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The relative uncertainty in the period of a satellite orbiting around the earth is 10^{–2}. If the relative uncertainty in the radius of the orbit is negligible the relative uncertainty in the mass of the earth is :
From kepler's Law
At some instant a radioactive sample S_{1} having an activity 5 μCi has twice the number of nuclei as another sample S_{2} which has an activity of 10 μCi. The half lives of S_{1} and S_{2} are :
Given : N_{1} = 2N_{2}
Hence 5years and 20 year
Two moles of helium are mixed with an moles of hydrogen. If C_{p}/C_{v} = 3/2 for the mixture then the value of n is
⇒
Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is θ, then
A and B have same alignment of transmission axis.
Lets assume c is introduced at θ angle
⇒
The deBroglie wav elength (λ_{B}) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (λ_{G}) by :
In the given circuit the current through zener diode is :
The end correction of a resonance column is 1cm. If the shortest length resonating with the tuning fork is 10cm, the next resonating length should be :
Given : e = 1 cm
For first resonance
For second resonance
⇒ ℓ_{2} = 3 × 11  1 = 32 cm
Two sitar strings A and B playing the note 'Dha' are slightly out of tune and produce beats of frequency 5Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease by 3Hz. If the frequency of A is 425 Hz. the original frequency of B is:
Frequency of B is either 420Hz or 430Hz As tension in B is increased its frequency will increase. If frequency is 430Hz, beat frequency will increase If frequency is 420 Hz beat frequency will decrease, hence correct answer is 420Hz
A power transmission line feeds input power at 2300V to a step down transformer with its primary windings having 4000 turns giving the output power at 230V. If the current in the primary of the transformer is 5A and its efficiency is 90% the output current would be :
A body of mass m starts moving from rest along xaxis so that its velocity varies as u = a√s where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :
Suppose that the angular velocity of rotation of earth is increased. Then as a consequence:
g' = g – ω^{2}Rcos^{2}φ
Where φ is latitude there will be no change in gravity at poles as φ = 90°
At all other points as ω increases g' will decrease.
Both the nucleus and the atom of some element are in their respective first excited states. They get deexcited by emitting photons of wavelengths λ_{N},λ_{A }respectively. The ratio λ_{N}/λ_{A } is closest to :
where E_{a} and E_{N} are energies of photons from atom and nucleus respectively. E_{N} is of the order of MeV and E_{a} in few eV.
So
A plane electromagnetic wave of wavelength λ has an intensity I. It is propagating along the positive Ydirection. The allowed expressions for the electric and magnetic fields are given by :
If E_{0} is magnitude of electric field then 1/2ε_{0}E^{2} x C = I
direction of
A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω with respect to normal axis then magnetic moment of the loop is :
A heating element has a resistance of 100Ω at room temperature. When it is connected to a supply of 220V a steady current of 2A passes in it and temperature is 500°C more than room temperature. What is the temperature coefficient of resistance of the heating element?
Resistance after temperature increases by 500°C = 220/2 = 110Ω
110 = 100 (1 + α500)
a = 2 × 10^{4} °C^{1}
A coil of crosssectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω the maximum e.m.f. induced in the coil will be :
A ray of light is incident at an angle of 60° on one face of a prism of angle 30°. The emergent ray of light makes an angle of 30° with incident ray. The angle made by the emergent ray with second face of prism will be :
δ = I + e  A
30 = 60 + e – 30°
⇒ e = 0
So angle with face = 90°
A galvanometer with its coil resistance 25Ω requires a current of 1mA for its full deflection. In order to construct an ammeter to read up to a current of 2A the approximate value of the shunt resistance should be :
An oscillator of mass M is at rest in the equilibrium position in a potential V = 1/2 k(x – X)^{2}. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1)
In first collision mu momentum will be imparted to system. In second collision when momentum of (M + m) is in opposite direction mu momentum of particle will make its momentum zero. on 13^{th} collision
⇒
⇒
One mole of an ideal monatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by :
Equation of line BC
In a circuit for finding the resistance of a galvanometer by half deflection method a 6V battery and a high resistance of 11kΩ are used. The figure of merit of the galvanometer produces a deflection of θ = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ/2 is
S = 110Ω
In the following circuit the switch S is closed at t = 0. The charge on the capacitor C_{1} as a function of time will be given by
q = C_{eq}E [1 – e^{–t/RC}eq)]
Both capacitor will have same charge as they are connected in series.
Let The magnitude of a coplanar vector is given by :
Solving equation (i) and (ii) we get
A particle executes simple harmonic motion and is located at x = a, b and c at times t_{0}, 2t_{0} and 3t_{0} respectively. The frequency of the oscillation is :
a = Asinωt_{0}
b = Asin2ωt_{0}
c = Asin3ωt_{0}
a + c = A[sinωt_{0} + sin3ωt_{0}] = 2Asin2ωt_{0}cosωt_{0
}
A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z' axes will be :
We know that, the moment of inertia of the circular disk about the center is
Using parallel axis theorem,
Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters and the force between them is F. A third identical conducting sphere C is uncharged. Sphere C is first touched to A then to B and then removed. As a result the force between A and B would be equal to :
when A and C are touched charge on both will be q/2
Then when B and C are touched
Two particles of the same mass m are moving in circular orbits because of force given by The first particle is at distance r = 1 and the second at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to
The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity The maximum percentage error in the value of A will be :
= 3 × 0.5 + 2 × 1 + 1/2 × 3 + 1.5
= 1.5 + 2 + 1.5 + 1.5
ΔA/A = 6.5%
A carrier wave of peak voltage 14V is used for transmitting a message signal given to achieve a modulation index of 80% will be :
A small soap bubble of radius 4cm is trapped inside another bubble of radius 6cm without any contact. Let P_{2} be the pressure inside the inner bubble and P_{0} the pressure outside the outer bubble. Radius of another bubble with pressure difference P_{2} – P_{0} between its inside and outside would be :
r = 2.4cm
For standardizing NaOH solution, which of the following is used as a primary standard?
Oxalic acid is used as a primary standard for NaOH standardizing.
Products A and B formed in the following reactions are respectively :
When XO_{2} is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO_{3} ; a dark green product is formed which disproportioates in acidic solution to afford a dark purple solution. X is :
The major product B formed in the following reaction sequence is :
In a complexometric titration of metal ion with ligand M (Metal ion) + L (Ligand) → C (Complex) end point is estimated spectrophotometrically (through light absorption). If 'M' and 'C' do not absorb light and only 'L' absorbs, then the titration plot between absorbed light (A) versus volume of ligand 'L' (V) would look like :
Initially ligand consumed by metal due to formation of complex. So absorbed light (A) remain constant, after complex formation is completed, extra volume of ligand solution increases ligand concentration and also increases absorbed light.
The major product of the following reaction is :
Among the following, the incorrect statement is :
In amylose1,4 αglycosidic linkage is present.
In the extraction of copper from its sulphide ore, metal is finally obtained by the oxidation of cuprous sulphide with :
Among the oxides of nitrogen :
N_{2}O_{3}, N_{2}O_{4} and N_{2}O_{5} ; the molecule(s) having nitrogennitrogen bond is/are :
Which of the following conversions involves change in both shape and hybridisation ?
The most polar compound among the following is :
the bond dipole vector of C–F bond is not subtractive.
In Wilkinson's catalyst, the hybridization of central metal ion and its shape are respectively :
Wilkinson catalyst
[RhCl(PPh_{3})_{3}]
At 320 K, a gas A_{2} is 20 % dissociated to A(g). The standard free energy change at 320 K and 1 atm in J mol^{–1} is approximately : (R = 8.314 JK^{–1} mol^{–1} ; ln 2 = 0.693 ; ln 3 = 1.098)
ΔGº = –2.303 × 8.314 × 320 log_{10} 0.2 = 4281 J/mole
Which of the following complexes will show geometrical isomerism?
Which of the following statements is false ?
When temperature is increased, black body emit high energy radiation, from higher wavelength to lower wavelength.
Rydberg constant has unit length^{–1} (i.e. cm^{–1})
When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of paminophenol produced is :
For which of the following processes, ΔS is negative ?
N_{2} (g, 1 atm) → N_{2} (g, 5 atm)
for isothermal process
ΔS < 0
An unknown chlorohydrocarbon has 3.55 % of chlorine. If each molecule of the hydrocarbon has one chlorine atom only ; chlorine atoms present in 1 g of chlorohydrocarbon are :
(Atomic wt. of Cl = 35.5 u ; Avogadro constant = 6.023 × 10^{23} mol^{–1})
C_{x}H_{y}Cl
% Cl = 3.55
The incorrect statement is :
Due to common ion effect, sufficient S^{2– }concentration is not produced and no ppt formed of NiS.
The mass of a nonvolatile, nonelectrolyte solute (molar mass = 50 g mol^{–1}) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75 %, is :
The incorrect geometry is represented by :
Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chlroide at same temperature and pressure is : (Atomic wt. of Cl 35.5 u)
The correct match between items of ListI and ListII is :
→ Phenelzine contains hydrazine
→ Chloroxylenol contains phenol
→ Uracil is the pyrimidine base
→ Ranitidine contains furan ring
The gas phase reaction 2NO_{2}(g) → N_{2}O_{4}(g) is an exothermic reaction. The decomposition of N_{2}O_{4}, in equilibrium mixture of NO_{2}(g) and N_{2}O_{4}(g), can be increased by :
2NO_{2}(g) → N_{2}O_{4}(g) ΔH = (–)
By addition of an inert gas at constant pressure, volume increases, so reaction moving in backward direction and decomposition of N_{2}O_{4} increases.
Which one of the following is not a property of physical adsorption?
Physical adsorption is multilayer adsorption.
A group 13 element 'X' reacts with chlorine gas to produce a compound XCl_{3}. XCl_{3} is electron deficient and easily reacts with NH_{3} to form Cl_{3}X → NH_{3} adduct; however, XCl_{3 }does not dimerize. X is :
The major product of the following reaction is :
Inversion takes place at the carbon containing bromine atom.
If 50 % of a reaction occurs in 100 second and 75 % of the reaction occurs in 200 second, the order of this reaction is :
First order reaction as half life is constant.
The major product of the following reaction is :
Which of the following compounds will most readily be dehydrated to give alkene under acidic condition?
will most readily be dehydrated to give conjugated alkene.
If then dy/dx is equal to :
Let N denote the set of all natural numbers. Define two binary relations on N as R_{1} = {(x, y) ∈ N × N : 2x + y = 10} and R_{2} = {(x, y) Î N × N : x + 2y = 10}. Then
The coefficient of x^{2} in the expansion of the product (2– x^{2})· ((1 + 2x + 3x^{2})^{6} + (1 – 4x^{2})^{6}) is :
coefficient of x2 = 2 coefficient of x^{2} in ((1 + 2x + 3x^{2})^{6} + (1 – 4x^{2})^{6}) – constant term
If the area of the region bounded by the curves, y = x^{2}, y = 1/x and the lines y = 0 and x = t (t > 1) is 1 sq. unit, then t is equal to :
If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is 3/2 units, then its eccentricity is:
The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is :
There are 4 places to be filled with the given digits. The thousands place can have only 2, 3 and 4 since the number has to be greater than 2000. For the remaining 3 places, we have pick out digits such that the resultant number is divisible by 3. The divisibility criteria for 3 states that sum of digits of the number should be divisible by 3. Case 1: if we pick 2 for thousand place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: 0,1 and 3 as 2 + 1 + 0 + 3 = 6 is divisible by 3. 0,3 and 4 as 2 + 3 + 0 + 4 = 9 is divisible by 3. In both the above combination, the remaining three digits can be arranged in 3! ways. Total number=2×3!=12 Case 2: If we pick 3 for thousands place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: 0,1and 2 as 3 + 1 + 0 + 2 = 6 is divisible by 3. 0,2 and 4 as 3 + 2 + 0 + 4 = 9 is divisible by 3. In both the above combinations, the remaining three digits can be arranged in 3! ways. Total number=2×3!=12
Case 3: if we pick 4 for thousand place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: 0,2 and 3 as 4 + 2 + 0 + 3 = 9 is divisible by 3. In the above combination, the remaining three digits can be arranged in 3! ways. Total number = 3! = 6.
Total number of numbers between 2000 and 5000 divisible by 3 are 12 + 12 + 6 = 30
Two different families A and B are blessed with equal number of children. There are 3 tickests to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is 1/12, then the number of children in each family is :
Let n number of children are there in each family
Let p, q and r be real numbers (p ≠ q, r ≠ 0), such that the roots of the equation are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :
(2x + p + q) r = (x + p) (x + q)
x^{2} + (p + q – 2r) x + pq – pr – qr = 0
p + q = 2r ……..(i)
α^{2} + β^{2} = (α + β)^{2}  2αβ
= 0 – 2 [pq – pr – qr] = – 2pq + 2r (p + q) = – 2pq + (p + q)^{2} = p^{2} + q^{2}
Let 1/x_{1} , 1/x_{2} ,..., 1/x_{n} (x_{i} ≠ 0 for i = 1, 2, &., n) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then is equal to :
The differential equation representing the family of ellipses having foci either on the xaxis or on the yaxis, centre at the origin and passing through the point (0, 3) is :
Equation of ellipse
since it passes through (0, 3)
⇒b^{2}=9
∴ Eqution of ellipse becomes
⇒
......(2)
The sum of the intercepts on the coordinate axes of the plane passing trhough the point (–2, –2, 2) and containing the line joining the points (1, –1, 2) and (1, 1, 1), is :
Equation of plane passing through three given points is
⇒  (x + 2) + 3 (y + 2) + 6 (z  2) = 0
⇒ x – 3y – 6z + 8 = 0
∴ sum of intercepts
Let A = and B = A^{20}. Then the sum of the elements of the first column of B is :
Sum of the elements of first column = 231
Let A, B and C be three events, which are pairwise independent and denotes the complement of an event E. If P(A∩B∩C) = 0 and P(C) > 0, then is equal to :
If p → (~p ν ~q) is false, then the truth v alues of p and q are respectively :
If the function f defined as f(x) = is continuous at x = 0, then the ordered pair (k, f(0)) is equal to :
Clearly k = 3 and f(0) = 1
If the angle between the lines, and is cos^{–1} then p is equal to :
Angle between both lines is
The locus of the point of intersection of the lines, √2 x – y + 4 √2 k = 0 and √2 kx + ky – 4 √2 = 0 (k is any nonzero real parameter), is
A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizonatal road. If it takes 18 min. for the angle of depression of the car to change from 30° to 45° ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :
Let length of tower = h
⇒ AC' = AB = h
and AC = AB cot 30º = √3 h
⇒ CC' = (√3 –1) h
Time taken by the car form C to C' = 18 min
⇒ time taken by the car to reach the foot of the tower =
= 9 (√3 + 1) min
If an angle A of a ΔABC satisfies 5 cosA + 3 = 0, then the roots of the quadratic equaiton, 9x^{2} + 27x + 20 = 0 are :
5cosA + 3 = 0 ⇒ cosA =  3/5 clearly A ∈ (90º, 180º)
Now roots of equation 9x^{2 }+ 27x + 20 = 0 are 5/3 and 4/3
⇒ Roots secA and tanA
If a circle C, whose radius is 3, touches externally the circle, x^{2} + y^{2} + 2x – 4y – 4 = 0 at the point (2, 2), then the length of the intercept cut by this circle C, on the xaxis is equal to :
Let P be a point on the parabola, x^{2} = 4y. If the distance of P from the centre of the circle, x^{2} + y^{2} + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :
Let P (2t, t^{2})
equation normal at P to x^{2} = 4y be
it passes through (–3,0)
Point P is (–2,1)
equation of tangent to x^{2} = 4y at (–2,1)
x(–2) = 2 (y + 1)
x + y + 1 = 0
If f(x) = dt then :
The number of v alues of k for which the system of linear equations,
(k + 2)x + 10y = k
kx + (k + 3) y = k – 1 has no soution, is :
For no solution
(k + 2) (k + 3) = 10 k
k^{2} – 5k + 6 = 0 ⇒ k = 2,3
k ≠ 2 for k = 2 both lines identical
so k = 3 only
so number of values of k is 1
If (C is a constant of integration), then the ordered pair (K, A) is equal to
The least positive integer n for which = 1, is
ω^{n} = 1
least positive integer value of n is 3.
The sum of the first 20 terms of the series ..... is:
Let and a vector be such that Then equals :
The mean and the standard deviation (s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is:
Here mean =
Now, standard deviation = 0
∴ all the five terms are same i.e., 9.
Now for changed observation
= 2
Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f (x) = 2x^{3}  9x^{2} + 12x + 5 in the interval [0, 3]. Then M  m is equal to:
For maxima or minima put f'(x) = 0
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