KIITEE Mock Test - 5 - JEE MCQ

If the length of capillary tube is l, the pressure difference is given by

where η is the coefficient of viscosity of water. If r becomes 2r and Q becomes Q/2, we have

Hence the correct choice is (a).

Loss in energy when two capacitors are connected in parallel is

When the frequency of the driving force is equal to the natural frequency i.e. at resonance, the amplitude is maximum.
At resonance, average power absorbed by the forced oscillator is also maximum.
So, both ω₁ and ω₂ are same at resonant frequencies.

Critical velocity of the liquid flowing through the pipe is , where r is the radius of the pipe and R is the Reynold's number.
Hence, 
The velocity of a liquid flowing through a tube is inversely proportional to the cross-sectional area of the tube.

Let A and B be the two extreme positions of the particle with O as the mean position. Displacements to the right of O are taken as positive while those to the left of O are taken as negative

Let the displacement of the particle is SHM be given by

Let us suppose that at time t = 0, the particle is at extreme position B. Setting x = A at t = 0 in Eq. (i)
we have

Now let us say that the particle reaches point C at t = t₁ and point D at t = t₂. At C, the displacement x(t₁) = + 12.5 cm and at D, it is x(t₂) = -12.5 cm (see Fig.).
So from (ii) we have

Notice that cos ωt₂ = -0.5 even for t₂ = 4π/3 .
This value of t₂ does not correspond to the minimum time because this is the time at which the particle, moving to left, reaches A and then returns to D.

Let T be temperature of source, T₁ be initial temperature of sink in the first case and T₂ be temperature of sink in second case.
For the first case,

T₁ = 
For the second case,


T₁ - T₂ = 50

= 50
T = 25 × 15 = 375 K

The fraction of the solid immersed in the liquid is given by:

 

where σ is the density of the solid, ρ is density of the liquid (upthrust = weight), V_i is volume immersed in the liquid and V is the total volume of the immersed body.

If the temperature increases, the fraction of the immersed solid becomes

Here,

Therefore, we have

 where M = mass of stone. If ρ is the density of the stone and V its volume, then m = ρV.
When the stone is wholly immersed in water of density ρ', the effective weight of the stone

Given l = 40 cm and l' = 30 cm.
Also v = v', which gives

Hence the correct choice is (2).

For uncatalysed reaction:
log k = log A - 
For catalysed reaction:
log k' = log A - 
Or, log k' = log A - 
log k' = log k + 
log k' - log k = 
  = 1.741 x 10^-3

In the reaction, two moles of C-H bonds and two moles of C-Cl bonds are broken.
Hence, energy absorbed = 2 x 416 + 2 x 325
= 1482 kJ

Among the group 15 hydrides, the stability decreases on moving down the group, hence the order of stabilities of the hydrides of group 15 is

NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃

This is due to the increasing bond length down the group.
All the other statements are correct.

i.e. NCl₃ + 3H₂O → NH₃ + 3HOCl

Since the P-H bond is greater than the N-H bond, phosphine is a stronger reducing agent than ammonia.
NO is paramagnetic in the gaseous state. However, in the solid state, it dimerises, which results in the pairing of the odd electron.

Oxymercuration-demercuration of alkenes yields a non-rearranged Markovnikov product.

The lone pair of electrons on the -NH₂ group is donated towards the benzene ring due to resonance effect. As a result, acidic character of -COOH group and basic character of -NH₂ group decrease. Therefore, the weakly acidic -COOH group cannot transfer an H⁺ ion to the weakly basic -NH₂ group. Thus, o- and p-aminobenzoic acids do not exist as zwitter ions.

From the Gibb's-Helmholtz equation

Hence,

ΔG= 0

W = nC_V(T₂ - T₁)
Work done by the gas = -3 kJ = -3000 J
C_V = 20 JK^-1 mol^-1
n = 1
T₁ = 273 + 27 = 300 K, T₂ = ?
Putting the values,
-3000 = 1  20(T₂ - 300)
-150 = T₂ - 300
300 - 150 = T₂
⇒  T₂ = 150 K

This option is correct because CrO₅ has a butterfly structure having two peroxo bonds as shown below:

In the peroxo bonds, the oxidation state of each oxygen is -1. Now, let the oxidation state of Cr in CrO₅ be x.
Therefore, x + 4(-1) + 1(-2) = 0
x = +6

According to anti-Markovnikov's rule, X is 1-bromobutane and Y is 2-bromobutane.

In this reaction, organic peroxide dissociates butene into free radicals. There is the formation of free radicals as intermediates.

More stable free radicals give major products according to anti-Markovnikov's rule.

The general equation of all non-vertical lines in a plane is: ax + by = 1, where b ≠ 0.
 a + b  = 0
 b = 0
 = 0
[ b  0]
This is the required differential equation.

Let 2n be the total number of terms

where T₁, T₃ ... T_{2n - 1} are terms occupying odd places

[Sum of G.P., where a is first term and r is common ratio]

Given that (cos θ + i sin θ)(cos 2 θ + i sin 2 θ) …. (cos n θ + i sin n θ) = 1
or, cos(θ + 2 θ + 3 θ + ….. n θ) + i sin(θ + 2 θ + ……… n θ) = 1

Since y = f(x) and the given differential equation is
y(1 + xy) dx - xdy = 0

Or, 
.
Integrating, we get
, which passes through (1, 2).
∴ c = 1
Then, the curve is

Or, 
.