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

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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|


Amomg the following, the paramagnetic compound is

(A)Na_2 O_2





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



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

The correct answer is A.

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

The correct answer is B.


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 )


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