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Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 1
If a point source is placed symmetrically from both the slits, then interference will be observed on the screen. if the source is placed unsymmetrically, then both slits will receive same type of input, Thus, interference will be observed in this case as well. If intensity of light is kept same at both the slits, the darker and brighter patterns will be more contrast. But if the two slits are illuminated by two independent sources, then the darker and the brighter pattern will not be observed because the interference will not be in proper sequence to project it on the screen.
The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statements is/are correct?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 2
(1) For an isothermal process, temperature is constant.
∴ T_{1} = T_{2}
(2) For an adiabatic process, temperature decreases.
∴ T_{3} < T_{1}
(3) In Wisothermal, ΔU = 0; so energy gets converted into work done, whereas in an adiabatic process, Q = 0.
∴ W_{isothermal} > Wadiabatic
(4) For an isothermal process, change in internal energy is constant, but not in an adiabatic process.
A tank of large base area is filled with water up to a height of 5 m. A hole of 2 cm^{2} cross section in the bottom allows the water to drain out in continuous stream. For the situation, mark the correct statement(s). [ρwater = 1000 kg/m^{3}, g = 10 m/s^{2}]
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 3
As the water falls down in the stream, the speed of the stream will increase due to gravity.
So according to equation of continuity (av = a constant), the area of the stream will decrease to keep the product of area and speed constant.
⇒ Option (a) is correct.
Velocity of water just coming out of the hole is v_{1} =
A cylindrical vessel of 90 cm height is kept filled up to the brim. It has four holes 1, 2, 3 and 4 which are, respectively, at heights of 20 cm, 30 cm, 40 cm and 50 cm from the horizontal floor PQ. The water falling at the maximum horizontal distance from the vessel comes from
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 5
The maximum horizontal distance form the vessel comes from hole numbers 3 and 4. Now, ν = √2gh, where h is the height of the hole from the top.
Horizontal distance,
Substituting the values we can see that whole three and four will have maximum range.
A driver in a stationary car blows a horn which produces monochromatic sound waves of frequency 1000 Hz normally towards a reflecting wall. The wall approaches the car with a speed of 3.3 m s^{−1}.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 6
The frequency received by the wall is
The reflected frequency will be same as the frequency received by the wall. Since the wall is moving towards the driver, the frequency heard by the driver will be,
A vertical glass capillary tube, open at both ends, contains some water. Which of the following shapes may not be taken by the water in the tube ?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 7
The two free liquid surface must provide a net upward force due to surface tension to balances the weight of the liquid column.
Usually, the water contact angle is smaller than 90^{o} with glass surface and so the surface will be hydrophilic. Shape of water surface on top will be concave. If gravity is taken into account, water will take the shape as shown in option D.
A long current carrying straight wire having current I_{1} = 10^{4} ampere is placed at the centre of an another current carrying loop having current I_{2} = 10^{4} ampere. Straight wire is perpendicular to the plane of the loop as shown in the figure. The torque (in N m) acting on the loop about its center is 10x. Find the value of x. (Radius R = 1 m)
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 9
A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances, etc. is shown in the figure. Calculate the energy (in mJ) stored in the capacitor C (4μF).
(Round off up to 2 decimal places)
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 11
Using Kirchhoff's first law at junction a and b, we have found the current in other wires of the circuit on which currents were not shown.
Now, to calculate the energy stored in the capacitor, we will have to first find the potential difference V_{ab} across it.
The figure below shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 2 Ω/m. The position of the conducting rod at t = 0 is shown. A time-dependent magnetic field B = 4t T is switched on at t = 0.
The rod moves towards right at 10 cm/s. The current (in A) in the loop at t = 3 s due to induced emf is
(Round off up to 2 decimal places)
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 12
Induced emf, e = Blv = 12 × 0.30 × 0.10 = 0.36 V
Resistance of U-shaped rail = 2 W, as its length is 100 cm
A sample of air is initially at 533 K and 700 N/m^{2}, and occupies a volume of 0.028 m^{3}. The air is expanded at constant pressure to three times. A polytropic process with n = 1.5 is then carried out followed by an isothermal process which completes the cycle.
The efficiency of the cycle (in %) is:
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 13
A sonometer wire resonates with a given turning fork forming standing waves with five antinodes between two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by M, the wire resonates with the same tuning fork forming three antinodes for the same positions of bridges. The value of M (in kg) is
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 14
Answer the following by appropriately matching the lists based on the information given:
Two wires, each carrying a steady current I are shown in four configurations in List I. Some of the resulting effects are described in List II.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 15
When P is situated midway between the wires, magnetic fields due to both the wires will be in opposite direction and cancel other.
(A) → (q), (r)
When point P is situated at the midpoint of line joining the centres of circular wires, which have the same radii, magnetic fields at point P will be in same direction.
(B) → (p)
In the given case, magnetic fields due to both wires will be in opposite direction and cancel out.
(C) → (q), (r)
In the above case, magnetic fields due to wires will be in opposite direction and wires will repel each other.
Answer the following by appropriately matching the lists based on the information given:
List I contains a list of processes involving expansion of an ideal gas and List II describes the thermodynamic change during this process.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 16
When P is situated midway between the wires, magnetic fields due to both the wires will be in opposite direction and cancel other.
(A) → (q), (r)
When point P is situated at the midpoint of line joining the centres of circular wires, which have the same radii, magnetic fields at point P will be in same direction.
(B) → (p)
In the given case, magnetic fields due to both wires will be in opposite direction and cancel out.
(C) → (q), (r)
In the above case, magnetic fields due to wires will be in opposite direction and wires will repel each other.
A 100 mL mixture of CO and CO_{2} is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 19
The structure of AZT (azidothymine) is given below. It is used to treat AIDS patients. It fights an AIDS infection but does not cure it.
Which of the following statements are correct about AZT?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 20
For (It shows acidic property), H atom attached to N is acidic due to the presence of the two electron-withdrawing (C = O) groups. Its ion is resonance stabilised.
It gives white precipitates with Anlyd. ZnCl_{2} + conc.HCl (Lucas reagent). 1^{o} ROH does not give Lucas test at room temperature.
(Azide ion (N_{3}^{–}) cannot replace Br^{⊖} by S_{N}2 reaction) as strong (C-N) bond is not broken by Br^{⊖}.
The decreasing order of hyperconjugative effect is
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 24
Hyperconjugation effect depends on the number of alpha-H-atoms.
Both options (a) and (b) are arranged in the order of decreasing number of alpha-hydrogen atoms and hence, in the decreasing order of the hyperconjugating effect.
III. NH_{3} > PH_{3} > AsH_{3} > SbH_{3} - Basic character
IV. NH_{3} > PH_{3} > AsH_{3} > SbH_{3} - Bond angle
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 25
All the orders regarding the hydrides of group 15 are correct.
The boiling point increases with the increase in the molecular mass, but the boiling point of NH3 is unexpectedly high because of H-bonding.
Thermal stability decreases down the group due to an increase in the bond length.
The basic character decreases down the group due to the decrease in the electronegativity of the central atom.
The bond angle decreases down the group as the size of the central atom increases and the electronegativity decreases. As a result the electron density around the central atom decreases which decreases the electronic repulsions and hence the bond angle.
In a quantitative determination of iron present in an ore, an analyst converted 0.42 g of the ore into its ferrous form. This required 42.0 ml of 0.1 N solution of KMnO_{4} (in the acidic medium) for titration. Choose the correct statement(s).
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 26
(1) Number of G. Eq. of Fe = Number of G. Eq. of KMnO_{4}
Number of G. Eq. of Fe = Number of G. Eq. of KMnO4 =
Hence, option (1) is correct.
(2) = 56%
Hence, option (2) is incorrect.
(3) Number of moles of KMnO_{4} = No. of moles of KMnO_{4 }/ 5
=
Hence, option (3) is incorrect.
(4)
So, one mole of KMnO_{4} will oxidise 5 moles of Fe^{+2} ions.
When two oppositely charged ions approach each other, the ion smaller in size attracts outermost electrons of the other ion and repels its nuclear charge. The electron cloud of anion no longer remains symmetrical, but is elongated towards the cation. Due to that, sharing of electrons occurs between the two ions to some extent and the bond shows some covalent character. The value of dipole moment can be used for determining the amount of ionic character in a bond. Thus, percentage ionic character
The dipole moment of LiH is 1.964 × 10^{-29} C.m. and the interatomic distance between Li and H in this molecule is 1.596 . What is the % ionic character in LiH?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 28
Assume that you are using an open-end manometer filled with mineral oil rather than mercury. What is the gas pressure in the bulb (in millimetre of Hg) if the level of mineral oil in the arm connected to the bulb is 237 mm higher than the level in the arm connected to the atmosphere and the atmospheric pressure is 746 mm Hg? (The density of mercury is 13.6 g/ml and that of mineral oil is 0.822 g/ml.)
(Round off up to 1 decimal place)
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 29
The gas pressure in the bulb equals the difference between the outside pressure and the manometer reading. The manometer indicates that the pressure of the gas in the bulb is less than the atmospheric pressure because the liquid level is higher on the side connected to the sample.
Mercury is more dense than mineral oil by a factor of
A given pressure will hold a column of mercury only 1/16.5 times the height of a column of mineral oil.
SO_{2}Cl_{2} and Cl_{2} are introduced into a 3 L vessel. Partial pressure of SO2Cl_{2} and Cl_{2} at equilibrium are 1 atm and 2 atm respectively. The value of K_{p} for the following reaction is 10.
The total pressure in atm at equilibrium would be :
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 30
'a' moles of NaNa reacts with sufficient water in an open vessel at 300 K. The work done by the liberated gas H_{2} is equivalent to 4988.4 J. What is aa?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 31
3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pd^{n+} was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 32
Complex compounds are molecular compounds which retain their identities even when dissolved in water. They do not give all the simple ions in solution but instead furnish complex ions. The complex compounds are often called co-ordination compounds because certain groups called ligands are attached to the central metal ion by co-ordinate or dative bonds. Co-ordination compounds exhibit isomerism, both structural and stereoisomerism. The structure, magnetic property, colour and electrical properties of complexes are explained by various theories.
Arrange the following compounds in order of their molar conductance :
out of (i) and (iv) K+ and Mg^{2+} , Mg^{2+} due to high solvation will have lower Ionic mobility & lower molar conductance. Molar conductance will have order (ii) < (iv) < (i) < (iii)
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 34
Step I is the chromyl chloride test, which is used to detect chloride ions in the qualitative analysis. If any alkali chloride salt is heated with acidified potassium dichromate, it produces red colour fumes of chromyl chloride. It confirms the presence of chloride ions in that salt.
However, for salts such as chlorides of antimony, lead and tin, chromyl chloride test is not applicable. This is because the chlorides of these metals are covalent and they do not generate Cl^{−} ions. The chromyl chloride test is applicable only for compounds having chloride ionic bonds. So, the chromyl chloride test is carried out mainly to detect chloride ions in ionic compounds.
When this gas is passed through a sodium hydroxide solution, a yellow solution is obtained due to the presence of sodium chromate.
Directions: The following question has four choices, out of which ONE or MORE is/are correct.
Let A and B be two distinct points on the parabola y^{2} = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 40
Let n be number of distinct objects, kk be number of like boxes and let P(n,k) be the number of ways of putting all the nn objects into the boxes so that no box is empty then which is/are correct ?
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 41
Given that the number of ways to put nn objects into kk like boxes = P(n,k)
Let us assume that we pick one object from nn distinct objects. Now there are two possibilities:
Case I: If this object is put into a box alone then the remaining n−1 objects need to be put in k−1 boxes which can be done in P(n−1,k−1) ways.
Case II: If this particular object is not kept in a box alone then the remaining n−1 objects are first distributed in kk boxes in P(n−1,k),k and then this object is put in any one of the k boxes in k ways.
The total number of ways is k⋅P(n−1, k)
Altogether the total number of ways is P(n, k) = P(n−1, k−1) + k⋅P(n−1, k)
For n = 6 & k = 4, we get P(6,4) = P(5,3) + 4⋅P(5,4)
Similarly, for n = 5 & k = 3 we get P(5,3) = P(4,2) + 3P(4,3)
Let an denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let b_{n} = the number of such n-digit integers ending with digit 1 and c_{n} = the number of such n-digit integers ending with digit 0.
The value of b_{6} is
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 45
To find b6, we have to find all 6 digit numbers ending with '1' such that no consecutive digits are '0'.
Some of the examples possible are:
1. 1 0 1 1 1 1
2. 1 0 1 0 1 1
3. 1 1 1 1 1 1
Three case possible:
1. One zero-It can be placed in any of the four places.
So, we get '4' such six digit numbers.
2. Two zeros-We get '3' such six digit numbers possible.
3. No zeros-We get only '1' such six digit number.
Let an ordered pair A be defined as A(x, y) where x∈ prime number, such that x < 10 and y∈ natural numbers and y≤10. If the probability that the ordered pair A satisfies the relation x^{2} − 3y^{2} = 1 is P then 60P equals.
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 48
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Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three times. If r_{1}, r_{2} and r_{3} are the numbers obtained on the die, then the probability that ω^{r1} + ω^{r2} + ω^{r3} = 0 is
Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 53
Total number of events = 6 × 6 × 6 = 216.
Given, ω^{r1} + ω^{r2} + ω^{r3} = 0
If one of r_{1},r_{2},r_{3} takes value from the set {3, 6} other takes values from the set {1, 4} and the third takes value from the set {2, 5}
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