Hi Brilliant! I recently wrote my National Science Olympiad and I really want to compare my answers with your answers.

Please look at my comments below and try solve the problems with clear explanations.

Update: Keep this note on reach, because there will be **more questions tomorrow!**

Update (2014/03/09): **Another 5 Questions** are added. Look for the questions that say "UPDATE 3". Chemistry forms an inextricable part of Science. Wait for the Chemistry problems from my Science Olympiad tomorrow!

Don't forget to **like or re-share** this note to keep getting more problems.

Thanks again to all the people that are contributing to this note. I'm learning so much!

**STAR SOLVER: Anish Puthuraya**

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TopNewest*UPDATE 3For a "p" electron (quantum number

I= 1) the possible values of the magnetic quantum numbersmare:Log in to reply

D

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D

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D

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D

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UPDATE 3Which one of the following statements is correct?

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D

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UPDATE 3A source emits electromagnetic waves of wavelength 3 m with intensity "l". One beam reaches the observer directly and the other after the reflection from a water surface, and so travelling an extra distance of 1.5 m whilst its intensity is reduced to 1/4 of the original intensity. The resultant intensity as seen by the observer is:

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D

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UPDATE 3Which one of following expressions will have the dimensions of time? Here "L" is inductance, "R" is resistance and "C" is capacitance.

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C

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Btw, would you make another separate note? This is getting too messy!

Or rather, just post these questions as problems (not notes)..

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UPDATE 2The age of wood can be found by comparing the amount of carbon-14 a sample contains to the amount of carbon-14 in a fresh piece of wood. Such a piece of wood contains 8 times the amount of carbon-14 as a sample from an ancient campfire. How many years ago was the campfire burning if the half-life of carbon-14 is 5600 years?

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C

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UPDATE 2Radiation of frequency \(10^{15}\) Hz shines on the surface of a metal whose work function is 1 eV (1.6 x \(10^{-19}\) J). The retarding potential which just prevents the ejection of photo-electrons is:

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Im getting the answer as 3.14375V...Could you check the problem again?

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I haven't made any mistake with the posting of the question.

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UPDATE 2A transformer has 100 turns in the primary coil and 2000 turns in the secondary coil. If an alternating potential difference of 12 V is applied across the primary, the potential difference across the secondary will be:

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B

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UPDATE 2Red light passes through a yellow filter, what colour is seen coming out of the filter?

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I suppose (im not

sosure, but im pretty sure), the answer is ALog in to reply

Why would it be orange? Yellow is a combination of green and red . So a yellow filter would only stop blue light. Red light would come out the same. i.e. Red. \( \boxed{C} \)

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My thinking was that the red light would be mixed with yellow and thus orange would be produced...I just used my logic, I didn't solve it rigorously..

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Filters allow the passing of light of specific frequencies(colors) and block all other light(colors).

Primary(RGB) color filters allow the passing of only the color they are of,e.g. a Red filter would only allow the passage of Red.

So if white light is passed through a red filter, only red light would come out. And if blue light is passed, no light would emerge.

Secondary color(Yellow, Magenta, Cyan) color filters allow the passing of the colors of which they're made up of. So a Yellow(Green + Red) filter will allow Green and Red light to pass. For Eg. if white light is passed through a yellow filter, then only Red and Green light will pass, which would appear Yellow to us. If Cyan(Blue + Green) light is passed through a yellow filter, then only the Green light would emerge. And if Red light is passed through a Yellow filter, then only Red light would pass through.

Note:- Filters don't add colors, only remove them.

Note II :- I didn't explain the bandwidth of filters which I felt would be too long.

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UPDATE 2Suppose you are standing on the platform of a railway station. As the train approaches the station, it gradually slows down. During this process of slowing down, the driver sounds a horn which emits sound waves at a constant frequency of 300 Hz. Which statement correctly describes the pitch, or changes in pitch, that you will hear as the train approaches you? It will:

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D

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I didn't know how to approach a problem like this. Can you explain the reasoning behind you answer?

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It is a very common phenomenon...When a train passes by you, you seem to hear a rise in the frequency of the train's horn and then a fall in the frequency when it moving away from you...Its quite a beautiful thing.

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UPDATETwo bodies, X and Y, of masses M kg and 2M kg respectively are moving towards each other due to their mutual gravitational attraction. If the velocity of X is "u" and the velocity of Y is "v", what is the velocity of their centre of mass?

(P.S.SQRT means to take the square root of a number. So SQRT(4) would be 2).

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\[v_{cm} = \frac{M_1v_1+M_2v_2}{M_1+M_2}\]

Thus,

\[v_{cm} = \frac{Mu-2Mv}{M+2M} = \frac{Mu-2Mv}{3M} = \frac{u-2v}{3}\]

The options are definitely wrong.

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Oh no! I think I made a type error with this question. It's not "mutual gravitational acceleration", but "mutual gravitational attraction". I've changed it now.

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relativevelocities, then the answer might be one of the options..Log in to reply

Since their is no external force, Vcm will be 0 so 'u' will be equal to '2v'

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accelerationis zero, but not the velocity..Log in to reply

onlydue to mutual gravitational attraction then the \(v_{cm}\) of the system would be zero.Yeah, that's true only if we assume the condition of being initially at rest. Only going by what has been given then your answer is absolutely correct.

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UPDATEThe RMS value of alternating current which produces heat in a given resistor at twice the rate as direct current of 3A is in amperes:

(P.S. SQRT means to take the square root of a number. So SQRT(4) would be 2)

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By definition,

\[P_{avg} = i_{rms}^2R\]

Using the given conditions,

\[P_{avg} = 2\times i^2R = 2\times (3)^2R\]

Equating the two equations,

\[i_{rms}^2R = 2\times (3)^2R = 2\times 9R = 18R\]

Thus,

\[i_{rms}^2 = 18\]

\[\Rightarrow i_{rms} = \sqrt{18} = 3\sqrt{2}A\]

Answer is D

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Hi Anish. Thanks for working through all these problems. I really need to acknowledge the help you've given me, so I made you the STAR solver. See the above note description.

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And thanks for that STAR badge...Its really like an Oscar to me.!

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Moreover, please continue solving my problems. Keep up your awesome work!

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here

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UPDATEA flash of lightning discharged 60 C of charge at a potential difference of \(10^{9}\) V in \(10^{-2}\) seconds. The current is:

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Current is defined as \(\displaystyle\frac{charge}{time}\), hence,

Current \(\displaystyle = i = \frac{q}{t} = \frac{60}{10^{-2}} = 6000A\)

Note that the current is independent of the P.D.

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UPDATEA galvanometer has a resistance of 3663 ohms. A shunt "S" is connected across it such that (1/34) of the total current passes through the galvanometer. The resistance of the shunt is:

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The potential difference across both the resistors must be the same (since they are in parallel)

Thus,

\[\frac{i}{34} 3663 = \frac{33i}{34} S\]

Solving this equation,

\[S = 111\Omega\] Answer is B

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UPDATEThree charges are placed along the X-axis. Charge A is a + 18 nC charge placed at the origin, 0 cm mark. Charge B is a - 27 nC charge placed at the 60 cm mark. Where along the axis must a negative charge C be placed in order to be in equilibrium. At the:

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I believe that the options are wrong (atleast in units)

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Thanks for bringing this to my attention. It's actually suppose to be "nC". Brilliant has this autocorrect feature.

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meters, not in cm)..I'll try it again later (though, I still think that my calculations were correct)

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Three resistors X, Y and Z are connected in parallel, with the resistance of X < Y < Z. The value of the equivalent resistance R of the parallel combination is:

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C.....In case of parallel combination, the equivalent resistance is less than even the smallest resistance while in case of series combination, the equivalent resistance will be greater than greatest resistance.

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It's pretty easy to show mathematically:

The equation is as follows: \(\frac{1}{R_{eq}} = \left(\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}\right) \). Without loss of generality, let \(0 < R_1 \leq R_2 \leq \dots \leq R_n\). If the only resistor was \(R_1\), when we have \(R_{eq} = R_1\) (of course, in that case, we could not hvae resistors

in parallel). By adding some other resistors \(R_2, R_3, \dots, R_n\), we increase the right hand side, thus we must increase the left hand side by decreasing \(R_{eq}\). Thus, \(R_{eq} < \min (R_1, R_2, \dots, R_n)\).Log in to reply

The world record for a high dive into deep water is 54 m. A diver of mass 64 kg dives into the water below him. If air resistance is insignificant, his velocity on entering the water is independent of his:

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Weight is such a quantity which is not considered at all when and

onlywhen there is no air resistance and other non conservative forces.Answer is B.

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I would disagree. B would be the best option if I had to chose (in an exam), but I don't think it is true.

We are already given his mass is 64 kg. Hence, if his weight is different (say compare 64N vs 640N), then it means that gravitational acceleration is different ( 1 m/s^2 vs 10 m/s^2), and hence his velocity will be different.

Note: The velocity is independent of his mass. However, weight is not equal to mass (most people do not distinguish between these two concepts).

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A variation in the weight can therefore be generated by either a variation in the mass

ora variation in g. If the mass varied, then the the velocity is unchanged as the mass cancels out of the dynamics. If g changes then the kinematics does change and so does the velocity. Hence, depending on what you choose to vary when the weight varies you get either the velocity changes or it doesn't. Ambiguity! Reading the question as holding the diver's mass fixed would mean that the weight change is due to a change in g, invalidating answer B. They messed up and should've put mass for option B.I hope you got it right anyways!

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Isn't that true? I mean to say that, If you were to calculate the velocity using Newton's Laws, then I believe that the answer would be independent of the weight of the diver, right?

I think it is not a matter of

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True, the only reason a feather takes longer to hit the ground is air resistance. With that gone, weight doesn't matter.

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B.......though now I too think it is just the best possible answer

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A stone of mass "m" is thrown from the top of a cliff "H" metres above the lake at a speed of "u" m/s. At what angles must it be thrown so that it hits the water at the maximum speed.

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The change in Potential Energy is the same for any angle, so the change in Kinetic Energy is independent of the angle of the stone.

Hence, the speed with which the stone hits the lake is constant (independent of angle).

Answer is D.

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D....final speed will be sqrt(2gh).

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D, vertical velocity is independent of horizontal velocity because gravity acts on every object the same which doesn't depend on the speed it's moving horizontally

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It takes a man 10s to ride down an escalator. The same man takes 15 s to walk back up the escalator against its motion. How long will the man take to walk down the escalator at the same rate he was walking before?

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Was None of the above an option? Let the speed of the escalator be "x" , spd of man be "y" and length of escalator "d". Then we have to find time taken when speed = x + y.

\( x = \frac{d}{10} \)

\( y - x = \frac{d}{15} \)

\( y + x = y - x + 2x = \frac{d}{15} + 2\frac{d}{10} = \frac{4d}{15} \).

Therefore Time = \( \frac{15}{4} \) = \( \boxed{3.75} \)

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Me too...I got the same answer as you. Something doesn't seem right with the options..Are they asking something else?

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A ball whose kinetic energy is E is thrown at an angle of 45 degrees with the horizontal. Its kinetic energy at the highest point of it flight will be:

(p.s. SQRT means square root, so SQRT(2) means the square root of 2)

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C

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I just have an area of concern with this problem. Why can't it be zero?

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A body of mass "M" is thrown horizontally with a velocity of 60 km/h from the ledge of a tower from height "h". It strikes the level ground a distance of 400 m from the foot of the tower. Next, a body of mass "2M" is thrown horizontally with a velocity of 30 km/h from another ledge of the tower, this time from a height of "4h". At what distance from the foot of the tower would it strike the level ground?

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B

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A boat can row with a speed of 10 km/h in still water. The river flows steadily at 5 km/h. In which direction, relative to the direction of the flow of the river, should the boatman row in order to reach a point on the other bank directly opposite to the point from where he started?

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Not a really good question. It's kind of obvious you would have to go against the current if you have to reach at a point directly opposite. The only one which satisfies that is D.

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A

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Can you work out the other problems?

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Can you please explain how you arrived to that answer? I really struggled to solve this problem. Thanks for the response!

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alt text

It is clear from the figure that the net velocity of the boat (relative to the ground) is the

vectoraddition of \(v_{boat}\) and \(v_{river}\)For the boat to reach a point directly opposite,

\(v_{net}\) should be along \(y-axis\)

In other words, the \(x\)-component of the net velocity of the boat must be

zero.The \(x\)-component of the net velocity is given by,

\[v_x = v_{river} + v_{boat}\cos\theta\]

As \(v_x = 0\),

\[v_{river} = -v_{boat}\cos\theta\]

Using the given values,

\[5 km/h = -10km/h\cos\theta\]

\[\Rightarrow\cos\theta = -\frac{1}{2}\]

Hence,

\[\theta = \boxed{120^o}\]

Im sorry..I didn't notice that the problem mentioned to take the angle relative to the direction of flow of the river..So, the answer is definitely \(120^o\).

Sorry for the confusion.

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A, B, C, D are four points on the same vertical line and are such that AB = BC = CD. If a particle falls freely from rest from point A, the times taken by it to move distances AB, BC, and CD are in the ratio of:

D \( \qquad {1}\ : \sqrt{2}\) - \({1}\ : \sqrt{3}\) - \(\sqrt{2}\)

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D

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It is D

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D

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UPDATE 3A train 100 m long travelling at 40 m/s overtakes another train 200 m long travelling at 30 m/s. The time taken by the first train to pass the second train is:

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A

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A

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