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Let there be a spherical symmetric charge distribution with charge density varying as ρ(r) = up to r = R and ρ(r) = 0 for r > R, where 'r' is the distance from origin. The electric field at a distance r (< r)="" from="" the="" origin="" is="" given="" />
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 1
Let dq be the small charge element on the spherical surface of radius r (< r)="" having="" thickness="" />
Hence, electric field can calculated using Gauss’s law.
A block of mass √3/10 kg is placed on a rough horizontal surface as shown in the figure.
A force of 1 N is applied at one end of the block, and the block remains stationary. The normal force exerted by the surface on block acts through (g = 10 m/s^{2})
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 2
For equilibrium, resultant force = 0 and resultant torque = 0
A ray of light is incident on the interface between water and glass at an angle i and refracted parallel at the water surface, then value of μ_{g} will be
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 4
A particle of mass m moves on the x-axis as follows:
It starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times (0 < t < 1). If denotes the instantaneous acceleration of the particle, then
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 5
Since the body is at rest at x = 0 and x = 1, α cannot be positive for all time in the interval 0 ≤ t ≤ 1.
Therefore, first the particle is accelerated and then retarded.
Now, total time t = 1 s (given)
Total displacement s = 1 m (given)
s = Area under v-t graph
∴ Height or v_{max} = 2s/t = 2 m/s is also fixed.
If height and base are fixed, area is also fixed.
In case 2: Acceleration = Retardation = 4 m/s^{2} (slope of the graph)
In case 1: Acceleration > 4 m/s^{2}, while retardation < 4="" />^{2}
In case 3: Acceleration < 4="" />^{2} and Retardation > 4 m/s^{2}
A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limits -ϕ and +ϕ. For an angular displacement θ (|θ| < ϕ), the tension in the string and the velocity of the bob are T and v, respectively. The relation(s) that hold(s) good under the above conditions is/are:
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 6
Motion of pendulum is part of a circular motion. In circular motion, resolving of forces occurs in two perpendicular directions; first along radius (towards centre) and second along tangent.
Along radius, Centripetal force,
Along tangent, Resultant force = ma_{T}, where a_{T} is the tangential acceleration.
A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2 v and v, respectively, strike the bar [as shown in the figure] and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity respectively by ω, E and v_{c}, we have after collision:
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 7
Using conservation of linear momentum. we have
Initial momentum = Final momentum
p_{i} = mi(2v) - 2m(v) = ⇒ 0 p_{f} = 0 hence velocity of the centre of mass of the combined system after collision is zero.
V_{cm} = 0
Using conservation of the angular momentum. we have
A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates. Two spheres of equal radii 1 unit, with their centres respectively at A (-2, 0, 0) and B (2, 0, 0), are taken out of the solid leaving behind spherical cavities as shown in the figure. Then,
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 8
Due to the solid sphere of a uniform density, gravitational field is zero at the centre of the sphere. The cavities A and B can be treated as negative masses and they are situated on opposite sides of the centre O. Hence, gravitational forces exerted by the cavity masses on a mass at O are opposite in direction. Hence, resultant force on mass at O is zero. Thus, gravitational force due to this object at the origin O is zero.
Consider a circle y^{2} + z^{2} = 36; the centre of the circle is (0,0,0) and the radius of the circle is 6 units. The circle lies in y-z plane. For the point outside the sphere or situated at the sphere, the mass of the sphere can be assumed to be situated at the centre. All the points of the circle y^{2} + z^{2} = 36 are equidistant from the centre; hence, gravitational potential is same at all the points in the given circle.
Consider the circle y^{2} + z^{2} = 4; its centre lies at (0, 0, 0) and the radius is 2 units. It lies in y-z plane perpendicular to x-axis. All the points of the circle y^{2} + z^{2} = 4 are equidistant from the centre; hence, gravitational potential is same at all the points in the given circle.
Let [ε_{0}] denote the dimensional formula of the permittivity of the vacuum, and [μ_{0}] that of the permeability of the vacuum. If M = mass, L = Length, T = time and I = electric current. Choose the correct option.
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 9
Two inductors having self inductances L_{1}, L_{2} and mutual inductance M are arranged in parallel combination as shown in the diagram. The sense of the helix is same for both the inductors.
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 11
To find the distance d over which a signal can be seen clearly in foggy conditions, a railway engineer uses dimensions and assumes that the distance depends on the mass density p of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to S^{1/n}. The value of n is
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 13
Three charges Q, q and q are placed at the vertices of a right-angled isosceles triangle such that the potential energy of the system is zero. The value of Q in terms of q is given by the expression, Then, the value of m is
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 14
A convex lens, when inserted between an object and a screen, which are at a fixed distance of 243 cm from each other, forms images of magnification 8. Then, the difference between the focal length of the lens and the object distance (in cm) is
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 15
A particle is suspended by a light vertical inelastic string of length ll from a fixed support. At its equilibrium position, it is projected horizontally with a speed . What is the ratio of the tension in the string in its horizontal position to that in string when the particle is vertically above the point of support ?
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 16
For vertical circular motion, student should always remember that
Two particles A and B are located at points (0, −10√3) and (0, 0) in xy plane. They start moving simultaneously at time t = 0 with constant velocities m s^{−1 } and m s^{−1 } respectively. Time when they are closest to each other is found to be K/2 seconds. Find K. All distances are given in meter.
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 17
As observed by B, motion of A is along AM and BM is the shortest distance between them. Relative displacement of A w.r.t to B is AM = AB cos 30^{o}
The equation of a standing wave propagating along a string fixed at both ends is given by y = (4 cm) sin(0.314 cm^{−1}) cos(3.14 s^{−1})t. The linear mass density of the string is 10 g cm^{−1}. The average power transmitted through the string is n × 10^{−4} W. Find n.
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 18
Net energy transmitted through a stationary wave is zero as stationary waves are basically just 2 equal waves with different phase difference moving in opposite directions
Directions: The following question is based on the paragraph given below.
P and Q are isomers of dicarboxylic acid C_{4}H_{4}O_{4}. Both decolorise Br_{2}/H_{2}O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U.
The compounds formed from P and Q respectively are:
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 20
Since both the dioic acids decolourise Br_{2}/H_{2}O, these are unsaturated dioic acids and display geometrical isomerism.
And also, 'P' forms a cyclic anhydride; therefore, it is a cis isomer.
'P' is maleic acid and 'Q' is fumaric acid.
On treatment with KMnO_{4} solution, both 'P' and 'Q' form tartaric acid which exists as three stereoisomers: S, T and U.
Maleic acid 'P' yields the meso isomer 'S', which is optically inactive.
Fumaric acid 'Q' yields the optically active isomer (T, U).
1 mole of an ideal gas A(C_{v, m} = 3R) and 2 moles of an ideal gas B are (C_{v,m} = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 21
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 25
(1) Replacement of -OH by a halogen in an alcohol is nucleophilic substitution reaction. It is the protonted alcohol which acts as a substrate.
(2) Alcohols are acidic enough to react with active metals to liberate hydrogen gas. They are basic enough to accept a proton from strong acids.
(3) Secondary alcohols, on oxidation, give a ketone containing the same number of carbon atoms. Further oxidation will give carboxylic acid containing lesser number of carbon atoms.
(4) Primary alcohols, on oxidation, give an aldehyde containing the same number of carbon atoms. Further oxidation will give carboxylic acid containing the same number of carbon atoms.
In which of the following arrangements, the order is not according to the property indicated against it?
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 29
Among the isoelectronic species, more the positive charge on the cation smaller is the size and more the negative charge on the anion larger is the size.
Al^{3+} < Mg^{2+} < Na^{+ }< F^{−}
The electron gain enthalpy decreases down the group, but the second element of the group has the highest negative value of electron gain enthalpy. Hence, I < Br < F < Cl is the correct order.
The metallic radius increases down the group. Hence, the order
Li < Na < K < Rb
The order should be B < C < O < N, as N has stable electronic configuration of 1s2, 2s^{2}, 2p^{3}.
On complete combustion of a 9 L mixture of ethane and propane, 21 L of CO_{2} at STP is produced. The molar ratio of ethane and propane in the mixture is
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 31
Avogadro's law states that under same conditions of temperature and pressure, equal moles of gases occupy equal volumes. Hence, in chemical reactions involving gaseous reactants and products, the volume ratio is equal to the molar ratio.
How many open chain structures are possible for N-Methly butanamine (including it) that are referred to as metamers?
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 32
Molecular formula of N-methyl butanamine is C_{5}H_{13}N. Primary, secondary and tertiary amines are different functional groups and metamerism cannot coexist with functional isomerism. Hence, we shall consider only those isomers of N-methyl butanamine which are secondary (2°) amines.
Every molecule of group A is a metamer of every molecule in group B.
In Goldschmidt aluminothermic process, the thermite mixture contains _____ parts of Fe_{2}O_{3} and 1 part aluminium by mass.
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 34
Aluminothermic process is an exothermic chemical reaction using AlAl as a reducing agent. In the reaction, the reactants are, Fe_{2}O_{3 }+ 2Al → 2Fe + Al_{2}O_{3}
So, 3 parts of Fe_{2}O_{3 } and one part Al by mass.
Prostaglandin E1 is a compound produced by the body to regulate a variety of processes including blood clotting, fever, pain and inflammation.
How many stereogenic centres are present in the given compound?
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 35
Asymmetric stereogenic carbons have 4 different groups attached to it. Carbons numbered 1, 4, 5 and 6 are such Carbons.
A stereogenic centre is that where interchanging the substituents results in a stereoisomer. Hence, the Carbons of the double bond (2 and 3) will be stereogenic centres.
Thus, the total number of stereogenic centres will be 6
10% of sites on a catalyst have been absorbed by H_{2}. On heating, H_{2} gas is evolved from the sites and collected at 0.03 atm and 300 K in a small vessel of 2.46 cm^{3}. Number of sites available is 6.0 x 10^{15} per cm^{2} and the surface area is 1000 cm^{2}. Find out the molecules of hydrogen occupied per site. (Given: N_{A} = 6 × 10^{23 }mol^{−1})
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 36
Adsorbed moles of
= 3 × 10^{−6}
∴ Number of absorbed molecules of H_{2} = 3 × 10^{−6 }× 6 × 10^{23 }= 18 × 10^{17 }
Total number of surface sites available = 6 × 10^{15 }× 1000 = 6 × 10^{18} cm^{2}
Number of surface sites that is occupied by adsorption of
f(x), g(x) and h(x) are all continuous and differentiable functions in [a, b], a < c < b and f(a) = g(a) = h(a). Point of intersection of the tangent at x = c with chord joining x = a and x = b is on the left of c in y = f(x), on the right in y = h(x) and tangent at x = c is parallel to the chord in the case of y = g(x).
If f’(x) > g’(x) > h’(x)
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 38
A point P moves such that the tangents PT_{1} and PT_{2} from it to the hyperbola 4x^{2 }− 9y^{2} = 36 are mutually perpendicular. Then the equation of the locus of P is
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 39
4x^{2 }− 9y^{2} = 36
The two tangent PT_{1} and PT_{2} will be perpendicular if the product of roots is −1.
(The point of intersection of perpendicular tangents lies on director circle)
For the hyperbola the equation of the director circle is x^{2 }+ y^{2 }= a^{2 }− b^{2}
∴ The equation of the locus of P(α,β) is x^{2 }+ y^{2} = 5
The tangent at any point P(a secθ, b tanθ) of the hyperbola makes an intercept of length p between the point of contact and the transverse axis of the hyperbola. If p_{1}, p_{2} are the lengths of the perpendiculars drawn from the foci on the normal at P, then
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 40
Let F_{1}(ae, 0) and F_{2}(–ae, 0) be the foci of the hyperbola.
Equations of the tangent and normal at P are
Q(a cosθ, 0) is the point of intersection of the tangent at P and the x-axis.
G(ae^{2 }secθ, 0) is the point where the normal at P meets xx-axis.
The foci of a hyperbola are the vertices of the ellipse and the directrices of the hyperbola pass through the foci of the ellipse. The radius of the auxiliary circle of the hyperbola is
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 41
Focus of hyperbola = Vertex of ellipse = (10, 0) and (-10, 0)
⇒ ae = 10 ...(1)
Directrix of hyperbola pass through focus of ellipse.
AB is a chord of the parabola y^{2} = 8x with vertex at A and BC is drawn perpendicular to AB meeting the axis of the parabola at C. The projection of BC on the axis of the parabola is equal to .....
Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 52
Let us assume that AB makes angle θ with positive x-axis.
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