A uniform magnetic field B exists in the region between x = 0 and x = 3R/2 (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge +Q and momentum p directed along xaxis enters region 2 from region 1 at point P_{1}(y = –R). Which of the following options(s) is/are correct ?
(A, B) For the charge +Q to return region 1, the radius of the circular path taken by charge should by 3R/2.
mv² ÷ (3R/2) = QvB
Therefore,
2p / 3R = Q
So,
B = 2p / 3QR
i.e., B should be equal or greater than 2p/2QR
'A' is the correct option.
When B = 8p / 13QR
mv² / r = Qv (8p / 13QR)
Therefore, v = 13R / 8
Also CP2² = CO² + OP2²
= [(5R/8)² + (3R/2)²]²
CP2 = 13R / 8
Thus the particle will enter region 3 through the point P1 on Xaxis 'B' is the correct option.
Change in momentum =√2p
Thus, 'C' is incorrect.
Further, mv² / r = qvB
Therefore, r = mv / qB
'D' is incorrect.
A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the sun and the Earth. The Sun is 3 × 10^{5} times heavier than the Earth and is at a distance 2.5 ×10^{4} times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is v_{e} = 11.2 km s^{–1}. The minimum initial velocity (v_{s}) required for the rocket to be able to leave the SunEarth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)
Given
From energy conservation
where r = distance of rocket from Sun
Three vectors and are shown in the figure. Let S be any point on the vector The distance between the points P and S is The general relation among vectors and is :
Let vector from point P to point S be
from triangle rule of vector addition
A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is :
The given points (1, 2, 3, 4, 5, 6) makes 360° angle at 'O'. Hence angle made by vertices 1 & 2 with 'O' is 60°.
Direction of magnetic field at 'O' due to each segment is same. Since it is symmetric star shape, magnitude will also be same.
Magnetic field due to section BC.
A photoelectric material having workfunction φ_{0} is illuminated with light of wavelength The fastest photoelectron has a deBroglie wavelength λ_{d}. A change in wavelength of the incident light by Δλ results in a change Δλ_{d} in λ_{d}. Then the ratio Δλ_{d }/ Δλ is proportional to
According to photo electric effect equation :
Assuming small changes, differentiating both sides,
A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT = 0.01 second and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 ms^{–2}and the velocity of sound is 300 ms^{–1}. Then the fractional error in the measurement, δL/L, is closest to
Total time taken
is also small, so taking binomial approximation
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density r remains uniform throughout the volume. The rate of fractional change in density is constant. The velocity v of any point on the surface of the expanding sphere is proportional to :
Density of sphere is
Velocity of any point on the circumfrence V is equal to
(rate of change of radius of outer layer).
Consider regular polygons with number of sides n = 3, 4, 5 ..... as shown in the figure. The center of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is Δ. Then Δ depends on n and h as :
OA = h
Initial height of COM = h
Final height of COM =
A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is q. Which of the following statements about its motion is/are correct ?
When the bar makes an angle q; the height of its COM (mid point) is
Since force on COM is only along the vertical direction, hence COM is falling vertically downward.
Instantaneous torque about point of contact is
Now;
Path of A is an ellipse.
Two coherent monochromatic point sources S_{1} and S_{2} of wavelength λ = 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is Δθ. Which of the following options is/are correct ?
At point P_{2}; Δ x = d = 1.8 mm = 3000 λ hence a (bright fringe) will be formed at P_{2}
_{Now,}
Δθ increases as θ decreases At P_{2}, the order of fringe will be maximum.
For total no. of bright fringes d = nλ
⇒ n = 3000
∴ total no. of fringes = 3000
A source of constant voltage V is connected to a resistance R and two ideal inductors L_{1} and L_{2} through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t = 0, the switch is closed and current begins to flow. Which of the following options is/ are correct?
Since inductors are connected in parallel
Current through resistor at any time t is given by
I_{1} + I_{2} = I ...(i)
L_{1}I_{1} = L_{2}I_{2} ...(ii)
From (i) & (ii) we get
(D) value of current is zero at t = 0
value of current is V/R at t = ∞
Hence option (D) is incorrect.
A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque τ about an axis normal to the plane of the paper passing through the point
Q. Which of the following options is/are correct ?
(A) is incorrect.
If force is applied normal to surface at point X
τ = F_{y} R sinθ
Thus τ depends on θ & it is not constant
(B) is incorrect
if force applied tangentially at S
but it will climb as mentioned in question.
If force is applied normal to surface at P then line of action of force will pass from Q & thus τ = 0
(D) is incorrect.
if force is applied at P tangentially the
Constant
The instantaneous voltages at three terminals marked X, Y and Z are given by
An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be:
Potential difference between X & Y = V_{X} – V_{Y}
Potential difference between Y & Z = V_{Y} – V_{Z }
Phasor of the voltages :
similarly
Also difference is independent of choice of two terminals.
A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct ?
Every point on circumference of flat surface is at equal distance from point charge
Hence circumference is equipotential.
Flux passing through curved surface = – flux passing through flat surface.
∴ Flux through curved surface =
Note : Flux through surface can be calculated using concept of solid angle.
∴ Solid angle subtended
φ for 4π solid angle
∴ φ for solid angle
PARAGRAPH–1
Consider a simple RC circuit as shown in figure 1.
Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V_{0} (i.e., charging continues for time T >> RC). In the process some dissipation (E_{D}) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is E_{C}.
Process 2 : In a different process the voltage is first set to and maintained for a charging time T >> RC. Then the voltage is raised to without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V_{0 }as in Process 1. These two processes are depicted in Figure 2.
Q. In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are related by :
When switch is closed for a very long time capacitor will get fully charged & charge on capacitor will be q = CV
Energy stored in capacitor
Work done by battery (w) = Vq = VCV = CV^{2}
dissipated across resistance ∈D = (work done by battery)  (energy store)
.........(ii)
from (i) & (ii)
∈_{D} = ∈_{C}
PARAGRAPH–1
Consider a simple RC circuit as shown in figure 1.
Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V_{0} (i.e., charging continues for time T >> RC). In the process some dissipation (E_{D}) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is E_{C}.
Process 2 : In a different process the voltage is first set to and maintained for a charging time T >> RC. Then the voltage is raised to without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V_{0 }as in Process 1. These two processes are depicted in Figure 2.
Q. In Process 2, total energy dissipated across the resistance E_{D} is :
For process (1)
Charge on capacitor =
energy stored in capacitor =
work done by battery =
Heat loss =
For process (2)
Charge on capacitor =
Extra charge flow through battery =
Work done by battery :
Final energy store in capacitor :
energy store in process 2 :
Heat loss in process (2) = work done by battery in process (2) – energy store in capacitor process (2)
For process (3)
Charge on capacitor = CV_{0}
extra charge flow through battery :
work done by battery in this process :
find energy store in capacitor :
energy stored in this process :
Now total heat loss (E_{D}) :
final energy store in capacitor :
so we can say that E_{D} =
PARAGRAPH 2
One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity w0. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is µ and the acceleration due to gravity is g.
Q. The total kinetic energy of the ring is :
PARAGRAPH 2
One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity w0. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is µ and the acceleration due to gravity is g.
Q.
The minimum value of w_{0} below which the ring will drop down is :
Which of the following combination will produce H_{2} gas ?
(A) Zn + 2NaOH → Na_{2}ZnO_{2} + H_{2}
(B) 4Au + 8NaCN + O_{2} + 2H_{2}O → 4Na[Au(CN)_{2}] + 4NaOH
(C) Cu + 4HNO_{3} → Cu(NO_{3})_{2} + 2NO_{2} + 2H_{2}O
(D) Formation of passive layer of Fe_{2}O_{3 }on the surface of Fe and NO_{2 }gas is evolved.
Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol^{–1}. The figures shown below represents plots of vapour pressure (V.P.) versus temperature (T).
[Molecular weight of ethanol is 46 g mol^{–1}]
Among the following, the option representing change in the freezing point is 
Ethanol should be considered non volatile as per given option
The order of the oxidation state of the phosphorus atom in H_{3}PO_{2} , H_{3}PO_{4} , H_{3}PO_{3} and H_{4}P_{2}O_{6} is
; oxidation state of P = +5
; oxidation state of P = +4
; oxidation state of P = +4
; oxidation state of P = +1
Hence Ans (A)
The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
Δ_{f}G° [C(graphite)] = 0 kJ mol ^{1}
Δ_{f}G° [C(diamond)] = 2.9 kJ mol ^{1}
The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10^{6} m^{3} mol^{1}. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is
[Useful information : 1 J = 1 kg m^{2} s^{–2} ; 1 Pa = 1 kg m^{–1} s^{–2} ; 1 bar = 10^{5 }Pa]
Ans. (A)
C(graphite) → C(diamond) ;
As dG_{T} = V.dP
P_{2} = 14501 bar
The major product of the following reaction is
The order of basicity among the following compounds is
Basic strength µ stability of conjugated acid. ∝ + M / +H / +I
(I)
Conjugated acid stabilized by 2 equivalent R.S.
(II)
(III)
(IV)
Conjugated acid stabilized by 3 equivalent R.S.
For the following cell :
Zn(s)  ZnSO_{4} (aq.)  CuSO_{4} (aq.)  Cu(s)
when the concentration of Z n^{2+} is 10 times the concentration of Cu^{2+} , the expression for ΔG ( in J mol^{–1}) is [F is Faraday constant , R is gas constant, T is temperature , Eº(cell) = 1.1V]
ΔG = ΔG_{0} + 2.303RT logQ
ΔG =  nFE_{0} + 2.303RT logQ
Given : Eº = 1.1 V and n = 2
ΔG = 2.2 F + 2.303RT
In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is(are):
K = P . A . e^{Ea / RT}
(A) If P < 1 A_{arr. }> A_{expt }
(D) If P > 1 A_{arr.} < A_{expt}
(C) If P is very small, then catalyst is required to carry out the reaction at measurable rate.
Compound P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C_{8}H_{8}O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction.
(i)
(ii)
The option(s) with suitable combination of P and R, respectively, is(are)
Ans. (A,C)
(B) Product of ozonolysis of R is having 9 carbon.
(D) Product of ozonolysis of R is having 9 carbon.
For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by
Ans. (BC)
Sol.
(Surrounding is unfavourable)
(Surrounding is favourable)
The option(s) with only amphoteric oxides is (are):
(C) Cr_{2}O_{3} , BeO , SnO , SnO_{2}
all are amphoteric oxides
(D) ZnO , Al_{2}O_{3} , PbO , PbO_{2}
all are amphoteric oxides
Among the following, the correct statement(s) is are
Ans. (ACD)
(A)
(D) Lewis acidic strength decreases down the group. The decrease in acid strength occurs because as size increases, the attraction between the incoming electron pair and the nucleus weakens. Hence Lewis acidic strength of BCl_{3} is more than AlCl_{3}.
The correct statement(s) about surface properties is (are)
(A) Emulsion is liquid in liquid type colloid.
(B) For adsorption, ΔH < 0 & ΔS < 0
(C) Smaller the size and less viscous the dispersion medium, more will be the brownian motion.
(D) Higher the T_{C }, greater will be the extent of adsorption.
For the following compounds, the correct statement(s) with respect of nucleophilic substitution reactions is(are);
(A) I and II follow S_{N}2 mechanism as they are primary
(B) Reactivity order IV > I > III
(C) I and III follows S_{N}1 mechanism as they form stable carbocation
(D) Compound IV undergoes inversion of configuration.
Upon heating KClO_{3} in the presence of catalytic amount of MnO_{2}, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO_{3} gives Y and Z.
Q. W and X are, respectively
W and X are respectively
W = O_{2} and X = P_{4}O_{10}
Upon heating KClO_{3} in the presence of catalytic amount of MnO_{2}, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO_{3} gives Y and Z.
Q. Y and Z are , respectively
Y and Z are respectively
Y = N_{2}O_{5 }and Z = HPO_{3}
The reaction of compound P with CH_{3}MgBr (excess) in (C_{2}H_{5})_{2}O followed by addition of H_{2}O gives Q, The compound Q on treatment with H_{2}SO_{4} at 0ºC gives R. The reaction of R with CH_{3}COCl in the presence of anhydrous AlCl_{3} in CH_{2}Cl_{2} f ollowed by treatment with H_{2}O produces compounds S. [Et it compounds P is ethyl group ]
Q.
The reactions, Q to R and S to S, are 
The reaction of compound P with CH_{3}MgBr (excess) in (C_{2}H_{5})_{2}O followed by addition of H_{2}O gives Q, The compound Q on treatment with H_{2}SO_{4} at 0ºC gives R. The reaction of R with CH_{3}COCl in the presence of anhydrous AlCl_{3} in CH_{2}Cl_{2} f ollowed by treatment with H_{2}O produces compounds S. [Et it compounds P is ethyl group]
Q. The product S is 
Three randomly chosen non negative int egers x, y and z are found to satisfy the equat ion x + y + z = 10. Then the probability that z is even, is
Let z = 2k, where k = 0, 1, 2, 3, 4, 5
∴ x + y = 10 – 2k
Number of non negative integral solutions
Let S = {1, 2, 3,.....,9}. For k = 1,2, ....., 5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N_{1} + N_{2} + N_{3} + N_{5} =
N_{1} + N_{2} + N_{3} + N_{5}
= Total ways – {when no odd}
Total ways = ^{9}C_{5} Number of ways when no odd, is zero (∵ only available even are 2, 4, 6, 8)
∴ Ans : 9C5 – zero = 126
If ƒ :R → R is a twice differentiable function such that f''(x) > 0 for all x ∈ R, and then
∵f'(x) is an increasing function
∴ f'(1) > 1
If y = y(x) satisfies the differential equation and y(0) = then y(256) =
How many 3 × 3 matrices M with entries from {0,1,2} are there, for which the sum of the diagonal entries of M^{T}M is 5 ?
Let =
∴ tr(M^{T}M) = a^{2} + b^{2} + c^{2} + d^{2} + c^{2} + f^{2 }+ g^{2} + h^{2} + i^{2} = 5,
where entries are {0,1,2} Only two cases are possible.
(I) five entries 1 and other four zero
∴^{ 9}C_{5} × 1
(II) One entry is 2, one entry is 1 and others are 0.
∴^{ 9}C_{2} × 2!
Total = 126 + 72 = 198.
Let O be the or igin and let PQR be an arbitrary triangl e. The point S issu ch that Then the triangle PQR has S as its
Let position vector of
with respect to
Now
Δ Triangle PQR has S as its orthocentre
∴ option (B) is correct.
The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is
The normal vector of required plane is parallel to vector
∴ The equation of required plane passing through (1, 1, 1) will be
–14(x – 1) – 2(y – 1) – 15(z – 1) = 0
⇒ 14x + 2y + 15z = 31
∴ Option (A) is correct
(Difference series)
.... (1)
For option (C) :
(on integration)
Hence option (C) is correct.
If ƒ : R → R is a differentiable function such that f"(x) > 2f(x) for all x ∈ R, and f(0) = 1, then
Given that,
∴ option (A) is correct
⇒ f(x) is strictly increasing on x ∈ (0, ∞)
⇒ option (C) is correct
As, we have proved above that
⇒ option (D) is incorrect
∴ options (A) and (C) are correct
If f(x) = then
Ans. (B,D)
Expansion of determinant
⇒ more than two solutions
Let a and b be nonzero real numbers such that 2(cosβ  cosα) + cosα cosβ = 1. Then which of the following is/are true ?
Ans. (A,C)
We get
Hence (A, C)
If g(x) =
No option matches the result
If the line x = a divides the area of region R = [(x, y) ∈ x^{3} < y < x, 0 < x < 1} into two equal parts, then
Area between y = x^{3} and y = x in x ∈ (0,1) is
Area of curve linear triangle
Let f(x) = Then
Let O be the origin, and be three unit vectors in the directions of the sides respectively, of a triangle PQR.
Q.
= sin R sin (P Q)
Let O be the origin, and be three unit vectors in the directions of the sides respectively, of a triangle PQR.
Q. If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R + P) is
–(cos P + cosQ + cosR) ≥ 3/2 as we know cosP + cosQ + cosR will take its maximum value when
P = Q = R = p/3
Let p,q be integers and let α, β be the roots of the equation, x^{2}  x  1 = 0, where α ≠ β. For n = 0,1,2, ........, let a_{n} = pα^{n} + qβ^{n}.
FACT : If a and b are rational numbers and then a = 0 = b.
Q. If a_{4} = 28, then p + 2q =
α^{2} = α + 1 ⇒ α^{4} = 3 α + 2
∴ α_{4} = 28 ⇒ pα^{4} + qβ^{4} = p(3α + 2) + q(3β + 2) = 28
⇒ p(3α + 2) + q(3  3α + 2) = 28
⇒ α(3p  3q) + 2p + 5q = 28 (αs α ∈ Q^{c}) ⇒ p = q, 2p + 5q = 28 ⇒ p = q = 4
∴ p + 2q = 12
Let p,q be integers and let α, β be the roots of the equation, x^{2}  x  1 = 0, where α ≠ β. For n = 0,1,2, ........, let a_{n} = pα^{n} + qβ^{n}.
FACT : If a and b are rational numbers and then a = 0 = b.
Q. a_{12 }=
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