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A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically.The tension in the string is equal to :
(Force diagram in the frame of the car)
Applying Newton's law perpendicular to strin
mg sin θ = ma cos θ
tan θ = a/g
Applying Newton's law along string
A uniform sphere has a mass M and radius R. Find the pressure p inside the sphere, caused by gravitational compression, as a function of the distance r from its centre. (γ is universal gravitational constant)
We consider a spherical concentric shell of radius x and thickness dx.
The mass of considered element is dm = (4π x^{2}dx)ρ.
Gravitational field at a point in the shell is
∴
The pressure in the element is
∴
But
By putting the value of ρ, we get,
A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
Here,
Input voltage, Vp = 220 V
Output voltage, V_{s} = 440 V
Input current, I_{p} = I_{p}
Output current, I_{s} = 2A
Efficiency of the transformer, η = 80%
Efficiency of the transformer,
A stone is dropped from a height of 45 m on a horizontal level ground. There is horizontal wind blowing due to which horizontal acceleration of the stone becomes 10 m/s^{2}.(Take g = 10 m/s^{2}).
The time taken (t) by stone to reach the ground and the net horizontal displacement (x) of the stone from the time it is dropped and till it reaches the ground are respectively
Taking motion in vertical direction,
u = 0, g = 10 m/s^{2 }, h = 45 m
⇒ h = 0 + (1/2) gt 2
⇒ t = √(2h/g) = √[(2× 45)/10]
⇒ t = 3 sec.
(b) Taking motion in horizontal direction,
The combination of 'NAND' gates shown here (in figure), are equivalent to
(if a NAND is given only 1 input it s behaves as NOT gate, as it is (one input is divided in two and then passed through NAND)
i.e. OR gate.
For second circuit C = Not of (A NAND B)
i.e. AND only
In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is
In balanced Wheatstone bridge, the galvanometer arm can be neglected so, equivalent resistance = R.
A neutron moving with speed v makes a headon collision with a hydrogen atom in ground state kept at rest. Find the minimum kinetic energy of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron ≈ mass of hydrogen = 1.67 x 10^{27} kg
Suppose the neutron and the hydrogen atom move at speeds v_{1} and v_{2} after the collision. The collision will be inelastic if a part of the kinetic energy is used to excite the atom. Suppose an energy ΔE is used in this way. Using conservation of linear momentum and energy,
mv = mv_{1} + mv_{2} ........(i)
and
From (i),
From (ii),
Thus,
Hence, (v_{1}  v_{2})^{2} = (v_{1} + v_{2})^{2}  4v_{1}v_{2} =
As v_{1}  v_{2} must be real,
or,
An electron in the ground state has an energy of 13.6 eV. The second energy level is 3.4 eV. Thus it would take E_{2} − E_{1} = 3.4 eV − 13.6 eV = 10.2 eV to excite the electron from the ground state to the first excited state.
The minimum energy that can be absorbed by the hydrogen atom in ground state to go in an excited state is 10.2 eV. Thus, the minimum kinetic energy of the neutron needed for an inelastic collision is
A transistoroscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be
In adiabatic process, ΔH = 0
Hence, zero is correct
Compute the bulk modulus of water if its volume changes from 100 litres to 99.5 litre under a pressure of 100 atmosphere.
By definition of bulk modulus,
Now as isothermal elasticity of a gas is equal to its pressure,
i e , bulk modulus of water is very large as compared to air This means that air is about 20,000 times more compressive than water, i.e , the average distance between air molecules is much larger than between water molecules
When a train is at a distance of 2km, its engine sounds a whistle. A man near the railway track hears the whistle directly and by placing his ear against the track of the train. If the two sounds are heard at an interval of 5.2 s, find the speed of the sound in iron (material of the rail track). given that velocity of sound in air is 330 ms^{1}.
Here, S = 2 km = 2,000 m ; v_{air} = 330 m s^{1}
Therefore, time taken by sound to travel through air,
As the two sounds (through air and iron rails) are heard at an interval of 5.2 s, time taken by sound to travel through iron rails,
t_{iron} = t_{air}  5.2
= 6.06  5.2 = 0.86 s
Therefore, velocity of sound in iron,
= 2,325.6 m s^{1}
Dispersive powers of materials used in lenses of an achromatic doublet are in the ratio 5:3. If the focal length of concave lens is 15 cm, then the focal length of the other lens will be
As per the diagram a point charge +q is placed at the origin O. Work done in taking another point charge Q from the point A [coordinates (0, a)] to another point B [coordinates (a, 0)] along the straight path AB is
Work done is equal to zero because the potential of A and B are the same
= 1 /4πε_{0} q/a
No work is done if a particle does not change its potential energy.
i.e., initial potential energy = final potential energy.
A man crosses the river perpendicular to river flow in time t seconds and travels an equal distance down the stream in Tseconds. The ratio of man's speed in still water to the speed of river water will be :
Let velocity of man in still water be v and that of water with repect to ground be u. Velocity of man perpendicular to river flow with respect to ground
Velocity of man downstream = v + u
⇒ ⇒
∴
The mean lives of a radioactive substance are 1620 years and 405 years for α  emission and β  emission respectively. Find the time during which threefourth of a sample will decay if it is decaying both by α – emission and β – emission simultaneously.
The decay constant λ is the reciprocal of the mean life τ
Thus,
and
∴ Total decay constant, λ = λ_{α} + λ_{β}
or
We know that
N = N_{0}e^{–λt}
When part of the sample has disintegrated, N = N_{0}/4
∴
or e^{λt} = 4
Taking logarithm of both sides, we get
or
= 2×324 × 0.693 = 449 year
An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it, How much does the spring elongate, when a 1 kg weight is suspended from it ? Take π = 3.14
Here. T = 0.314 s ; m = 1 kg
or
When spring is loaded with a weight 1 kg:
m g = k I
or
or
A point object is moving with a speed v in front of an arrangement of two mirrors as shown in figure. If the velocity of image in mirror M_{1} with respect to image in mirror M_{2} is n V sinθ, n =
Angle between Their magnitudes is v.
A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity k. Calculate the thickness of the shell if temperature difference between the outer and inner surfaces of the shell in steady state is T.
Consider a concentric spherical shell of radius r and thickness dr as shown in diagram. The radial rate of flow of heat through this shell in steady state will be,
(Negative sign is used as with increase in r. θ decreases]
Now as for spherical shell A =4πr^{2}
which on integration and simplification gives
Now in steady state as no heat is absorbed, rate of loss of heat by conduction is equal to that of supply i.e., H = P and here
So Equation (i) becomes,
i.e., thickness of shell (b  a)
A body falling freely from a given height ‘H’ hits an inclined plane in its path at a height ‘h’. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of (h/H) the body will take maximum time to reach the ground?
For finding the maximum time using the concept of differentiation
Two radio antennas radiating waves in phase are located at point A and B, 200 m part (Figure). The radio waves have a frequency of 6 MHz. A radio receiver is moved out from point B along a line perpendicular to the line connecting A and B (line BC shown in figure). At what distances from B will there be destructive interference?
There is a long solid conductor of resistivity ρ=1x10^{−6}Ωm, surrounded by air. There is a steady electric current along the length of the conductor. At point ‘P’ just outside the cylinder the electric field strength E=10^{−4}V/m is directed at an angle of α to the normal to the surface. Find the current density in the conductor in the vicinity of point ‘P’ (see figure)
A parallel beam of light falls normally on the first face of a prism of small angle A. At the second face it is partly transmitted and partly reflected. Then the reflected beam strike the first face again and emerges from first surface in a direction making an angle 6^{0}30′ with the normal at the first surface. The refracted beam is found to have undergone a deviation of 1
Calculate the angle of prism in degree.
The speed of sound in a mixture of n_{1}=2 moles of He, n_{2}=2 moles of H_{2} at temperature is Find (Take )
Six identical parallel metallic large plates are located in air at equal distances d to neighbouring plates. The area of each plate is A. Some of the plates are connected by conducting wires to each other. The capacitance of the system of plates between two points P and Q in pF is:
(Take A = 0.05 m^{2}, d = 17.7 mm, ε_{0}=8.85x10^{−12}F/mε_{0}=8.85x10^{−12}F/m).
The equivalent capacitance between points PQ is capacitance between two neighboring plates by wheat stone bridge.
A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has a length 4.8 m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of wave pulse. Find amplitude (in mm) of reflected pulse from junction Q.
If helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is .
For a monatomic gas kinetic energy = E, the relation with rms velocity is
For a monatomic gas with only Translational degrees of freedom
E_{a} = 3/2 k_{}_{b}T (Avg K E per molecule)
eq^{n} (2) multiply with 3 and adding with eq^{n} (1) we get
So x = 1
y = 3
z = 6
The terminal hydrogen is acidic in
H_{3}C  C ≡ CH
(propyne) and it reacts with ammoniacal AgNO_{3}. In propene, CH_{3}CH = CH_{2} there is no acidic hydrogen.
A piston is cleverly designed so that it extracts the maximum amount of work out of a chemical reaction, by matching P_{external} to the P_{internal} at all times. This 8cm diameter piston initially holds back 1 mol of gas occupying 1 L, and comes to rest after being pushed out a further 2 L at 25^{o}C .After exactly half of the work has been done, the piston has travelled out a total of
Piston change in volume
V = of cylinder = πr^{2}h
A monatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equal to 1. What is the molar heat capacity of the gas?
C is molar heat capacity for process (given)
Integrate,
Which of the following carboxylic acids undergoes decarboxylation easily ?
Complete octet & Extended conjugation
Cu_{2}S + 2Cu_{2}O ⟶ 6Cu + SO_{2}
This is an example of auto  reduction.
A solution which is 10^{–3} M each in Mn^{2}^{ +}, Fe^{2}^{ +}, Zn^{2}^{ +} and Hg^{2}^{ +} is treated with 10^{16} M sulphide ion. If K_{sp }of MnS, FeS, ZnS and HgS are 10^{13}, 10^{–18}, 10^{–24} and 10^{–53} respectively, which one will precipitate first ?
[S2^{}] need for precipitation of
Thus minimum [S^{2}] for precipitation is for HgS it will be precipitated first
In which of the following arrangements, the sequence is not strictly according to the property written against it?
Drago's Rule :
In case of NH_{3}, PH_{3},    bond angle decrease ,NH_{3} (≈107) & PH_{3} (≈90) So, %S charactor of lp in NH_{3} is less & %s charactor of lp top to bottom increase so basic charactor decrease.
A crystal made up of particles X, Y, and Z. X forms fcc packing. Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. It all particles along one body diagonal are removed, then the formula of the crystal would be
For fcc, number of X atoms = 4/unit cell
Number of Tetrahedral Voids = Z = 8
Number of Octahedral Voids = Y = 4
Number of atoms removed along one body diagonal = 2X (corner) and 2Z (TVs) and 1 Y (OV at body centre)
∴ Number of X atoms left = 4  [2 × (1/8) ]= (15/4)
Number of Y atom left = 4  (1 × 1) = 3
Number of Z atom left = 8  (2 × 1) = 6
The simplest formula
⇒ X_{5} Y_{4}Z_{8}
In the formal charge on each oxygen atom and the P – O bond order are respectively
There are Four Resonating Structures B.O = 1.25
= 0.75
π Bond of P = 0 is delocalised due to resonance on four Oatom
bond order
Based on following information determine the value of x and y
atomic masses of Al,Cl and Ag are respectively 27,35.5,108.
Aqueous ammonia is used as a precipitating reagent for Al^{3}^{+} ions as Al(OH)_{3} rather than aqueous NaOH, because
Al(OH)_{3} formed with NaOH dissolves in excess of NaOH to form aluminate ion.
which is stable.
Bakelite is made from phenol and formaldehyde. The initial reaction between them is the example of :
The reaction of phenol and formaldehyde, is called Lederer Mannase reaction. It follows electrophilic substitution and rearrangement reaction.
For an exothermic reaction, following two steps are involved.
Step 1. A + B → I (slow)
Step 2. I → AB (fast)
Q. Which of the following graphs correctly represent this reaction ?
In any reaction mechanism the no. of activated complexes showing maxima of potential energies is the no. of steps in that mechanism and in between these maxima is a valley where lies a more stable intermediate. There is one reaction intermediate lying at valley of two maxima representing two steps. E_{a} for first step should be higher than second step as first is slow step. These requirements fulfilled by choice (ii) & not choice (i) the choice (iv) is not possible because that is for endothermic reaction (Δ H + ve) & (ii) is single step equation so not possible.
An electron moving with velocity 'v' is found to have a certain value of deBroglie wavelength. The velocity to be possessed by the neutron to have the same deBroglie wavelength is
Chlorobenzene on heating with NaOH at 300^{o}C and high pressure followed by reaction with dil. HCl gives
A sample of an ideal gas is expanded from original volume of 1m^{3} to twice its volume in a reversible process for which p = αV^{2} (α = 5 atm m^{ 6})
Calculate (i) Work done on P  V diagram
(ii) Calculate ΔS_{m} ifC _{V}_{m} = 20J K ^{1 }m 1 , R = 8JK^{–1} m^{–1} and ln2 = 0.7
The values of work done & ΔS_{m} may be
8.21dm^{3} of NH^{3} at 500K and 1 atm expands adiabatically upto 0.5 atm against a constant external pressure of 0.5 atm. Calculate magnitude of change in internal energy (in calories). Assume that NH^{3} behaves as an ideal gas and vibration degrees of freedom are inactive.
(R=2 calmol^{−1}K^{−1}=0.0821atm−litK^{−1}mol^{−1}).
The total number of sigma and pi bonds in tricyclometaphosphoric acid, (HPO_{3})_{3} is:
It has 15 sigma bonds, and 3 Pi bonds.
If the dipole moment of AB molecule is given by 1.2 D and A–B the bond length is 4.8 D then % covalent character of the bond is:
Hence a unit vector perpendicular to
Hence the required vector is
If x = 9 is the chord of contact of the hyperbola x^{2}  y^{2} = 9, then the equation of the corresponding pair of tangents is
Let (h, k) be point whose chord of contact with respect to hyperbola x^{2}  y^{2} = 9 is x = 9.
W e know that, chord of contact of (h, k) with respect to hyperbola x^{2}  y^{2} = 9 is T = 0
But it is the equation of the line x = 9.
This is possible when h = 1, k = 0 (by comparing both equations).
Againe quation of pair of tangents is T^{2} = SS_{1}
Roots of the corresponding equation are
∴ by sign  scheme :
If α, β and γ are the angles which a directed line makes with the positive directions of the coordinates axes, then find the value of sin^{2}α + sin^{2}β + sin^{2}γ.
The direction cosines of the line are
In an acute  angled triangle ABC, points D, E and F are the feet of the perpendiculars from A, B and C onto BC, AC and AB, respectively. H is the orthocentre of ΔABC. If sin A = 3/5 and BC = 39, then the length of AH is
The line x = c cuts the triangle with corners (0, 0), (1, 1) and (9, 1) into two regions. For the area of the two regions to be the same c must be equal to :
The area of the triagle formed by three given vertices is
Now the line x = c bisect the area of triangle in two equal parts so the area of the both the parts of the triangle must be four
(or c = 15 which is not possible)
The number of solutions of the equation a ^{f(}^{x)} + g(x) = 0, where a > 0, g (x) ≠ 0 and has least value 1/2 is
Which is impossible.
Hence, no solutions.
∴ Number of solutions = zero.
The area of the curvilinear triangle bounded by the yaxis & the curve y = tan x & y = (2/3) cos x is
The equations of the tangents of the parabola y^{2}^{ }= 12x, which passes through the point (2,5).
y^{2 }= 12 x
4a = 12 (∵ y^{2 }= 4ax)
⇒ a = 3
Equation of tangent in slope form is
It passes through (2,5)
⇒ 5=2m+3/m
⇒ 5m=2m^{2}+3
⇒ 2m^{2}5m+3=0
⇒ (m1)(2m3)=0
⇒ m = 1,3/2
So equation of tangents are
The focal chord to y^{2} = 16xis tangent to (x  6)^{2} + y^{2 }= 2, then the possible values of the slope of this chord, are
The focal chord of y^{2} = 16 x is tangent to circle (x  6)^{2} + y^{2} = 2 .
Focus of parabola is (a; 0) i.e., (4, 0) is the Focus
Now, tangents are drawn from (4, 0) to (x  6)^{2} + y^{2} = 2 Since, PA is tangent to circle.
Since, tanθ = slope of tangent
∴ Slope of focal chord as tangent to circle
= ±1
If P(1) = 0 and then value of x for which is P(x) > 0 satisfy interval
If 5a + 4b + 20c = t, then the value of t for which the line ax + by + c  1 = 0 always passes through a fixed point is
Equation of given line is ax + by + c  1 = 0
From given relation, substuting value of c
Clearly for t = 20, the given line will pass through the point for all values of a & b
If f(x) is a polynomial satisfying f (x) f (1/x) = f (x) + f (1/x) and f (2) > 1,
Then by comparing the coefficients of like powers, we get
The equation of the pair of straight lines parallel to the y  axis and which are tangents to the circle x^{2} + y^{2}  6x  4y  12 = 0 is :
If a line parallel to yaxis, x = λ touches the given circle, then
The number of common tangents that can be drawn to the circles x^{2} + y^{2}  4x  6y  3 = 0 and x^{2} + y^{2} + 2x + 2y + 1 = 0 is :
Centre and radius of S_{1} are C_{1} (2, 3) and r_{1} = 4 and centre and radius of S_{2} are C_{2} (1, 1) and r_{2} = 1.
Hence the given circles are externally touching each other Hence, number of common tangents = 3
The distance between the circumcentre and orthocentre of the triangle whose vertices are (0, 0), (6, 8) and (4, 3) is
ΔOBC is right angled at O
Circum centre = mid point of hypotenuse BC
Orthocentre = vertex O (0, 0)
If A is a square matrix of order n such that adj adj(adjA)=[A]^{27}, then find the possible value of n.
Normals are drawn from the point P with slopes m_{1},m_{2},m_{3} to the parabola y^{2}=4x, if locus of P with m_{1}m_{2}=α is a part of the parabola itself then find the value of a.
The equation of the normal to the parabola y^{2} = 4ax. is y = mx – 2am – am^{3}.
Hence, the equation of the normal to the parabola y^{2} = 4x is y = mx 2m – m^{3}
Now, if it passes through (h, k), then we have k = mh  2m  m^{3}
This can be written as m^{3} + m(2h) + k = 0 .... (1)
Here, m_{1} + m_{2} + m_{3} = 0
m_{1}m_{2} + m_{2}m_{3}+ m_{3}m = 2  h
Also, m_{1}m_{2}m_{3} = k,
where m_{1}m_{2} = α
Now, this gives m^{3} = k/α and this must satisfy equation (1).
Hence, we have (k/α)3  (k/α)(2h) + k = 0
Solving this, we get k^{2} = α^{2}h  2α^{2} + α^{3}
And so we have y^{2} = α^{2}x  2α^{2}  α^{3}
On comparing this equation with y^{2} = 4x we get
α^{2} = 4 and 2a^{2} + a^{3} = 0
This give α = 2
Line segments drawn from the vertex opposite to the hypotenuse of a right angles triangle to the point trisecting the hypotenuse have length sin x and cos x where 0 < x < π/2. If the length of hypotenuse is where p and q are coprime, then find 2p+q.
If the equation of the curve on reflection of the ellipse about the line x−y−2=0 is
16x^{2}+6y^{2}+k_{1}x−36y+k_{2}=0, then find the value of 2k_{1}+k_{2}2k_{1}+k_{2}
Let p(4(1+cos θ)), 3(1+sin θ) is any point on the curve its reflection about the line x  y  2= 0
If f(x) is a polynomial of degree 4 with leading coefficient 1 satisfies the relation f(1)=1,f(2)=2, f(3)=3 and f(12)+f(−8) is equal to λ. Then find the digit at unit place of λ.
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