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In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9. This implies that
Detailed Solution for JEE Advanced Practice Test- 17 - Question 1
Ratio of maximum to minimum intensity of the interference pattern
A particle of charge +q and mass m moving under the influence of a uniform electric field and uniform magnetic field follows a trajectory from P to Q as shown in figure. The velocities at P and Q are Which of the following statement(s) is/are correct?
Detailed Solution for JEE Advanced Practice Test- 17 - Question 2
Magnetic force does not do work. From work-energy theorem
An incompressible and non-viscous liquid in a V-shape tube is shown in the figure. When the liquid is slightly depressed, ('A' is the area of cross-section.)
Detailed Solution for JEE Advanced Practice Test- 17 - Question 3
The liquid will execute S.H.M.
When the liquid in the tube is slightly depressed and released, a restoring force is developed due to hydrostatic pressure difference.
The tube is of uniform cross-section.
Initially, the level of liquid in two limbs is at the same height 'h' from the ground.
If the liquid is depressed by 'y' in one limb, it will rise by 'y' along the length of tube in the other limb.
Therefore, restoring force developed by hydrostatic pressure difference, F = -△p x A = (h_{1} + h_{2})gdA = -Agd(sin θ_{1} + sin θ_{2})y
As the restoring force is linear, the motion will be linear simple harmonic with force constant
The unbalanced force is F = -Agd(sin θ_{1} + sin θ_{2})y
A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision,
Detailed Solution for JEE Advanced Practice Test- 17 - Question 4
We know that 0.1 × 2.0 = 2 kg m/s
Friction with ground gives no impulse.
FΔt = mΔv
= 0.1(20 - 0) = 2 N-s
Using: angular impulse = change in angular momentum, we get:
Note that we have assumed that direction of angular velocities is same before and after and since LHS of the above equation is positive, ω_{2} > ω_{1}.
Thus, the ring must be slipping to the right; hence, the friction will be to the left as it will be opposite to the direction of motion.
A projectile is fired at an angle θ with the horizontal. The condition under which it lands perpendicular on the fixed inclined plane with inclination α as shown in figure.
Detailed Solution for JEE Advanced Practice Test- 17 - Question 5
Applying equation of motion perpendicular to the incline for y = 0
If in a hydrogen atom, radius of nth Bohr orbit is r_{n}, frequency of revolution of electron in nth orbit is f_{n}, and area enclosed by the nth orbit is A_{n} , then which of the following graphs is/are correct?
Detailed Solution for JEE Advanced Practice Test- 17 - Question 6
(i) Since in hydrogen atom r_{n} ∝ n^{2}, therefore graph between r_{n} and n will be a parabola through origin and having increasing slope.
(ii) It means, graph between log(r_{n}/r_{1}) and log n will be a straight line passing through and having positive slope (tanθ = 2)tanθ = 2.
(iii) If radius of an orbit is equal to r, then area enclosed by it will be equal to A = πr^{2}.
Since r_{n} ∝ n^{2}, therefore A_{n }∝ n^{4}
It means, graph between log(A_{n}/A_{1}) and log n will be a straight line passing through origin and having positive slope (tanθ = 4).
(iv) If frequency of revolution of electron is f, then its angular velocity will be equal to ω = 2π. Hence, its angular momentum will be equal to Iω = mr^{2}. But according to Bohr's theory, it is equal to nh/2π, therefore,
It means, graph between log (f_{n}/f_{1}) and log n will be a straight line passing through origin and having negative slope, tan θ = −3. Hence, it will be as shown in diagram.
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 f_{1} = 800 Hz to f_{2} = 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
Detailed Solution for JEE Advanced Practice Test- 17 - Question 9
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
Detailed Solution for JEE Advanced Practice Test- 17 - Question 10
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
Detailed Solution for JEE Advanced Practice Test- 17 - Question 11
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
Detailed Solution for JEE Advanced Practice Test- 17 - Question 12
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
When you throw a ball in air with some velocity at some angle with horizontal, vertical component of velocity at highest point is zero and horizontal component of velocity remains unchanged.
Velocity of a projectile at height 15 m from ground is Here, is in horizontal direction and vertically upwards. Then
Angle of projectile with ground is
Detailed Solution for JEE Advanced Practice Test- 17 - Question 13
Answer the following by appropriately matching the lists based on the information given:
Six point charges, each of the same magnitude q, are arranged in different manners as shown in List II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.
Detailed Solution for JEE Advanced Practice Test- 17 - Question 15
List - II
1. By symmetry, E = 0, V = 0 and B = 0, μ = NIA = 0 but I = 0
Since total charge is zero, current is zero.
2. E ≠ 0, V = 0, B = 0
Since I = 0, μ = NIA = 0
3. E = 0 (By symmetry)
V ≠ 0, since distances are different.
B ≠ 0, since radii are different, μ ≠ 0
4. E = 0 (By symmetry)
V ≠ 0, since distances are different,
B ≠ 0, μ ≠ 0
5. E ≠ 0, V = 0, B = 0, since each rotating charge is equivalent to a steady current.
And μ = 0, since μ for pairs of opposite charges will cancel each other.
Angular frequency in SHM is given by Maximum acceleration in SHM is ω^{2}A and maximum value of friction between two bodies in contact is μN, where N is the normal reaction between the bodies.
In the diagram shown, what can be the maximum amplitude of the system so that there is no slipping between any of the blocks ?
Detailed Solution for JEE Advanced Practice Test- 17 - Question 16
Maximum friction between 1 kg and 2 kg blocks can be 0.6 x 1 x 10 = 6 N. Therefore, maximum acceleration of 1 kg block can be 6/1 = 6 m/s^{2}. Maximum force of friction between 2 kg and 3 kg blocks can be 0.4 x 3 x 10 = 12 N.
Therefore, maximum acceleration of 1 kg and 2 kg blocks jointly can be 12/3 = 4 m/s^{2}.
So, maximum acceleration of the whole system, so that there is no slipping between any of blocks is 4 m/s^{2}.
A.DC circuit consisting of two cells of emf V and 2V having no internal resistance are connected with two capacitors of capacity C and 2C and four resistors R, R, 2R and 2R as shown in figure. The ammeter and voltmeter used in the circuit are ideal.
The reading of the ammeter as soon as the switch is closed is
Detailed Solution for JEE Advanced Practice Test- 17 - Question 17
Capacitors will offer no resistance and hence the circuit becomes
An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy.
The molar specific heat of the gas in this process is given by C whose value is
Detailed Solution for JEE Advanced Practice Test- 17 - Question 18
It is given that there is decrease of the internal energy while the gas expands absorbing some heat, which numerically equal to the decrease in internal energy.
Hence, dQ = −dU
Thus, dQ = dW + dU = −dU
⇒ dW = −2dU (remembering that both dW and dU are negative)
For the Balmer series of hydrogen spectrum, the wave number for each line is given by: where R_{H} is Rydberg Constant and n_{1} and n_{2} are intergers
Choose the correct statement(s) about the given information.
Detailed Solution for JEE Advanced Practice Test- 17 - Question 19
Only statements (1), (2) and (4) are correct as
(1) Beyond a certain wavelength the line spectrum become band spectrum.
(2) For Balmer series n_{1} = 2.
(4) For the calculation of longest wavelength, we use nearest value of n_{2}.
Hence, for the longest wavelength in Balmer series of hydrogen spectrum is n_{2} and n_{3}.
Detailed Solution for JEE Advanced Practice Test- 17 - Question 24
In an isocyanide, an electrophile attacks first followed by a nucleophile at the same carbon atom bearing negative charge. on partial hydrolysis in acidic medium, will give N-methylmethanamide while on complete hydrolysis in acidic medium it gives CH_{3}NH_{2} and HCOOH.
One mole of an ideal gas is subjected to a two-step reversible process (A→B and B→C). The pressure at A and C is the same. Mark the correct statement(s).
Detailed Solution for JEE Advanced Practice Test- 17 - Question 25
At A
∵ V_{B} > V_{A}, so, the expansion of gas takes place.
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.)
Detailed Solution for JEE Advanced Practice Test- 17 - Question 27
As we know according to Boyle's law, P_{1}V_{1 }= P_{2}V_{2}.
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 = P_{1} and that of CD = P_{2}.
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,
Substituting values of P_{1} and P_{2} and P_{2} from (1) and (2) in (3),
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 'P_{2}' and column AB = 20 − x having pressure P_{1}'.
According to Boyle's law,
Substituting values of P_{1}' and P_{2}' from (4) and (5) in (6)
Putting value of ''P' in the above equation,
Solving for x,
Therefore, length of gas column AB = 20 − x = 20 − 6.115 = 13.885 cm
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/sec^{2})
Detailed Solution for JEE Advanced Practice Test- 17 - Question 28
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.
.0 L each of CH_{4} (g) at 1.00 atm, and O_{2} (g) at 4.00 atm, at 300^{o}C are taken and allowed to react by initiating the reaction with the help of a spark.
Detailed Solution for JEE Advanced Practice Test- 17 - Question 29
Number of moles of CH4 (g) used = = 0.0425 mol
Number of moles of O2 (g) taken = = 0.1700 mol
Here, methane is the limiting reactant. Thus, according to the balanced equation, the number of moles of CO^{2} formed is same as the number of moles of methane reacted, i.e. 0.0525 mol.
Hence, mass of CO^{2} (g) formed = 0.0425 mol × 44 g mol-1 = 1.87 g
In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface, surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer, so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'.
The number of atoms in this HCP unit cell is
Detailed Solution for JEE Advanced Practice Test- 17 - Question 30
Atom at the face = 2
Atom at the centre = 3
Atom at corners = 12
Total number of atoms = (2 × 1/2) + (3 × 1) + (12 × 1/6) = 6
In the analysis it is found that 1 molecule of H_{2} is yielded per 100 eV of energy absorbed. A nuclear power reactor of 200 kW capacity has been installed based on this reaction.
The volume of H_{2} produced per minute in the above reactor is
Detailed Solution for JEE Advanced Practice Test- 17 - Question 31
Energy required for one molecule = 100 eV = 100 x 1.6 x 10^{-19} = 1.6 x 10^{-17} J
Arrange the following by appropriately matching the lists based on the information given in the paragraph.
List - 1 includes starting materials and reagents of selected chemical reactions.
List - 2 gives the types of reactions of List - 1.
For the synthesis of ortho- and para-hydroxyacetophenone, the only correct combination is ____________.
Detailed Solution for JEE Advanced Practice Test- 17 - Question 33
The Fries rearrangement is an organic reaction used to convert a phenyl ester to an ortho- and para-hydroxy aryl ketone using a Lewis acid catalyst and Bronsted acid, such as HF, AlCl_{3}, BF_{3}, TiCl_{4} or SnCl_{4}. The mechanism begins with the coordination of the ester to the Lewis acid, followed by a rearrangement which generates an electrophilic acylium cation.
The aromatic compound then attacks the alkyl cation (both the ortho and para attacks are allowed) via an electrophilic aromatic substitution (SEAr). The deprotonation to regenerate aromaticity and Bronsted acid work ups to regenerate the Lewis acid catalyst providing the two hydroxy aryl ketone products. The complete reaction is shown as below.
wo beaker are placed in a sealed flask. Beaker A initially contained 0.15 mol of naphthalene (non-volatile) in 117 g of benzene and beaker B initially contained 31 g of an unknown compound (non-volatile, non-electrolytic) in 117 g of benzene. At equilibrium, beaker A is found to have lost 7.8 g of weight. Assume ideal behaviour of both solutions to answer the following question.
The molar mass of solute in solution B is closest to:
Detailed Solution for JEE Advanced Practice Test- 17 - Question 35
Two beaker are placed in a sealed flask. Beaker A initially contained 0.15 mol of naphthalene (non-volatile) in 117 g of benzene and beaker B initially contained 31 g of an unknown compound (non-volatile, non-electrolytic) in 117 g of benzene. At equilibrium, beaker A is found to have lost 7.8 g of weight. Assume ideal behaviour of both solutions to answer the following question.
Now beaker B, at equilibrium, is replaced by another beaker C which contain 31 g of same solute as in B, but in different solvent X. Also moles of X in solution C is same as mol of benzene in solution B at equilibrium. When the container containing beakers A and C is closed again and allowed to attain equilibrium, beaker C has lost some weight. Which of the following can be said with guarantee ?
Detailed Solution for JEE Advanced Practice Test- 17 - Question 36
for beaker A,
= 0.0967 (before equilibrium)
for solvent x
= 0.0965 (before equilibrium)
Relative lowering in vapour pressure is less over beaker C i.e., vapour pressure of X is more than vapour pressure of benzene.
Let L_{1} be a straight line passing through the origin and L_{2} be the straight line x + y = 1. If the intercepts made by the circle x^{2} + y^{2} − x + 3y = 0 on L_{1} and L_{2} are equal, then which of the following equation can represent L_{1} ?
Detailed Solution for JEE Advanced Practice Test- 17 - Question 41
Let equation of line L_{1} be y = mx. Intercepts made by L_{1 }and L_{2} on the circle will be equal ie, L_{1 }and L_{2} are at the same distance from the centre of the circle.
Centre of the given circle is (1/2,−3/2).
Therefore,
Thus, two chords are x + 7y = 0 and y − x = 0.
Therefore, x − y = 0 and x + 7y = 0 are correct answers.
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)
Detailed Solution for JEE Advanced Practice Test- 17 - Question 45
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.
The order of the differential equation whose general solution is given by y = (c_{1} + c_{2}) cos(x + c_{3}) - c_{4}ex + c_{5}, where c_{1}, c_{2}, c_{3}, c_{4} and c_{5} are arbitrary constants is
Detailed Solution for JEE Advanced Practice Test- 17 - Question 46
Phosphorus forms a number of oxoacids which differ in their structures and oxidation state of phosphorus. All the acids contain phosphorus atom/atoms linked tetrahedrally to four other atoms or groups. Each of them has at least one P = O or P → O unit and one P - OH unit. The -OH group is ionisable but H atom linked directly to P is non-ionisable. Structures of all the acids are considered to be derived either from phosphorus acid or phosphoric acid.
Which one is mono-basic acid?
Detailed Solution for JEE Advanced Practice Test- 17 - Question 53
A monobasic acid is an acid that has only one hydrogen ion to donate to a base in an acid-base reaction. Therefore, a monobasic molecule has only one replaceable hydrogen atom.
And the structure can be graphically represnted as follows:
Disaccharides are carbohydrates those contain two monosaccharides molecules, each in the hemiacetal form, joined together by the elimination of a water between two hydroxyl groups. Dehydration involves the anomeric carbon of one monosaccharide and may or may not involve the anomeric carbon of the other monosaccharide when the hemiacetal hydroxyl group on an anomeric carbon is involved in a dehydration, the resulting product is an acetal (in common) and glycoside (in carbohydrate).
Sucrose is a non-reducing sugar (while its hydrolysis products glucose and fructose are reducing sugars) because :
Detailed Solution for JEE Advanced Practice Test- 17 - Question 54
Sucrose is having structure as
In above structure there is no free carbonyl centre of an aldehydic group present ; It is a non-reducing sugar.
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