54 Questions MCQ Test JEE Main & Advanced Mock Test Series - JEE Advanced Practice Test- 15
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A particle of charge 'q' and mass 'm' moves rectilinearly under the action of an electric field E = α - βx. Here, α and β are positive constants, and 'x' is the distance from the point where the particle was initially at rest.
Chose the correct option(s).
Detailed Solution for JEE Advanced Practice Test- 15 - Question 1
As we have, ma = qE so acceleration of the particle is a = q(α - β)/m
At mean position a = 0
Hence velocity of the particle is zero at x = 0 and x = 2α/β
We can conclude that particle oscillated between these two limits and have amplitude A = α/β
Acceleration of the particle is maximum at x = 0 and x = 2α/β , a_{max} = qα/m
Let Z_{1} and Z_{2} complex numbers such that Z_{1} ≠ Z_{2} and |Z_{1}| = |Z_{2}|. If Z_{1} has positive real part and Z_{2} has negative imaginary part, then may be
Detailed Solution for JEE Advanced Practice Test- 15 - Question 3
A particle of mass m moves along a curve y = x^{2}. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s^{−1} then :
Detailed Solution for JEE Advanced Practice Test- 15 - Question 5
On the curve
y = x^{2} ⇒ at x = 1/2
y = 1/4
Hence the coordinate
Differentiating : y = x^{2 }⇒ v_{y }= 2xv_{x}
Which satisfies the line
4x − 4y − 1= 0 (tangent to the curve)
and magnitude of velocity :
As the line 4x - 4y = 1 does not pass through the origin, therefore The magnitude of angular momentum of particle about origin at that position is 0 is not correct.
A particle is moving in a straight line according to the graph given below, assume that at = 0 particle is located at x = 1 m and have initial velocity 1 m s^{−1}.
then :
Detailed Solution for JEE Advanced Practice Test- 15 - Question 6
In v − x graph,
v = (tan 45^{∘})x ⇒ v = x ⇒ dv/dx = 1 ,and we know that acceleration, a = dv/dt = vdv/dx
⇒ a = x × 1 ⇒ a = x, so the graph between acceleration and displacement is,
Now we know that rate of change of linear displacement is called velocity,
In the circuit shown, resistance R = 100 Ω, inductance L= 2π and capacitance C = 8/π µF are connected in series with an AC source of 200 volt and frequency f. If the readings of the hot wire voltmeters V_{1} and V_{2} are same, then,
Detailed Solution for JEE Advanced Practice Test- 15 - Question 7
A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.
The focal length of the lens is (in cm)
Detailed Solution for JEE Advanced Practice Test- 15 - Question 10
An 820-turn wire coil of resistance 24 Ω is placed on a top of a-12,500 turn, 7 cm long solenoid as shown in the figure. Both coil and solenoid have a cross-section area of 10^{-4} m^{2}. The magnetic field produced by the solenoid at the location of the coil is one half as strong as the field at the centre of the solenoid. The solenoid is connected to a battery of emf 60 V and resistance 14 Ω with the help of a switch. The switch is closed at t = 0.
Determine the average emf (in V) induced in the circuit which opposes the current in the circuit to grow for a time interval 0 to t_{0}.
Detailed Solution for JEE Advanced Practice Test- 15 - Question 11
Average emf induced in interval 0-t_{0} sec,
Therefore, the average emf induced in the circuit is 37.9 V.
In YDSE, if incident light consists of two wavelengths λ_{1 }= 4000 Å and λ_{2 }= 5600 Å and is parallel to line SO. The minimum distance yy upon screen, measured from point O, will be where the bright fringe due to two wavelengths coincide is nλ_{1}D/d. Find n.
Detailed Solution for JEE Advanced Practice Test- 15 - Question 12
An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. If the order of accelerating voltage (for electrons) is required in such a tube is (10x) kV. Then the value of x(integer) is .......
Detailed Solution for JEE Advanced Practice Test- 15 - Question 13
In X - ray tube, accelerating voltage provides the energy to the electrons which produce X - rays. For getting X - rays, photons of 27.61 keV is required that the incident electrons must process kinetic energy at length 27.61 keV.
Gravitational acceleration on the surface of a planet is √6g/11 , where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km s^{−1}, the escape speed on the surface of the planet in km s^{−1} will be:
Detailed Solution for JEE Advanced Practice Test- 15 - 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- 15 - Question 15
When path difference increases from 0 to λ/2 , intensity will decrease from maximum zero. Hence, in this case, I(P_{2}) > I(P_{1}) > I(P_{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 between μ_{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- 15 - 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.
Four identical uniform rods of mass m = 6 kg each are welded their ends to form a square and then welded to a uniform ring having mass m = 4 kg and radius R = 1m. The system is allowed to roll down he incline of inclination θ = 30^{∘}
Detailed Solution for JEE Advanced Practice Test- 15 - Question 17
Four capacitors C_{1}(= 1 μF), C_{2}(= 2 μF), C_{3}(= 3 μF), and C_{4}(=4 μF), are connected in a network as shown in the diagram. The e.m.f of the battery is E = 12 V and its internal resistance is negligible. The keys S_{1} and S_{2} can be independently put on or off. Indicate the charge on the capacitors C_{1}, C_{2}, C_{3} and C_{4} by q_{1}, q_{2}, q_{3} and q_{4} respectively and the potential drops across them by V_{1}, V_{2}, V_{3} and V_{4} respectively.
Initially both the keys are open. The the key S_{1} is closed. Then the charges on the capacitors are
Detailed Solution for JEE Advanced Practice Test- 15 - Question 18
With both keys S_{1} ans S_{2} initially open, if now S_{1} is closed, the capacitors C_{1} and C_{3} are in series as also the capacitors C_{2} and C_{4}.
The compound formed upon combustion of sodium metal in the excess of air is
Detailed Solution for JEE Advanced Practice Test- 15 - Question 19
When sodium metal is burnt in excess of air, mainly sodium peroxide (Na_{2}O_{2}), with little sodium oxide (Na_{2}O), is formed.
If the oxidation is carried at high pressure, then sodium superoxide (NaO_{2}) may be formed. The question does not mention the condition of increased pressure.
Select from following molecules or ions, which is/are aromatic.
Detailed Solution for JEE Advanced Practice Test- 15 - Question 21
For aromatic compounds
For a compound to be aromatic, it should follow Huckle's (4n + 2)πe^{− }rule, i.e., 2, 6, 10, 14....πe^{−}
They should show delocalised, conjugate π system with alternate single and double bonds. In addition to this, the compound should be cyclic and planar or almost planar.
The number of πe^{−} is even, but not a multiple of 4.
For anti-aromatic compounds:
4nπe^{−}, i.e., 4, 8, 12....πe^{−} should be delocalized in a cyclic, planar, or almost planar system with alternate single and double bonds.
Planar conjugated carbocyclic polyene, carbocyclic polyene structure is less stable than its open structure.
Detailed Solution for JEE Advanced Practice Test- 15 - Question 22
All the above statements are true according to the kinetic theory of gases.
The collisions of the gas molecules are elastic and the energy lost by one molecule is equal to the energy gained by the other.
Hence, at a given temperature, the average energy of the gas is constant.
Heavier molecules transfer more momentum to the walls of the container. But this is not the postulate of the kinetic theory of gases.
Only a small fraction of the molecules possess high velocity (V_{r.m.s}), whereas the maximum number of molecules possess a lower velocity called the most probable velocity (V_{mp}).
The molecules move in random, zig-zag straight lines. The motion is referred to as Brownian motion.
Choose the correct reason(s) for the stability of the lyophobic colloidal particles.
Detailed Solution for JEE Advanced Practice Test- 15 - Question 23
Lyophobic colloids are stable due to preferential adsorption of ions common in their lattice from the solution.
Thus, all the colloidal particles acquire a charge and repel the other colloidal particles as they have like charges.
And potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles stablises the charges acquired by adsorption and makes the lyophobic sol stable.
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- 15 - Question 24
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.
Consider a reaction: aG + bH → products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is
Detailed Solution for JEE Advanced Practice Test- 15 - Question 27
Since the rate doubles when [G] is doubled, the order of the reaction with respect to the reactant G is one.
Since the rate becomes 8 times when both [G] and [H] are doubled, the order of the reaction with respect to the reactant H is two.
Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is
Detailed Solution for JEE Advanced Practice Test- 15 - Question 28
One mole of potassium dichromate reacts with 6 moles of ferrous sulphate. Thus, the number of moles of Mohr's salt required per mole of dichromate is 6.
Find the number of correct statements about the given series.
It is one of the natural series found on earth.
The number of alpha particles obtained on disintegration of one _{90}Th^{232} atom to get the stable product is 6.
The number of beta particles obtained on disintegration of one _{90}Th^{228} atom to get the stable product is 2.
_{90}Th^{232} and _{88}Ra^{228} are isodiaphers.
The nuclear isomers of _{83}Bi^{212} gives _{84}P^{212}4 on beta decay and _{81}Tl^{208} on alpha decay.
Detailed Solution for JEE Advanced Practice Test- 15 - Question 30
The correct statements will be-
(A) It is one of the natural series found on earth.
(B)
(C)
(D) Isodiaphers are atoms, having the different atomic number and mass number but have the same difference between a number of neutron and number of a proton.
When 20 g of naphthoic acid (C_{11}H_{8}O_{2}) is dissolved in 50 g of benzene (K_{f} = 1.72 K kg mol^{-1}), a freezing point depression of 2 K is observed. The van't Hoff factor (i) is
(Round off up to 1 decimal place)
Detailed Solution for JEE Advanced Practice Test- 15 - Question 31
Molar mass of naphthoic acid (C_{11}H_{8}O_{2}) = 172 g mol^{-1}
20% of the surface sites adsorbed N_{2}. On heating, N_{2} gas evolves from sites and is collected at 0.001 atm and 298 K, in a container of the volume 2.46 cm^{3}. The density of surface sites is 6.023 × 10^{14 }cm^{−2}. Area of the surface site is 1000 cm^{2}. Find out the number of surface sites occupied per molecule of N_{2}?
Detailed Solution for JEE Advanced Practice Test- 15 - Question 32
For adsorbed N_{2} on surface sites,
Now, to find the moles of N_{2} gas adsorbed using the ideal gas equation:
Molecules of adsorbed,
Surface site density = 6.023 × 10^{14} cm^{−2}Area of surface site = 1000 cm^{3}Total surface sites available = Number of sites per cm^{2} × Area
Total surface sites available = 6.023 × 10^{14} × 1000 = 6.023 × 10^{17}
Surface sites on which N_{2} is adsorbed = 20% × Available sites
Valence shell electron-pair repulsion theory predicts the shape of a molecule by considering the most stable configuration of the bond angles in the molecule. The main points of the theory are
(i) Electron pairs in the valence shell of central atom of a molecule, whether bonding or lone pairs are regarded as occupying localised orbitals. These orbitals arrange themselves so as to minimize the mutual electronic repulsions.
(ii) The magnitude of the different types of electronic repulsions follow the order given below:
Lone pair-lone pair> lone pair-bonding pair > bonding pair-bonding pair. These repulsive forces alter the bond angles of the molecule or ion.
(iii) The electron repulsion between two pairs of electrons will be minimum if they stay as far as possible. On this basis, several studied. have been done
Which of the following statements is correct ?
Detailed Solution for JEE Advanced Practice Test- 15 - Question 36
This problem includes the concept of bond angle the basis of Pπ − dπ back bonding and bond stifness, read all statements carefully and answer according to question using above concept.
Due to presence of Pπ − dπ back bonding, NCl_{3} is sp^{2} hybridised but due to the absence of vacant d orbital in NH_{3} it can't form Pπ − dπ back bonding. So bond anlge of NCl_{3} is greater than NH_{3}. Due to Pπ − dπ back bonding (d-orbitals available in Cl). The lone pair on N not available for exerting repulsion and hence bond angle in NCl_{3} > NH_{3}.
In BF_{3} there is Pπ − dπ back bonding from filled orbitals of F creating a partial double bond character making it a poor Lewis acid than BCl_{3}.
Bond multiplicity leads to more repulsion - double bond - lone pair repulsion is too high and hence bond angle increases.
SiCl_{4} and CCl_{4} bond angles are both tetrahedral (same).
Consider the quadratic equation (a + c − b)x^{2} + 2cx + (b + c − a) = 0, where a, b and c are distinct rational numbers and a + c − b ≠ 0. Then,
Detailed Solution for JEE Advanced Practice Test- 15 - Question 37
For the given quadratic equation (a + c − b)x^{2} + 2cx + (b + c − a) = 0, discriminant = 4c^{2} − 4(a + c − b)(b + c − a) = 4(a − b)^{2}, which is greater than equal to zero and is a perfect square of a rational number.
The tangent PT and the normal PN to the parabola y^{2} = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose
Detailed Solution for JEE Advanced Practice Test- 15 - Question 40
Equation of tangent and normal at point P(at^{2}, 2at) is
The smallest value of k, for which both the roots of the equation x^{2} - 8kx + 16(k^{2} - k + 1) = 0 are real, distinct and have values at least 4 , is
Detailed Solution for JEE Advanced Practice Test- 15 - Question 45
The equation is x^{2} - 8kx + 16(k^{2} - k + 1) = 0
A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then what is the value of k – 20?
Detailed Solution for JEE Advanced Practice Test- 15 - Question 47
If the area bounded by the curve y = sin^{−1}(sin x), y = {[|x|]} from x = 0 to x = π is A sq. units, where [⋅] & {⋅} are the greatest integer function and fractional part function respectively, then the value of 8A/π^{2} is
Detailed Solution for JEE Advanced Practice Test- 15 - Question 49
y = {[|x|]} = 0
The graph of y = sin^{−1}(sin x)can be shown as following
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