JEE Main Mock Test - 6 - JEE MCQ

Binding energy per nucleon of a deuteron 

Binding energy per nucleon of helium  

We have to find released energy if helium is formed from the fusion of two deuterons. Rest mass energy of deuteron,

Given, force, F = 5 N
Mass, m = 500 g = 5 × 10⁻¹ kg
Displacement, s = 5 m
Work done is given by
Let's assume the angle between force and displacement to be zero. Then work will be,

W = Fs = 5 × 5 = 25 J
Let the time taken to travel the displacement of 5 m be t s.
Applying Newton's second law of motion,
F = ma


The average power is given by

In 1 - > small portion from top left corner is removed there will be shift in x-co ordinate of COM in right direction

In 2 - > small portion from top left corner and bottom left corner also are removed there will be shift in x-co ordinate of COM in right direction (greater shift in right direction than 1)

In 3 - > small portion from top left corner and bottom left corner also are removed there will be shift in x-co ordinate of COM in right direction (greater shift in right direction than 1) and one portion from right top corner is than removed so a shift in leftward direction is attained which is comparable to 1. but is still in rightward direction to 1.

Excess of Pressure inside liquid bubble P₁ = 4T/R
electrostatic pressure due to charge P₂ = σ²/2ε₀
Comparing them 4T/R = σ²/2ε0
Therefore T = σ²R/8ε0

Due to frictional force (which acts in a direction opposite to the direction of acceleration) on the rear face, the pressure in the rear side will be increased. Hence the pressure in the front side will be lowered. Hence the correct choice is (b).

Range of vision for healthy eye is 25 cm (near point) to ∞ (far point). If the person can see clearly only upto a maximum distance of 50 cm he is suffering from myopia (short sightedness). A shortsighted eye can see only nearer objects. This defect can be removed by using a concave lens of suitable focal length f = 50 cm.

An oscillating LC-circuit having a maximum charge on the capacitor as Q.
We have to find the charge on this capacitor when the energy is stored equally between the electric and magnetic fields.

Magnetic energy stored by the inductor at any instant is 

Electrical energy stored by the capacitor at any instant is  [where, q is the charge at that instant]

Maximum electrical energy stored by the capacitor is 

We are given that energy is equally stored between the electric and magnetic fields 

By the principle of conservation of energy, 

Dynamic resistance,

Net force on particle towards centre of circle is,

This force will act as centripetal force. Distance of particle from centre of circle is a/√2

Let initial volume V₁ be V

Given: Final volume,  

Initial temperature, 

We have to find the rise in temperature, say T₂.

For an adiabatic process: 

Intensity of sunlight = I = 0.092 Wm^-2

The peak value of electric and magnetic field are related as:

Using constraint

Common glass consists of silica (SiO₂). When HF reacts with silica it will form H₂SiF₆ or SiF₄ Both can soluble in water which means that a fresh surface of SiO₂ will be exposed to attack by HF.

BF₃ is least acidic among boron halide.

Molar conductivity at infinite dilution of KCl is

= (73.52 + 76.34)Ω⁻¹ cm² mol⁻¹
= 149.86 Ω−cm² mol⁻¹
Conductivity of 0.01 M KCl is

(0.01 × 10⁻³ mol cm⁻³) = 1.4986 × 10⁻³Ω⁻¹ cm⁻¹
Cell constant is
K = k₁R = (1.4986 × 10⁻³Ω⁻¹ cm⁻¹)(525 Ω) = 0.7868 cm⁻¹
Conductivity of 0.1 M NH₄OH is

Molar conductivity of NH₄OH at infinite dilution is

= (73.4+197.6)Ω⁻¹ cm² mol⁻¹
= 271.0 Ω⁻¹ cm² mol⁻¹
Degree of dissociation of NH₄OH is

Equilibrium constant of NH₄OH is

After electrolysis, aqueous NaCl is converted into aqueous NaOH.
The quantity of electricity passed  
The number of equivalents of OH−ion formed = 5.44 × 10⁻³
∴ Morality of NaOH =

∴ pOH = −log(1.36 × 10⁻²) = 1.87
∴ pH = 12.13

The reaction of CO + HCl in the presence of AlCl₃ with benzene to form benzaldehyde is called Gatterman-Koch reaction.

Addition across C=O group occurs via nucleophilic mechanism.

Let, PQ is a focal chord and S is the focus
Now, OS = 1
⇒ In ΔOSR (R is the foot of the perpendicular from O on PQ)
⇒ cosecθ = 1/0.4 = 2.5
Length of PQ is equal to the length of latus rectum×cosec2θ
⇒ PQ = 4(2.5)² = 4 × 6.25
= 25

Let, the breadth of the rectangle be B, then the length will be 2B, since the ratio of length and breadth is 2:1,
Now, the area of rectangle is L × B = 2B × B = 2B²  ...(1)

Also, the ellipse is inscribed in the rectangle, hence, length of major axis of the ellipse is equal to the length of the rectangle i.e., 2a = L = 2B, ⇒ a = B, 

And, the length of minor axis of the ellipse is equal to the breadth of the rectangle, hence 2b = B, ⇒ b = B/2,

On putting this value in equation (1), we get,
Area of rectangle 

Let, the end points of the diameter of circle are
(a sec θ, b tan θ) & (−a sec θ, b tan θ)
⇒ Equation of circle is
(x − a sec θ)(x + a sec θ)+(y − b tan θ)² = 0
⇒ x² − a²sec²θ + y² + b² tan²θ − 2b tan θ y = 0
Because (2,0) satisfies above equation,
hence, 4 − a²(1 + tan²θ) + 0 + b² tan 2θ = 0
⇒(4 − a²) + (b² −a²)tan²θ = 0
Which is always true if |a| = 2 & |b| =2

Putting u = x^x and using logarithmic differentiation, we get log u = x log x.