Crack the Problem

(30) What is the e.m.f. of a cell at room temperature containing two hydrogen electrodes, the negative one in contact with 10^-6 M OH^- ions and the positive one in contact with 0.05 M H⁺?

Submission: Closed

Solution provided by Jithin edited by INDIANET :

It is a concentration cell as both the electrodes are made of hydrogen.

\(E_{cell}=\frac {0.059}{n}log \frac {c_1}{c_2}\); (at room temperature)

n = 1;

c₁= 0.05 M .......(Given that cathode has 0.05 concentration of H⁺ which is the positive side)

The ionic product of water = K_w = 10^-14 = [ OH^-] x [H⁺]

or, 10^-6x [H⁺] = 10^-14;

Therefore, c₂= [H+]= 10^-14/10^-6=10^-8M;

\(E_{cell}=\frac {0.059}{1}log \frac {0.05}{10^{-8}}\)

=0.059log(5x10⁶)

=0.059 x 6.6989

=0.3952V

Cracked by:

1. Aman Verma, Hans Raj Model School, Punjabi Bagh, New Delhi.
2. Jithin Prakash, 12, Army School, Deolali.

(29) a, b, c are sides of a triangle, then prove that

.

Submission: Closed

Solution (.pdf) provided by Jathin

Cracked by:

1. Umang Merwana, Bhopal (M.P.).
2. Jathin Das, Class XII, Abu Dhabi Indian School, Abu Dhabi, UAE.

(28) Evaluate:

Where [ ] denotes the greatest integer function.

Submission: Closed

Solution Provided by Nachiket:

We shall first prove one general result (A)
\(S=[x]+[x+\frac{1}{n}]+[x+\frac{2}{n}]+............+[x+\frac{n-1}{n}]=[nx]\)

Proof::
Let p be the first number for which \([x+\frac{p}{n}]=[x]+1\)
So {x} i.e. fractional part of x =\(\frac{n-p}{n}\) + (a number less than 1/n)
Also, [x +(n-1/n)]<=[x]+1
As n-1/n is a fraction less than 1
So [x]+....[x+(p-1/n)]=p[x]
& [x+p/n] +...+[x+(n-1/n)]=(n-p)[x]+(n-p)
Hence S=n[x]+(n-p).......(1)
Now n{x}=n* (n-p/n) +n*( a number less than 1/n)
= n-p + a number less than 1 ........(2)
Hence [nx]=[ n[x] + n{x} ] =[ n[x] + n-p + a number less than 1 ] from (2)
=n[x] +(n-p) ........................(3)
Hence proved from (2) & (3)

Now returning to the given problem From result (A) the fraction in problem is transformed to
Z={[x] +[2x] +......+[nx]}/n^2

We know that x-1<[x] < x+1
:::::::::::::::::::::::::::::
:::::::::::::::::::::::::::::
:::::::::::::::::::::::::::::
Similarly nx-1<[nx]<nx+1
So Adding All these terms and dividing each side by n^2 we get
{(x-1)+....(nx-1)}/n^2<Z<{(x+1)+....(nx+1)}/n^2
So{ x(1+...+n)-n(1)} /n^2<Z<{ x(1+...+n)+n(1)} /n^2
So x(n )(n+1)/2 n^2 -1/n<Z< x(n )(n+1)/2 n^2 +1/n
So x/2+(x-2/2n)< Z< x/2+(x+2/2n)
Taking limit on both sides as n-->infinity we get
X/2<= Required Limit <= x/2
Hence By Sandwich Theorem We Get The Required Limit = x/2

Cracked by:

Nachiket Vaidya, 12th, Sathaye College, Vile Parle, Mumbai.

(27)

Submission: Closed

Jithin's Solution (*.pdf)

Cracked by:

1. Yash Agrawal, Bhopal (M.P.).
2. Jithin Prakash, 12, Army School, Deolali.

(26)

Submission: Closed

Answer: .

Cracked by:

1. Umang Merwana, Bhopal (M.P.).
2. TusharJain, Babu Banarsidas Institute of Technology, Gaziabad, U.P.
3. Yash Agrawal, Bhopal (M.P.).