Engineering Entrance Questions: Punch of the Week(19-Feb-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

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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

Two particle of mass m each are tried at the ends of light string of length 2a. the whole system is kept is kept on frictionless horizontal surfacewith string held tight so that each mass is at a distance 'a' from the center P (as shown in the figure). Now, the mid point of string is pulled vertically upwards with small but constant force F. as a result, the particles move towards each other on the surface. the magnitude of acceleration between them becomes 2x is.

(A) $\cfrac{F}{2m} \cfrac{a}{\sqrt{a^2 - x^2}}$

(B) $\cfrac{F}{2m} \cfrac{x}{\sqrt{a^2 - x^2}}$

( C) $\cfrac{F}{2m} \cfrac{x}{a}$

(D) $\cfrac{F}{2m} \cfrac{\sqrt{a^2 - x^2}}{x}$

Chemistry

The regent(s) for the following conversion

(A)alcoholic KOH

(B)alcoholic KOH follow by NaNH_2

( C)aqueous KOH followed by NaNH_2

(D) Zn/CH_3OH

Math

The number of solutions of the pair of equations
$2 \sin^2 \theta - \cos 2\theta = 0$
$2 \cos^2 \theta - 3\sin \theta = 0$
in the interval $[0, 2\pi]$ is

(A) zero

(B) one

( C) two

(D) four

Click here for Previous week's Punch of the Week

Posted on 19-Feb-10

So, folks here are the correct answers B

Physics

Solution:

$2T \sin \theta = F$

$2T \cos \theta = mA$
$2T \tan \theta = \cfrac{F}{mA}$
$A = \cfrac{F}{2m} \Big ( \cfrac{x}{\sqrt{a^ - x^2}}\Big )$

Chemistry
The correct answer is B.

Solution:

Because CH_2 = CH - Br has partial C-Br double bond character, its require more stronger base to remove HBr
Hence(B) is the charcter answer

Math
The correct answer is B.

Solution:

$2 \sin ^2 \theta - \cos 2\theta = 0 \Rightarrow \sin^2 \theta = \cfrac{1}{4}$
also $2\cos ^2 \theta = 3\sin \theta \Rightarrow \sin \theta = \cfrac{1}{2}$
$\Rightarrow$ two solution in $[0, \pi]$