JEE Advanced Practice Test- 1 - JEE MCQ

During time 't₂', capacitor is discharging with the help of three resistors of resistance 'R' each.

Therefore, q = q0-3t/RC

[∵Q = CV]

V = V₀e^-3t/RC

As

t₂ = (RC/3).log_e2

During time 't₁', capacitor is charging with the help of battery.

Therefore, q = q₀(1 - e^-3t/RC) or V = V₀(1 - e^-3t/RC)

As

t₁ = (RC/3).log_e2

T = t₁ + t₂ = (2RC/3).log_e2

As PV = αT²

PdV + VdP = 2 αTdT

For constant pressure process, dP = 0

PdV = 2 α TdT

W = 3αT₀²

We know ΔS = q_rev / T

But

It is given n = 1,V₂ / V₁ = 10

q_rev = 2.303 RT log 10 = 2.303 RT

Hence

= 2.303 R

Fe²⁺ + KSCN → No colouration

If charges are of opposite signs then the two fields are along the same direction. So, they cannot be zero. Hence, the charges should be of same sign.

Further, work done per unit charge by external force = change in potential energy

∴

=

or

1 N × cos θ acts towards Q along the inclined plane, and mgsinθ acts towards P along inclined plane

At θ = 45^o, mgsinθ = 1 × cosθ

At θ > 45^o, mgsinθ > 1×cosθ (friction acts upward)

At θ < 45^o, mgsinθ < 1×cosθ (friction acts downward)

Slope ∝ ρ

Mg − T = Ma ........(i)

T = ma ........(ii)

Solving (i) and (ii)

FBD of man

Mg − N = Ma

On applying the Fleming's right-hand rule, the induced current is from left to right on the crossbar.

Current in the loop

Forces acting on the cross bar

Its weight mg acts downward.

Magnetic force B₀iL acts in upward direction.

When rod attains terminal velocity, resultant force on the rod is zero

We have, x = A sin²ωt + B cos²ωt + C sinωtcosωt = A/2 (1 - cos2ωt) + B/2(1 + cos2ωt) + C/2sin2ωt

For A = -B; C = 2B: x = B cos2ωt+ B sin2ωt; which is an SHM, whose amplitude is B√2.

For A = B, C = 0:

We get x = A, which in not an SHM.

For A = B, C = 2B:

x = B + B sin2ωt; which is an SHM, whose amplitude is B.

For any values of A, B and C (except C = 0):

;

which is an SHM.

When switch S₁ is closed, the equation that describes the charging of the capacitor is given by:

Maximum charge on the capacitor, Q = CE.

Energy stored in the capacitor, when it is fully charged, is

When switch S₂ is closed, according to the law of conservation of energy, the energy in the circuit cannot exceed 4.05 J, as the loss of energy is not considered; the energy in the circuit is 4.05 J.

Charge across the capacitor and current in the circuit are given by:

Magnitude of the rate of change of current is

At t = 0,

Using Fourier law,

Q = UA△T

Q_A = 2K/L

Q_B = 3K/4L

Q_C = K/L

Q_D = 5K/4L

Q_E = 6K/L

Also, heat flow across the system remains same.

∴ Option 1 is correct.

And since U_E is maximum and heat flow across the system remains constant, option 3 is also correct.

Now,

∵ U_C = K\L

∴ Q_C = KA△T/L

Also,

∵ U_B + D = 5K/4L + 3K/4L + 4K/L

∴ Q_B + D = KA△T/L

∴ Q_C = Q_B + D

Let the heat capacity for the polytropic process be C.

Here, μ is the number of moles.

Hence,

For the polytropic process,

1 + n = 6

⇒ n = 5

Using the quantum condition: 2π = nλ

Putting x⁴ = N, so the bullet loses the velocity by v/n on passing through each plank.

As the total number of planks = 8, therefore

change in K.E. in passing through 8 planks is 100%.

Solving this, we get:

N = 15.5

or approximately N = 16

x⁴ = 16

x = 2

M × 5 × 10⁻² = 0.5

m = 10

2n =10

n = 5

From momentum conservation, Velocity of B just after impact = v/2

Angular velocity, ω = v / 2l

So, n = 2

Velocity of first block before collision,

By conservation of momentum,

Also v'₂−v'₁ = v₁ for elastic collision

It gives V'₂ = 0.4m/s

v'₁ = −0.2m/s

Now distance moved after collision

∴ s = s₁ + s₂ = 0.05m = 5 cm

(I) ⇒ Aromatic 6πe⁻ system in conjugation

(II) ⇒ Non - aromatic, 8πe⁻ system but ring is not planar, so not anti - aromatic but non - aromatic.See

(III) ⇒ anti - aromatic 8πe^- in conjugation

Stability order: Aromatic > Non - aromatic > Anti - aromatic.

(I) > (II) > (III).

2NaI + Cl₂ ⟶ 2NaCl + I₂(violet colouration)

The reaction proceeds through the formation of a carbocation intermediate, which re-arranges by a hydride shift followed by ring expansion in order to reduce the Bayer strain.

Thus, the major product formed is 1,2-dimethyl cyclohexene.

Step 1:

Step 2:

Hence, option (a), (c) and (d) are correct.

Benzoquinone is highly stable due to conjugation and does not exhibit tautomerism.

Option (a) is incorrect as cold alkaline solution of KMnO₄ would produce vicinal diols.

Option (b) is correct as KMnO₄ is stronger oxidising agent and in the acidic medium, it oxidises the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds.

Option (c) is correct as the process is ozonolysis by ozone, which is arrested at the aldehyde stage by adding zinc water.

Option (d) is incorrect as decarboxylation, with soda lime, would produce ethane. Ethene is obtained by Kolbe's electrolysis. Hence, 'S' represents the process of electrolysis and not a reagent.

1) Number of Ba ions per unit cell = 1/8 x 8 = 1

Number of Ti ions per unit cell = 1 x 1 = 1

Number of O ions per unit cell = 1/2 x 6 = 3

Hence, the empirical formula of the substance should be BaTiO₃.

Thus, (a) is correct.

2) Titanium ions can be assumed to occupy a hole in the lattice formed by barium and oxide ions. The titanium ions occupy the body centres of the face centred cubic and this is one of the octahedral holes in the fcc lattice.

So, (b) is also true.

3) The octahedral holes at the centres of unit cell constitute just one-fourth of all the octahedral holes in a face centred cubic lattice. Thus, (c) is also true.

4) An octahedral hole at the centre of a unit cell, occupied by titanium ion, has 6 nearest neighbour oxide ions. The other octahedral holes located at the centres of the edges of unit cells have 6 nearest neighbours each, as in case with any octahedral hole, but 2 of the 6 neighbours are barium ions (at unit cell corners terminating the given edge) and 4 oxide ions. The proximity of two cations Ba²⁺ and Ti⁴⁺ would be electrostatically unfavourable. So, titanium cannot occupy all of the octahedral holes and restrictions as stated above exist.

So, (d) is wrong.

Tin is obtained by reducing the ore cassiterite with coal in a reverberatory furnace. Limestone is added to produce a slag with the impurities, which can be separated.

SnO₂ + 2C → Sn + 2CO

Crude tin so obtained is contaminated with iron, copper, lead and other metals. It is, therefore, remelted on an inclined furnace. The process is called liquation.

The easily fusible tin melts away and the less fusible impurities are left behind. Molten tin is finally stirred with green poles of wood in contact with air. In this process, any remaining metal impurities are oxidised forming a scum, which rises to the surface and is removed. This process is called poling.

In NO (15 electrons), the last electron is unpaired in

NO + (BO = 3.0) & O₂⁺(BO = 2.5)

In OF⁺ (Electrons = 16), 2 unpaired electrons in

In Ne₂⁺ (Electrons = 19), 1 unpaired electron in

ie.

π -bond has inter-nuclear axis coincident with the nodal plane.

Hence, answer ​is Face plane in FCC and ​Body diagonal plane in BCC.

Body diagonal plane in FCC the atom at body centre will not be touching four corner atoms. Face plane in BCC shall not contain the atom at body centre.

The dichlorocarbene formed acts as an electrophile and attacks the para methyl phenol molecule at ortho and para position which are the sites where the negative charge is localised on the ring.

There are two products formed out of which the para substitution product is minor due to steric hindrance caused by the methyl group while the ortho substitution product is the major as the site has no effect of steric hindrance.

H₃P₃O₉ is a cyclic tri-metaphosphoric acid.