A simple pendulum is oscillating in a vertical plane. If resultant acceleration of bob of mass m at a point A is in horizontal direction, find the tangential force at this point in terms of tension T and mg.
When the acceleration of bob is horizontal, net vertical force on the bob will be zero.
T cos θ – mg = 0
The tangential force at that instant is
Hailstones falling vertically with a speed of 10 m/s, hit the wind screen (wind screen makes an angle 30° with the horizontal) of a moving car and rebound elastically. The velocity of the car if the driver finds the hailstones rebound vertically after striking is :
For the driver to observe the rain move vertically upward after the elastic collision, rain should come at angle 30^{º }with the horizontal (as clear from figure).
Let, velocity of rain w.e.t. car be V_{m//c}
But , since rain fall vertically down.
[Since ; V_{R/G(y)} = – 10 m/s ; V_{C/G(y) }= 0
⇒ V sin 30° = 10 ⇒ V = 20 m/s.
Substituting V = 20 m/s in equation (i)
A water tank stands on the roof of a building as shown. Then the value of 'h' for which the distance covered by the water 'x' is greatest is 
a
the roots of x are (0,4) and the maximum of x is at h = 2.
The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval.
Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint ,lower the range.Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m.
A Utube of base length “l” filled with same volume of two liquids of densities ρ and 2ρ is moving with an acceleration “a” on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by:
For the given situation, liquid of density 2ρ should be behind that of ρ. From right limb :
.......(1)
But from left limb : P_{c} = P_{atm} + (2ρ) gh ....(2)
From (1) and (2) :
A particle A of mass 10/7 kg is moving in the positive direction of x. Its initial position is x = 0 & initial velocity is 1 m/s. The velocity at x = 10 is: (use the graph given)
Area under Px graph
from graph ; area
v = 4m/s
ALITER : from graph
or
or
Now integrate both sides,
The string of a step rolling wheel is pulled by applying force F with different lines of action in two situations as shown. The wheel starts rolling without slipping due to application of the force :
(Pure rotation about instantaneous point of contact)
Note : If line action passes through point of contact, it only spins.
Two particles start together from a point O and slide down straight smooth wires inclined at 30º & 60º to the vertical & in the same vertical plane as in figure. The relative acceleration of second with respect to first will be (in magnitude & direction) as :
Resultant Acceleration
α = 30°
i.e. Resulting acceleration is in vertical direction.
Two points A & B on a disc have velocities v_{1} & v_{2} at some moment. Their directions make angles 60° and 30° respectively with the line of separation as shown in figure. The angular velocity of disc is :
For rigid body separation between two point remains same.
In the figure shown ADB & BEF are two fixed circular paths. A block of mass m enters in the tube ADB through point A with minimum velocity to reach point B. From there it moves on another circular path of radius R'. There it is just able to complete the circle.
For minimum velocity. at A ;
For looping the loop ;
The displacement of a body from a reference point is given by, (√x) = 2t3 where ' x ' is in metres and t in seconds. This shows that the body :
Vel c = dx/dt = 2(2t  3) x 2 = 4 (2t  3)
if v = 0 t = 3/2
acceleration
so correct ans is (A) and (B)
If the resultant force on a system of particles is nonzero, then :
The resultant force can be accelerating or decelerating, hence the momentum can increase or decrease. Hence (A) is wrong.
Since F_{net }= M a_{cm}
∴ a_{cm } ≠ 0 ;
hence v_{cm} must change
Hence (B) is correct.
In case of a circular motion of centre of mass about a point the distance of centre of mass will remain constant. Hence (C) is correct.
Kinetic energy of some particles may increase and of some particles may decrease at the same time.
A painter is applying force himself to raise him and the box with an cceleration of 5 m/s^{2} by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s^{2}, then :
For the whole system, 2T – 1500 = 150 x 5 ⇒ T = 1125 N
For the person, T – 1000 + N = 100 x 5 N = 1500 – 1125 = 375 N
Statement1 : For a disc undergoing fixed axis rotation, the magnitude of angle between velocity and acceleration vector of any moving point on disc at a particular instant of time are same.
Statement2 : Each moving point on a disc undergoing fixed axis rotation has same angular speed and same angular acceleration at an instant of time. Hence the ratio of magnitude of tangential acceleration and magnitude of centripetal acceleration is same for all moving points at an instant of time.
Statement1 is True, Statement2 is True; Statement2 is a correct explanation for Statement1
Statement1 : The equation of distance travelled by a particle moving in a straight line with constant acceleration in nth second is where letters have usual meaning, is dimensionally incorrect.
Statement2 : For every equation relating physical quantities to be true, it must have dimensional homogenity.
The equation of distance travelled in nth second is dimensionally correct because the interval of time 1 second has already been substituted into the equation and its dimension should be taken into account. Therefore statement1 is false.
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm^{–1}. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π^{2} = 10; g = 10 m/s^{2}). Assume that the spring does not affect the collision.
Mass of the block B is
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm^{–1}. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π^{2} = 10; g = 10 m/s^{2}) Assume that the spring does not affect the collision.
Amplitude of the SHM as being executed by block Bspring system, is 
Immediately after the collision, suppose velocities of the blocks are V_{1} and V_{1} as shown 1/2 vel. of approach = velocity of separation.
⇒ 5 = V_{2}  V_{1} .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V_{1} + V_{2}
or 10 = V_{1} + 5V_{2} .....(2)
On solving V_{2} = 2.5 m/s; V_{1} =  2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm^{–1}. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π^{2} = 10; g = 10 m/s^{2}) Assume that the spring does not affect the collision.
The distance 'd' will be equal to 
Matrix Match Type
This section contains 1 questions. Each question has four statements (A, B, C and D) given in ColumnI and five statements (p,q,r, s and t) in ColumnII. Any given statement in ColumnI can have correct matching with ONE OR MORE statement(s) in ColumnII. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are Ap, Ar, Bp, Bs, Cr, Cs, Dq and Dt then the answer should be written as : A→ p,r ; B→ p, s ; C → r, s ; D → q, t
Q. A uniform disc of mass M and radius R lies on a fixed rough horizontal surface at time t = 0. Initial angular velocity ω_{o} of each disc (magnitude and sense of rotation) and horizontal velocity v0 of centre of mass is shown for each situation of columnI. Match each situation in columnI with the results given in columnII.
In all four situation of columnI, angular momentum of the disc about a point on ground is conserved. Take angular momentum out of the paper as positive
(A) Initial angular momentum about its point of contact on ground = 1/2 {mR^{2}ω_{o}–mR(2Rω_{o} )} = negative. Hence final state of the disc is as shown if figure B.
Hence angular velocity shall first decrease and then increase in opposite sense. The velocity of centre shall decrease till the disc starts rolling without slipping.
(B) The initial angular momentum about its point of contact on ground =0.
Hence angular speed and velocity of centre simultaneously reduce to zero without a change in direction.
(C) Because v_{0} > Rω_{0}, velocity of centre of mass will decrease and angular velocity will increase without a change in direction till disc starts rolling without slipping.
(D) Because v_{0} < Rω_{0}, velocity of centre of mass will increase and angular velocity will decrease without a change in direction till disc starts rolling without slipping.?
Two particles are projected simultaneously with the same speed vv in the same vertical plane with angles of elevation θ, and 2θ, where θ<45^{∘}. At what time will velocities be parallel?
A man can swim in still water with a speed of 3 m/s. x and y axis are drawn along and normal to the bank of river flowing to right with a speed of 1 m/s. The man starts swimming from origin O at t = 0 second. Assume size of man to be negligible. Locus of all the possible points where man can reach at t = 1 sec. is (x–a)^{2} + y^{2} = c^{2} Find value of ac^{2}.
Method 1
If the river is still, the man will be at a distance 3 metres from origin O after 1 second. The locus of all the point where man can reach at t = 1 second is a semicircle of radius 3 and centre at O (dotted semicircle shown in figure).
The river flows to right with a speed 1 m/s. Hence there shall be additional shift in position by 1 m/s x 1 sec = 1m towards right. Hence the locus of all points giving possible position after one second will be the dotted semicircle shifted to right by 1 m as shown in figure.
Hence locus all the points where the man can be at t = 1 sec. is a semicircle of radius 3 and centre at 0' (1 m, 0 m)
∴ Equation of locus of all the points is
(x  1)^{2} + (y  0)^{2} = 3^{2}
or (x  1)^{2} + y^{2} = 9
Method  2
Let the relative velocity of the man make angle 'θ' with the xaxis.
Then at time 't' :
x = (3 cosθ + 1) t
and y = 3 sinθ t
⇒(x  t)^{2} + y^{2} = (3 cosθ)^{2} t^{2} + (3 sinθ)^{2} t^{2}
(x  t)^{2} + y^{2} = 9t^{2}
at t = 1 sec. the required equation is
(x  1)^{2} + y^{2} = 9.
In the figure shown a small block ‘B’ of mass ‘m’ is released from the top of a smooth movable wedge ‘A’ of the same mass ‘m’. ‘B’ ascends another movable smooth wedge ‘C’ of the same mass. Neglecting friction any where the maximum height attained by ‘B’ on ‘C’ is h/2x Find the value of x .
Let u and v be the speed of wedge A and block B at just after the block B gets off the wedge A. Applying conservation of momentum in horizontal direction, we get. mu = mv ..............(1) Applying conservation of energy between initial and final state as shown in fig (1), we get
solving (1) and (2) we get
.......(3)
At the instant block B reaches maximum height h' on the wedge C (figure 2),the speed of block B and Wedge C are v'. Applying conservation momentum in horizontal direction,we get mv= (m+m)v'................(4)
Applying conservation of energy between initial and final state
......... (4)
Solving equations (3) (4) and (5) we get h' = h/4.
A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?
fR = lα1 fR = lα_{2}
α_{1} = α_{2} = 2p red/sec^{2}
For A cylinder : ω = ω_{0}  αt ω = 100π  2πt ...(i)
For B cylinder ω = ω_{0} αt ω_{0} = 0
ω = αt ω = 27πt ....(ii)
From (i) and (ii) ω= 100 π  ω
2ω = 100π
ω = 50π
From (ii) euqation 50 π = 2 πt
t = 25 sec
The volume occupied by 2.0 mole of N_{2 }at 200K and 8.21 atm pressure, if is
So
So Volume of 2 moles = 3.6 L
If 10 gram of V_{2}O_{5} is dissolved in acid and is reduced to V^{2+} by zinc metal, how many mole of I_{2} could be reduced by the resulting solution if it is further oxidised to VO^{2+} ions ?
[Assume no change in state of Zn^{2+}ions] (V = 51, O = 16, I = 127) :
________________________________________
....(1)
Now
_____________________________________________
So we have 1 moles of V_{2}O_{5} will reduce 2 moles of iodine
So moles of I_{2} will be reduced by given amount of V_{2}O_{5} = 0.11 moles of I_{2}
0.5 mole each of two ideal gas A(C_{v }= 3/2 R) and B(C_{v }= 5/2 R_{) }are taken in a container and expanded reversibley and adiabatically from V = 1 litre to V = 4 litre starting from initial temperature t = 300K. ΔH for the process (in cal/mol) is
When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :
Initially solution will have low buffer capacity,will have maximum buffer capacity at half neutralisation and will be close to zero at equivalence point.After equivalence point the buffer capacity will keep on increasing with increase in concentration of base
One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :
[R = 0.082 L atm mol^{–1} K^{–1} = 8.3 J mol^{–1}K^{–1}].
An electron in a hydrogen like atom makes transition from a state in which its deBroglie wavelength is λ_{1} to a state where its deBroglie wavelength is λ_{2} then wavelength of photon (λ) generated will be : where m is mass of the electron, c is speed of light in vaccum.
⇒
⇒
Given
Find out the negative of logarithm of the solubility of solid Zn(OH)_{2 }at 25^{°}C ,at pH=6.Consider Zn(OH)_{2 }makes saturated solution at 25^{°}C.
Dissolved [Zn(OH)_{2}] = [Zn^{+2}]_{aq} + [Zn(OH)^{+}]_{aq }+ (Zn(OH)_{2})aq + [Zn(OH)_{3}^{}] + [Zn(OH)_{4}]^{2}
Now, [Zn(OH)_{2}]_{aq} = 10^{6} M in saturated solution.
so.
Dissolved Zn (OH)_{2} = + 10^{6} + 10^{3} [OH^{}] + 10^{2} [OH^{}]^{2}
+ 10^{6} + 10^{3} x 10^{8} = 10^{18} = 10^{1} + 10^{5} + 10^{6} + 10^{11} = 10^{1}
The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is / are incorrect (in the low pressure region):
If Z > 1 positive deviation
Z < 1 negative deviation
2CaSO_{4}(s) 2CaO(s) + 2SO_{2}(g) + O_{2}(g), Δ H > 0
Above equilibrium is established by taking some amount of CaSO_{4(s) }in a closed container at 1600K.Then which of the following may be correct options?
 As reaction is endothermic therefore with increase in temperature it will go in the forward direction hence moles of CaO will increase.
 With increase or decrease of volume partial pressure of the gases will remain same.
 Due to the addition of the insert gas at constan pressure reaction will proceed in the direction in which more number of gases moles are formed.
The normal boiling point of a liquid `X` is 400 K. Which of the following statement is true about the process X (l) ———? X(g)?
Boiling of a liquid at normal boiling point is a equilibrium process and on decreasing the pressure equilibrium will go forward and ΔG will be negative and vice versa.
The variation of pH during the titration of 0.5 N Na_{2}CO_{3 }with 0.5 N HCl is shown in the given graph.The following tables indicates the colour and pH ranges of different indicators:
Based on the graph and table ,Which of the following statement are true?
(A) First equivalence point pH is in between 810 and also this point is inbetween 8.29.8 (indicator range of cresolphthalein).(B), (C) second equivalence point (complete neutralisation).
(D) 50/2 = 25.
In a Hlike sample electrons make transition from 4^{th} excited state to 2^{nd} state then
Transition is taking place from 5 → 2
⇒ so Δn = 3
Hence maximum number of spectral line obsrved
(C) Number of lines belonging to the balmer series = 3
(D) Number of lines belonging to paschen series is 2.
Statement1 : A reaction which is spontaneous and accompanied by decrease of randomness must be exothermic.
Statement2 : All exothermic reaction are accompanied by decrease of randomness.
A reaction that is spontaneous and accompanied by a decrease of randomness must be exothermic.
A spontaneous reaction has a negative value of ΔG. Also, the decrease in randomness means a negative value of ΔS. Since, ΔG=ΔH−TΔS, the enthalpy change will be negative. Hence, the reaction will be exothermic.
Those exothermic reactions in which gaseous products are formed from a liquid or solid reactants are accompanied by an increase of randomness.
Statement1 : The titration curve for (W_{A} + W_{B}) is as follows :
For this titration no suitable indicator is present.
Statement2 : Indicator should change its colour sharply for indication of reaction to be complete so its pHrange should lie perfectly with in sharp change to avoid experimental error.
When there is no sharp increase (change) in pH of titration curve within pH range of indicator,there will be experimental error
Amphoteric oxides, such as aluminium oxide, are soluble both in strongly acidic and in strongly basic
solutions :
Dissolution of Al(OH)_{3} in excess base is just a special case of the effect of complexion formation on solubility. Al(OH)_{3} dissolves because excess OH¯ ions convert it to the soluble complex ion Al(OH)_{4}¯ (aluminate ion). The effect of pH on the solubility of Al(OH)_{3} is shown in figure.
Other examples of amphoteric hydroxides include Zn(OH)_{2}, Cr(OH)_{3}, Sn(OH)_{2} and Pb(OH)_{2}, which react with
excess OH¯ ions to form the soluble complex ions Zn(OH)_{4}^{2} (zincate ion), Cr(OH)_{4}¯ (chromite ion), Sn(OH)_{3}¯
(stannite ion), and Pb(OH)_{3}¯ (plumbite ion), respectively. By contrast, basic hydroxides, such as Mn(OH)_{2},
Fe(OH)_{2},and Fe(OH)_{3}, dissolve in strong acid but not in strong base.
Which of the following curves best represents the variation of solubility of ferrous hydroxide Fe(OH)_{2} with the concentration of [H^{+}] ions in the solution :
On increasing concentration of [H^{+}] ions the solubility of basic hydroxide, Fe(OH)_{2 }will increase.
Amphoteric oxides,such as aluminium oxide, are soluble both in strongly acidic and in strongly basic solutions.
Dissolution of Al(OH)_{3 }in excess base is just a special case of the effect of complexion formation on solubility.Al(OH)_{3 } dissolves because excess OH^{ }ions convert it to the soluble complex ion (aluminate ion). The effect of pH on the solubility of Al(OH)_{3 }is shown in figure.
Other examples of amphoteric hydroxides include Zn(OH)_{2, }Cr(OH)_{2, } Sn(OH)_{2 }and Pb(OH)_{2, }which react with excess OH^{ }ions to from the soluble complex ions Zn(OH)_{4}^{2} (zincate ion), Cr(OH)_{4}¯ (chromite ion), Sn(OH)_{3}¯
(stannite ion), and Pb(OH_{3})¯ (plumbite ion),respectively.By contrast, basic hydroxides, such as Mn(OH)_{2},
Fe(OH)_{2},and Fe(OH)_{3}, dissolve in strong acid but not in strong base.
Aluminium is mined as bauxite (Al_{2}O_{3}.xH_{2}O) a hydrated oxide that is always contaminated with Fe_{2}O_{3} and SiO_{2}. In Bayer process, Al_{2}O_{3} is purified by some reagents, which of the following reagents will be the best for this purification (Note : Strong heating of precipitate of Al(OH)_{3} produces Al_{2}O_{3})
First heat bauxite with NaOH so, Al_{2}O_{3} get dissolved in it (SiO_{2} will also get dissolved as it is an acidic oxide but not Fe_{2}O_{3} (basic acidic),then filter lt. To filterate add a weak acid like CO_{2} so Al(OH)_{3} forms Al^{3+} ions). On heating the ppt. of Al(OH)_{3} we will get pure Al_{2}O_{3}.
Amphoteric oxides, such as aluminium oxide, are soluble both in strongly acidic and in strongly basic
solutions :
Dissolution of Al(OH)_{3} in excess base is just a special case of the effect of complexion formation on solubility. Al(OH)3 dissolves because excess OH¯ ions convert it to the soluble complex ion Al(OH)_{4}¯ (aluminate ion).The effect of pH on the solubility of Al(OH)_{3 }is shown in figure.
Other examples of amphoteric hydroxides include Zn(OH)_{2}, Cr(OH)_{3}, Sn(OH)_{2} and Pb(OH)_{2}, which react with
excess OH^{} ions to form the soluble complex ions Zn(OH)_{4}^{2} (zincate ion),Cr(OH)_{4}^{} (chromite ion),Sn(OH)_{3}¯(stannite ion), and Pb(OH)_{3}¯ (plumbite ion), respectively. By contrast, basic hydroxides, such as Mn(OH)_{2}, Fe(OH)_{2},and Fe(OH)_{3}, dissolve in strong acid but not in strong base.
Zn(OH)_{2} is a amphoteric hydroxide and is involved in the following two equilibria in aqueous solutions
At what pH the solubility of Zn(OH)2 be minimum ?
Let solubility of Zn(OH)_{2} = s, some of will go in Zn^{2+} form and some in complex Zn(OH)_{4}^{2}, of = 10^{x} M,
........(1)
[S_{1} + S_{2} total solubility]
So, S_{1} S_{2} = 1.2 x 1.2 10^{18}
Now, we want S = S_{1} + S_{2} to be minimum we will have S_{1} = S_{2}
So, = S_{1} = S_{2} = 1.2 x 10^{9} M
Hence from 1" equation we get
So,
Hence x = 10
Matrix Match Type
This section contains 1 questions. Each question has four statements (A, B, C and D) given in ColumnI and five statements (p,q,r, s and t) in ColumnII. Any given statement in ColumnI can have correct matching with ONE OR MORE statement(s) in ColumnII. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are Ap, Ar, Bp, Bs, Cr, Cs, Dq and Dt then the answer should be written as : A→ p,r ; B→ p, s ; C → r, s ; D → q, t
Q. Match the following :
0.0333 M KMnO_{4} is used for the following titration. What volume of the solution in ml will be required to react with 0.158 g of Na_{2}S_{2}O_{3} ? (mol. wt. of Na_{2} S_{2} O_{3} = 158)
S_{2}O_{3}^{2} + MnO_{4}^{} +H_{2}O →MnO_{2}(s) + SO_{4}^{2 }+ OH^{ }(not balanced)
V = 80 ml
Change in oxidation num ber of sulphur per molecule of
Change in oxidation number of Mn per molecule of
No. of moles in 0.158 g of
No. of equivalents = 8 x 10^{3}
Normality of 0.0333 M KMnO_{4} solution = 0.0333 x 3 = 0.1
Let V mL of volume of KMnO_{4} be required; then
or V = 80 ml
Heat of hydrogenation of cyclohexene to cyclohexane is –28.6 kcal/mol. The observed heat of hydrogenation of benzene to cyclohexane is –49.8 kcal/mol calculate the resonance energy of benzene in kcal/mole.
Resonance energy = 49.8  (3 x 28.6) = 36 k cal/mol.
The equilibrium constant for the reaction Br_{2}(l) + Cl_{2}(g) 2Br Cl(g) at 27°C is kp = 1 atm. In a closed container of volume 164 L initially 10 moles of Cl_{2} are present at 27ºC. What minimum mass of Br_{2}(l) (in grams) must be introduced into this container so that above equilibrium is maintained at total pressure of 2.25 atm. Vapour pressure of Br_{2}(l) at 27ºC is 0.25 atm. Assume that volume occupied by liquid is negligible. [R = 0.082 L atm mole^{1} K^{1}, Atomic mass of bromine = 80].
To maintain the mentioned equilibrium there must be some (negligible) mass of Br_{2}(l) at equilibrium so there will be an equilibrium
so some Br_{2}(ℓ) will be required for this conversion (into vapour) moles of Br_{2}(ℓ) required for above equilibrium
Now for equilibrium
otal gaseous moles = (10 + x) {Let x be moles of Br_{2} (ℓ) required just to maintain above equilibrium}
we have
So,
So
⇒ ⇒ 8x^{2} = 100  x^{2} ⇒ 9x^{2} = 100
so x = 10/3
Hence total moles of Br_{2}(ℓ) required to maintain both of above equilibria
so mass of Br_{2} (ℓ) required = (80 g/mole) (5 mole) = 400 gm
The equilibrium constants for amino acids are given in terms of succesive ionization constants of the protonated form. For example, equilibrium constants for Glycine (NH_{2}CH_{2}COOH) are K_{a1} = 5 × 10^{3} M and K_{a2} = 2 × 10^{10 }M. What will be the pH at the Isoelectric point for this amino acid and pH of 0.02 M protonated Glycine in pure water respectively ? [Take : log 2 = 0.30]
Isoelectric point
If (1 + 2x) (1 + x + x^{2})^{n}
Integratin both sides of
If all the distinct roots of the equation x^{47} + 2x^{46} + 3x^{45} + ........+ 24x^{24} + 23x^{23} + ...... + 2x^{2} + x = 0 are z_{1}, z_{2} ...... z_{k} and imaginary part of z_{k}^{2} is b_{k} . Then the value of b1 + b2 ..... + bk is
x^{4}7 + 2x^{46} + 3x^{45} +..... + 24x^{24} + 23x^{23} +.... + 2x^{2} + x ≡ x (x^{23} + x^{22} + ...... + x + 1)^{2}
∴ roots of the given equation are 0 and roots of x^{23} + x^{22} + .... + x + 1 = 0 each repeated once.
Now roots of x^{23} + x^{22} + .....+ x + 1 = 0 are cos where k = 1, 2, ......... 23
the distinct roots of the given equation are 0, cos where k = 1, 2, .... . . . . 23
If all the words formed from the letters of the word "HORROR" are arranged in the opposite order as they are in a dictionary, then the rank of the word "HORROR" is
RRROOH
For complex number z, the minimum value of z + z – cos α – i sin α + z – 2(cos α + i sin α) is
z + z  cos α  i sin α + z  2(cos α + i sin α)
As shown in figure A and B are the points cos α + i sin α and 2(cos α + i sin α) respectively and let z be the affix of P. Then z + z  cos α  i sin α + z  2 (cos a + i sin α.) is minimum only when P coincides with the point A
∴ OP + AP + BP = 2
If a, b, c are in H.P. and a, b, c ≠ 9/4 , then are in
since a,b,c are in H.P
∴
If a determinant of order 3 × 3 is formed by using the numbers 1 or – 1, then minimum value of determinant is
so mimimum value = 4
Range of value of 'θ' for which Arg (z + 2√3 – 2i) = θ and z 2 have atleast one z common is
From the figure range of θ is
If (1 + x + x^{2} + x^{3})^{n} = a_{0} + a_{1}x + a_{2}x^{2} + .......... + a_{3n} . x^{3}^{n} and
A = a_{0} + a_{4}+ a_{8} + ......., B = a_{2} + a_{6} + a_{10} + ............., C = a_{3} + a_{7} + a_{11} + .........., then
As A = B = C
A + C = 2B
If Arg, then which of the following is (are) correct :
z lies on major arc of circle with z_{0} as centre, where
If P(x) = ax^{2} + bx + c satisfy a, b > 0 and  P’(0)_{max} = α, then
∵
∴ maximum occurs at x = 1 and minimum occurs at x = 0
∴ P(1) < 1 and P(0) >  1 ⇒ a + b + c < 1 and c >  1
⇒ a + b < 2 ∴ b < 2
P'(0)_{max} = 2
if b = 2, then c = 1
⇒
Product of 22 integers is equal to 1, then their sum cannot be
Integers can be either + 1 or  1
(+1)^{r} (1)^{22r} = 1
⇒ r = even
as r is given
2r  22 can not be , 0, 1 or 4
If each root of x^{2} – ax – b = 0, a, b R has absolute value < 1, then
Statement1 : Let m, n, a, b and c are nonzero real numbers such that a, b, c are in H.P., then , , are also in H.P.
Statement2 : If a, b, c are in G.P., then a – b/2 , b/2, c – b/2 are in H.P.
Statment 1 is true :
⇔ ⇔
⇔ ⇔ a,b,c are in H.P.
Statment2 is false.
α, β are the real roots of a_{1}x^{2} + b_{1}x + c_{1} = 0; γ, δ are the real roots of a_{2}x^{2} + b_{2}x + c_{2} = 0; D_{1} = b_{1}^{2} – 4a_{1}c_{1}, and D_{2} = b_{2}^{2} – 4a_{2}c_{2.}
Statement1 : If α, β, γ, δ are in H.P., then D_{1} : D_{2} : : c_{12} : c_{22}
Statement2 : If α, β, γ, δ are in A.P., then D_{1} : D_{2} : : a_{12} : a_{22}
Statment 2 ⇒
⇒
⇒ ⇒
Statment1
∴ ⇒
⇒
Statement1 and statement2 both are correct.but not each other's correct explanations.
Let z = e^{iθ} and a = e^{iα}. The given figure consists of a unit circle and two parallelograms OAHE and OBDC. AOB and OEF are two line segments and G is the mid point of FH. ΔEFH is isosceles where EF = EH.
If OH is perpendicular to OC, then the value of θ is
∴ [∵ OAHE is a rhombus]
Let z = e^{iθ} and a = e^{iα}. The given figure consists of a unit circle and two parallelograms OAHE and OBDC. AOB and OEF are two line segments and G is the mid point of FH. ΔEFH is isosceles where EF = EH.
The affix of the point G is
affix of E is az, affix of H is z + az and affix of F is 2az
∴
Let z = e^{iθ} and a = e^{iα}. The given figure consists of a unit circle and two parallelograms OAHE and OBDC. AOB and OEF are two line segments and G is the mid point of FH. ΔEFH is isosceles where EF = EH.
The length of AF is
Matrix Match Type
This section contains 1 questions. Each question has four statements (A, B, C and D) given in ColumnI and five statements (p,q,r, s and t) in ColumnII. Any given statement in ColumnI can have correct matching with ONE OR MORE statement(s) in ColumnII. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are Ap, Ar, Bp, Bs, Cr, Cs, Dq and Dt then the answer should be written as : A→ p,r ; B→ p, s ; C → r, s ; D → q, t
(A)
m is of the from (i + 2)^{2}  2, i = 1, 2, 3 ..., n
(B)
∵
∴ required value = 16  2(8) = 0
(C) ⇒
(D)
If z_{1} + z_{2} = z_{1} – z_{2} = 2 and 2z_{2} + 2i(z_{3} – z_{2}) = z_{2} + 2iz_{3} + (1 – 2i) z_{2} = 10 where z_{1} = 3 + 4i, then find the value of z_{1}^{2} + z_{2}^{2} + z_{3}^{2}
∵
⇒ Z_{1}, Z_{2} and origin are collinear and origin lies between Z_{1} and Z_{2} but Z_{1}  Z_{2} = 2
∵ 2Z_{2} + 2i (Z_{3}  Z_{2})[ = Z_{2} + 2iZ_{3} + (1  2i) Z_{2}
let (1  2i) Z_{2} + 2iZ_{3} = Z_{4}
∴ Z_{2} + Z_{4} = Z_{2} + Z_{4}
∴ Z_{2}, Z_{4} and origin are collinear and Z_{2}, Z_{4} lies on the same side of origin
∵ Z_{4} = Z_{2} + 2i (Z_{3}  Z_{2})
∴ By rotating it about x=π/2 in anticlockwise direction we get Z_{4} and Z_{4}  Z_{2} = 2 Z_{3}  Z_{2}
Z_{4} = 10  3 = 7 [∵ Z_{2} + Z_{4} = 10] Z_{4}  Z_{2} = 4
Z_{3}  Z_{2} = 2
∴ Z_{1}^{2} + IZ_{2}I^{2} + IZ_{3}I^{2} = 25 + 9 + 13 = 47
.......(1)
Also (x + 1)^{n} = C_{0}X^{n} + C_{1}X^{n1} + C_{2}X^{n1} + .............. + C_{n} ( 2 )
Multiply (1) & (2) and compare coefficient of x^{n+1}
Each digit has to be use only once then 6 digit numbers are there whose 3 digits are even and 3 are odd
is λ, then find exponent of 10 in λ.
Even : 0, 2, 4, 6, 8 odd : 1, 3, 5, 7, 9
^{5}C_{3} . ^{5}C_{3}. 6!  ^{4}C_{2} ^{5}C_{3} 5! = 64800 (Total)  (when zero occupies first place)
^{5}C_{3} x ^{5}C_{3}. 6!: Selection of 3 even from 5 even numbers and selection of 3 odd from 5 odd number.
^{4}C_{2} x ^{5}C_{3}. 5 ! ; Selection of 2 even from remaining four numbers and selection of 3 odd from 5 odd numbers.
In order to find numbers divisible by 5, Consider two cases.
In order to find numbers divisible by 5, Consider two cases.
(i) fixing 0 at last digit
^{4}C_{2} x ^{5}C_{3} x 5!
(i) fixing 5 at last digit
^{4}C_{2} x ^{5}C_{3} x 5!  ^{4}C_{2} . ^{4}C_{2}. 4!
Number divisible by 5 = ^{4}C_{2 }x ^{5}C_{3} x 5! + ^{4}C_{2 }x ^{5}C_{3} x 5!  ^{4}C_{2}. ^{4}C_{2}. 4! = 13536
If n € odd integers such that (1 + x + x^{2})^{n} = a_{0} + a_{1}x + a_{2} x^{2} + ........+ a_{2n}x^{2n }and (a_{1})^{2} – (2 a_{2})^{2} + (3 a_{3})^{2} – (4 a_{4})^{2} + ......–(2na_{2n})^{2} = – λn^{2} a_{n1} then find the value of λ.
∵ (1 + x + x^{2})^{n }= a_{0} + a_{1}x + a_{2}x^{2} + ....+ a_{2n}x^{2n} ........... (1)
differentiate w.r.t. x, we get.
⇒ n (1 + x + x^{2})^{n1} (1 + 2x) = a_{1} + 2a_{2} x + 3a_{3}x^{2} + ...... 2na_{2n}x^{2n1} ...... (2)
⇒ n(x^{2}  x + 1)^{n 1} (x  2) = a_{1}x2^{n1}  2a_{2}x^{2n2} + 3a_{3}x^{2n3}  ..............  2n a_{2n} ........ (3)
multiplying equation (2) and (3) and compare coefficient of x^{2n1} of both sides.
⇒ (a_{1})^{2}  (2a_{2})^{2} + (3a_{3})^{2}  ....  (2n a_{2n})^{2} = coefficient of x^{2n1} in n^{2} (1 + x^{2} + x^{4})^{n1} (2x^{2} 3 x 2)
= 2n^{2} {coefficient of x^{23} in (1 + x^{2} + x^{4})^{n 1}}  3n^{2} {coefficient of x^{2n2} in (1 + x^{2} + x^{4})^{n1}}  2n^{2} {coefficient of x^{2n1} in (1 + x^{2} + x^{4})^{n1}}
= 2n^{2} (0)  3n^{2} {coefficient of x^{n1} in (1 + x+ x^{2})^{n1}}  2n^{2}(0).
= 3n^{2} a_{n1}
Note : ∵ n is odd ⇒ (2n  3) and (2n  1) both will be odd.
∴ in the expansion of (1 + x^{2} + x^{4})^{n1} all the powers of x will be even
∴ coefficient of x^{2n1} and x^{2n3} will be zero.
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