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**

## Comments

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

– Mark Mottian · 2 years, 10 months agoI= 1) the possible values of the magnetic quantum numbersmare:Log in to reply

– Anish Puthuraya · 2 years, 10 months ago

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– Pankaj Joshi · 2 years, 10 months ago

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– Srinivas Kola · 1 year ago

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– Abhijeet Dhakane · 2 years, 8 months ago

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UPDATE 3Which one of the following statements is correct?

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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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.

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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Or rather, just post these questions as problems (not notes).. – Anish Puthuraya · 2 years, 10 months ago

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– Mark Mottian · 2 years, 10 months ago

Alright Anish. I'll post the chemistry problems (and maybe some of the other physics problems in a separate note tomorrow). I can't post these questions as problems because I don't know some of the answers and I need to discuss the problems I don't know how to solve.Log in to reply

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?

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

Im getting the answer as 3.14375V...Could you check the problem again?Log in to reply

– Mark Mottian · 2 years, 10 months ago

I haven't made any mistake with the posting of the question.Log in to reply

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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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

– Mark Mottian · 2 years, 10 months agoLog in to reply

sosure, but im pretty sure), the answer is A – Anish Puthuraya · 2 years, 10 months agoLog in to reply

– Siddhartha Srivastava · 2 years, 10 months ago

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. \( \boxed{C} \)Log in to reply

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.. – Anish Puthuraya · 2 years, 10 months ago

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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. – Siddhartha Srivastava · 2 years, 10 months ago

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– Anish Puthuraya · 2 years, 10 months ago

That makes sense, ThanksLog in to reply

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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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– Mark Mottian · 2 years, 10 months ago

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

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. – Anish Puthuraya · 2 years, 10 months ago

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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). – Mark Mottian · 2 years, 10 months ago

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Thus,

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

The options are definitely wrong. – Anish Puthuraya · 2 years, 10 months ago

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– Mark Mottian · 2 years, 10 months ago

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.Log in to reply

– Mark Mottian · 2 years, 10 months ago

Does your answer work out now? The options remain the same.Log in to reply

– Anish Puthuraya · 2 years, 10 months ago

But, the options are still wrong.! The answer is \(\displaystyle\frac{u-2v}{3}\)..Im pretty sure that it is correct.Log in to reply

– Mark Mottian · 2 years, 10 months ago

Well that's weird. I'm not sure what happened over here. That explains why I couldn't work out this question. At the end, I just guessed one the answers.Log in to reply

relativevelocities, then the answer might be one of the options.. – Anish Puthuraya · 2 years, 10 months agoLog in to reply

– Kalpit Jain · 2 years, 10 months ago

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

accelerationis zero, but not the velocity.. – Anish Puthuraya · 2 years, 10 months agoLog 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. – Sudeep Salgia · 2 years, 10 months ago

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– Anish Puthuraya · 2 years, 10 months ago

Yeah.Log in to reply

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) – Mark Mottian · 2 years, 10 months ago

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

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 – Anish Puthuraya · 2 years, 10 months ago

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– Mark Mottian · 2 years, 10 months ago

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.Log in to reply

And thanks for that STAR badge...Its really like an Oscar to me.! – Anish Puthuraya · 2 years, 10 months ago

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Moreover, please continue solving my problems. Keep up your awesome work! – Mark Mottian · 2 years, 10 months ago

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here – Anish Puthuraya · 2 years, 10 months ago

Check it outLog in to reply

– Mark Mottian · 2 years, 10 months ago

Those IIT-JEE problems are not straightforward. They'll be useful material to me. Thanks! (P.S. After looking at that paper, the level of Science in India looks much higher than South Africa).Log in to reply

– Aabhas Mathur · 2 years, 10 months ago

Well yes the answer is D....and great going Anish . I got a bit late today and yes our IIT JEE problems are lot harder than these.Log in to reply

– Anish Puthuraya · 2 years, 10 months ago

Why thanks!Log in to reply

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

– Mark Mottian · 2 years, 10 months agoLog in to reply

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

Note that the current is independent of the P.D. – Anish Puthuraya · 2 years, 10 months ago

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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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

Thus,

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

Solving this equation,

\[S = 111\Omega\] Answer is B – Anish Puthuraya · 2 years, 10 months ago

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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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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

– Mark Mottian · 2 years, 10 months ago

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

meters, not in cm)..I'll try it again later (though, I still think that my calculations were correct) – Anish Puthuraya · 2 years, 10 months ago

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– Sudeep Salgia · 2 years, 10 months ago

Yeah it is mostly \(-2.7 m\).Log in to reply

– Anish Puthuraya · 2 years, 10 months ago

mostly?Log in to reply

– Sudeep Salgia · 2 years, 10 months ago

It is \(-2.7m\). Sorry for the uncertainty.Log in to reply

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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Aabhas Mathur · 2 years, 10 months ago

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.Log in to reply

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, \(R_{eq} < \min (R_1, R_2, \dots, R_n)\). – Michael Tong · 2 years, 10 months agoLog 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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

onlywhen there is no air resistance and other non conservative forces.Answer is B. – Anish Puthuraya · 2 years, 10 months ago

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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). – Calvin Lin Staff · 2 years, 10 months ago

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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! – David Mattingly Staff · 2 years, 10 months ago

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– Mark Mottian · 2 years, 10 months ago

Hi David. Yeah, I also found this question a bit sceptical. I was working through this problem and I also noticed that grey area. My answer was "B", but I was a bit unsure, that's why I posted it here. Thanks for clearing things up.Log in to reply

– Parth Thakkar · 2 years, 10 months ago

Well, I think it is just a matter of not wording the question properly. I think they shouldn't have given weight as an option if mass was already given.Log in to reply

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

changinghis weight. – Anish Puthuraya · 2 years, 10 months agoLog in to reply

– Parth Thakkar · 2 years, 10 months ago

What he is saying is that if the mass is already given (fixed) and weight is a variable, then it would mean that \( g \) is a variable. When I read that comment first, I too thought he was saying it depends upon the mass/weight.Log in to reply

– Robert Fritz · 2 years, 10 months ago

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

– Aabhas Mathur · 2 years, 10 months ago

B.......though now I too think it is just the best possible answerLog in to reply

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.

– Mark Mottian · 2 years, 10 months agoLog in to reply

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

Answer is D. – Anish Puthuraya · 2 years, 10 months ago

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– Aabhas Mathur · 2 years, 10 months ago

D....final speed will be sqrt(2gh).Log in to reply

– Robert Fritz · 2 years, 10 months ago

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 horizontallyLog in to reply

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?

– Mark Mottian · 2 years, 10 months agoLog in to reply

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

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

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

Therefore Time = \( \frac{15}{4} \) = \( \boxed{3.75} \) – Siddhartha Srivastava · 2 years, 10 months ago

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– Anish Puthuraya · 2 years, 10 months ago

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

– Mark Mottian · 2 years, 10 months ago

Hmmm, I can see your reasoning. I double checked the question and I think they may have made a mistake with the options.Log in to reply

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) – Mark Mottian · 2 years, 10 months ago

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– Aabhas Mathur · 2 years, 10 months ago

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– Mark Mottian · 2 years, 10 months ago

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

– Aabhas Mathur · 2 years, 10 months ago

The horizontal velocity in case of projectile motion is always constant=u cos(theta). So kinetic energy cannot be zero.Log in to reply

– Anish Puthuraya · 2 years, 10 months ago

True..Your answer is also correct.Log in to reply

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?

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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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?

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Siddhartha Srivastava · 2 years, 10 months ago

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.Log in to reply

– Anish Puthuraya · 2 years, 10 months ago

ALog in to reply

– Mark Mottian · 2 years, 10 months ago

Can you work out the other problems?Log in to reply

– Anish Puthuraya · 2 years, 10 months ago

Done, I solved every problem except for the first one..It should be \(3.75 sec\)Log in to reply

– Mark Mottian · 2 years, 10 months ago

Geez, you're hard-core! Keep up this fantastic physics problem solving!Log in to reply

– Mark Mottian · 2 years, 10 months ago

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

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. – Anish Puthuraya · 2 years, 10 months ago

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– Mark Mottian · 2 years, 10 months ago

Yeah that makes a lot more sense. Thank you for your concise explanation! This solution is awesome!Log in to reply

– Kou$htav Chakrabarty · 2 years, 10 months ago

Yea really awesome! Thanks!Log in to reply

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 \( \qquad {1}\ : \sqrt{2}\) - \({1}\ : \sqrt{3}\) - \(\sqrt{2}\) – Mark Mottian · 2 years, 10 months ago

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– Anish Puthuraya · 2 years, 10 months ago

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– Kalpit Jain · 2 years, 10 months ago

It is DLog in to reply

– Pankaj Joshi · 2 years, 10 months ago

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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:

– Mark Mottian · 2 years, 10 months agoLog in to reply

– Anish Puthuraya · 2 years, 10 months ago

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– Pankaj Joshi · 2 years, 10 months ago

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