## Hurried Trip to Avoid a Bad Star M. Lilla and C. Bishop Barry

Posted: July 30, 2013 in Questions of class-11

Moved to New site

## Travelling Through the Dark

Posted: July 30, 2013 in Questions of class-11

Observe the central idea of the poem.

## Travelling Through the Dark

Posted: July 30, 2013 in Questions of class-11

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Posted: July 29, 2013 in Questions of class-11 ## IOE

Posted: July 28, 2013 in Questions of class-11

GET IOE TEST Q. SOLVE here like this

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• ### Q.1

The value of C0 / 1 +  C1 / 2  +  C2 / 3  + …….. +  Cn / (n + 1)  =

• 2n / (n+1)

• (2n – 1) / (n+1)

• (2n+1 – 1) / (n+1)

• 2n+1 / (n+1)

### Solution

(1+x)n =C0  +  C1 x +  C2x2  + …….. + Cnxn
Integrating wrt x,
(1+x)n+1 / (n + 1) + k =
C0x + C1x2 / 2 + C1x3 / 3 + ………….. +  Cnxn+1 / (n + 1)
Put x = 1, 2n+1/ (n + 1) + k = C0/1 + C1/2 + C2/3 + ……….. + Cn/(n + 1)
Again, put  x = 0, hen k = – 1/1+n
Therefore, C0/1 + C1/2 + C2/3 + ……….. + Cn/(n + 1) = (2n -1) / (n + 1)

• ### Q.2

If a, b, c are in A. P., then (a-c)2 / (b2-ac) =

• 1

• 2

• 3

• 4

### Solution

a, b, c are in A. P.    ==> b = (a+c) /2
So, (a-c)2 / (b2-ac) = (a-c)2 / [{(a+c)/2}2 -ac]
= 4 (a-c)2 / (a-c)2 = 4

• ### Q.3

How many roots of the equation x – 2/(x-1) = 1 – 2/(x-1) are there

• one root

• two root

• infinite

• no root

### Solution

It seems x = 1. But this solution makes given equation indeterminate. So, the equation have no root.

• ### Q.4

If  α, β, γ are angles which a directed line makes with the positive directions of the co-ordinate axes, then sin2α + sin2β + sin2γ =

• 2

• 1

• 3

• None

### Solution

cos2α +cos2β +cos2γ = 1
=>(1-sin2α) +(1- sin2β) + (1-sin2γ) = 1
=>sin2α + sin2β +sin2γ = 2

• ### Q.5 • 1/2

• 1

• 0

• 1/4

### Solution • ### Q.6

The equation x2 / (2-a) + y2 / (a-5) + 1 = 0 represents an ellipse if

• a > 2

• a > 5

• 2 < a < 5

• none

### Solution

x2 / (2-a) + y2 /(a-5) + 1 = 0 ==> x2 / (a-2) + y2 /(5-a) = 1.
To represent ellipses, a-2 > 0 & 5-a > 0 i.e. 2 < a < 5

• ### Q.7

If 2x + 2y = 2x+y, then value of dy/dx at x = y = 1 is

• 1

• 2

• 0

• -1

### Solution

2x + 2y = 2x+y Differentiating wrt x,
2x log2 + 2y log2 dy/dx  = 2x+y log2 (1 + dy/dx)
At x = y = 1, 2 log2 + 2log2 dy/dx = 22log2 (1 + dy/dx)
Therefore, dy/dx = -1

• ### Q.8

The area of loop between the curve y = a sinx-a and x-

• a

• 2a

• 3a

• none

### Solution

Loop is formed between x = 0 and x = π.
Therefore, Area =πʃ0 a sinx dx = [ -a cosx] π0 = 2a.

• ### Q.9

If the equation 2x2 – 2hxy + 2y2 = 0 represents two coincident straight line passing through origin, then h =

• ± 6

• √6

• -√6

• ± 2

### Solution

2x2 – 2hxy + 2y2 = 0 condition of co-incident line is h2 = ab => h2 = 2 (2)
Therefore, h = ± 2

• ### Q.10

ʃ 1 / {x√(1+logx)} dx =

• 2/3 (1+log x)3/2 + c

• (1+log x)3/2 + c

• 2√(1+logx) + c

• √(1+logx) + c

### Solution

ʃ 1 / x√(1+logx) dx = 2√(1+logx) + c
[Therefore, ʃ f(x) / √f(x) dx = 2√f(x) + c]

• ### Q.11

If the lengths of the sides of a triangle be 6, 4√3 and √13, then the smallest angle is

• 150

• 300

• 600

• 450

### Solution

a = 7, b = 4√3, c = √13. So smallest side = c => smallest angle = C
Therefore, cos C = (72 + 48 – 13) / (56√3 )
= 84 / 56√3
= √3 /2
= cos 300
=>  C =  300

• ### Q.12

CCl3COH is commonly known as

• Maleic aldehyde

• chloral

• Gloyxal

• Ethylene glycol

### Solution

CCl3COH is commonly called as Chloral.

• ### Q.13

Anion of BeO is

• BeO

• BeO– –

• BeO2

• BeO2 – –

### Solution

2NaOH + BeO  ———–> Na2BeO2 + H2O
i.e. anion is BeO2– –

• ### Q.14

When KMnO4 is passed through acidic H2S, the product is obtained. In this rxn. The equivalent weight of KMnO4 is

• Mol. wt/5

• Mol wt. /3

• Mol. wt. /2

• Mol. wt.

### Solution

KMn+7O4 + H2SO4 + H2S ———-> Mn+2SO4 + K2SO4 + H2O + S
i.e. change in O.S. of  Mn = 5
So, equivalent wt. of KMnO4 = Mol.wt / 5

• ### Q.15

Which of these both Bronsted and Lewis acid but not Arrhenius acid?

• H2O

• H3O+

• NH2HCl

• HCl

### Solution

H3O+ right answers as it gives H+ ion and can also donate proton and can accept electron.

• ### Q.16

Which of the following is strongest reducing agent.

• F2

• Cl2

• Br2

• I2

### Solution

The correct order of reducing agent is,
F2< Cl2< Br2 < I2
So, strongest reducing agent is I2.

• ### Q.17

The types of bond present in NH4Cl is

• Ionic and covalent

• Ionic, covalent and co-ordinate covalent

• Covalent and coordinate covalent bond

• Ionic only

### Solution

NH4+ Cl
Due to this two ions it contains Ionic bond, and in formation of  NH4 contains covalent and coordinate covalent bond so, it contains all ionic, covalent and coordinate covalent bond.

• ### Q.18

In Ostwald’s process the catalyst used is

• Pt

• Fe

• V2O5

• Mo

### Solution

In Ostwald’s process,
4NH3 + SO2 => NO + 6H2O (here temperature should be 8500C and catalyst should be Pt)
So, Pt is the required catalyst.

• ### Q.19

The salt of Na2CO3 is ———- in nature

• acidic

• basic

• netural

• Amphoteric

### Solution

Na2CO3 when dissolves in water it gives NaOH and Na2COas NaOH is strong base and Na2CO3 is weak acid so, no. of OH–  ion always exceeds i.e. it is basic in nature.

• ### Q.20

The copper obtained from blast furnace is known as

• Blister copper

• Poling

• Matte

• Pure copper

### Solution

Blister copper is obtained from blast furnace.

• ### Q.21

Bayer’s test is not given by

• Ethene

• Ethyne

• Benzene

• None

### Solution

Not given by  benzene as it contains delocalized π – bond.

• ### Q.22

If X and Y are the sets, then X ∩ (X U Y) is

• X

• Y

• φ

• None

### Solution

X ∩ (X U Y) = X.

• ### Q.23

The value of sin-1(4/3) + 2 tan-1(1/3)  is equal to

• π / 2

• π / 3

• π / 4

• None of above

### Solution

sin-1(4/3) + 2 tan-1(1/3)
= sin-1(4/3) +  sin-1{(2/3) / (1-1/9)}
= sin-1(4/3) +  sin-1(3/4)
= sin-1(4/3) +  cos-1(4/3)
= π / 2

• ### Q.24

If A is a matrix of order 3×3 and │A │= 2, then │Adj A│ =

• 1

• 2

• 23

• 22

### Solution

For nth order matrix A, │Adj A│ = │A│n-1
Here, n = 3

: . │Adj A│= │A│3-1 = 22

• ### Q.25 • abc

• 1/abc

• ab + bc + ca

• 0

### Solution • ### Q.26

1.4/ 0! + 2.4/1! + 3.6/2! + 4.7/3! + 5.8/4! + ———— =

• 11 e

• 5 e

• e-4

• e+4

### Solution • ### Q.27

One gram of ice is mixed with one gram of steam. After thermal equilibrium, the temperature of the mixture is

• 0o C

• 100o C

• 55o C

• 80o C

### Solution

One gram of ice should receive 80  calorie of heat for melting and them it receives 100 calories of heat to be heated upto 100o C. But one gram of steam releases 536 calrorie of heat if it is converted into water at 100o C.Since total heat received by ice is less than 536 cal. So temperature of mixture is 100o C.

• ### Q.28

A Carnot engine working between 300 K and 600 K has a work output of 800 J per cycle. What is the amount of heat energy supplied to the engine from source per cycle

• 800 J

• 1200 J

• 1600 J

• 2000 J

### Solution

Efficiency Ƞ = 1 – T2/T1 = 1 –  3000/600 = 1/2
W/Q = Ƞ = 1/2
==> Q = 2W
= 2 * 800 J
=  1600 J

• ### Q.29

The speed of sound in a gas at N.T.P. is 300 m/s. If the pressure is increased four times without change in temperature, the velocity of sound will be

• 150 m/s

• 300 m/s

• 600 m/s

• 1200 m/s

### Solution

Velocity of sound is independent of pressure at constant temperature because
p/ρ = PV/M = RT/M (constant).

• ### Q.30

When the temperature increases the frequency of a tuning fork

• increases

• decreases

• remains same

• increases or decreases depending on the material.

### Solution

As the temperature increases, the length of prongs increases due to thermal expansion and hence frequency decreases.

• ### Q.31

A light beam ( red + blue ) is incidence on one face of a glass slab. After emerging from glass slab through second face, they.

• ( red + blue ) will separate

• appear at the same point and move n the same direction

• appear at the different point and move in the same direction

• will not emerge through the second face

### Solution

Refractive index is different for different colors of light so they appear at different points. There is no deviation for a glass slab so they move in same direction.

• ### Q.32

When white light passes through a dispersive medium, it breaks up into various colours. Which of the following statements is true?

• velocity of light for violet is greater than the velocity of light for red colour

• velocity of light for violet is less than the velocity of light for red

• velocity of light  is the same for all colours

• velocity of light for different colours has nothing to do with the phenomenon of dispersion

### Solution

Refractive index for violet is maximum given by Cauchy’s relation.

• ### Q.33

For achromatic combination of lenses if we use two lenses of focal lengths f and f’, dispersive powers ω and ω’, respectively, then

• ω = ω0, ω’ = 2ω0, f’ = 2f

• ω = ω0, ω’ = 2ω0, f’ = – 2f

• ω = ω0, ω’ = 2ω0, f’ = f / 2

• ω = ω0, ω’ = 2ω0, f’ = – f /2

### Solution

For achromatic combination of lens.
f / f’ =  – ω / ω’ which is possible if ω = ω0, ω’ = 2ω0 and f’ = – 2f

• ### Q.34

A charge q is place at the centre of the joining two equal charges. Q The system of three charges will be in equilibrium if q is equal to

• 4Q

• -Q/4

• -2Q

• – (Q/2)

### Solution

System of charge be equilibrium if potential energy of the system is zero
U = 0
Let r be the distances between two equal charges Q, then
U = (Q.Q)/(4πЄ0r) +  2 [(Q.q) /{4πЄ0 (r/2)}]
0 = Q/ 4πЄ0r  (Q + 4q)
Q + 4q
Therefore,  q = -Q/4

• ### Q.35

n-equal capacitors are first connected in series and then in parallel. The ratio of capactitances in series and parallel arrangement will be:

• n

• n2

• 1/n

• 1/n2

• ### Q.36

A wire has resistance 6 ohm. It is cut in two parts and both half values are connected in parallel. The new resistance is

• 6 Ω

• 3 Ω

• 12 Ω

• 1.5 Ω

### Solution

Rp = 6/22 = 1.5 Ω

• ### Q.37

A galvanometer has a resistance 100 Ω and gives a full scale deflection for a current of 0.1 A. It is to be converted into a voltmeter of range 50 V. The resistance of voltmeter  so formed is

• 400 Ω

• 500 Ω

• 200 Ω

• 80 Ω

### Solution

Resistance of the voltmeter   = V/ig = 50/0.1 = 500 Ω

• ### Q.38

The frequency of AC is 50 Hz. How many times the current becomes zero in one second?

• 50 times

• 100 times

• 200 times

• 25 times

### Solution

The current in AC circuit becomes zero twice in a cycle. So if frequency of AC is 50 Hz then current becomes zero 100 times in a second.

• ### Q.39

A proton and an α-particle are injected into a uniform magnetic field at right angles to the direction of the field with equal speeds. Then ratio of radii of their circular paths in the field will be

• 1 : 1

• 2 : 1

• 1 : 2

• √2:1

### Solution

r = mv/qB
So for equal speed in same field   r α m/q
rp/rα = (mp/mα) (qα/qp) = mp/4mp * 2e/e = 1:2

• ### Q.40

The dimensions for light year are

• T

• T-1

• L

• L-1

• ### Q.41

A block of mass 2 kg is placed on a horizontal surface. The coefficient of static friction is 0.4. If a force of 5 N is applied on the block parallel to the surface, the force of friction between the block and the surface (taking g = 10 m/s2) is

• 5 N

• 2 N

• 8 N

• zero

### Solution

Limiting frictional force f = μmg = 0.4 * 2 * 10 = 8 N
Applied force on the block F = 5 N
Since applied force < limiting frictional force. So block is at rest and there exists a static frictional force, which should be equal to applied force. Frictional force = 5 N

• ### Q.42

A balloon is rising upward at constant velocity 15 m/s. At height 50 m above the ground, a stone is released from that balloon. The stone reaches the ground in

• 3 sec

• 5 sec

• 7.5 sec

• 10 sec

### Solution

H  = – ut + 1/2 gt2
50 = – 15t + 1/2 * 10 * t2
5t2 – 3t – 10 = 0
5t2 – 5t + 2t – 10 = 0
(5t + 2) (t – 5) = 0
t = – 2/5 sec or 5 sec.
Since time must be positive
So,  t = 5 seconds

• ### Q.43

At what depth x from earth’s surface the value of g will become one-fourth of its value at earth’s surface? (R = radius of earth).

• R

• R/2

• R/4

• 3/4 R

### Solution

At depth x inside earth, the value of g is
g’  = GM/R3 (R – x) = g/R (R – x)
Hence g’/g = (R – x) / R
1/4 = (R – x) / R
x = 3/4 R

• ### Q.44

In S.H.M, the acceleration of the particle is zero when

• its velocity is zero

• its velocity is maximum

• its velocity is half its maximum value

• its velocity isone – forth its maximum value

### Solution

Acceleration is zero at mean position where v = ω √(A2 – 02) = Aω i.e.  maximum

• ### Q.45

A body is just floating in a liquid. If the body is presses down and released then it will

• sink

• come back to initial

• oscillate S.H.M

• remain in new position