Crack the Problem

Find the area of the region consisting of the set of points given by,
{(x, y)|x²+y²-6|x|-6|y|+9\(\le\) 0, |x|+|y|\(\le\)3}.

Submission Closed

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Cracked By:

• Praveen Gorakavi, 12th Pass, Narayana Junior College, Hyderabad
• T.R. Manoj, 12th, Vijaya Ratna Junior College, Hyderabad
• Ankul Garg, 12th, Bal Bharati Public School, Rohini, Delhi (Ankul has qualified for Indian Physics Olympiad-2007)
• Chandni Bhatia, 12th, Kendriya Vidyalaya No.2, Bhopal
• Anit Kumar Sahu, 12th, Ispat English Medium School, Sector-20, Rourkela (Anit has topped Orissa in regional maths Olympiad)
• Ravi Tej, 12th, Vijaya Ratna Junior College, Hyderabad

Answer: \(9(\pi-2)\)

An organic compound C₆H₄N₂O₄ is insoluble in both dilute acid and base. Dipole moment of the compound is zero. Write the IUPAC name of the compound.

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Submission Closed

Cracked By:

Ankul Garg, 12th, Delhi
T.R. Manoj, 12th, Hyderabad
Sooraj Prakash, 12th Pass, Bhopal
Harikrishna Devarajan, Chennai
Virgaonkar Waman, 12th, Nashik
Chandni Bhatia, 12th, Bhopal
Akshay Hattiangadi, 12th, Mumbai
Praveen Gorakavi, 12th Pass, Hyderabad

Answer: 1,4-dinitrobenzene

The following answer is the edited version of the explanations provided by Manoj & Sooraj:

As the D.U. is 6 and hydrogen to carbon ratio is less than 1, we expect the benzene ring (Huckel's rule satisfies this as well). As the above compound is insoluble in base, it means there is no acidic proton donor site, hence no phenolic or carboxylic acid group is expected in the side chain (in fact looking at the chemical formula, the side chain must have nitrogen and oxygen). Now as the above compound is insoluble in acid it means there is no basic site .i.e there is no lone pair on nitrogen available for donation. All these are possible in the case of nitro group. Also, intermolecular hydrogen bonding is ruled out as well (further cause for solubility). Finally, as the dipole moment is zero it means the molecule is symmetric.

Match the following (more than one and non-unique matches are possible). Answer will be considered only if all matches are correct:

(a) K electron capture	(p) excess protons
(b) positron emission	(q) x-rays
(c) crystallography	(r) \(\gamma\)-rays
(d) electron-positron annihilation	(s) photons

Submission Closed

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The following explanation is the edited version of Ankul Garg's submission.

a) K electron capture is a decay mode for those isotopes which have too many protons in their nuclei. It happens by the capture of a K-shell electron in the nucleus leading to formation of a positron and neutrino. Also, after K electron capture, the atom in its excited state emits X-rays due to electron transition from outer shell to inner shell. X-rays are made of photons. Thus, a) matches with p), q) & s).

b) Positron emission is also another mode of decay in isotopes having too many protons in their nuclei. It is Beta plus decay in which a proton gets converted into neutron, positron and a neutrino.
So b) matches with p).

c) In crystallography, X-rays are used to analyse the diffraction pattern of the crystal. X-rays are made of photons. So c) matches with q) & s).

d) Electron-Positron annihilation occurs when an electron collides with a positron to create gamma rays photons. In fact, the conservation of linear momentum and total energy in the process forbids the emission of a single gamma ray but leads to emission of a large no. of gamma ray photons.
So d) matches with r) and s).

Hence correct match is: pqs,p,qs,rs.

Evaluate, \(\int\limits_0^{\pi /2}{\ln [1+2e(e-1)(1+cos 2x)]}dx\).

Type your final answer in numbers writing the method in brief.

Cracked By:

Chandni Bhatia
Praveen Gorakavi
Sanchit Sharma

Answer: 3.14 (\(\pi\))

Take each side of the pyramid having resistance 1 ohm. Find the resistance between a and b. Type the numerical answer up to at least two decimal places assuming the unit to be ohm. Treat the dashed line just like any other line, i.e. having resistance 1 ohm.

Cracked by*:

Garg, Ankul
vs, karthik
Mukherjee, Soumik
Poolla, Chaitanya
Tirukodi Radhakrishn, Manoj

*Some students have written non-numerical answer. Since the question mentioned numerical answer, these answers are not considered.

Answer: 0.53

Sea-water is found to contain 5.85% NaCl and 9.50% MgCl₂ by weight of solution. Assuming 80% ionisation of NaCl and 50% ionisation of MgCl₂ calculate the boiling point of sea-water in ºC. K_b for water=0.51 K.Kg/mol. Type only the numerical value up to at least one decimal place.

Cracked By:

Garg, Ankul
BOBADE, PARAG

Answer: 101.94

10^sinx-10^-sinx-y=0 has at least one real solution for \(x\in[-\frac{\pi}{2},\frac{\pi}{2}]\) then [|y|] cannot exceed (type only the smallest numerical value as a natural number, [ ] denotes the greatest integer function):

Cracked By:

vs, karthik
Garg, Ankul
patil, rahul
Zafar, Asif
Narsapur, Sameer
Saxena, Vishesh
Thanki, Parashar
gorthi, chandu

Answer: 9

In an experiment 31.26 ml of a 0.165 M solution of Ba(OH)₂ solution is required to just neutralise 25 ml of citric acid. What is the concentration of the citric acid (H₃C₆H₅O₇) solution?

Type in brief your solution along with the final answer in molarity.

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Cracked By:

vs, karthik
Tirukodi Radhakrishn, Manoj

Answer:

0.138 M

Consider two situations:

1. A point charge q is placed at a distance d from a large conducting uncharged plate. The magnitude of the force experienced by q is R.
2. A point charge q is placed at a distance d each from the two perperdicular conducting uncharged large plates meeting at right angle. The magnitude of the force experienced by q is S.

Enter the numerical value of (S/R) where ( ) denotes the least integer function in the space provided at http://courses.manishverma.site/mod/quiz/view.php?id=433 .

Note: Only number(s) should be typed.

Cracked by:

• karthik vs
• Soumik Mukherjee

Answer: 1

Hint:

These types of problems are solved by the image method. A system of charge q at distance d from a plane metal sheet is replaced by a system of charge q with -q charge at distance 2d. In this method when the system is substituted by the charges, the potential at the plane originally occupied by the sheet must be zero.

When charge q is at distance d each from the two perpendicular metal sheets and if the system is replaced by q, -q and -q then the zero potential condition is not met. In this case the system should be replaced by q, -q, -q and a fourth charge q. The location of these three image charges is the same as the location of images when an object is placed in front of two mutually perpendicular mirrors and three images are formed.

(Edited by Manish Verma - Friday, 1 September 2006, 11:38 AM)

30 ml of a solution containing 9.15 gm/l of an oxalate K_xH_y(C₂O₄)_z.nH₂O are required for titrating 27 ml of 0.12 N NaOH and 36 ml of 0.12 N KMnO₄ separately. Assume all H atoms are replaceable and x, y, z are in the simple ratio of g atoms.

Calculate the following:

1. x
2. y
3. z
4. n

Submission Closed

Cracked by:

1. Aditya Gopi, 12'th, C.M.R N.P.S, Bangalore.

2. Ajay N. Nandoriya, 12'th, School of Science, Rajkot.

3. Udit Khanna, 12'th, St. Anselm’s Pink City School, Jaipur.

4. Ankul Garg, 12'th, Bal Bharti Public School, Rohini, Delhi.

Udit's Solution:

Let the no. of millimoles of the salt in 30 ml soln = a

Then, the no. of meqts of H⁺ = ay (let meqts = milliequivalents)

Since meqts of H⁺ = meqts of OH^- in NaOH

Therefore, ay = 27 * 0.12 = 3.24 ………….(i)

No. of meqts of C₂O₄^2- = 2az

Again since meqts of C₂O₄^2- = meqts of MnO₄^- in KMnO₄

Therefore, 2az = 36 * 0.12 = 4.32 ………….(ii)

Since salt must be electrically neutral,

Therefore, x + y = 2z ……………………(iii)

From (i), (ii) and (iii) , we get ,

x : y : z = 1 : 3 : 2

Since, x, y and z are the simplest ratio’s, therefore,

x = 1

y = 3

z = 2

From (i), a * 3 = 3.24

=> a = 1.08 …………………..(iv)

Molecular wt. of the salt = (39 * 1) + (1 * 3) + (88 * 2) + (18 * n)

= 218 + 18n

wt. of salt in 1 Lt. soln = 9.15 gm

=> wt. of salt in 30 ml soln = 9.15 * 30 milligram = 274.5 mgm

=> moles of salt in 30 ml soln = mmoles

=> a =

=> n = 2

Hence, the answers are,

x = 1

y = 3

z = 2

n = 2

And the salt is : KH₃(C₂O₄)₂.2H₂O

Ankul's Solution is attached (.pdf).