JEE Main Part Test - 2 - JEE MCQ

Carbonic acid is a weak acid it dissociates as follows 

Since both acids are weak cannot be 0.034 M as there is no complete dissociation. 

Total hydrogen ion concentration is approximately equal to concentration of hydrogen ion in first reaction as K₁ is larger. This is approximately equal to concentration of 

The concentration of  depends on  K₂ and that of H⁺ on K₁. But . So The concentration of H⁺ is not double that of 

  (orbital speed υ₀ of a satellite)

Near the earth’s surface is equal to 1/√2
times the escape velocity of a particle on earth’s surface) Now from conservation of mechanical energy: Decrease in kinetic energy = increase in potential energy

By the principle of superposition of fields

Here  = net field at the centre of hole due to entire mass

 = field due to remaining mass

 = field due to mass in hole = 0

From principle of continuity,

Let ρ = density of the liquid.

 Let α = area of a cross-section of each hole.

 Volume of liquid discharged per second at a hole = αv.

 Mass of liquid discharged per second = αvρ.

 Momentum of liquid discharged per second = αv²ρ.

∴ the force exerted at the upper hole (to the right) = αρv2²

 and  the force exerted at the lower hole (to the left) = αρv1²

Net force on the tank = 2αρgh. 

Let p0 = atmospheric pressure,

p1 and p2 = pressures inside the two bubbles

Let r = radius of curvature of the common surface

The thermal resistance

Fe²⁺ is added asFeSO₄
Fe⁺ is formed by charge transfer from NO to Fe²⁺ 



Fe⁺ has three unpaired electrons (N).

Volume of the original solution = 1 L

pH of the new solution = 2

Volume of the new solution = ?
As per volumetric principle

That means the volume of water added to the original solution of 1 L = 10 −1 = 9 L

= 37.22

Since the acid is only 0.059% ionized, therefore the concentration of ions in solution = 0.1 x 0.059 / 100 = 0.000059

Ka = [H⁺] [X^–] / [HX] = (0.000059)² / 0.1 = 3.5 x 10^-8

When ammonia is added, solubility of AgCl increases due to formation of complex salt which decreases the concentration of radicals in the product side and thus drives the reaction in forward direction.
When we add KCl common ion effect is applied in presence of common ion solubility decreases and reaction goes in backward direction.

First off, since NaOH is a strong base, it will dissociate completely into Na+ and OH-. Thus, we know that we have 0.01 M OH-.

However, we do not know anything about the concentration of H+. Fortunately, we do not need to, as pH + pOH = 14. So, if we find pOH, we can solve for pH. p is a mathematical function equivalent to -log. So, pH actually means -log[H+] (Note that brackets indicate concentration of).

pOH = -log 0.01M OH-

pOH = 2

pH + 2 = 14

pH = 12

This result makes sense, since a solution of strong base should have a high pH.

 (1+x)ⁿ= nC0+ nC1x+ nC2x²+...+ nC0xⁿ ....(1)
(1−x)ⁿ= nC0 − nC1x+ nC2x²−...+ nC0 xⁿ....(2)
⇒(1+x)ⁿ−(1−x)ⁿ = nC0+ nC1x+ nC2x²+...+ nC0xⁿ− nC0+ nC1x− nC2x²+...+ nC0xⁿ
 =2(nC1x+nC3x³+....+xⁿ)

Given: n=60 and put x=1
⇒(1+1)⁶⁰−(1−1)⁶⁰
 =2⁶⁰
⇒2⁶⁰=2(nC1x+nC3x³+....+xⁿ)

∴Sum of odd powers of x in the expansion is = nC1x+nC3x³+....+xⁿ =2⁶⁰−1 =2⁵⁹

Middle term in the expansion of (1+x)²ⁿ; is 
=tn+1 = 2nCn.1⁽²ⁿ⁻ⁿ⁾.xⁿ
= {(2n)!.xⁿ}/(2n−n)!n!
= {{2n(2n−1)(2n−2)(2n−3)....4×3×2×1}/n! n!}/xⁿ
= {{2ⁿ[n(n−1)(n−2)....×2×1][(2n−1)(2n−3)....3×1]}/n! n!}xⁿ
= [[(2n−1)(2n−3)....3×1]/n!] 2ⁿxⁿ

Let the point be (h, k)

Now equation of tangent to the parabola y² = 4ax whose slope is m is

as it passes through (h, k)]

 Parabola is x² – 8x + 2y + 7 = 0 
∴   (x – 4)² = – 2y – 7 + 16 
∴   (x – 4)² = – 2[y – (9/2)]
 ∴   x² = – 4ay  
 ⇒ x = x – 4, y = y – (9/2) and 2 = 4a
 i.e. a = (1/2) 
 Its focus is given by x = 0 and y = 0 i.e. x – 4 = 0   and     y – (9/3) = 0 
∴    x = 4    and y = (9/2) 
∴ focus [4, (9/2)].

Using the diagram, OA⊥BC

⇒ Slope of OA × Slope BC = −1

Solving Equations(i)i and (ii)ii, we get

α = −1, β = 6

∴ The third vertex is (−1, 6)