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Engineering Entrance Questions: Punch of the Week(24-Dec-09)

deepa mittal
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What is "Punch Of The Week"?
  • Every week Learnhub's team will post one question each on Physics, Chemistry and Mathematics of the IIT-JEE, AIEEE and BITSAT Test Preparation
  • Members can post their answers along with explanations for the next one week.
  • After one week, Learnhub's team will post the official answer along with the explanation.Besides answering the previous week's questions, Learnhub's Team will come up with the next set of questions for that week as well.
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This Week's Punches

Physics

A length of wire caries a steady current. It is bent fist to form a circular plane coil of one turn. The same circular length is now bent more sharply to give a double loop of small radius. when the same current is passed. find the ratio of the magnetic field at the center with it fist value

(A) 1:1
(B) 1:2
(C) 2:1
(D) 4:1

Chemistry

Elecctrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.001 mol of H_2 gas at the cathode is (1 Faraday = 96500 c mil)

(A) 9.65 \times 10^4 sec
(B) 19.30 \times 10^4 sec
(C) 25.95 \times 10^4 sec
(D) 38.6 \times 10^4 sec

Math

In geometric progression consisting of positive terms, each term equals the sum of the next two term, then the common ratio of this progression equals

(A) \sqrt{5}
(B) \cfrac{1}{2} \; (\sqrt{5} - 1)
(C) \cfrac{1}{2} \; (1 - \sqrt{5})
(D) \cfrac{1}{2} \; (\sqrt{5})

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Posted on 11-Jan-10

So, folks here are the correct answers

Physics

Solution:

(i) Let l be the length of the wire. When the wire is bent in the from of one tern circular coil,
l = 2\pi r_1orM = \cfrac{l}{2\pi},N = 1
B_1 = \cfrac{\mu_0 NI}{2r}=\cfrac{\mu_0 * 1 * 1}{2* (\cfrac{l}{2\pi})}\cfrac{\mu_0 \pi I}{l}
When the wire is bent in from of two - turn coil
l = 2*2\pi r_2 or r^2 = \cfrac{l}{4/pi}, N=2
\thereforeB_2 = \cfrac{\mu_0 * 2 * I}{2*(\cfrac{l}{4\pi})}=\cfrac{4\mu_0 \pi I}{l}
Hence \cfrac{B_2}{B_1} = \cfrac{4}{1}


Chemistry
The correct answer is B.

Solution:

Q = i * t
Q = 10*10^{-3} * t
2H_2O + 2e^- \rightarrow H_2 + 2OH^-
To liberate 0.001 mole of H_2, 0.002 Faraday charge is required
Q = 0.002 * 96500 C
\therefore 0.02 * 96500 = 10^{-2} * t
t = 19.30 * 10^4sec


Math
The correct answer is B.

Solution:

Q = i * t
a = ar + ar^2
\Rightarrowr^2 + r - 1 = 0
\Rightarrowr = \cfrac{1}{2}(\sqrt{5} - 1)


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4 Comments

  1. gaurav311092 said – Wed, 30 Dec 2009 19:01:05 -0000 ( Link )

    answer to physics question is 4:1

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  2. sai_ram said – Wed, 06 Jan 2010 18:54:26 -0000 ( Link )

    answer to the physics problem is 1:1 in this question might be we need to take care that we cannot use a solonoid formula for the magnetic field in the second part.

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  3. prathameshb said – Thu, 07 Jan 2010 18:23:26 -0000 ( Link )

    math answer is b…..its simple calculation yaar……….

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  4. g_a_r_i_m_a said – Sun, 10 Jan 2010 13:37:55 -0000 ( Link )

    ans to phy is 1:2 and ans to maths is option b….

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