/0.24solve itF = 5208 N |
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| 13266. |
the average depth of IndiarnOcean is 3 kilometre calculateDelta be divided by we givethat bulk modulus of elasticityfor water is 2.2 into 10 raise topower 9 Newton -2 gravitationequals to 10 M per secondsquare |
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| 13267. |
The surface tension of a liquid is 5 Newton per metre. If a film of this liquid is held on a ring of area0.02 metres2, its surface energy is about(A) 5 x 10-2J(B) 2.5 x 10-2 J(C) 2 x 10-1J(D) 3 x 10-1J |
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| 13268. |
in the circuit shown, value of R in ohm whisresult in no current through the 30 V battery is30 V| 50 v LZR2003102W |
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| 13269. |
The displacement of a particle from a point having position vector 2i + 4) to another point having position vector 5 i+\hat{j} is(1) 3 units(3) 5 units(2) 3/2 units(4) 5/3 units |
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| 13270. |
what is gravitational potential energy |
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Answer» Gravitational potential energy is the energy an object has due to its position above Earth, energy due to its height. The equation for gravitational potential energy is 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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| 13271. |
xample 14 An electric dipole of length 4 cm, whenplaced with its axis making an angle of 60° withauniform electric field, experiences a torque of43 N-m. Calculate the potential energy of theDelhi 2014dipole, if it has charget 8 nC. |
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| 13272. |
32.Obtain an expression for gravitational potential energy. |
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| 13273. |
.A block of wood of mass 0.5kg is placed onplane making angle 30° with the horizontal Ifizontal.Ifthe coefficient of friction between the surfaceof contact of the body and the plane is 0.2the force required to keep the body slidingdown with uniform velocity is |
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| 13274. |
The potential energy (in joule) of a body moving ina force field is given by, U=4+ 2x, where x is thedisplacement in metre. The force acting on the bodyat x= 1 metre is(1) -6 newton(2) - 10 newton13) -2 newton(4) zerotantino |
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Answer» 1) is the correct answer 2 option is correct the potential energy of a body moving in a force field is given by U= 4+ 2x^3 , where x is the displacement in meter. the force acting on body at x=1 meter option a is the correct answer of the given question 3) is the right answer of the following option 1 is the correct answer me |
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| 13275. |
le of R in ohm which willin the circuit shown, value of R in ohm wresult in no current through the 30 V battery is130 V| 50 v ,ER3202 }102 |
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| 13276. |
2ale fan the boto stap |
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Answer» u = 0.5m/sa = - 0.5m/s²t = ?v= 0a= v-u t0.5= 0-0.5 ——— tt= 0.5*0.5t= 0.25 sec |
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| 13277. |
40. In the figure given below, the position-time graphof a particle of mass 0.1 kg is shown. The impulseat t-2 sec is -(1) 0.2 kg-m/s 6(2) -0-2 kg-m/s 2(3) 0.1 kg-m/s(4)-0.4 kg-m/sX(m)4t(seconds |
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| 13278. |
) Find the position vector of the centre of mass ofl kg, 2 kg and 3 kg at points (31 + 2. (+d2i + k respectively |
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| 13279. |
What are celestial bodies? Name any three celestial bodies |
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Answer» An astronomical object or celestial object is a naturally occurring physical entity, association, or structure that exists in the observable universe. In astronomy, the terms object and body are often used interchangeably. |
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| 13280. |
5. Infinite number of bodies, each of mass 2 kg aresituated on x-axis at distances 1 m, 2 m, 4 m,8 m,...., respectively, from the origin. The resultinggravitational potential due to this system at theNEET-2013]origin will be84(2)G(3)- 4G(4) - G |
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| 13281. |
2. A force of 10 Newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum is(a) 5 kg m/s(b) 100 kg m/s(c)200 kg m/s(d)1000 kg m/s |
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Answer» F = mv-mu/t10 = mv-mu/10Change in momentum = 100 Ns or 100kgm/s |
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| 13282. |
TU36. Calculate the work done by the brakes of a car of mass 1000 kg when its speed is reduced from 20 m/s to10 m/s?1111 |
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| 13283. |
121You find your mass to be 42 kg. on a weighing matching. Is your mass more or less than42 kgQ.7121 |
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| 13284. |
Two otjocts of masses I kg and 2 kg respectively are placed at a distance of 1 m apart. Find theaoceleration of the bodies procbced due to gravitational force between them. |
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Answer» Gravitational Force between two objects, F = GMm/(d^2) where F is force, G is gravitational constant, M is mass of object 1, m is mass of object 2, d is the distance between them. Formula of Force, by Newton's second law. F = ma Where F is force, m is mass and a is acceleration. So ma = GMm/(d^2) => a = GM/(d^2) Cancelling the smaller mass involved. => a = 6.67*(10^(-11))*2/(1^2) = 13.34*10^-11 m/s^2 So, that's the final answer 1.334*10^-10 m/s^2. |
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| 13285. |
Calculate the binding energy of an artificialsatellite of mass 1000 kg orbiting at height of3600 km above the earth's surface. Also find thekinetic energy, potential energy and total energyof the satellite.(R= 6400 km, |
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| 13286. |
The moon is experiencing a gravitational force due to the earth and is revolving around the earthin a circular orbit. How much work is done by the moon |
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Answer» Work done by forces and not by a mass either Moon or Earth. Here the average work done by gravitational force between Moon and Earth is zero. Reason is that displacement is along the tangential direction to the circular orbit. The force is always along the radial line inwards. So the dot product is always 0. As work done is always zero , the kinetic energy remains same. Potential energy remains the same. |
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| 13287. |
What will be velocity of a satellite revolving arounthe earth at a height h above surface of earth ifradius of earth is R:59.\begin{array}{l}{\text { (1) } R^{2} \sqrt{\frac{g}{R+h}}} \\ {\text { (3) } R \sqrt{\frac{g}{R+h}}}\end{array} |
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| 13288. |
What are geostationary satellites? Find the expression of total energy ofa satellite revolving around the surface of the earth. What is thesignificance of negative sign in the expression? Write the expression ofangular momentum of a satellite. |
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Answer» Ageostationary satelliteis an earth-orbitingsatellite, placed at an altitude of approximately 35,800 kilometers (22,300 miles) directly over the equator, that revolves in the same direction the earth rotates (west to east). ... |
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| 13289. |
Two satellite of masses 100kg and 200kg are revolving the earth at attitudes 6400 kmand 44800 km. The ratio of orbital velocity of the satellites is46. |
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| 13290. |
11. If the earth is a point mass of 6 x 1024 kg revolving aroundthe sun at a distance of 1.5 x 108 km and in time,T 3.14 x 10's, then the angular momentum of the eartharound the sun is |
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Answer» the mass of the earth, m = 6×10²⁴ kgAssume the mass of the sun is Mdistance between earth and sun , r = 1.5×10¹¹mforce between sun and earth , F = 3.5×10²² N use gravitational force formula, F = GMm/r²3.5× 10²² = 6.67× 10⁻¹¹× 6× 10²⁴ M/(1.5× 10¹¹)²3.5× 10²²× 2.25× 10²²/(6.67× 6× 10¹³) = MM = 0.196 × 10³¹ kg hence,mass of the sun is 1.96× 10³⁰ kg |
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| 13291. |
21.) Deduce an expression of orbital velocity of a satellite revolving around theearth at a height 'h' above the surface of earth. |
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| 13292. |
11. If the earth is a point mass of 6 x10^24 kg revolving aroundthe sun at a distance of 1.5 x 10^8 km and in time,T =3.14 x 10^7 s, then the angular momentum of the eartharound the sun is |
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| 13293. |
28. Which one does secrete bile ?(A) Stomach(B) Pancreas(C) Liver(D) Buccal cavity |
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Answer» Liver secretes bile juicehence option c |
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| 13294. |
a. You have three cells. Draw a diagram to show the connection of the thPee elsb. What are the safety measures adopted while handling electric devices?electricity |
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Answer» The safety measures adopted are:-1)Disconnect the power source before servicing or repairing electrical equipment. 2) Use only tools and equipment withnon-conducting handles when working on electrical devices. 3) Never use metallic pencils or rulers, or wear rings or metal watchbands when working with electrical equipment. |
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| 13295. |
9.A potential difference of 220 V is appliedacross a rheostat AB of 12000 ohm. Thevoltmeter V has a resistance of 6000 ohm andCB is one fourth of the distance of AB. Theerror in the reading of voltmeter isapproximately -126220Vwwwwwwwwww600092(1) 27%(2) 10%(3) 40%(4) 15% |
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| 13296. |
1. A river 2 km wide is flows at the rate of 2km/h. Aboatman who can row a boat at a speed o4 km/h in still water, goes a distance of 2 kmupstream and then comes back. The time taken byhim to complete his journey is(1) 60 min (2) 70 min (3) 80 min (4) 90 min |
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Answer» Speed of water is 2 Km/hr, Speed of Boat is 4 Km/hr, Up stream Speed = 4-2 = 2 Km/hr Down Stream Speed = 2+4 = 6 Km/hr So Total time = 2/6 + 2/2 =4/3 hr =4/3*60 =80 min |
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| 13297. |
Three cells of EMF 1.5 volts each andinternal resistance of 1 ohm are connectedin parallel. This combination is used to sendcurrent through an external resistance of0.001 ohm. What is the value of the current?1.5 A0.5 A4.5 A3.0 A |
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| 13298. |
A voltmeter is connected acrossa battery of emf 12V and internalresistance of 10 ohm. If the voltmeterresistance 230 ohm, calculatethe current flowing through theresistance and reading shown by thevoltmeter12:24 AM |
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| 13299. |
4. A battery of e.m.f. 15 V and internal resistance3 ohm is connected to two resistors of resistances3 ohm and 6 ohm in series. Find(a) the current through the battery(b) the p.d. between the terminals of the batteryilsAns. (a) 1-25 A (b) 11-25 V |
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Answer» The current through the battery is = (15)/(3+3+6) = 1.25A b) P.D between the terminal of the battery is = 1.25*(9) = 11.25 v |
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| 13300. |
Rain is falling vertically witha speed of 35 m st, Winds starts blowingafter sometime with a speed of 12 m/sI ineast to west direction. In which directionshould a boy waiting at a bus stop holdhis umbrella ? |
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Answer» The drops which are falling a little ahead of the current position of the man, will fall on him, because he moves in that direction. relative velocity of the rain drops wrt cyclist = velocity of woman - velocity of rain the magnitude of relative velocity the rain drops wrt man = √(12² + 35²) = 37 m/s direction of relative velocity of the rain drops = = tan⁻¹ (12/35) = 18.92 deg. from the vertical. So the boy should hold the umbrella at 18.92 deg to the vertical in the forward direction. |
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