GET IOE TEST Q. SOLVE here like this
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GET IOE TEST Q. SOLVE here like this
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The value of C_{0} / 1 + C_{1} / 2 + C_{2} / 3 + …….. + C_{n} / (n + 1) =
2^{n} / (n+1)
(2^{n} – 1) / (n+1)
(2^{n+1} – 1) / (n+1)
2^{n+1} / (n+1)
Given Answer :none Status :Not Answered Right Answer : C
(1+x)^{n} =C_{0} + C_{1} x + C_{2}x^{2} + …….. + C_{n}x^{n}
Integrating wrt x,
(1+x)^{n+1} / (n + 1) + k =
C_{0}x + C_{1}x^{2} / 2 + C_{1}x^{3 }/ 3 + ………….. + C_{n}x^{n+1} / (n + 1)
Put x = 1, 2^{n+1}/ (n + 1) + k = C_{0}/1 + C_{1}/2 + C_{2}/3 + ……….. + C_{n}/(n + 1)
Again, put x = 0, hen k = – 1/1+n
Therefore, C_{0}/1 + C_{1}/2 + C_{2}/3 + ……….. + C_{n}/(n + 1) = (2^{n} -1) / (n + 1)
If a, b, c are in A. P., then (a-c)^{2} / (b^{2}-ac) =
1
2
3
4
Given Answer :none Status :Not Answered Right Answer : D
a, b, c are in A. P. ==> b = (a+c) /2
So, (a-c)^{2} / (b^{2}-ac) = (a-c)^{2} / [{(a+c)/2}^{2} -ac]
= 4 (a-c)^{2} / (a-c)^{2} = 4
How many roots of the equation x – 2/(x-1) = 1 – 2/(x-1) are there
one root
two root
infinite
no root
Given Answer :none Status :Not Answered Right Answer : D
It seems x = 1. But this solution makes given equation indeterminate. So, the equation have no root.
If α, β, γ are angles which a directed line makes with the positive directions of the co-ordinate axes, then sin^{2}α + sin^{2}β + sin^{2}γ =
2
1
3
None
Given Answer :none Status :Not Answered Right Answer : A
cos^{2}α +cos^{2}β +cos^{2}γ = 1
=>(1-sin^{2}α) +(1- sin^{2}β) + (1-sin^{2}γ) = 1
=>sin^{2}α + sin^{2}β +sin^{2}γ = 2
1/2
1
0
1/4
Given Answer :none Status :Not Answered Right Answer : A
The equation x^{2} / (2-a) + y^{2 }/ (a-5) + 1 = 0 represents an ellipse if
a > 2
a > 5
2 < a < 5
none
Given Answer :none Status :Not Answered Right Answer : C
x^{2} / (2-a) + y^{2 }/(a-5) + 1 = 0 ==> x^{2} / (a-2) + y^{2 }/(5-a) = 1.
To represent ellipses, a-2 > 0 & 5-a > 0 i.e. 2 < a < 5
If 2^{x} + 2^{y} = 2^{x+y}, then value of dy/dx at x = y = 1 is
1
2
0
-1
Given Answer :none Status :Not Answered Right Answer : D
2^{x} + 2^{y} = 2^{x+y} Differentiating wrt x,
2^{x} log2 + 2^{y} log2 dy/dx = 2^{x+y} log2 (1 + dy/dx)
At x = y = 1, 2 log2 + 2log2 dy/dx = 2^{2}log2 (1 + dy/dx)
Therefore, dy/dx = -1
The area of loop between the curve y = a sinx-a and x-
a
2a
3a
none
Given Answer :none Status :Not Answered Right Answer : B
Loop is formed between x = 0 and x = π.
Therefore, Area =^{π}ʃ_{0} a sinx dx = [ -a cosx] ^{π}_{0} = 2a.
If the equation 2x^{2} – 2hxy + 2y^{2} = 0 represents two coincident straight line passing through origin, then h =
± 6
√6
-√6
± 2
Given Answer :none Status :Not Answered Right Answer : D
2x^{2} – 2hxy + 2y^{2} = 0 condition of co-incident line is h^{2} = ab => h^{2} = 2 (2)
Therefore, h = ± 2
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
Given Answer :none Status :Not Answered Right Answer : C
ʃ 1 / x√(1+logx) dx = 2√(1+logx) + c
[Therefore, ʃ f(x) / √f(x) dx = 2√f(x) + c]
If the lengths of the sides of a triangle be 6, 4√3 and √13, then the smallest angle is
15^{0}
30^{0}
60^{0}
45^{0}
Given Answer :none Status :Not Answered Right Answer : B
a = 7, b = 4√3, c = √13. So smallest side = c => smallest angle = C
Therefore, cos C = (7^{2} + 48 – 13) / (56√3 )
= 84 / 56√3
= √3 /2
= cos 30^{0}
=> C = 30^{0}
CCl_{3}COH is commonly known as
Maleic aldehyde
chloral
Gloyxal
Ethylene glycol
Given Answer :none Status :Not Answered Right Answer : B
CCl_{3}COH is commonly called as Chloral.
Anion of BeO is
BeO
BeO^{– –}
BeO_{2}^{–}
BeO_{2}^{ – –}
Given Answer :none Status :Not Answered Right Answer : D
2NaOH + BeO^{ }———–> Na_{2}BeO_{2} + H_{2}O
i.e. anion is BeO_{2}^{– –}
When KMnO_{4} is passed through acidic H_{2}S, the product is obtained. In this rx^{n}. The equivalent weight of KMnO_{4} is
Mol. wt/5
Mol wt. /3
Mol. wt. /2
Mol. wt.
Given Answer :none Status :Not Answered Right Answer : A
KMn^{+7}O_{4} + H_{2}SO_{4} + H_{2}S ———-> Mn^{+2}SO_{4} + K_{2}SO_{4} + H_{2}O + S
i.e. change in O.S. of Mn = 5
So, equivalent wt. of KMnO_{4 }= Mol.wt / 5
Which of these both Bronsted and Lewis acid but not Arrhenius acid?
H_{2}O
H_{3}O^{+}
NH_{2}^{–}HCl
HCl
Given Answer :none Status :Not Answered Right Answer : B
H_{3}O^{+ }right answers as it gives H^{+} ion and can also donate proton and can accept electron.
Which of the following is strongest reducing agent.
F_{2}
Cl_{2}
Br_{2}
I_{2}
Given Answer :none Status :Not Answered Right Answer : D
The correct order of reducing agent is,
F_{2}< Cl_{2}< Br_{2} < I_{2}
So, strongest reducing agent is I_{2}.
The types of bond present in NH_{4}Cl is
Ionic and covalent
Ionic, covalent and co-ordinate covalent
Covalent and coordinate covalent bond
Ionic only
Given Answer :none Status :Not Answered Right Answer : B
NH_{4}^{+ }Cl^{–}
Due to this two ions it contains Ionic bond, and in formation of NH_{4}^{+ } contains covalent and coordinate covalent bond so, it contains all ionic, covalent and coordinate covalent bond.
In Ostwald’s process the catalyst used is
Pt
Fe
V_{2}O_{5}
Mo
Given Answer :none Status :Not Answered Right Answer : A
In Ostwald’s process,
4NH_{3} + SO_{2} => NO + 6H_{2}O (here temperature should be 850^{0}C and catalyst should be Pt)
So, Pt is the required catalyst.
The salt of Na_{2}CO_{3} is ———- in nature
acidic
basic
netural
Amphoteric
Given Answer :none Status :Not Answered Right Answer : B
Na_{2}CO_{3} when dissolves in water it gives NaOH and Na_{2}CO_{3 }as NaOH is strong base and Na_{2}CO_{3} is weak acid so, no. of OH^{– }ion always exceeds i.e. it is basic in nature.
The copper obtained from blast furnace is known as
Blister copper
Poling
Matte
Pure copper
Given Answer :none Status :Not Answered Right Answer : A
Blister copper is obtained from blast furnace.
Bayer’s test is not given by
Ethene
Ethyne
Benzene
None
Given Answer :none Status :Not Answered Right Answer : C
Not given by benzene as it contains delocalized π – bond.
If X and Y are the sets, then X ∩ (X U Y) is
X
Y
φ
None
Given Answer :none Status :Not Answered Right Answer : A
X ∩ (X U Y) = X.
The value of sin^{-1}(4/3) + 2 tan^{-1}(1/3) is equal to
π / 2
π / 3
π / 4
None of above
Given Answer :none Status :Not Answered Right Answer : A
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
If A is a matrix of order 3×3 and │A │= 2, then │Adj A│ =
1
2
2^{3}
2^{2}
Given Answer :none Status :Not Answered Right Answer : D
For nth order matrix A, │Adj A│ = │A│^{n-1}
Here, n = 3
: . │Adj A│= │A│^{3-1} = 2^{2}
abc
1/abc
ab + bc + ca
0
Given Answer :none Status :Not Answered Right Answer : D
1.4/ 0! + 2.4/1! + 3.6/2! + 4.7/3! + 5.8/4! + ———— =
11 e
5 e
e-4
e+4
Given Answer :none Status :Not Answered Right Answer : A
One gram of ice is mixed with one gram of steam. After thermal equilibrium, the temperature of the mixture is
0^{o} C
100^{o} C
55^{o} C
80^{o} C
Given Answer :none Status :Not Answered Right Answer : B
One gram of ice should receive 80 calorie of heat for melting and them it receives 100 calories of heat to be heated upto 100^{o} C. But one gram of steam releases 536 calrorie of heat if it is converted into water at 100^{o} C.Since total heat received by ice is less than 536 cal. So temperature of mixture is 100^{o} C.
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
Given Answer :none Status :Not Answered Right Answer : C
Efficiency Ƞ = 1 – T_{2}/T_{1 }= 1 – 3000/600 = 1/2
W/Q = Ƞ = 1/2
==> Q = 2W
= 2 * 800 J
= 1600 J
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
Given Answer :none Status :Not Answered Right Answer : B
Velocity of sound is independent of pressure at constant temperature because
p/ρ = PV/M = RT/M (constant).
When the temperature increases the frequency of a tuning fork
increases
decreases
remains same
increases or decreases depending on the material.
Given Answer :none Status :Not Answered Right Answer : B
As the temperature increases, the length of prongs increases due to thermal expansion and hence frequency decreases.
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
Given Answer :none Status :Not Answered Right Answer : C
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.
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
Given Answer :none Status :Not Answered Right Answer : B
Refractive index for violet is maximum given by Cauchy’s relation.
For achromatic combination of lenses if we use two lenses of focal lengths f and f’, dispersive powers ω and ω’, respectively, then
ω = ω_{0}, ω’ = 2ω_{0}, f’ = 2f
ω = ω_{0}, ω’ = 2ω_{0}, f’ = – 2f
ω = ω_{0}, ω’ = 2ω_{0}, f’ = f / 2
ω = ω_{0}, ω’ = 2ω_{0}, f’ = – f /2
Given Answer :none Status :Not Answered Right Answer : B
For achromatic combination of lens.
f / f’ = – ω / ω’ which is possible if ω = ω_{0}, ω’ = 2ω_{0} and f’ = – 2f
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)
Given Answer :none Status :Not Answered Right Answer : B
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)/(4πЄ_{0}r) + 2 [(Q.q) /{4πЄ_{0} (r/2)}]
0 = Q/ 4πЄ_{0}r (Q + 4q)
Q + 4q
Therefore, q = -Q/4
n-equal capacitors are first connected in series and then in parallel. The ratio of capactitances in series and parallel arrangement will be:
n
n^{2}
1/n
1/n^{2}
Given Answer :none Status :Not Answered Right Answer : D
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
Given Answer :none Status :Not Answered Right Answer : D
R_{p} = 6/2^{2 }= 1.5 Ω
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
Given Answer :none Status :Not Answered Right Answer : B
Resistance of the voltmeter = V/i_{g} = 50/0.1 = 500 Ω
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
Given Answer :none Status :Not Answered Right Answer : B
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.
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
Given Answer :none Status :Not Answered Right Answer : C
r = mv/qB
So for equal speed in same field r α m/q
r_{p}/r_{α} = (m_{p}/m_{α}) (q_{α}/q_{p}) = m_{p}/4m_{p} * 2e/e = 1:2
The dimensions for light year are
T
T^{-1}
L
L^{-1}
Given Answer :none Status :Not Answered Right Answer : C
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/s^{2}) is
5 N
2 N
8 N
zero
Given Answer :none Status :Not Answered Right Answer : A
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
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
Given Answer :none Status :Not Answered Right Answer : B
H = – ut + 1/2 gt^{2}
50 = – 15t + 1/2 * 10 * t^{2}
5t^{2} – 3t – 10 = 0
5t^{2} – 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
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
Given Answer :none Status :Not Answered Right Answer : D
At depth x inside earth, the value of g is
g’ = GM/R^{3} (R – x) = g/R (R – x)
Hence g’/g = (R – x) / R
1/4 = (R – x) / R
x = 3/4 R
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
Given Answer :none Status :Not Answered Right Answer : B
Acceleration is zero at mean position where v = ω √(A^{2} – 0^{2}) = Aω i.e. maximum
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
Given Answer :none Status :Not Answered Right Answer : A
On pressing down, the body experiences additional pressure due to water column above resulting in slight compression so that weight of the body now exceeds upthrust.
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