54 Questions MCQ Test JEE Main & Advanced Mock Test Series - JEE Advanced Practice Test- 16
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Three identical coils A, B and C are placed coaxially as shown with their plane parallel to one another. Coil B is fixed in position. Coil A is moved towards B with constant speed of 3 m/s and coil C is moved away from B with constant speed of 5 m/s. Then, the
Detailed Solution for JEE Advanced Practice Test- 16 - Question 1
The coil A increases the flux in the coil B therefore the direction of induced current in coil B is opposite to the direction of the current in A .The coil C decreases the flux in coil B therefore the direction of the induced current in B is same as the direction of the current in C.Rate of increase in the magnetic flux due to coil A is less compared to the rate of decrease in the magnetic flux due to the coil C. Due to the movement of the coil A and C direction of the induced current in coil B is anticlockwise.
If a particle is projected at an angle 'α' with the horizontal from the foot of a plane, whose inclination to the horizontal is 'β', then it will strike the plane at right angle
Detailed Solution for JEE Advanced Practice Test- 16 - Question 2
If a projectile hits at Q at an angle of 90º to PQ, the velocity parallel to PQ is zero.
If 'T' be the time of flight from P to Q, component of the initial velocity of the projectile along the plane PQ is u cos (α - β), the component of the final velocity along the PQ plane = 0 and the component of the acceleration due to gravity = -g sin β
A cubical box of copper 6 cm on each side is floating on mercury in a beaker. The portion of the box is inside the mercury. Now, water is poured in to the beaker so that it just covers the box. Density of copper = 5.6 gm/cc, density of mercury = 13.6 gm/cc. Then,
Detailed Solution for JEE Advanced Practice Test- 16 - Question 3
As the upthrust acting on the cube = Weight of cube
Upthrust = Weight of the mercury displaced
Hence, V_{in} ρ_{Hg}g = Vcu ρ_{cu}g
V_{in} ρ_{Hg} = V_{cu} ρ_{cu}
Or,
or,
, where Hin is the height of the box inside and HCu is the height of the box given
When water is poured in the beaker, then up-thrust required for the floatation of the box is provided by both mercury and water.
Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string. They are fully immersed in a fluid of density dF. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if
Detailed Solution for JEE Advanced Practice Test- 16 - Question 4
F = upthrust = Vd_{Fg}
Equilibrium of A
At d_{A} < />_{F}B d_{B} > d_{F}D d_{A} + d_{B} = 2d_{F} for sphere A to be at the top in liquid, the density of sphere A should be less than the density of liquid and for sphere B to be at the bottom in liquid, the density of sphere B should be greater than the density of liquid. The FBD of the top sphere gives us Vd_{F} g = T + V dAg and of the bottom sphere gives us T + V d_{F} g = V d_{B} g.
Substituting T from the first equation in the second, we get
2V d_{F} g = Vd_{A} g + V d_{B}g or 2d_{F} = d_{A} + d_{B}
A composite block is made of slabs A,B,C,D and E of different thermal conductivities ( given in terms of a constant K ) and size ( given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in steady state
Detailed Solution for JEE Advanced Practice Test- 16 - Question 5
Thermal resistance R = ℓ / KA
∴
(Here w = width )
=
R_{A} : R_{B} : R_{C} : R_{D} : R_{E}
15 : 160 : 60 : 96 : 12
So, let us write, R_{A} = 15 R, R_{B} = 160 R etc and draw a simple electrical circuit as shown in figure
H = Heat current = Rate of heat flow.
H_{A} = H_{E} = H
∴ Heat flow through A and E slabs are same, is correct.
In parallel current distributes in inverse ratio of resistance.
A vernier callipers has 1 mm marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is
Detailed Solution for JEE Advanced Practice Test- 16 - Question 8
Least count of vernier callipers
LC = 1MSD − 1VSD
= Smallest division on main scale / Number of divisions on vernier scale
20 divisions of vernier scale = 16 divisions of main scale
One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dotted lines as shown in the graph below. If the work done along the solid line path is w_{s} and that along the dotted line path is w_{d}, then the ratio w_{d}/w_{s} is
(Round off up to 2 decimal places)
Detailed Solution for JEE Advanced Practice Test- 16 - Question 9
Solid line path work done (w_{s}) is isothermal because PV is constant (Boyle's law) and dotted line (horizontal) path work done w_{d} is isobaric. Work done in vertical line is zero as ΔV = 0.
Total work done on solid line path (w_{s}) = 2.303 nRT log V_{2}/V_{1}
= 2.303 PV log V_{2}/V_{1} = 2.303 x 4 x 0.5 log 5.5/0.5 = 4.8 L atm.
Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m^{2}. The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio n_{A}/n_{B}, where n_{A} and n_{B} are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.]
(Round off upto 2 decimal places)
Detailed Solution for JEE Advanced Practice Test- 16 - Question 10
Excess pressure above atmospheric pressure, due to surface tension in a bubble = 4T/r
Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 mm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that by B?
Detailed Solution for JEE Advanced Practice Test- 16 - Question 11
According to Wien's displacement law,
According to Stefan Boltzmann law, rate of energy radiated by a black body is
In the figure shown below, the maximum possible unknown resistance X (in Ω), that can be measured by the post office box is given by a × 10^{b} Ω in scientific notation, then find the value of a−b?
(in this experiment, we take out only one plug in arm AB and only one plug in arm BC, but in arm AD we can take out many plugs).
Detailed Solution for JEE Advanced Practice Test- 16 - Question 12
A cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10^{−3} m^{2} enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.
Detailed Solution for JEE Advanced Practice Test- 16 - Question 13
During heating, the process was isobaric, so, ΔW_{1} = PΔV
And internal energy, ΔU_{1} = nC_{v}ΔT = 50nC_{v}
Thus, heat transfer, ΔQ_{1} = 50nC_{v} + 55
During cooling, the process is isochoric, so, ΔW_{2} = 0 and ΔU_{2} = −50nC_{v}
Consider a frictionless plane of height h = 2 m m and angle α=30°. One end of an electric string is connected to point B, the other end is passed through an orifice at point D on the plane and connected to a body of mass m = 2 kg initially stationary at point A. This body comes to rest at point C at the bottom of the inclined plane. The line drawn from point B to point D is perpendicular to side AC. Knowing that the length of the string is unstretched state is equal to the length of segment BD. Determine the elastic constant in N m^{−1}. Take g = 10 m s^{−2}
Detailed Solution for JEE Advanced Practice Test- 16 - Question 14
Answer the following by appropriately matching the lists based on the information given:
List I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S_{1} and S_{2}. In each of these cases S_{1}P_{0} = S_{2}P_{0}, S_{1}P_{1} - S_{2}P_{1 }= λ /4 and S_{1}P_{2} - S_{2}P_{2} = λ/3, where λ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index μ and thickness t is pasted on slit S_{2}. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by δ (P) and the intensity by I(P). Match each situation given in List I with the statement(s) in List II valid for that situation.
Detailed Solution for JEE Advanced Practice Test- 16 - Question 15
(A) → (p, s) Intensity at P_{0} is maximum. It will continuously decrease from P_{0} toward P_{2}.
(B) → (q) Path difference due to slap will be compensated by geometrical path difference. Hence, δ(P_{1}) = 0.
Answer the following by appropriately matching the lists based on the information given:
Two transparent media of refractive indices μ_{1} and μ_{3} have a solid lens shaped transparent material of refractive index μ_{2} between them as shown in figures in List II. A ray traversing these media is also shown in the figures. In List I different relationship betweenμ_{1} , μ_{2} and μ_{3} are given. Match them to the ray diagrams shown in List II.
Detailed Solution for JEE Advanced Practice Test- 16 - Question 16
In the above equation, as sine is an increasing function in 0 to π/2 , if μ_{2} > μ_{1}, i > r and if μ_{2} < />_{1}, i < />
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.
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.
The difference in maximum kinetic energy of photoelectrons from A and from B
Detailed Solution for JEE Advanced Practice Test- 16 - Question 18
Directions: The following question has four choices, out of which one or more is/are correct.
Which of the following statements about some of the observations during qualitative salt analysis is/are correct?
Detailed Solution for JEE Advanced Practice Test- 16 - Question 23
Gelatenous white crystals of Al(OH)_{3} are soluble in NaOH(aq) because of the formation of soluble sodium aluminate.
Al(OH)_{3} + NaOH → Na[Al(OH)4]
However, Al(OH)_{3} is insoluble in aqueous NH_{3} because Al(OH)_{3} precipitates at higher pH, but dissolves when the pH is low.
Hence, option (b) is incorrect.
An aqueous solution of Co(II) thiocyanate (10% freshly prepared) and mercuric nitrate solution, when taken in equal volumes, produces deep-blue precipitate on scratching the wall of the vessel with a glass rod due to the formation of mercuric tetrathiocyanatocobaltate(II) and Hg_{2}[Co(CNS)_{6}] (deep blue ppt).
Green precipitate of Cr(OH)_{3} readily dissolves in excess of sodium hydroxide, forming a green solution.
Chromium (III) salts give green coloured borax beads in both oxidising and reducing flame.
The maximum number of isomers (including stereoisomers) that are possible on monochlorination of the following compound is:
Detailed Solution for JEE Advanced Practice Test- 16 - Question 27
The monochlorination involves free radical substitution reaction. In order to find the maximum number of isomers, we should count the number of different hydrogens present [3°,2°,1°].
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of the dimer. At 25°C, the equilibrium constant for the dimerisation is 1.3 × 10^{3} (pressure in atmatm). What is ΔS^{∘} for the reaction? Assume that ΔH does not vary with temperature.
Detailed Solution for JEE Advanced Practice Test- 16 - Question 28
Additional bond forms in product = ( 2 hydrogen bonds)
A blue colour complex is obtained in the analysis of Fe^{+3} having formula Fe_{4}[Fe(CN)_{6}]_{3}
Let a = oxidation number of iron in the coordination sphere, b = no. of secondary valencies of central iron ion. C = Effective atomic number of iron in the coordination sphere.
Then find the value of (c + a − 2b).
Detailed Solution for JEE Advanced Practice Test- 16 - Question 29
The arrangement of X^{–} ions around A^{+} ion in solid AX is given in the figure (not drawn to scale). If the radius of X^{–} is 250 pm, what is the radius (in pm) of A^{+}? (Round off up to 1 decimal place)
Detailed Solution for JEE Advanced Practice Test- 16 - Question 30
According to the given figure, A^{+} is present in the octahedral void of X^{–}.
The limiting radius in octahedral void is related to the radius of sphere as:
In a vessel gaseous dimethyl ether at initial pressure 'p0' atm. Dimethyl ether decomposes on heating as per the reaction by first order kinetics.
It is observed that the half-life period for this reaction is 0.2 h. After a very long time, pressure in the vessel is observed to be 1.2 atm. Assume ideal behaviour of all gases with constant V-T conditions.
The initial rate of formation of CH_{4} gas is
Detailed Solution for JEE Advanced Practice Test- 16 - Question 35
The initial rate of a reaction is the instantaneous rate at the start of the reaction (i.e., when t = 0). The initial rate is equal to the negative of the slope of the curve of reactant concentration versus time at t = 0.
Which of the following statement(s) is/are incorrect regarding chloride of iron?
Detailed Solution for JEE Advanced Practice Test- 16 - Question 36
This problem is based on various properties of iron chlorides. To solve this problem, students must have the knowledge of solubility of iron chloride in aqueous solution, the existence of iron chloride in ether solution, oxidising properties of iron chloride.
Properties of iron chloride
(i) FeCl_{3} (Iron chloride) is back coloured in anhydrous conditions. But its hydrated form is yellowish-brown.
(ii) FeCl_{3} sublimes at 300°C3 to give a dimer.
(iii) FeCl_{3} is water-soluble as well as ether soluble due to solvation.
(iv) It is an oxidising agent, during oxidation. The yellow colour of aqueous Fe (III) changes to light green aqueous Fe (II).
Let PMPM be the perpendicular from the point (1, 2, 3) to xy-plane. If OP makes an angle θθ with the positive direction of the z-axis and OM makes an angle ϕϕ with the positive direction of x-axis, where O is the origin, θ and ϕ are acute angles, then
Detailed Solution for JEE Advanced Practice Test- 16 - Question 39
Here, P be (x, y, z)
Then, x = rsin θ ⋅ cos ϕ, y = rsin θ sin ϕ, z = rcos θ
A consignment of 15 record players contains 4 defective ones. The record players are selected at random, one by one, and examined. The ones examined are not put back.
What is the probability that the 9^{th} one examined is the last defective?
Detailed Solution for JEE Advanced Practice Test- 16 - Question 44
The first 8 record players examined contain the remaining 3 detectives. (as 9th one examined is the last defective)
The number of ways in which we can have 3 defectives in a total of 8 objects is C(8, 3)
= 8!/3!(8 − 3)! = 56.
The number of ways in which we can have 4 defectives in a total of 15 objects is C(15, 4)
= 15!/4!(15 − 4)!
= 1365.
The probability that the ninth one examined is the last defective = 56/1365 = 8/195
Therefore, the probability that the 9th one examined is the last defective is 8/195
Ten different letters of an alphabet are given. Four-lettered words are formed using these given letters. Then, the number of words which have at least one letter repeated is
Detailed Solution for JEE Advanced Practice Test- 16 - Question 47
Number of words which have at least one letter repeated = Total number of words formed - Total number of words in which no letter is repeated
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