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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.
| 9051. |
What is converging and diverging lens |
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| 9052. |
What is the significance of the area of closed curve on a P V diagram? |
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Answer» During an adiabatic process no heat is transferred to the gas, but the temperature, pressure, and volume of the gas change as shown by the dashed line. As described on the work slide, thearea undera process curve on ap-V diagramis equal to the work performed by a gas during the process. |
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| 9053. |
Q6 A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens is the prescribed lensdiverging or converging? |
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Answer» Power of lens= 1/focal length ⇒ focal length = 1/power of lens =1/1.5 =0.6667 m Since the focal length is positive, the prescribed lens is convex and is converging lens. |
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| 9054. |
A magnetic needle free to rotate in vertical plane is taken to a magneticpole of the Earth. How will it orient itself ? oppredelos |
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| 9055. |
whichThe reflection surface of a plane mirror is vertical. A particle is projected in a vertical planeis also perpendicular to the mirror. The initial velocity of the particle is 10 m/s and the angle ofprojection is 60. The point of projection is at a distance 5 m from the mirror. The particle movestowards the mirror. Just before the particle touches the mirror the velocity of approach of theparticle and its image is: |
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| 9056. |
0 sec such that its horizontal and vertical components ofA stone is projected from level ground at tinitial velocity are 10 m/s and 20 m/s respectvely. Then the instant of time at which magnitude oftangential and magnitude of normal components of acceleration of stone are same is (neglect airresistance) g 101.m/s |
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Answer» there are 2answer 1 and 3 second |
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| 9057. |
(2) Assuming that the earth performsde uniform circular motion around the Sun, findthe centripetal acceleration of the earth. [Speedce of the earth - 3 x 10 m/s, distance between theearth and the Sun 1.5 1011 m)Solution : Data: v = 3 x 10 m/s, r = 1.5 x 10 mre |
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| 9058. |
11. A fort is provided with food for 80 soldiers t8 lal12. 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldwants to reach school IIILthe food last if 20 additional soldiers join after 15 days?sent to another fort and thus, the food lasted for 32 more days. How manleft the fort? |
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Answer» Total period = 60 daysAfter 15 days this period will be = 60 - 15 = 45 daysNumber of soldiers in the beginning = 80Number of soldiers join = 20New number of soldiers = 80 + 20= 100 Soldiers.Number of Soldiers Days 80 45 100 xMore men, the food will last lesser days.This is an inverse proportion.80 : 100 : : 45 : xx = (80*45)/100x = 36 days15 days has already gone, so, the food will last for 15 + 36 = 51 days instead of 60 days. |
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| 9059. |
A projectile is thrown with velocity v at an angle θwith horizontal. When the projectile is at a heightequal to half of the maximum height, the verticalcomponent of the velocity of projectile is((1) v sin θ x 33vsin θN2 |
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Answer» option 3 is the answer. |
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| 9060. |
What is polygon law of vectors?Ans: Polygon law of vector addition:If a number of vectors are represented inmagnitude and direction by the sides of anincomplete polygon taken in order, thenresultant is represented in magnitude anddirection by the remaining side of the polygondirected from the starting point (tail) of the firstvector to the end point (head) of last vector |
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Answer» Polygon law of vector additionstates that if a number ofvectorscan be represented in magnitude and direction by the sides of apolygontaken in the same order, then their resultant is represented in magnitude and direction by the closing side of thepolygontaken in the opposite order. |
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| 9061. |
A projectile is thrown with velocity v at an angle θ withhorizontal. When the projectile is at a height equal tohalfof the maximum height, the vertical component ofthe velocity of projectile is -vsín θ(A) v sin θ3(B) 3v sin θvsín θ(C) 2(D) 3 |
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| 9062. |
Twe vectors A and B make angles of 40° and120° resepctively with the X-axis. If IA I-6 andBI=5 unit, then resultant vector of these two |
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| 9063. |
1. एक वस्तु 10 किमी उत्तर की ओर तथा 20 किमी पूर्व की ओर जाती । (d) दूसरी गेंद पहै। प्रारम्भिक स्थिति (initial point) से विस्थापन क्या होगा ? | 7, 19.6 मीटर गहरी(a) 22.36 किमी(b) 2 किमीगिरने की ध्वनि :(c) 5 किमी(d) 20 किमी(a) 342 मी/से । |
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Answer» option (a) 22.36 km. is right answer विस्थापन =√ (400 - 100) =5 km approx. =10× root 5=10×2.236=22.36 km 2) is the correct answer a) 22.36 is the correct answer the answer could be option c |
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| 9064. |
Two billiard balls are rolling on a flat table. Onehas velocity components v 1m/s, v, /3 m/sand the other has components v 2m/s andv, 2 m/s. If both the balls start moving from thesame point, the angle between their path is(1) 60° (2) 45 (3) 22.5° (4) 15 |
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| 9065. |
A body is projected horizontally from the top of a hillwith a velocity of 9.8 m/s. The time elapses beforethe vertical velocity is twice the horizontal velocity is32.)(1) 0.5 sec(3) 2 sec(2) se(4) 1.5 sec |
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| 9066. |
projectile is thrown from a point in a horizontal placesuch that its horizontal and vertical velocity componentare 9.8 mls and 19.6 m/s respectively. Its horizontal range is(a) 4.9 m(c) 19.6 m(b) 9.8 m(d) 39.2 m |
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Answer» thank you so much... |
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| 9067. |
படம்) 3-I vector F is egnal to aſ 24+4kin which ügik belongs to x-axis,Y axis, 2-axis.Lo Find the work done due to 3i along ty4-axis.undity.enlete |
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| 9068. |
15. A projectile is thrown from a point in a horizontal place such that their horizontal and vertical velocitycomponents are 9.8 m/s and 19.6 m/s respectively. Its horizontal range is(A) 4.9 m(8) 9.8 m(C) 19.6 m(D) 39.2 mm |
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| 9069. |
A particle performs uniform circularmotion with an angular momentum L. If the frequencyparticle's motion is doubled and its K.E. is halved, whathappens to its angular momentum? |
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| 9070. |
A particle performs uniform circular motion withan angular momentum L. If the frequency ofparticle's motion is doubled and its kinetic energyhalved, the angular momentum becomes(1) 2L (2) 4L (3)24 |
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| 9071. |
A car is moving with a speed of 72 Km/hour.The diameter of its wheels is 50cm. If itswheels come to rest after 20 rotations as aresult of application of brakes, then theangular retardation produced in the car willbeQ.5(A) 25.5 Radians/sec2(B) 0.25 Radians/sec2(C) 2.55 Radians/sec2(D) 0 |
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Answer» thanku |
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| 9072. |
16. Find the angular dispersion between red and blue raysof light produced by a flint glass prism of refractingangle 6ă. The refractive indices of flint glass for redand blue light are 1.644 and 1.664 respectively.Ans. 0.12 |
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| 9073. |
Se WIL LILLIUc) 15b) 10 sa) 5sA bus is moving with a constant speed of 75 kmph. The driver observes the redSignal and applies the brakes. If the speed of the bus reduces uniformly to 21kmph in 3s find the retardation produced in the bus.d) 6m/s2a) 3m/s2b) 4m/s2c) 5m/s2hill celerates uniformly with 0.5 m/s2 to |
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Answer» Usingv = u + at v = 21 km/hr = 21*5/18 = 35/6 m/su = 75 km/hr = 75*5/18 = 25*5/6 = 125/6 m/st = 3 s 35/6 = 125/6 + a*33a = 35/6 - 125/63a = - 90/6 = - 15a = - 15/3 = - 5 m^2/s -ve sign show retardation (c) is correct option |
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| 9074. |
SICS VOL-IRMass 1 kg is divid(1-x) m for aLaw of Gravitation:1 kg is divided into two parts 'x m and16a given seperation, the value of x'ich the gravitational attraction betweenieces becomes maximum is (2001 E)the two1) 1/22) 3/53)14) 2 |
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| 9075. |
) For the circuit shown in the diagram given below5Ω10Ω30Ω6 VCalculate:(a) the value of current through each resistor(b) the total current in the circuit.(c) the total effective resistance of the circuit. |
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| 9076. |
←Time-Time←TimeA wheel has a constant angular acceleratio3.0 rad/s2, During a certain 4.0 s interval, it tthrough an angle of 120 rad. Assuming tht - 0, angular speed op 3 rad/s how lonmotion at the start of this 4.0 second inter(1) 7 sec.(2) 9 sec.(3) 4 sec.(4) 10 sec. |
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| 9077. |
A particle performs S.H.M. of amplitude 10cm, its maximum velocity during oscillatios is 100cm/s. What is its displacement, when the velocity is 60 cm/s2 |
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| 9078. |
4. The equation of motion of a projectile is:y = 12x-4x2The horizontal componenet of velocity is 3 ms-1. Given that gA) 12.4 m10 ms2, what is the range of the projectile?B) 21.6 mC) 30.6 mD) 36.0 m |
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| 9079. |
A man going east in a car with a velocity of 40 km/hr. A train appears to move towards north with a velocity of 40-/3km/hrthen the velocity and direction of motion of train is(1) 80 km/hr 60° N ofE (2) 80/2 km/hr 609 E of N (3) 80/2 km/hr 60° N of E (4) of thesebollo tbe is in hic hand The water is fallina vertically |
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| 9080. |
2. A ball is thrown from the top of a tower with an initial velocity of 10 m s1 at an angle of 30° with the horizontal. If ithits the ground of a distance of 17.3 m from the back of the tower, the height of the tower is (Take g 10 ms2A) 5 mB) 20 mC) 15 mD) 10 m |
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| 9081. |
A car is moving at a speed of 72 km hr. The radiusits whels is 0.25 m. If the wheels are stopped in 20rotations by applying brakes, then angular retardationproduced by the brakes is(a) 25.5 rad s-28.(b) -29.5 radsd)-45.5 rad s33.5 rad s |
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| 9082. |
A rocket is fired vertically from the ground. It moves upwards with a constant acceleration 10 ms2 for30 seconds, after which the fuel is consumed. After what time from the instant of firing, the rocket will attairnthe maximum height ? (Take g = 10 ms-2 )g102)?A)75 s (taFZ)(B) 60 säťăŽre)(C) 45 s (torg)(D) 30 s (m) |
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Answer» S=ut+1/2(at²) s=4500m Putting in v²-u²=2as We get v=300m/s Now a=-10m/s² after the fuel is consumed So v=u-gt We get t=30s Total time for which rocket is ascending=30s+30s=60s Like my answer if you find it useful! |
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| 9083. |
what is mobility? derive the equation of mobility with respect to drift velocity and electric field. |
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Answer» The charge carrier in most metals is the negatively charged electron (see electron scattering).Mobilityis formally defined as the value of the drift velocity per unit of electric field strength; thus, the faster the particle moves at a given electric field strength, the larger themobility. in solid state physics, electron mobility is characterizes how quickly an electron can move through a metal or semiconductor, when pulled by an electric field. In semiconductors, there is an analogous quantity for holes called hole mobility. when an electric field is applied across a piece of material, the electrons respond by moving with an average velocity called drift velocity. |
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| 9084. |
A particle is projected vertically upwardswith a velocity of 20 m/sec. Find thetime at which thedistance travelled is twice thedisplacement. |
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Answer» The maximum height H = v^2/ 2g = 20*20/(2*10) = 20m.Time to reach maximum height is t = v/g = 20/10 = 2 s.If distance = H + x, displacement = H-x.Given H+ x = 2*(H-x)=> x = H/3 = 20/3 m.Time to fall a distance of 20/3 m from rest is found fromt^2 = 2h/ g = 2*(20/3) / 10 = 4/3t = 2/ root3Total time = 2+ 2/ root3 s. |
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| 9085. |
A car is travelling at 36 kmph on a road. If u = 0.5 between the tyres and theroad, the minimum turning radius of the car is (g 10 ms2)a) 20 mb) 25 mc) 30 md) 35 m |
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| 9086. |
is moving in a circular horizontal track ofmovin20. A carradius 10 m with a constant speed of 10 msllumb bob is suspended from the roof of the car bya string of length 1 m. The angle made by the stringwith vertical is (g 10 ms2(a) 0o(b) 30°(d) 60°45° |
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Answer» can we take mv^v/r on left side |
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| 9087. |
A projectile has the same range RGiven g 10 ms2ectile has the same range R 40 m for two angles of projection. If Ti and Th are the times of flight, find T T40 m for two angles of pr |
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Answer» For same rangle... the angles should be complimentary to each other.. and the relationship between R , T1 and T2 is given by R = (T1)(T2)g/2 here R = 40 so , T1*T2 = 40*2/10 = 8 |
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| 9088. |
49. Acar of mass 1500 kg is mevirng with a velocity oi 60 kmtr1, The work done by ts brakes to brinrg t torest is(1) 208.42 k(3) 112,42 Jwork done by its brakes to ring it o e l(2) 198,52(4) 212.52 |
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Answer» Velocity = 60km/h = 60*5/18 = 16.66 work done = change in K.E = 1/2*(1500)*(16.66)² = 208166.7 J = 208.2 KJ thanks please answer my one more question which I have apploaded |
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| 9089. |
A charge of 7 C is carrying from one point to anotherhaving a potential difference of 25 volt. The work donewill be(1) 125 J(2Y 175 J(3) 225 J(4) 275 J |
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Answer» As work done W is equal to the product of potential difference V and charge carried q hence W=V×qW=25 V × 7 CW=175 J |
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| 9090. |
The maximum and minimum distances of a comet from the sun are 1.4 x 1012 m and 7 x 100m Iris velocity nearestto the sun is 6 x 10 ms, what is the velocity in the farthest position? Assume that path of the comet in both theinstantaneous positions is circular25¡3000 m/s] |
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| 9091. |
2A planet of mass m moves arounM in an elliptical orbit. The maximumdistances of the planet from the sunthe sun of masse maximum and minimum2.respectively. The time period oproportional to(2) r 3/2(4) (r1- r2)372(1) r,373/2(3) (r, + 2)3/2t the centre of sun is |
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Answer» Given the perihelion = r2 = shortest distance of a planet from Sun aphelion = r1 = longest distance of planet from Sun mass of planet = m. mass of Sun = M So the length of the major axis of the Elliptical orbit of the planet = r1 + r2. Semimajor axis = R = (r1 + r2)/2. According to Kepler's laws: The square of time period T of a planet revolving around Sun is proportional to the cube of semi major axis R of the elliptical orbit of the planet. T² ∞ R³ T∞R³/² T ∞ [r1 + r2] ³/² Answer. option 3 |
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| 9092. |
1.A particle is projected vertically upwards from ground with velocity 10 m/s. Find the time taken byreach at the highest point ? |
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Answer» Given:velocity=10m/s Time take will be the ascent given by u/gwhere u=initial velocity=10m/sg=acceleration due to gravity=9.8m/s² now t=10/9.8=1.02 second what is your question |
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| 9093. |
A block of mass 2 kg is resting on a horizontaltable. A force of 10 N is applied for 4 s on theblock in the horizontal direction. If g 10 ms-1and the cofficient of kinetic friction between theblock and the table 0.2, then thethe net force is.(a) 144J (b)96J (c) 72J (d) 36 Jwork done by |
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Answer» Answer: a)144 JExplanation: Friction force= 4 Napplied force=10 NNet force=applied - friction=10-4=6 Nacceleration=Force/ mass= 6 N/2 kg=3 m/s^2distance= 24 mwork done by net force= 6 N* 24 m=144 J Distance travelled =s=ut+1/2(ať^2)here u =initial speed =0s=0+1/2(3*4*4)=3*4*2=24 |
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| 9094. |
27. A block of mass 2 kg is resting on a horizontaltable. A force of 10 N is applied for 4 s on theblock in the horizontal direction. If g 10 ms2,and the cofficient of kinetic friction between theblock and the table- 0.2, then the work done bythe net force is.(a) 144J (b)96J (c) 72J (d) 36J |
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Answer» Friction force= 4N APPLIED FORCE= 10 N NET FORCE= 10N-4N= 6N acceleration of block= 6N/2kg= 3m/s2 distance travelled in 4 second= 24 m WORK DONE BY APPLIED NET FORCE= 6N X 24m=144JOption (A) |
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| 9095. |
. When an arrow is shot, where from the arrow will acquire its kinetic energy? |
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Answer» A stretched bow has potential energy due to the change in its shape. To shot the arrow, the bow has to be released.Therefore, its potential energy is converted into the kinetic energy of the arrow. |
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| 9096. |
If two people are pulling a box in the samedirection with 85 N and 65 N force. Netdisplacement of the box is 18m. Find the network done. |
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| 9097. |
4. (a) How does the fusion point and solidificationpoint of ice and water compare numerically? |
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Answer» Melting of a pure substance occurs at a particular constant temperature called melting point. The change of state from a liquid to a solid is called solidification or freezing or casting. A pure substance freezes at a temperature equal to its melting point. it ask to compare numerically |
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| 9098. |
5. The trajectory of a projectile in a vertical plane is yaz -ba2 where a and b are constants and and y arerespectively horizontal and vertical distances of the projectile from the point of projection. The maximum heightattained by the particle and the angle of projection from the horizontal are:A) tan-1 (b)C) , tan-1 (a)B) tan (2)D)2, )tan-1 (a4b' |
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| 9099. |
hackhadin toangula20 an s-1, Calcola4ÂŁ94.21 |
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Answer» it's not angular speed it's angular velocity |
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| 9100. |
How to Hack Wifi password using cmd |
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Answer» Wifi password are usually based on 128bit and 256-bit encryption. Today mostly encryption is done using RSA algorithm so it is quite difficult for anyone to hack. Moreover RSA algorithm is bases on factorization of prime number and cracking needs super computer to run all the possibility. netsh wlan show profiles |
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