JEE Advanced Practice Test- 9 - JEE MCQ

At mean position a = 0

Hence velocity of the particle is zero at x = 0 and x = 2α/β

We can conclude that particle oscillated between these two limits and have amplitude A = α/β

Acceleration of the particle is maximum at x = 0 and x = 2α/β , a_max = qα/m

Current through the 2Ω resistance

Current through the 2Ω resistance

Z₁ has positive real part and

Z₂ has negative imaginary part

Then there can be following cases

When a, b are positive real numbers

Also,

When a, b are non negative real numbers

Similarly in all the other cases

is always purely imaginary

is either zero or purely imaginary.

Using Gauss's law, we have

Hence, Gauss law is invalid and option 1 is wrong.

Electric field due to a dipole is not symmetric; hence, a Gaussian surface cannot be used to find the electric field due to a dipole using

Gauss law. Hence, option 2 is wrong.

Electric field on the line joining the charges is opposite in direction if they are of the same sign only. Hence, option 3 is correct.

By definition of potential, work done by external force to move a unit positive charge from point A to point B is given by:

W = q(V_B -V_A) = V_B - V_A

Hence, option 4 is correct.

y = x² ⇒ at x = 1/2

y = 1/4

Hence the coordinate

Differentiating : y = x² ⇒ v_y = 2xv_x

Which satisfies the line

4x − 4y − 1= 0 (tangent to the curve)

and magnitude of velocity :

As the line 4x - 4y = 1 does not pass through the origin, therefore The magnitude of angular momentum of particle about origin at that position is 0 is not correct.

v = (tan 45^∘)x ⇒ v = x ⇒ dv/dx = 1 ,and we know that acceleration, a = dv/dt = vdv/dx

⇒ a = x × 1 ⇒ a = x, so the graph between acceleration and displacement is,

Now we know that rate of change of linear displacement is called velocity,

at t = 0, x₀ = 1 (given) ⇒ x = e^t

∵ v = x ⇒ x = v ⇒ x = (tan 45^∘) v = x = v,

= 1000 volt

I₁ = I_o(I - e^-t/(L/R)), where I_o = E/R

I₂ = I_oe ^-t/RC

Constant current means current I through the battery is constant.

I = I₁ + I₂ is constant with time.

This is possible, if

L/R = RC

u = -15 m

v = mu

v = (-2) × (-15) = 30 cm

Hence,

f = 10 cm

Therefore, the average emf induced in the circuit is 37.9 V.

Energy = eV = E

eV = 27.61 keV

V = 27.61 kV

So, the order of accelerating voltage is 30 kV.

or g ∝ ρR

Now escape velocity,

= 3 km s⁻¹

When path difference increases from 0 to λ/2 , intensity will decrease from maximum zero. Hence, in this case, I(P₂) > I(P₁) > I(P₀)

In this case I(P₁) = 0.

Thus light rays going from rarer medium to denser medium will bend towards the normal and vice versa.

It can be seen from the image how the light rays bend when they refract at a curved surface. (Line along the radius is the normal here as shown in the figure)

If the ray goes undeviated, then the refractive indexes are same.

So in 1, μ₁ < />₂ and μ₂ = μ₃

So in 2, μ₁ > μ₂​ and μ₂ > μ₃

So in 3, μ₁ < />₂ and μ₂ = μ₃

So in 4, μ₁ > μ₂​ and μ₂ > μ₃

So in 5, μ₁ > μ₂ and μ₂ = μ₃

With both keys S₁ ans S₂ initially open, if now S₁ is closed, the capacitors C₁ and C₃ are in series as also the capacitors C₂ and C₄.

When sodium metal is burnt in excess of air, mainly sodium peroxide (Na₂O₂), with little sodium oxide (Na₂O), is formed.

If the oxidation is carried at high pressure, then sodium superoxide (NaO₂) may be formed. The question does not mention the condition of increased pressure.

For a first order reaction, [A] = [A]₀e^-kt , concentration of A decreases with time exponentially.

Half-life for the first order reaction implies that half-life is independent of initial concentration.

From Arhennius Law, k =

As temperature increases value of k increases, as can be seen from the equation.

As the value of k increases we conclude that the half-life decreases.

For 99.6% completion, [A] = 0.4 and [A]₀ = 100

Therefore, number of half-lives used

For a compound to be aromatic, it should follow Huckle's (4n + 2)πe⁻ rule, i.e., 2, 6, 10, 14....πe⁻

They should show delocalised, conjugate π system with alternate single and double bonds. In addition to this, the compound should be cyclic and planar or almost planar.

The number of πe⁻ is even, but not a multiple of 4.

For anti-aromatic compounds:

4nπe⁻, i.e., 4, 8, 12....πe⁻ should be delocalized in a cyclic, planar, or almost planar system with alternate single and double bonds.

Planar conjugated carbocyclic polyene, carbocyclic polyene structure is less stable than its open structure.

The collisions of the gas molecules are elastic and the energy lost by one molecule is equal to the energy gained by the other.

Hence, at a given temperature, the average energy of the gas is constant.

Heavier molecules transfer more momentum to the walls of the container. But this is not the postulate of the kinetic theory of gases.

Only a small fraction of the molecules possess high velocity (V_r.m.s), whereas the maximum number of molecules possess a lower velocity called the most probable velocity (V_mp).

The molecules move in random, zig-zag straight lines. The motion is referred to as Brownian motion.

Thus, all the colloidal particles acquire a charge and repel the other colloidal particles as they have like charges.

And potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles stablises the charges acquired by adsorption and makes the lyophobic sol stable.

Hence, options (1) and (4) are correct.

In above structure there is no free carbonyl centre of an aldehydic group present ; It is a non-reducing sugar.

2. Decomposition of acidified KMnO₄ is catalysed by sunlight.

3.

4. KMnO₄ also acts as an oxidizing agent in alkaline medium:

Second electron gain enthalpy always be +ve

Na⁺ (g) + Cl⁻ (g) ⟶ Na⁺Cl⁻(s); ΔH_L.E. = (−)ve

Lattice energy is -ve when two ion combine & from 1 mole solid

N(g) ⟶ N⁻(g); ΔH_gain = (+)Ve

On electron addition in half filled (more stable configuration)energy will be added +ve

Al²⁺ (g) ⟶ Al³⁺ (g); ΔH_I.E. =(+)Ve

to electron remove energy have to gives [+ve]

Since the rate becomes 8 times when both [G] and [H] are doubled, the order of the reaction with respect to the reactant H is two.

Rate of reaction = k[G][H]²

Overall order of the reaction = 1 + 2 = 3

The equation can be written as:

K₂SO₄ + 6FeSO₄ + 7H₂SO₄ → Cr₂(SO₄)₃ + 3Fe₂(SO₄)₃ + K₂SO₄ + 7H₂O

One mole of potassium dichromate reacts with 6 moles of ferrous sulphate. Thus, the number of moles of Mohr's salt required per mole of dichromate is 6.

CH₃ - CH₂ - CH₂ - CH₂ - CH₂ - CH₃

(A) It is one of the natural series found on earth.

(B)

(C)

(D) Isodiaphers are atoms, having the different atomic number and mass number but have the same difference between a number of neutron and number of a proton.

₉₀Th²³² and ₈₈Ra²²⁸ are isodiaphers,