#Wishing you all the best guys n gals who are appearing for IIT JEE 2017 MAIN 2017 exam…… Please do well and come out with flying colours……

No documents or writing material is permitted inside exam hall other than the admit card…..

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# Tag: IIT JEE

## IIT JEE MAIN 2017

## IIT JEE Main 2016 Maths Question 32

## IIT JEE Main 2016 Maths Question 32

#### Solution

#### Answer

## IIT JEE Main 2016 Maths Question 31

## IIT JEE Main 2016 Maths Question 31

#### Solution

#### Answer

## IIT JEE Main 2016 Physics Question 30

## IIT JEE Main 2016 Physics Question 30

#### Solution

#### Answer

## IIT JEE Main 2016 Physics Question 29

## IIT JEE Main 2016 Physics Question 29

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#### Answer

## IIT JEE Main 2016 Physics Question 28

## IIT JEE Main 2016 Physics Question 28

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#### Answer

## IIT JEE Main 2016 Physics Question 27

## IIT JEE Main 2016 Physics Question No 27

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#### Answer

## IIT JEE Main 2016 Physics Question 26

## IIT JEE Main 2016 Physics Question No 26

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#### Answer

## IIT JEE Main 2016 Physics Question 25

## IIT JEE Main 2016 Physics Question No 25

#### Solution

#### Answer

## IIT JEE Main 2016 Physics Question 24

## IIT JEE Main 2016 Physics Question No 24

#### Solution

#### Answer

#Wishing you all the best guys n gals who are appearing for IIT JEE 2017 MAIN 2017 exam…… Please do well and come out with flying colours……

No documents or writing material is permitted inside exam hall other than the admit card…..

If f(x) + 2f(1/x) = 3x, x<> 0 and S = {x ∈ R : f(x)=f(-x)}; then S :

(1) contains exactly two elements | (2) contains more than two elements | (3) is an empty set | (4) contains exactly one element |

Solution to the problem requiring to find out the set S as per the relation specified is as follows:

f(x) + 2f(1/x) = 3x …………. (1)

Replace x->(1/x)

=>f(1/x) + 2f(x) = 3/x

=> 2f(1/x) + 4f(x) =6/x ……(2)

=> from (1) and (2)

3f(x) = 6/x – 3x

=> f(x) = (2/x) – x

=> f(-x) = (-2/x) + x

=> f(x) = f(-x)

=> (2/x) – x = (-2/x) + x

=> 4/x = 2x

=>x^{2} = 2

=>x= +/- √2

Hence set S contains two elements, i.e. +√2 , -√2.

Correct answer to given problem No 32 of IIT JEE Main Maths paper of 2016 is option number (1) that is set S contains exactly two elements.

The area (in sq. units) of the region {(x, y) : y^{2} >=2x and x^{2} + y^{2} <= 4x, x>= 0, y >= 0} is:

(1) Π – {(4√2)/3} | (2) (Π/2) – {(2√2)/3} | (3) Π – (4/3) | (4) Π – (8/3) |

Solution to the problem requiring to find out the area under a region bounded by given curves, is obtained by integration as follows:

y^{2} = 2x, x^{2} + y^{2} = 4x

=> x^{2} + 2x = 4x

=> x^{2} = 2x

=> x = 0, 2

Required area = ∫_{0}^{2} [√(4x – x^{2}) – √(2x)] dx

= ∫_{0}^{2} [√(4x – x^{2}) – √2√x] dx

=[{(x-2)/2}√(4x – x^{2}) + 2 sin^{-1}{(x-2)/2} -{(2/3)√2}x^{3/2}]]^{2}_{0}

= 0 + 0 – (8/3) – [0 + 2(-Π/2)]

=Π – (8/3)

Correct answer to given problem No 31 of IIT JEE Main Maths paper of 2016 is option number (4).

A pendulum clock loses 12 s a day if the temperature is 40ºC and gains 4s a day if the temperature is 20ºC. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively:

(1) 30ºC; α = 1.85 × 10^{-3}/ºC |
(2) 55ºC; α = 1.85 × 10^{-2}/ºC |
(3) 25ºC; α = 1.85 × 10^{-5}/ºC |
(4) 60ºC; α = 1.85 × 10^{-4}/ºC |

(ΔT/T) =(ΔL/L)

(ΔT/T) = (αΔT/2)

=> (ΔT/T) = (αΔT_{1}/2)

=> (1/2)(αΔT_{1})x24x3600

=> 12 = (αΔT_{1}) x 12 x 3600

=> 4 = (αΔT_{2}) x 12 x 3600

=> 3 = (ΔT_{1}/ΔT_{2}) = (40 – T)/(T-20)

=> 3T – 60 = 40 – T

=> T = 25ºC

α = 1.85 × 10^{-5}/ºC

Correct answer to given problem No 30 of IIT JEE Main Physics paper of 2016 is option number (3).

In an experiment for determination of refractive index of glass of a prism by i – δ, plot, it was found that a ray incident at angle 35º, suffers a deviation of 40º and that it emerges at angle 79º. In that case which of the following is closest to the maximum possible value of the refractive index?

(1) 1.7 | (2) 1.8 | (3) 1.5 | (4) 1.6 |

∠i = 35º

δ = 40º

e = 79º

i = 35º

e = 79º

δ = 40º

δ + 4 = i + e

=> A = 74º

with above set of data, as refractive index increases emerging ray will have tendency to suffer TIR.

For i = 35º

μ = sin 35º/ sin r_{1}

=> μ = sin 90º / sin (74º – r_{1})

Using reversibility principle

For i = 79º

μ = sin 79º/ sin r_{2} = sin 90º / sin (74º – r_{2})

From 2nd case r_{2} = 37º

=> μ =1.63

Correct answer to given problem No 29 of IIT JEE Main Physics paper of 2016 is option number (4).

An observer looks at a distance tree of height 10m with a telescope of magnifying power of 20. To the observer the tree appears:

(1) 20 times taller | (2) 20 times nearer | (3) 10 times taller | (4) 10 times nearer |

By definition, for telescopes, magnifying power = 20 means that the object appears 20 times nearer.

Correct answer to given problem No 28 of IIT JEE Main Physics paper of 2016 is option number (2).

An ideal gas undergoes a quasi static reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PV^{n} constant, then n is given by (Here C_{p} and C_{v} are molar specific heat at constant pressure and constant volume, respectively):

(1) n= (C_{p} – C)/(C- C_{v}) |
(2)n= (C- C_{v})/(C – C_{p}) |
(3) n= C_{p} / C_{v} |
(4) (C- C_{p})/(C – C_{v}) |

PV^{n} = constant

Then C_{process} for the above polytropic process is known

C_{process} = C = C_{v} + R/(1-n)

C- C_{v} = R/(1-n)

(1-n) = R / (C- C_{v})

So after solving we get,

n = (C- C_{v} – R)/(C -C_{v})

=> n = (C- C_{p})/(C – C_{v})

Correct answer to given problem No 27 of IIT JEE Main Physics paper of 2016 is option number (4) that is (C- C_{p})/(C – C_{v}).

A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take 2 g 10 ms^{-2} ).

(1) 2√2s | (2) √2s | (3) 2Π√2s | (4) 2s |

ν = √[(μxg)/μ]

= √(xg)

dx/dt = √(xg)

∫_{0}^{-1} [dx/√(x)] = t√(g)

2x^{1/2} = t√(g)

[2/√(g)]√(20) = t

2√2 = t

Correct answer to given problem No 26 of IIT JEE Main Physics paper of 2016 is option number (1) that is 2√2.

The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say b_{min}) when:

(1) a=√(λL) and b_{min}=√(4λL) |
(2) a=λ^{2}/L and b_{min}=√(4λL) |
(3) a=λ^{2}/L and b_{min}=(2λ^{2}/L) |
(4) a=√(λL) and b_{min}=(2λ^{2}/L) |

b sin θ = nλ

(b-a)/L = tan θ ≈ sin θ

a(b-a)/L = λ

ab – a^{2} = λL

a^{2} -ab + λL = 0

b^{2} >= 4λL

b_{min} = √(4λL)

a(√(4λL)) – a^{2} = λL

a = λ^{2}/L

Correct answer to given problem No 25 of IIT JEE Main Physics paper of 2016 is option number (2).

Hysteresis loops for two magnetic material A and B are given below:

These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:

(1) A for transformers and B for electric generators

(2) B for electromagnets and transformers

(3) A for electric generators and transformers

(4) A for electromagnets and B for electric generators

Option no 2 as Hysteresis loss for electromagnets and transformers are required to be less, so that, it can magnetize or demagnetize easily.

Correct answer to given problem No 24 of IIT JEE Main Physics paper of 2016 is option number (2).