COMEDK Mock Test - 9 - JEE MCQ

P = constant

By integration,

Consider the equilibrium of wooden block.
Forces acting in the downward direction are

(a) Weight of wooden cylinder

(b) Force due to pressure (P₁) created by liquid of height h₁ above the wooden block is
= P₁ × π (2r)² = [P₀ + h₁rg] × π (2r)²
Force acting on the upward direction due to pressure P₂ exerted from below the wooden block and atmospheric pressure is

At the verge of rising

[Negative sign is used as with increase in r, θ decreases.]

Now as for spherical shell A = 4πr²

which on integration and simplification gives

Now in steady state as no heat is absorbed, rate of loss of heat by conduction is equal to that of supply, i.e., H = P, and here

θ₁ - ​θ₂ = T and a ≃ b = Ra ≃ b = R

So Equation (i) becomes,

i.e., thickness of shell

Given,

Now we see that the velocity is perpendicular to the magnetic field so the magnitude of velocity will not change due to magnetic field.
The magnitude of velocity will change in 'i' direction due to  only

Let us say that the speed of particle doubles after time 't'. in y direction the speed will be V₀ only in y - z plane always perpendicular to x-direction so net speed at time 't'.

Also by given condition V_net = 2V₀

Length of the antenna = 25 m
 = λ / 4
⇒ λ = 100m

Initial Surface Area = 4πR²
= 4 π (D/2)²
= 4 π D² / 4 = π D²
Surface Area at ΔT
= 4π(D^'/2)²
Now
D^'=D(1+α ΔT)
So Surface Area at ΔT
= 4π(D²/4 )(1+α ΔT)²
= π D² (1+ α²ΔT² + 2α ΔT)
Change in surface Area
= π D² (α²ΔT² + 2α ΔT)
= π D²α ΔT (α ΔT+ 2)

According to first law of thermodynamics, for the path iaf,

Q_iaf = ΔU_iaf + W_iaf

or, ΔU_iaf = Q_iaf - W_iaf

= 50 - 20 = 30 cal

For the path ibf,

Q_ibf = ΔU_iaf + W_ibf

Since ΔU_iaf = ΔU_ibf, change in internal energy is path independent.

Q_ibf = ΔU_iaf + W_iaf

∴ W_ibf = Q_ibf - ΔU_iaf = 36 - 30 = 6 cal

The reaction of sodium carbonate with slaked lime to produce caustic soda is as follows:

Also, Ca(OH)₂ when bubbled with CO₂ gives milkiness due to the formation of calcium carbonate.

Bicarbonates of Ca and Mg present in the water result in temporary hardness. These bicarbonates react with slaked lime to form insoluble calcium and magnesium carbonates, which can be easily filtered off.

This is known as Clark's method to remove temporary hardness.

From balanced chemical reaction

Optically active compounds rotate the plane of polarised light either towards the right or left side. And the compound which has at least one chiral centre will be optically active.

Since only the given compound contains a chiral carbon, it will rotate the plane of polarised light.

Option D also has chiral carbons but it also contain plane of symmetry. So, it will be meso compound.

For a NaCl crystal, Cl- ions form the FCC lattice and Na+ ions lie in the octahedral voids.
The minimum value of radius ratio (radius of cation/radius of anion) in an octahedral void is 0.414.

Base BOH is dissociated as follows.

So, dissociation constant of BOH base,

At equilibrium, [B⁺] = [OH]^-

In general, higher the oxidation state, more is the covalent character of the oxide. Therefore, the only covalent oxide of iodine is I₂O₅, whereas I₂O₇ and I₂O₉ do not exist and I₂O₄ is iodyl iodate, i.e. [IO]⁺[IO₃]^-, which is ionic in nature.

Given 125 mL of 1 M AgNO₃ solution.

1000 mL of AgNO₃ solution contains 108 g Ag.

 125 mL of AgNO₃ solution contains

Balanced equation :

For finding limiting reagent, moles of both the reactants are divided by their coefficients and the least value gives L.R.

∴ TiCl₄ is the limiting reagent So, on the basis of L.R., product amount will be calculated.

∵ One mole of TiCl₄ produces one mole of Ti.

 Number of moles of Ti produced by 1.88 moles of 

TiCl₄ = 1.88 moles.

Mass of Ti= moles of Ti x molar mass of Ti

Lyophilic colloids have very less or no charge and their particle size is within colloidal range.

So, charge and size do not show any effect on stability.

But they are highly solvated in the dispersion medium, they exist and remain stable. So, dispersion medium is the reason behind the existence of lyophilic colloids in stable form.

According to Faraday' first law of electrolysis:

The euivalent weight of Ag is maximum among these hence, AgNO₃ solution will deposit the maximum mass of the metal at the cathode on electrolysis.

y = e^cx
dy/dx = c. e^cx
y' = cy