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Engineering Entrance Questions: Punch of the Week(12-Feb-10)

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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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"Punch Of The Week" Team Members
Learnhub's Punch
This Week's Punches
Physics

Water is filled up a height h in a beaker of radius R as shown in the figure. The density of water is \rho, the surface tension of water is T and the atmospheric pressure is P_0. Consider a vertical section ABCD of the water column through a diameter of the beaker. The force on water on the other side of this section has magnitute.

(A) |2P_0 Rh + \pi R^2 \rho gh - 2RT|

(B) |2P_0 Rh + R\rho gh^2 - 2RT|

(C) |P_0 \pi R^2 + R\rho gh^2 - 2RT|

(D) |P_0 \pi R^2 + R\rho gh^2 + 2RT|

Chemistry

Amomg the following, the paramagnetic compound is

(A)Na_2 O_2

(B)O_3

(C)N_2O

(D)KO_2

Math

Let V_t denote the sum of the fist r terms of arithmetic progression(A.P.) whose fist term is r and the common (2r-1). Let
T_1 = V_{r+1} - V_r - 2 and Q_r = T_{r+1} - T_r for r = 1, 2, \cdots
The sum V_1 + V_2 + \cdots + V_n is

(A) \cfrac{1}{12} n(n+1)(3n^2 - n +1)

(B) \cfrac{1}{12} n(n+1)(3n^2 + n +2)

(C) \cfrac{1}{2} n(2n^2 - n + 1)

(D) \cfrac{1}{3} n(2n^3 - 2n + 3)

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:


\overset{h}{\underset{0}{\int}}(P_0 + \rho g x)2Rdx - 2RT = F
2P_0 Rh + R \rho gh^2 - 2RT = F

Chemistry
The correct answer is A.

Solution:
O^{2-} _2 = \sigma 1s^2 \sigma * 1s^2, \sigma 2 s^2 \sigma* 2s^2, \sigma 2p ^2 _z, \pi 2p ^2 _x = \pi 2p^2 _y, \pi * 2p ^2 _x = \pi * 2p^2 _y

Number of unpaired electrons = 0

N=N \Rightarrow O Number of unpaired electrons = 0

Number of unpaired electrons = 0

O^- _2 = \sigma 1s^2, \sigma * 1s^2 \sigma 2s^2, \sigma * 2s^2, \sigma 2p ^2 _z,\pi 2 p^2 _x = \pi 2 p^2 _y, \pi * 2 p ^2 _x= \pi * 2p ^1 _y
Number of unpaired electrons = 1

Thus O ^- _2 is a paramagnetic. hence (D) is Correct

Math
The correct answer is B.

Solution:

V_r = \cfrac{r}{2}[2r + (r-1)(2r+1)] = \cfrac{1}{2}(2r^3 - r^2 + r)

\Sigma V_r = \cfrac{1}{12} n(n+1)(3n^2 + n + 2)

Click here for Next week's Punch of the Week

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  1. saikrishna123456 saidWed, 17 Feb 2010 16:03:17 -0000 ( Link )

    chemistry-d

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