InterviewSolution
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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.
| 201. |
Define 1 W of power. |
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Answer» Power of an agent is said to be 1 W if it does 1 joule of work in 1 s, i.e., `1 watt (W) = (1 joule(J))/(1 second (s)) =1 J//s` In other words, we say that power is 1 W when the rate of consumption of energy is 1 J//s. |
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| 202. |
A lamp consumes 1000 J of electrical energy is 10 s. What is its power ? |
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Answer» Here, electrical energy consumed, W = 1000 j time in which this energy is consumed, t =10 s Power fo the lamp, `P =(W)/(t) = (1000J)/(10s) = 100W` |
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| 203. |
How can you visualize a joule of energy ? |
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Answer» A body of mass 1 kg dropped from a height of 10cm possesses a kinetic energy of 1J. `[ as upsilon^2 =2gh = 2xx10xx0.1 = 2, KE = (1)/(2) m upsilon^2 = (1)/(2) (1) (2) = 1J]` |
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| 204. |
What is the relation between a watt (W) and a joule (J) ? |
| Answer» `1 W = 1 J//s`. | |
| 205. |
How is horse power (hp) related to a watt (W) ? |
| Answer» `1 hp = 746 W`. | |
| 206. |
If an electric iron of 1200 W is used for 30 minutes everyday, find electric energy consumed in the month of Aprill. |
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Answer» We are given that power of the electric iron, `P = 1200 W =1.2 kW`, total time for which the electric iron is used in the month of April `(30 days)` =` (30 min//day) (30day)` `=((1)/(2)h//day) (30day) = 15 h` Energy consumed, `E = Pt = (1.2 kW) (15h) = 18kWh (as P = "work done//time" = E//t)` |
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| 207. |
A light body and a heavy body have the same momentum, which of the two bodies will have greater kinetic energy ? |
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Answer» As `E-k = (p^2)/(2m), E_k prop (1)/(m) ("P is the saem for both the bodies")` Thus, the lighter body has more kinetic enegy then the heavier body through both have the same momentum. |
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| 208. |
A light and a heavy body have the same kinetic energy. Which one of the two will have greater momentum ? |
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Answer» Kinetic energy, `E_k = (1)/(2)m upsilon^2 = ((m upsilon)^2)/(2m) = (P^2)/(2m) (momentum, P = m upsilon)` Thus, `P^2 = 2 m E_k or p = sqrt(2m E_k)` Since `E_k` is the same for both the bodies, `p prop sqrtm.` Thus, the heavier body (with more m) will have more momentum then the ligheter body (with less m), though both have the same kinetic energy. |
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| 209. |
which has a greater kinetic energy : a supertanker berthed at a pier or a motor boat pulling a water skier? |
| Answer» Motor boat pulling a water skier as the supertaker is at rest (Pier is a structure leading out to sea and used as a landing stage for boats). | |
| 210. |
A single conservative `f_((x))` acts on a `m = 1 kg` particle moving along the x-axis. The potential energy `U_((x))` is given by : `U_((x)) = 20 + (x - 2)^(2)` where x is in metres. At `x = 5 m`, a particle has kintetic energy of 20 J The maximum speed of the particle is:A. `1 m//s`B. `29 m//s`C. `sqrt(29) m//s`D. `sqrt(58) m//s` |
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Answer» Correct Answer - D |
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| 211. |
A spring of spring constasnt 50 N/m is compressed from its natural position through 1 cm. Find the work done by the work done by the spring force on the agency compressing the spring. |
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Answer» The magnitude of the work is `1/2 kx^2=1/2xx(50N/m)xx(1cx)^2` `=(25N/m)xx(1xx10^-2)^2=2.5xx10^-3J` As the compressed spring will push the agency, the force will be opposite to the displacement of the point of application and the work will be negative. Thus, the work done by the spring force is -2.5mJ. |
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| 212. |
A single conservative `f_((x))` acts on a `m = 1 kg` particle moving along the x-axis. The potential energy `U_((x))` is given by : `U_((x)) = 20 + (x - 2)^(2)` where x is in metres. At `x = 5 m`, a particle has kintetic energy of 20 J The least value of x (position of particle is ) will be:A. zeroB. `-2`C. `- sqrt(29) + 2`D. `sqrt(29) + 2` |
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Answer» Correct Answer - C::D |
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| 213. |
A single conservative `f_((x))` acts on a `m = 1 kg` particle moving along the x-axis. The potential energy `U_((x))` is given by : `U_((x)) = 20 + (x - 2)^(2)` where x is in metres. At `x = 5 m`, a particle has kintetic energy of 20 J The largest value of x will be:A. zeroB. `-2`C. `- sqrt(29) + 2`D. `sqrt(29) + 2` |
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Answer» Correct Answer - D |
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| 214. |
A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to `(-K//r^(2))`, where k is a constant. The total energy of the particle is -A. `KE = ((k)/(2r)), PE = - ((k)/(r)), ME = - ((k)/(2r))`B. `KE = ((k)/(2r)), PE = - ((k)/(2r)), ME =` zeroC. `KE =` zero, PE = zero, ME = zeroD. `KE = ((k)/(r)), PE = - ((k)/(2r)), ME = ((k)/(2r))` |
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Answer» Correct Answer - A |
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| 215. |
A ball P is projected vertically up. Another similar ball Q is projected at ari angle `45^(@)`. Both reach the same height during their motion. Then, at the starting point, ratio of kinetic energy of P and Q is?A. 0.5B. 0.25C. 2D. 4 |
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Answer» Correct Answer - A |
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| 216. |
A ball is projected vertically upwards. Air resistance and variation in g may be neglected. The ball rises to its maximum height H in a time T, the height being h after a time t : (1) The graph of kinetic energy `E_(k)` of the ball against height h is shown in figure 1 (2) The graph of height h against time t is shown in figure 2 (3) The graph of gravitational energy `E_(k)` of the ball against height h is shown in figure 3 (1) (2) In the above situation the block wili have maximum velocity when:A. the spring force becomes zeroB. the frictional force becomes zeroC. the net force becomes zeroD. the acceleration of block becomes zero |
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Answer» Correct Answer - C::D |
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| 217. |
A ball is projected vertically upwards. Air resistance and variation in g may be neglected. The ball rises to its maximum height H in a time T, the height being h after a time t : (1) The graph of kinetic energy `E_(k)` of the ball against height h is shown in figure 1 (2) The graph of height h against time t is shown in figure 2 (3) The graph of gravitational energy `E_(k)` of the ball against height h is shown in figure 3 (1) -Which the figure shows the correct answers ?A. 3 onlyB. 1, 2C. 2, 3D. 1 only |
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Answer» Correct Answer - A |
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| 218. |
A boy of mass 40 kgruns up a flight of 50 steps,each of 10 cm high, in 5s, Find the power developed by the boy, |
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Answer» Here,mass of the boy, `m =40 kg` total height gained, `h =50xx10cm 500cm =5m` time taken to climb, t =5 s Work done by the boy, `W = mgh =(40kg)(10m//s^2)(5m) =2000J` power developed `P = (W)/(t) =(2000J)/(5s) = 400W` |
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| 219. |
What is the change in gravitational potentia energy of a 50 kg person whon climbs a flight of staris with a height of 3m and a horizontal extent of 5m ? |
| Answer» The change in gravitational potential energy `= mgh = 50 xx 10 xx 3 = 1500 J`. The horizontal extent of stairs has no effect on the answer. | |
| 220. |
A force `F=-K(yhati+xhatj)` (where K is a positive constant) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point `(a, 0)`, and then parallel to the y-axis to the point `(a, a)`. The total work done by the force F on the particle isA. `-2Ka^(2)`B. `2 Ka^(2)`C. `-Ka^(2)`D. `Ka^(2)` |
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Answer» Correct Answer - C |
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| 221. |
A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. At point away from the origin, the particle is in unstable equilibriumB. For any finite non-zero ,value of x, there is force directed away from the originC. If its total mechanical energy if K/2, it. has its minimum ri at the originD. None of these |
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Answer» Correct Answer - D |
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| 222. |
Find the work done of move an object of mass 3 kg to a height of 20 m from the ground (Take g `=10ms^(-2)`). |
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Answer» Work done=mgh `=3xx10xx20=600J` |
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| 223. |
Calculated the amount of force (in N) respuird to displace an object by 2 m in the direction of force. Work done in this process is 5kJ. |
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Answer» `"Work =Force"xx"Displacement"` `5xx1000=Fxx2` `:.F=(5000)/(2)=25000N` force is applied. |
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| 224. |
What amount of work is done in reading this book in terms of scientific definition of work? |
| Answer» Zero, as we are exerting only mentally and not physically to cause a displacement of the book. | |
| 225. |
A block of mass m moving with a velocity `v_(0)` on a smooth horizontal surface strikes and compresses a spring ofstiffuess k till mass coines to rest as shown in the figure. This phenomenon is observed by two observers : (a) Standing on the horizontal surface (b) standing on the block To an observer A, the work done by the normal reaction N between the block and the spring on the block is :A. zeroB. `- (1)/(2) mv_(0)^(2)`C. `+ (1)/(2) mv_(0)^(2-)`D. none of these |
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Answer» Correct Answer - B |
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| 226. |
A ball of mass 5.0 gm and relative density 0.5 strikes the surface of the water with a velocity of 20 m/sec. It comes to rest at a depth of 2m. Find the work done by the resisting force in water: (take `g = 10 m//s^(2)`)A. `-6J`B. `+ 7.5 J`C. `- 9 J`D. `- 10 J` |
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Answer» Correct Answer - C |
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| 227. |
Many people debate about whether there is a God, or what God is. Is God and individual personality, sitting up in heaven listening to and greating out request, or is God simply all the energy that exists in the Universe, or is God simply nature or the laws of nature ? |
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Answer» (a) The Bible says that God is light. And what is light ? It is energy. (b) When a person is brain dead, no electrical activity registers on an ECG test. Why Because the electrical energy which powerd the body is no longer present. (c ) Everything that exists in the Universe is made up of energy and God is the sum total of universal energy. |
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| 228. |
A body of mass 5kg falls from a height of 5m. How much energy does it possess at any instant? |
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Answer» Correct Answer - B Energy at any instant = Potential energy `(E_p)` at a height of 5m i.e., `E_p = mgh = (5kg) (10m//s^2)(5m) = 250J` |
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| 229. |
An object of mass m is moving with a constant velocity `upsilon` How much work should be done on the object in order to bring the object to rest ? |
| Answer» An object of mass m moving with a constant velocity v has a kinetic energy equal to `(1)/(2) mv^(2)`. An equal amount of work (that is, `(1)/(2)mv^(2)` work) should be done on this moving object to make its kinetic energy zero and bring it to rest. | |
| 230. |
A rocket is moving up with a velocity `upsilon` if the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies ? |
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Answer» `Initial KE = (1)/(2) mu^2` `"Final" KE = (1)/(2)m upsilon^(2) = (1)/(2)m (3u)^2 = (9)/(2)mu^2 (as upsilon - 3u)` Thus, `("initial" KE)/("final"KE) = (1/2"mu"^(2))/(1/9"mu"^(2))=1/9=1:9`. |
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| 231. |
When a ball is hung from a vertical spring, it stretches the spring. As it drops, it loses gravitational potential energy , but this does not at all show up as kinetic energy. What happens to the gravitational potential energy ? |
| Answer» The gravitational potentail energy is converted to kinetic energy and the elastic potentic eneergy of the spring. At the bottom, it is all elastic potential energy. | |
| 232. |
Tarzan, who weighs 875 N, swings from a vine through the jungle. How much work is done by the vine as he drops, through a vertical distance of 4 m ? |
| Answer» The moment Tarzan drops from the vine, tension (F) in the vine becomes zero i.e., F= 0. Therefore, work done by the tension in the vine, `W = F xx s = 0 xx 4m =0`. | |