JEE Advanced Mix Test 58

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JEE Advanced Mix Test 58

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Weekly Quiz Competition

Question 1
4 / -1

Two trains A and B are moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle.

Assume that the sound of the whistle is composed of components varying in frequency from f₁ = 800 Hz to f₂ = 1120 Hz, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.

The spread of frequency as observed by the passengers in train B is

A
310

B
315

C
320

D
325

Solution
Question 2
4 / -1

A resistance of 2Ω is connected across one gap of a metre-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is

A
3

B
4

C
5

D
6

Solution
Question 3
4 / -1

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s^-2).

The magnitude of the normal reaction that acts on the block at the point Q is

A
7.0

B
7.7

C
7.8

D
7.5

Solution

Given:

m = 1 kg

R = 40 m

θ = 30°

Work against friction = 150 J

g = 10 m/s²

Initial potential energy at top = m g R = 1 × 10 × 40 = 400 J

Height of point Q from base = R sin(30°) = 40 × 0.5 = 20 m

Potential energy at Q = m g h = 1 × 10 × 20 = 200 J

Apply work-energy theorem:

400 = KE + 200 + 150

KE = 50 J

KE = (1/2) m v² ⇒ 50 = (1/2) × 1 × v² ⇒ v² = 100 ⇒ v = 10 m/s

Centripetal force = m v² / R = 100 / 40 = 2.5 N

Component of weight towards centre = m g sin(30°) = 10 × 0.5 = 5 N

N - 5 = 2.5 ⇒ N = 7.5 N

Answer: 7.5
Question 4
4 / -1

Work function of metal A is equal to the ionization energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionization energy of He+ ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.

Value of E ( in eV) is

A
21.8

B
24

C
23.8

D
20.8

Solution

W_A = ionization energy of electron ion 2nd orbit of hydrogen atom = 3.4 eV

W_B = ionization energy of electron in 2nd orbit of He+ ion

= 13.6 eV.

Now, given that

K_A= 2K_B

W_A = 3.4 eV, W_B = 13.6 eV.

K_A = E − W_A, K_B = E − W_B,

E − 3.4 = 2(E − 13.6)

E − 3.4 = 2E − 27.2

E = 23.8 eV.
Question 5
4 / -1

A glass tube AD of uniform cross-section of length 100 cm100 cm sealed at both ends contains two columns of ideal gas AB and CD separated by a column of mercury of length 20 cm20 cm. When the tube is held horizontally, AB = 20 cm cm and CD = 60 cm. When the tube is held vertically with the end A up, the mercury column moves down 10 cm10 cm. What will be the length of gas column AB when the tube is held vertically with the end D up? (Give answer upto two digits after decimal point.)

A
9.88

B
8.88

C
10.88

D
13.88

Solution

As we know according to Boyle's law, P₁V₁= P₂V₂.

In horizontal position the pressure of gases in column AB and CD is equal and let it be ''P'. When the tube is held vertically with end AA up, let's consider pressure in column AB = P₁ and that of CD = P₂.

Volume of gas is equal to the volume of column, and volume of a cylinder is the product of its height and area of base ('A' as cross-section is uniform). As mercury column moves down by 10 cm, therefore, length of column AB becomes \30 cm and that of column CD becomes 50 cm. So,

Now let's consider when the tube is held vertically with end D up, then column CD moves down by ‘x’ cm, so that length of CD = 60 + x having pressure 'P₂' and column AB = 20 − x having pressure P₁'.

According to Boyle's law,

Therefore, length of gas column AB = 20 − x = 20 − 6.115 = 13.885 cm
Question 6
4 / -1

A gas is enclosed in a cylinder with a piston. Weights are added to the piston, giving a total mass of 2.20 kg. As a result, the gas is compressed and the weights are lowered 0.25 m. At the same time, 1.50 J of heat is evolved from the system. What is the change in internal energy of the system? (Take g = 9.8 m/sec²)

A
3.89

B
2.89

C
1.89

D
5.89

Solution

From the first law of thermodynamics,

△U = △q + w ....(i)

Given:

△q = −1.50 J (Heat evolved from the system)

Mass =2.20 kg

Work and heat are path functions. Work done on the system,

w = −Pdv = Force × displacement

w = −F.ds

F = Mass × Gravity = 2.2 × 9.8

ds = 0.25 m

w = 2.2 × 9.8 × 0.25

= 5.39 J

Work will be positive since volume of system is decreasing.

According to the first law of thermodynamics,

Internal energy can be determined as given below. Internal energy is a state function and an extensive property.

△U = △q + w

= −1.50 + 5.39

△U = 3.89 J

Alternatively:

Given:

Mass added = 2.20 kg

Distance lowered = 0.25 m

g = 9.8 m/s²

Heat evolved (Q) = -1.50 J (negative because heat is released)

Work done on the gas (W) = force × distance = m × g × h = 2.20 × 9.8 × 0.25 = 5.39 J

Using the first law of thermodynamics:

ΔU = Q + W

ΔU = -1.50 + 5.39 = 3.89 J

Change in internal energy = 3.89 J
Question 7
4 / -1

.0 L each of CH₄ (g) at 1.00 atm, and O₂ (g) at 4.00 atm, at 300^oC are taken and allowed to react by initiating the reaction with the help of a spark.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), ΔH = -802 kJ

Mass of CO₂ (g) produced in the reaction is

(Round off up to 2 decimal places)

A
1.87

B
2.88

C
4.87

D
5.87

Solution

Here, methane is the limiting reactant. Thus, according to the balanced equation, the number of moles of CO₂ formed is same as the number of moles of methane reacted, i.e. 0.0425 mol.

Hence, mass of CO₂ (g) formed = 0.0425 mol × 44 g mol-1 = 1.87 g
Question 8
4 / -1

The probability of India winning a test match against West Indies is 1/2. Assuming independence from match to match, the probability that in a 5-match series, India's second win occurs in the third test, is (Round off up to 2 decimal places)

A
0.25

B
0.30

C
0.35

D
0.20

Solution

Given: P(India wins) = p = 1/2

P(India loses) = p' = 1/2

Out of 5 matches, India's second win occurs in the third test.

⇒ India wins the third test and simultaneously it has won one match from the first two and lost the other.
Question 9
4 / -1

The order of the differential equation whose general solution is given by y = (c₁ + c₂) cos(x + c₃) - c₄ex + c₅, where c₁, c₂, c₃, c₄ and c₅ are arbitrary constants is

A
1

B
2

C
3

D
4

Solution

The given equation is

y = (c₁ + c₂) cos(x + c₃) - c₄e^x + c₅

Let y = A cos(x + B) - Ce^x; where A = c₁ + c₂, B = c₃ and C = c₄e^c5

dy/dx = -A sin(x + B) - Ce^x

Differentiating again,

d²y/dx² = -A cos(x + B) - Cex

Or d²y/dx² + y= -2 Ce^x

Or d³y/dx³ + dy/dx = -2Cex = [d²y/dx²] + y

Or d³y/dx³ - d²y/dx² + dy/dx - y = 0; which is a differential equation of order 3.

Alternate Solution:

y = (c₁ + c₂) cos(x + c₃) + (-c₄ e^c5)ex

Put c₁ + c₂ = k₁, c₃ = k₂, and -c₄ e^c5 = k₃

Then, y = k₁ cos (x + k₂) + k3e^x;where k₁, k₂ and k₃ are constants.

Now, y contains 3 independent constants; hence, the order of the differential equation is 3.
Question 10
4 / -1

If z is any complex number satisfying |z - 3 - 2i| ≤ 2, then the minimum value of |2z - 6 + 5i| is

A
8

B
10

C
5

D
6

Solution