Engineering Entrance Questions:Punch of the Week(14-Mar-10)

by Deepa Mittal

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.

What are the rules of "Punch Of The Week"?

Every user must refrain from posting sarcastic comments on the answers posted by others.

Once the Learnhub's team posts the official answer along with the explanation, no comments must be made regarding those questions. In case of further clarification please send a personal message to the respective team member

If you think that there is something wrong with the explanation of the Learnhub's Team, which seldom happens, please send a personal message to the respective team member informing him/her the same. Posting the same here will not be considered and will be deleted.

Please bear in mind that only Learnhub's Team members have the privilege of posting questions here. Questions posted by other members will not be entertained and hence will be deleted immediately.

"Punch Of The Week" Team Members

This Week's Punches

Physics

The resistance of bulb filament is 100 $\Omega$ at a temperature of $100^{\circ}C$ . If its temperature coefficient of resistance be 0.005 per $^{\circ}C$ , its resistance be 0.005 per $^{\circ}C$ , its resistence will become 200 $\Omega$ at temperature of.

(A) $200^{\circ} N$

(B) $300^{\circ} N$

( C) $400^{\circ} N$

(D) $500^{\circ} N$

Chemistry

Given the data at $25^{\circ}c$ ,

$Ag + I^{-} \Rightarrow Agl + e^{-}$ ; $E^{\circ} = 0.152 V$

$Ag \Rightarrow Ag^{+} + e^{-}$ ; $E^{\circ} = -0.800 V$

What is the value of log $K_{sp}$ for Agl

$(2.303 \cfrac{RT}{F} = 0.059)$

(A) -8.12

(B) +8.612

( C) -37.83

(D) -16.13

Math

If the straight lines $\cfrac{k-1}{k} = {y-2}{2} = {z-3}{3}$ and $\cfrac{x-2}{3} = \cfrac{y-3}{k}= {z-1}{z}$ integer k equal to

(A) -5

(B) 5

( C) 2

(D) -2

Click here for Previous week's Punch of the Week

Posted on 8-March-10

So, folks here are the correct answers B

Physics

Solution:

$200 = 100 [1 + (0.005 * \Delta t)]$

T - 100 = 200

$T - 300^{\circ} C$

Chemistry
The correct answer is D.

Solution:

Math
The correct answer is A.

Solution:

$\cfrac{x-1}{k} = \cfrac{y-2}{2} = \cfrac{z-3}{3} \And \cfrac{x-2}{3} = \cfrac{z-1}{2}$

since lines intersect in point

$\therefore$ 2k^2 + 5k - 25 = 0

k = -5, 5/2

Click here for Next week's Punch of the Week

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mps_delhi – Thu, 29 Apr 2010 03:23:05 -0000

How is BITSAT different from IIT-JEE and AIEEE in terms of syllabus and types of questions? Someone who has already appeared in these exams, what specific additional preparation is required? Thanks!!

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