1 Crore+ students have signed up on EduRev. Have you? 
Four point charges –Q, –q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is :
Let the side length of square be 'a' then potential at centre O is
= – Q – q + 2q + 2Q = 0 ⇒ Q + q = 0 (Given) Q
= – q
Two long parallel wires carry equal current i flowing in the same direction are at a distance 2d apart. The magnetic field B at a point lying on the perpendicular line joining the wires and at a distance x from the midpoint is –
The magnetic field due to two wires at P
Both the magnetic fields act in opposite direction.
In the circuit shown, the symbols have their usual meanings. The cell has emf E. X is initially joined to Y for a long time. Then, X is joined to Z. The maximum charge on C at any later time will be
Current in inductor = E/R
∴ its energy = 1/2 LE^{2}/R^{2}
Same energy is later stored in capacitor
A point object O is placed in front of a glass rod having spherical end of radius of curvature 30 cm. The image would be formed at
Using, μ/v  1/u = μ  1/R
Or 1.5/v  1/15 = 1.5  1/+30
∴ v =  30cm
In Young’s double slit exper iment, λ = 500n m, d = 1mm, D = 1m. Minimum distance from the central maximum for which intensity is half of the maximum intensity is
What is the voltage gain in a common emitter amplifier, where input resistance is 3 Ω and load resistance 24 Ω, b = 0.6 ?
Voltage gain, A_{v} = β R_{L}/R_{i} = 0.6 x 24/3 = 4.8
The acceleration due to gravity on the surface of the moon is 1/6 that on the surface of earth and the diameter of the moon is onefourth that of earth. The ratio of escape velocities on earth and moon will be
A particle of mass m executes simple harmonic motion with amplitude a and frequency n. The average kinetic energy during its motion from the position of equilibrium to the end is
The kinetic energy of a particle executing S.H.M. is given by
K = 1/2 ma^{2} ω^{2} sin^{2}ωt
= 1/2 mω^{2}a^{2}(1/2) (∵ < sin^{2} θ > = 1/2)
= 1/4 mω^{2}a^{2} = 1/4 ma^{2 }(2πv)^{2} (∵ ω = 2πv)
or, <K> = π^{2 }ma^{2}v^{2}
The dipole moment of the given charge distribution is
At a place, if the earth's horizontal and vertical components of magnetic fields are equal, then the angle of dip will be
tan θ = B_{y}/B_{H }= 1 ∴ θ = 45°
The third line of Balmer series of an ion equivalnet to hydrogen atom has wavelength of 108.5 nm.The ground state energy of an electron of this ion will be
For third line of Balmer series n_{1 }= 2, n_{2} = 5
On putting values Z = 2 From
E
The binding energy per nucleon of ^{10}X is 9 MeV and that of ^{11}X is 7.5 MeV where X represents an element. The minimum energy required to remove a neutron from ^{11}X is
If C, the velocity of light, g the acceleration due to gravity and P the atmospheric pressure be the fundamental quantities in MKS system, then the dimensions of length will be same as that of
Figure shows a capillary rise H. If the air is blown through the horizontal tube in the direction as shown then rise in capillary tube will be
Due to increase in velocity, pressure will be low above the surface of water.
A boy running on a horizontal road at 8 km/h finds the rain falling vertically. He increases his speed to 12 km/h and finds that the drops makes 30° with the vertical. The speed of rain with respect to the road is
A hunter aims his gun and fires a bullet directly at a monkey on a tree. At the instant the bullet leaves the barrel of the gun, the monkey drops.
Pick the correct statement regarding the situation.
t = OC/u cos θ = x/u cos θ
AC = x tan q
BC = distance travelled by bullet in time t, vertically.
y = u sin θ t – 1/2 gt^{2}
AB = x tan θ – (u sin θt – 1/2 gt^{2})
= x tan θ – (usinθ × x/u cos θ – 1/2 gt^{2})
(∴ bullet will always hit the monkey)
A particle of mass m1 moving with velocity v collides with a mass m2 at rest, then they get embedded. Just after collision, velocity of the system
The ratio of the specific heats of a gas is C_{p}/C_{v} = 1.66, then the gas may be
Let ‘n’ be the degree of freedom
⇒ n = 3 ⇒ gas must be monoatomic.
Two oscillators ar e started simultaneously in same phase. After 50 oscillations of one, they get out of phase by p, that is half oscillation. The percentage difference of frequencies of the two oscillators is nearest to
Phase change π in 50 oscillations.
Phase change 2π in 100 oscillations.
So frequency different ~ 1 in 100.
A juggler keeps on moving four balls in the air throwing the balls after intervals. When one ball leaves his hand (speed = 20 ms^{–1}) the position of other balls (height in m) will be (Take g = 10 ms^{–2})
Time taken by same ball to return to the hands of juggler = 2u/g = 2 x 20/10 = 4 S. So he is throwing the balls after each 1 s. Let at some instant he is throwing ball number 4. Before 1 s of it he throws ball. So height of ball 3:
h_{3} = 20 × 1 – 1/2 10(1)^{2} = 15
Before 2s, he throws ball 2. So height of ball 2:
h_{2} = 20 × 2 – 1/2 10(2)^{2} = 20 m
Before 3 s, he throws ball 1. So height of ball 1:
h_{1} = 20 × 3 – 1/2 10(3)^{2} = 15 m
If a stone of mass 0.05 kg is thrown out a window of a train moving at a constant speed of 100 km/ h then magnitude of the net force acting on the stone is
After the stone is thrown out of the moving train, the only force acting on it is the force of gravity i.e. its weight.
∴ F = mg = 0.05 × 10 = 0.5 N.
A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is
Impulse experienced by the body = MV – (–MV) = 2MV.
A hoop rolls down an inclined plane. The fraction of its total kinetic energy that is associated with rotational motion is
Infinite number of masses, each 1 kg are placed along the xaxis at x = ± 1m, ± 2m, ± 4m, ± 8m, ± 16m ..... the magnitude of the resultant gravitational potential in terms of gravitational constant G at the orgin (x = 0) is
Water of volume 2 litre in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at rate of 160 J/s.
In how much time temperature will rise from 27°C to 77°C? [Given specific heat of water is 4.2 kJ/kg]
Heat gained by the water = (Heat supplied by the coil) – (Heat dissipated to environment)
⇒ mc Δθ = P_{Coil} t – P_{Loss} t
⇒ 2 × 4.2 × 10^{3} × (77 – 27) = 1000 t – 160 t
⇒ t = 4.2x10^{5}/840 = 500 s = 8 min 20 s
In the following PV diagram of an ideal gas, two adiabates cut two isotherms at T_{1} = 300K and T_{2} = 200K. The value of V_{A} = 2 unit, V_{B} = 8 unit, V_{C} = 16 unit. Find the value of V_{D}.
The mass of H_{2} molecule is 3.32 × 10^{–24} g. If 10^{23} hydrogen molecules per second strike 2 cm^{2} of wall at an angle of 45^{o} with the normal, while moving with a speed of 10^{5} cm/s, the pressure exterted on the wall is nearly.
The wavelength of two waves are 50 and 51 cm respectively. If the temperature of the room is 20°C then what will be the number of beats produced per second by these waves, when the speed of sound at 0°C is 332 m/s?
λ_{1}= 50 cm. λ_{2}= 51 cm.
⇒ v_{2} = 319.23.
v_{1 }= v_{2}/λ_{1 }= 319.23/0.50 = 640 Hz.
v_{2} = v_{2}/λ_{2} = 319.23/51x10^{2} = 625.94 = 62 HZ
No. of beats = v_{2}  v_{1 }= 14 Hz
The figure shows the interference pattern obtained in a doubleslit experiment using light of wavelength 600nm. 1, 2, 3, 4 and 5 are marked on five fringes.The third order bright fringe is
Electric potential at any point is V = –5x + 3y + √15z , then the magnitude of the electric field is
E_{x} =  rv/rx = r/rx(5x + 5y + √15z) = 5
E_{y} = rv/ry = 3, E_{z} = √15
Now E =
Seven resistan ces, each of value 20 Ω , are connected to a 2 V battery as shown in the figure.
The ammeter reading will be
The variation of magnetic susceptibility (c) with temperature for a diamagnetic substance is best represented by
A copper rod of length l rotates about its end with angular velocity ω in uniform magnetic field B. The emf developed between the ends of the rod if the field is normal to the plane of rotation is
A 10V battery with internal resistance 1W and a 15V battery with internal resistance 0.6 W are connected in parallel to a voltmeter (see figure).
The reading in the voltmeter will be close to:
As the two cells oppose each other hence, the effective emf in closed circuit is 15 – 10 = 5 V and net resistance is 1 + 0.6 = 1.6 Ω (because in the closed circuit the internal resistance of two cells are in series.
Current in the circuit, I = effective emf/total resistance = 5/1.6A
The potential difference across voltmeter will be same as the terminal voltage of either cell.
Since the current is drawn from the cell of 15 V
∴ V_{1} = E_{1} – Ir_{1} = 15 – 5/1.6 x 0.6 = 13.1 V
10 for ks are arranged in in creasing or der of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies (in Hz) are
Using n_{Last} = n_{First} + (N – 1)x
where N = Number of tuning forks in series
x = beat frequency between two successive forks
⇒ 2n = n + (10 – 1) × 4 ⇒ n = 36 Hz
A charged particle enters in a uniform magnetic field with a certain velocity. The power delivered to the particle by the magnetic field depends on
Power = work done/time
As no work is done by magnetic force on the charged particle because magnetic force is perpendicular to velocity, hence power delivered is zero.
A r esistor an d an inductor a re connected to an ac supply of 120 V and 50 Hz. The current in the circuit is 3 A. If the power consumed in the circuit is 108 W, then the resistance in the circuit is
In an ac cir cuit, a pure indcutor does not consume any power. Therefore, power is consumed by the resistor only.
∴ P = I^{2}_{v}R
or 108 = (3)^{2} R or R = 12 Ω
In an electron gun, the poten tial differ en ce between the filament and plate is 3000 V. What will be the velocity of electron emitting from the gun?
V = 3000 volt.
1/2 mv^{2} = eV ⇒ v =
= 32.6 × 10^{6} = 3.26 × 10^{7} m/s.
A radioactive substance with decay constant of 0.5s^{–1} is being produced at a constant rate of 50 nuclei per second. If there are no nuclei present initially, the time (in second) after which 25 nuclei will be present is
N = (100 (1 – e^{–t/2})) = 25
t = 2 ln (4/3)
The 25 mL of a 0.15 M solution of lead n itrate, Pb(NO_{3})_{2} reacts with all of the aluminium sulphate, Al_{2}(SO_{4})_{3}, present in 20 mL of a solution. What is the molar concentration of the Al_{2}(SO_{4})_{3}?
3Pb(NO_{3})_{2} (aq) + Al_{2}(SO_{4})_{3} (aq) → 3PbSO_{4}(s) + 2Al(NO_{3})_{3} (aq)
Molar mass of Pb(NO_{3})_{2} = 25 × 0.15 = 3.75 m. moles Molar mass of Al_{2} (SO_{4})_{3} = 1/3 x 3.75 M x 20
M = 0.0625 = 6.25 × 10^{–2} M
100 mL O_{2} and H_{2} kept at same temperature and pressure. What is true about their number of molecules
This is Avogadro’s hypothesis.
According to this, equal volume of all gases contain equal no. of molecules under similar condition of temperature and pressure.
If the Planck’s constant h = 6.6×10^{–34} Js, the de Broglie wavelength of a particle having momentum of 3.3 × 10^{–24} kg ms ^{–1} will be
Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy?
The smaller the atomic size, larger is the value of ionisation potential. Further the atoms having half filled or fully filled orbitals are comparatively more stable, hence more energy is required to remove the electron from such atoms.
Wh ich of t h e followin g is th e cor r ect a n d increasing order of lone pair of electrons on the central atom?
The number of lone pairs of electrons on central atom in various given species are
Thus the correct increasing order is
According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding O^{+}_{2}
O_{2} : σ1s^{2}, σ*1s^{2 }, σ2s^{2}, σ*2s^{2}, σ2p^{2}_{z},
Bond order 10  6/2 = 2
(two unpaired electrons in antibonding molecular orbital)
Bond order 10 5/2 = 2.5
(One unpaired electron in antibonding molecular orbital).
Hence O_{2} as well as O^{+}_{2} both are paramagnetic, and bond order of O^{+}_{2} is greater than that of O_{2}.
If V is the volume of one molecule of gas under given conditions, the van der Waal’s constant b is
van der Waals’s constant b = 4 times the actual volume of 1 mole molecules = 4VN_{0}
For vapor ization of water at 1 atmosph er ic pressure, the values of DH and DS are 40.63 kJmol^{–1} and 108.8 JK^{–1} mol^{–1}, respectively. The temperature when Gibbs energy change (DG) for this transformation will be zero, is:
ΔH = 40630 J mol^{ –1}
ΔS = 108.8 JK^{–1} mol^{–1}
ΔG = ΔH  TΔS When ΔG = 0, ΔH  TΔS = 0
T = ΔH/ΔS
For the react ion taking place at certain temperature
NH_{2} COONH_{4} (s) ⇌ 2 NH_{3} (g) + CO_{2} (g), if equilibrium pressure is 3X bar then ΔrG° would be
ΔG° = – RT ln K_{P}; K_{P} = (2X)^{2} X = 4X^{3}
ΔG° = – RT ln (4X^{3})
ΔG° = – RT ln 4 – 3RT ln X
The pH of 0.1 M solution of the following salts increases in the order :
(i)
∴ [H^{+}] = 0.1 M
pH = – log [H^{+}] = – log 0.1 = 1
(ii) NaCl is a salt of strong acid and strong base so it is not hydrolysed and hence its pH is 7.
(iii) NH_{4}Cl + H_{2}O ⇌ NH_{4}OH + HCl
∴ The solution is acidic and its pH is less than that of 0.1 M HCl.
(iv) NaCN + H_{2}O ⇌ NaOH + HCN
∴ The solution is basic and its pH is more than that of 0.1 M HCl.
∴ Correct order for increase in pH is
HCl < NH4Cl < NaCl < NaCN.
When N2O5 is heated at certain temperature, it dissociates as N_{2}O_{5} (g) ⇌ N_{2}O_{3} (g) + O_{2} (g); K_{c} = 2.5. At the same time N_{2}O_{3} also decomposes as:
N_{2}O_{3} (g) ⇌ N_{2}O (g) + O_{2} (g).
If initially 4.0 moles of N_{2}O_{5} are taken in 1.0 litre flask and allowed to dissociate. Concentration of O_{2} at equilibrium is 2.5 M. Equilibrium concentration of N_{2}O_{5} is :
Q [O_{2}] = x + y = 2.5
for N_{2}O_{5}, Kc = [N_{2}O_{5}] [O_{2}]/ [N_{2}O_{5}]
and 2.5 = (x +y (x y)/4x
∴ x = 2.166
[N_{2}O_{5}] = 4 – x = 1.846
Consider the reactions
(A) H_{2}O_{2} + 2HI → I_{2} + 2H_{2}O
(B) HOCl + H_{2}O_{2} → H_{3}O^{+} + Cl^{–} + O_{2}
Which of the following statements is correct about H_{2}O_{2} with reference to these reactions?
Hydrogen peroxide is ______.
Following are colours shown by some alkaline earth metals in flame test. Which of the following are not correctly matched?
Calcium gives brick red colour and barium gives apple green colour in flame test.
Ber yllium shows diagonal relationship with aluminium. Which of the following similarity is incorrect ?
The Be(OH)_{2} and Al(OH)_{3} are amphoteric in nature.
An elemen t X occurs in shor t period having configuration ns2 np1. The formula and nature of its oxide is
ns^{2} np^{1} is the electronic configuration of III A period. X^{2}O^{3}(Al^{2}O^{3}) is an amphoteric oxide.
Which of the following is strongest nucleophile
The strength of nucleophile depends upon the nature of alkyl group R on which nucleophile has to attack and also on the nature of solvent.
The order of strength of nucleophiles follows the order:
CN^{–} > I^{–} > C_{6}H_{5}O^{–} > OH^{–} > Br^{–} > Cl^{–}
IUPAC name – 3, 3Dimethyl 1cyclohexanol
Which of the following will have a mesoisomer also?
The compound has two similar assymmetric Catoms. It has plane of symmetry and exists in meso form.
Meso  2, 3 dichlorobutane
In a set of reactions, ethylbenzene yielded a product D.
Identify the incorrect statement from thefollowing:
The ozone layer, existing between 20 to 35 km above the earth’s surface, shield the earth from the harmful U. V. radiations from the sun.
Depletion of ozone is caused by oxides of nitrogen
N_{2}O + hn → NO + N
NO + O_{3} → NO_{2} + O_{2}
O_{3} + hn → O_{2} + O
NO_{2} + O → NO + O_{2}
2O_{3} + hn → O_{2} (Net reaction)
The presence of oxides of nitrogen increase the decomposition of O_{3}.
Each edge of a cubic unit cell is 400 pm long. If atomic mass of the element is 120 and its density is 6.25 g/cm^{3}, the crystal lattice is : (use N_{A} = 6 × 10^{23})
Chloroform, CHCl_{3}, boils at 61.7 °C. If the K_{b} for chloroform is 3.63°C/molal, what is the boiling point of a solution of 15.0 kg of CHCl_{3} and 0.616 kg of acenaphthalene, C_{12}H_{10}?
ΔT_{b} = K_{b}.m ⇒
T_{b} = 61.7 + 0.968
= 62.67° C
pH of a 0.1 M monobasic acid is found to be 2.Hence, its osmotic pressure at a given temperature TK is
pH = 2
[H^{+}] = 0.01 M = Cx = 0.1x
x = 0.1
i = 1 + x = 1.1
π = i n/V RT = iMRT = 1.1 x 0.1RT = 0.11RT
On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl_{2}, all copper of the solution was deposited at cathode. The strength of CuCl_{2} solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 Cmol^{–1})
By Faraday's Ist Law, W/e = q/96500
(where q = it = charge of ion)
we know that no of equivalent
(where i = 1 A, t = 16×60+5 = 965 sec.) Since, we know that
Normality = no. of equivalent/Volume (in litre)
A 100.0 mL dillute solution of Ag^{+ }is electrolysed for 15.0 minutes with a current of 1.25 mA and the silver is removed completely. What was the initial [Ag^{+}]?
No. of moles of
The accompanying figure depicts a change in concentration of species A and B for the reaction A → B, as a function of time. The point of inter section of the two curves represents
The intersection point indicates the half life of the reactant A when it is converted to B.
The rate constant of a reaction is 1.5 × 10^{–3} at 25°C and 2.1 × 10^{–2} at 60°C. The activation energy is
T_{1} = 273 + 25 = 298 K
T_{2} = 273 + 60 = 333 K
Freundlich equation for adsorption of gases (in amount of x g) on a solid (in amount of m g) at constant temperature can be expressed as
According to Freundlich equation,
x/m ∝ p^{1/n} or x/m = Kp^{1/n}
or log x/m = log Kp^{1/n} or log x/m = log K + 1/n log K + 1/n log p
Which of the following feature of catalysts is described in reactions given below?
(i) CO (g) + 2H_{2} (g) CH_{3}OH (g)
(ii) CO (g) + H_{2} (g ) HCHO (g)
(iii) CO (g) + 3H_{2} (g) CH_{4} (g)+ H_{2}O (g)
Given reactions shows that the selectivity of different catalysts for some reactants is different.
Which of the following is not a member of chalcogens?
chalcogens are defined as oreforming elements.
Catenation tendency is higher in
phosphorus when compared with other elements of same group.
Which of the following element do not form complex with EDTA?
Be is the only group 2 element that does not form a stable complex with [EDTA]^{4–}. Mg^{2+} and Ca^{2+} have the greatest tendency to form complexes with [EDTA]^{4–}.
Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour ?
CN^{–} is a strong field ligand and it causes pairing of electrons; as a result number of unpaired electrons in Co^{3+} becomes zero and hence it has lowest value of paramagnetic behaviour.
When an aqueous solution of copper (II) sulphate is saturated with ammonia, the blue compound crystallises on evaporation. The formula of this blue compound is:
Following compounds are given:
(1) CH_{3}CH_{2}OH
(2) CH_{3}COCH_{3}
(3)
(4) CH_{3}OH
Which of the above compound(s), on being warmed with iodine solution and NaOH, will give iodoform?
Among the given compounds only CH_{3}OH does not give iodoform reaction.
Arrange the following alcohols in increasing order of their reactivity towards the reaction with HCl.
(CH_{3})_{2}CHOH (1), (CH_{3})_{3}COH (2), (C_{6}H_{5})_{3}COH (3)
Alkylhalide formation in the reaction of alcohol with HCl undergoes SN_{1} reaction in which formation of the carbocation as intermediate occurs. Stability of carbocation is greatest for (C_{6}H_{5})_{3}C^{+} due to resonance effect, and stability of tertiary carbocation is greater than the secondary carbocation hence the option (a) shows the correct order.
Thirty percent of the bases in a sample of DNA extracted from eukaryotic cells is adenine. What percentage of cytosine is present in this DNA?
If 30 percent of DNA is adenine, then by Chargaff’s rule 30 percent will be thymine. The remaining 40 percent of the DNA is cytosine and guanine. Since the ratio of cytosine to guanine must be equal, then each accounts for 20 percent of the bases.
The blue colour of snail is due to presence of
Most snail blood is blueish in color. This is because their blood cells use haemocyanin, which gets its blue color from the copper that is part of its structure.
Diamines are those compounds which contain two amino groups.
Choose the word which is most similar in meaning to the word 'Optimistic'.
Optimistic means hopeful and confident about the future.
Choose the word which is most opposite in meaning to the word 'Drowsy'.
Drowsy means sleepy and lethargic.
Therefore, option (d) is the correct antonym of it.
Rest of the options are its synonyms.
Direction: Which of the following phrases (I), (II), and (III) given below each sentence should replace the phrase printed in bold letters to make the sentence grammatically correct? Choose the best option among the five given alternatives that reflect the correct use of phrase in the context of the grammatically correct sentence. If the sentence is correct as it is, mark (d) i.e., "No correction required" as the answer.
He is really feeling under the weather today; he has a terrible cold.
(I) feeling like the weather
(II) feeling over the weather
(III) feeling in the weather
The phrase used in the sentence is grammatically correct hence, it doesn't require any correction. The meaning of the phrase' under the weather' is to feel ill.
Direction: Which of the following phrases (I), (II), and (III) given below each sentence should replace the phrase printed in bold letters to make the sentence grammatically correct? Choose the best option among the five given alternatives that reflect the correct use of phrase in the context of the grammatically correct sentence. If the sentence is correct as it is, mark (d) i.e., "No correction required" as the answer.
By working parttime and looking after his old mother, he managed to get the best for both worlds.
(I) the best at both worlds
(II) the best of both worlds
(III) the best on both worlds
The correct phrase to be used is 'get the best of both worlds' which means a winwin situation.
Direction: Which of the following phrases (I), (II), and (III) given below each sentence should replace the phrase printed in bold letters to make the sentence grammatically correct? Choose the best option among the five given alternatives that reflect the correct use of phrase in the context of the grammatically correct sentence. If the sentence is correct as it is, mark (d) i.e., "No correction required" as the answer.
Hey, Nanny, speak about the devil and you are here.
(I) speak at the devil
(II) speak on the devil
(III) speak of the devil
The correct phrase to be used is 'speak of the devil'. This phrase is said when a person appears just after being mentioned.
DIRECTION: Read the following passage carefully and answer the questions given below it.
The likelihood of at least 600,000 deaths being caused annually in India by fine particulate matter pollution in the air is cause for worry, even if the data released by the World Health Organisation are only a modelled estimate. The conclusion that so many deaths could be attributed to particulate matter 2.5 micrometres or less in size is, of course, caveated, since comprehensive measurement of PM2.5 is not yet being done and the linkages between pollution, disease and deaths need further study. What is not in doubt is that residents in many urban areas are forced to breathe unhealthy levels of particulates, and the smallest of these  PM10 and less  can penetrate and get lodged deep in the lungs.
The WHO Global Burden of Disease study has been working to estimate pollutionlinked health impacts, such as stroke and ischaemic heart disease, acute lower respiratory infection and chronic obstructive pulmonary disease. Data on fine particulates in India show that in several locations the pollutants come from burning of biomass, such as coal, fuel wood, farm litter and cow dung cakes. In highly builtup areas, construction debris, road dust and vehicular exhaust add to the problem. The Prime Minister launched an Air Quality Index last year aimed at improving pollution control. The new data, which the WHO says provide the best evidence available on the terrible toll taken by particulates, should lead to intensified action. A neglected aspect of urban air pollution control is the virtual discarding of the Construction and Demolition Waste Management Rules, notified to sustainably manage debris that is dumped in the cities, creating severe particulate pollution.
The Environment Ministry has highlighted the role that debris can play as a resource. Municipal and government contracts are, under the rules, required to utilise up to 20 per cent materials made from construction and demolition waste, and local authorities must place containers to hold debris. This must be implemented without delay. Providing cleaner fuels and scientifically designed cookstoves to those who have no option but to burn biomass, would have a big impact on reducing particulate matter in the northern and eastern States, which are the worsthit during winter, when biomass is also used for heating.
Greening the cities could be made a mission, involving civil society, with a focus on landscaping open spaces and paving all public areas to reduce dust. These measures can result in lower PM10 and PM2.5 levels.
Comprehensive measurement of these particulates is currently absent in many cities, a lacuna that needs to be addressed.
Q. According to the WHO Global Burden of Disease study which of the following is/are pollution linked health impacts?
(I) Infection of the lower respiratory system
(II) Chronic obstructive pulmonary disease
(III) Stroke and ischaemic heart disease
DIRECTION: Read the following passage carefully and answer the questions given below it.
The likelihood of at least 600,000 deaths being caused annually in India by fine particulate matter pollution in the air is cause for worry, even if the data released by the World Health Organisation are only a modelled estimate. The conclusion that so many deaths could be attributed to particulate matter 2.5 micrometres or less in size is, of course, caveated, since comprehensive measurement of PM2.5 is not yet being done and the linkages between pollution, disease and deaths need further study. What is not in doubt is that residents in many urban areas are forced to breathe unhealthy levels of particulates, and the smallest of these  PM10 and less  can penetrate and get lodged deep in the lungs.
The WHO Global Burden of Disease study has been working to estimate pollutionlinked health impacts, such as stroke and ischaemic heart disease, acute lower respiratory infection and chronic obstructive pulmonary disease. Data on fine particulates in India show that in several locations the pollutants come from burning of biomass, such as coal, fuel wood, farm litter and cow dung cakes. In highly builtup areas, construction debris, road dust and vehicular exhaust add to the problem. The Prime Minister launched an Air Quality Index last year aimed at improving pollution control. The new data, which the WHO says provide the best evidence available on the terrible toll taken by particulates, should lead to intensified action. A neglected aspect of urban air pollution control is the virtual discarding of the Construction and Demolition Waste Management Rules, notified to sustainably manage debris that is dumped in the cities, creating severe particulate pollution.
The Environment Ministry has highlighted the role that debris can play as a resource. Municipal and government contracts are, under the rules, required to utilise up to 20 per cent materials made from construction and demolition waste, and local authorities must place containers to hold debris. This must be implemented without delay. Providing cleaner fuels and scientifically designed cookstoves to those who have no option but to burn biomass, would have a big impact on reducing particulate matter in the northern and eastern States, which are the worsthit during winter, when biomass is also used for heating.
Greening the cities could be made a mission, involving civil society, with a focus on landscaping open spaces and paving all public areas to reduce dust. These measures can result in lower PM10 and PM2.5 levels.
Comprehensive measurement of these particulates is currently absent in many cities, a lacuna that needs to be addressed.
Q. The conclusion regarding the deaths attributed to particulate matter 2.5 micrometers is considered to be caveated because
DIRECTION: Read the following passage carefully and answer the questions given below it.
The likelihood of at least 600,000 deaths being caused annually in India by fine particulate matter pollution in the air is cause for worry, even if the data released by the World Health Organisation are only a modelled estimate. The conclusion that so many deaths could be attributed to particulate matter 2.5 micrometres or less in size is, of course, caveated, since comprehensive measurement of PM2.5 is not yet being done and the linkages between pollution, disease and deaths need further study. What is not in doubt is that residents in many urban areas are forced to breathe unhealthy levels of particulates, and the smallest of these  PM10 and less  can penetrate and get lodged deep in the lungs.
The WHO Global Burden of Disease study has been working to estimate pollutionlinked health impacts, such as stroke and ischaemic heart disease, acute lower respiratory infection and chronic obstructive pulmonary disease. Data on fine particulates in India show that in several locations the pollutants come from burning of biomass, such as coal, fuel wood, farm litter and cow dung cakes. In highly builtup areas, construction debris, road dust and vehicular exhaust add to the problem. The Prime Minister launched an Air Quality Index last year aimed at improving pollution control. The new data, which the WHO says provide the best evidence available on the terrible toll taken by particulates, should lead to intensified action. A neglected aspect of urban air pollution control is the virtual discarding of the Construction and Demolition Waste Management Rules, notified to sustainably manage debris that is dumped in the cities, creating severe particulate pollution.
The Environment Ministry has highlighted the role that debris can play as a resource. Municipal and government contracts are, under the rules, required to utilise up to 20 per cent materials made from construction and demolition waste, and local authorities must place containers to hold debris. This must be implemented without delay. Providing cleaner fuels and scientifically designed cookstoves to those who have no option but to burn biomass, would have a big impact on reducing particulate matter in the northern and eastern States, which are the worsthit during winter, when biomass is also used for heating.
Greening the cities could be made a mission, involving civil society, with a focus on landscaping open spaces and paving all public areas to reduce dust. These measures can result in lower PM10 and PM2.5 levels.
Comprehensive measurement of these particulates is currently absent in many cities, a lacuna that needs to be addressed.
Q. Which of the following is/are not true in the context of the passage?
DIRECTION: Read the following passage carefully and answer the questions given below it.
The likelihood of at least 600,000 deaths being caused annually in India by fine particulate matter pollution in the air is cause for worry, even if the data released by the World Health Organisation are only a modelled estimate. The conclusion that so many deaths could be attributed to particulate matter 2.5 micrometres or less in size is, of course, caveated, since comprehensive measurement of PM2.5 is not yet being done and the linkages between pollution, disease and deaths need further study. What is not in doubt is that residents in many urban areas are forced to breathe unhealthy levels of particulates, and the smallest of these  PM10 and less  can penetrate and get lodged deep in the lungs.
The WHO Global Burden of Disease study has been working to estimate pollutionlinked health impacts, such as stroke and ischaemic heart disease, acute lower respiratory infection and chronic obstructive pulmonary disease. Data on fine particulates in India show that in several locations the pollutants come from burning of biomass, such as coal, fuel wood, farm litter and cow dung cakes. In highly builtup areas, construction debris, road dust and vehicular exhaust add to the problem. The Prime Minister launched an Air Quality Index last year aimed at improving pollution control. The new data, which the WHO says provide the best evidence available on the terrible toll taken by particulates, should lead to intensified action. A neglected aspect of urban air pollution control is the virtual discarding of the Construction and Demolition Waste Management Rules, notified to sustainably manage debris that is dumped in the cities, creating severe particulate pollution.
The Environment Ministry has highlighted the role that debris can play as a resource. Municipal and government contracts are, under the rules, required to utilise up to 20 per cent materials made from construction and demolition waste, and local authorities must place containers to hold debris. This must be implemented without delay. Providing cleaner fuels and scientifically designed cookstoves to those who have no option but to burn biomass, would have a big impact on reducing particulate matter in the northern and eastern States, which are the worsthit during winter, when biomass is also used for heating.
Greening the cities could be made a mission, involving civil society, with a focus on landscaping open spaces and paving all public areas to reduce dust. These measures can result in lower PM10 and PM2.5 levels.
Comprehensive measurement of these particulates is currently absent in many cities, a lacuna that needs to be addressed.
Q. As per the given passage, which of the following is/are the measures for lowering particulate matter in the atmosphere?
(I) Making cleaner fuels available
(II) Landscaping open areas
(III) Providing cooking stoves designed scientifically
If sentence (B) "The Finance Ministry's warning to potential investors in bitcoin and other cryptocurrencies has come at a time when a new, seemingly attractive investment area has opened up that few have enough information about." is the first sentence, what is the order of other sentences after rearrangement?
(A) One of the main reasons for this volatility is speculation and the entry into the market of a large number of people lured by the prospect of quick and easy profits.
(B) The Finance Ministry's warning to potential investors in bitcoin and other cryptocurrencies has come at a time when a new, seemingly attractive investment area has opened up that few have enough information about.
(C) A number of investors, daunted by the high price of bitcoin, have put their money into less wellestablished and often spurious cryptocurrencies, only to lose it all.
(D) Investment in bitcoin and other cryptocurrencies increased tremendously in India over the past year, but most new users know close to nothing of the technology, or how to verify the genuineness of a particular cryptocurrency.
(E) The price of bitcoin, the most popular of all cryptocurrencies, not only shot up by well over 1000% over the course of the last year but also fluctuated wildly.
(F) The government's caution comes on top of three warnings issued by the Reserve Bank of India since 2013.
The first sentence talks about the fact that only few investors have idea about bitcoins and other cryptocurrencies (which seems an attractive investment area), so, the finance ministry has warned the potential investors about it. Sentence E will follow the first sentence because it says that 'bitcoin not only shot up well over by 1000%......' which justifies 'attractive investment area' and forms a link. Now, we are left with only option (b) and (d) to choose from.
When we consider the sentence F, we can see that this line seems to be a part somewhere in the middle of the paragraph, also, the first line starts with a warning, therefore, it must justify the consequences of the investment in bitcoins and other cryptocurrencies which is justified by sentence C. Hence, option (b) is the correct choice.
If sentence (C) "Clinical trials involving human subjects have long been a flashpoint between bioethicists and clinical research organisations (CROs) in India." is the first sentence, what is the order of other sentences after rearrangement?
(A) Such overvolunteering occurs more frequently in bioequivalence studies, which test the metabolism of generics in healthy subjects.
(B) Landmark amendments to the Drugs and Cosmetics Act in 2013 led to better protection of vulnerable groups such as illiterate people, but more regulation is needed to ensure truly ethical research.
(C) Clinical trials involving human subjects have long been a flashpoint between bioethicists and clinical research organisations (CROs) in India.
(D) The big problem plaguing clinical research is an overrepresentation of lowincome groups among trial subjects.
(E) While CROs have argued that more rules will stifle the industry, the truth is that ethical science is often better science.
(F) Sometimes CROs recruit them selectively, exploiting financial need and medical ignorance; at other times people overvolunteer for the money.
After reading all the sentences carefully, we see that sentence A and F should go one after another as both talk about 'overvolunteer'.
Moreover, sentence A will follow sentence F because of the presence of the word 'such' which signifies that the subject of the sentence has already been discussed in the previous sentence.
So, we have option (c), (d) and (e) to choose from. Considering sentence D which talks about 'a big problem', we find that it can't be the second sentence as no problem of any sort has been dealt in the first sentence, so, option (c) and (e) gets eliminated. Hence, by elimination method, we can conclude that option (d) is the correct choice.
DIRECTION: Read the sentence to find out whether there is any grammatical error or idiomatic error in it. The error, if any, will be in one part of the sentence.
The letter of that part is the answer. If there is no error, the answer is (d). (Ignore errors of punctuation, if any.)
Despite being (a)/ a good teacher, (b)/ he has no influence on his pupil. (c)/ No error (d)
Replace the preposition 'on' with 'over' to make the sentence grammatically correct.
DIRECTION: Read the sentence to find out whether there is any grammatical error or idiomatic error in it. The error, if any, will be in one part of the sentence.
The letter of that part is the answer. If there is no error, the answer is (d). (Ignore errors of punctuation, if any.)
Yesterday, when we were returning from the party, (a)/ our car met with an accident, (b)/ but we were fortunate to reach our home safely. (c)/ No error (d)
Replace the adverb 'safely' with the adjective 'safe' to make the sentence.
Grammatically correct.
In a code language, if REGAINS is coded as QDFZHMR, then the word PERIODS will be coded as 
According to question,
If 5 # 6 = 121 and 10 # 8 = 324, then find the value of 23 # 14 = ?
According to question, (5 + 6)2 = 121
(10 + 8)2 = 324
(23 + 14)2 = 1369
Which of the following cube in the answer figure cannot be made based on the unfolded cube in the question figure?
Which one of the following diagram represents the correct relationship among Professor,
Male and Female.
Select the related letter/word/ number from the given alternatives.
Distance : Odometer :: ? : Barometer
Distance is measured by odometer. Similarly, Pressure is measured by Barometer.
Find the odd word/letters/ number pair/number from the given alternatives.
According to question, 24 = 1 × 6 ×1 × 4
270 = 5 × 6 × 9
120 = 4 × 3 × 2 × 5
162 ¹ 6 × 9 × 3 × 0 = 0
Choose the correct alternatives from the given ones that will complete the series.
L_N O_ _MLLM_ OO_ML
According to question, L M N O /O N M L/ L M N O/ O N M L
Choose the correct alternatives from the given ones that will complete the series.
22, 26, 53, 69, 194, ?
The pattern is :
Select the missing number from the given responses.
The pattern is : (1 × 2 × 3 × 5) + (1 + 2 + 3 + 5) = 41
(3 × 4 × 2 × 6) + (3 + 4 + 2 + 6) = 159
(9 × 8 × 3 × 4) + (9 + 8 + 3 + 4) = 888
Identify the figure that will complete the pattern.
The domain of the function f (x) = , where [x] denotes the greatest integer less than or equal to x, is
f(x) is defined if x^{2}  [x^{2}] ≥ 0 ⇒ x^{2} ≥ [x]^{2}, which is true for all positive real x and all negative integers x.
If m sin θ = n sin(θ + 2α) then tan(θ + α) is
= tan(θ + α) cot α
Number of solutions of equation sin 99 = sin θ in the interval [0,2π] is
sin 9θ = sin θ ⇒ 9θ = nπ + (1)^{n} θ
If n = 2m then 9θ = 2mπ + θ ⇒ θ = mπ/4
If n = 2m +1 then 9θ = (2m +1)π  θ
⇒ θ = (2m +1) π/10
The values belonging to [0, π] are
A pole stands vertically inside a triangular park ABC. If the angle of elevation of the top of the pole from each corner of the park is same, then the foot of the pole is at the
The foot of the pole is at the centroid.
Because centroid is the point of intersection of medians AD, BE and CF, which are the lines joining a vertex with the mid point of opposite side.
Let A, B, and C are the angles of a plain triangle and tan A/2 = 1/3, tan B/2 = 2/3 Then tan C/2 is equal to
A + B + C = π
∴ tan C/2 = 7/9.
If the amplitude of z – 2 – 3i is p/4, then the locus of z = x + iy is
z  23i = x+ iy  23i = (x  2)+i(y 3)
⇒ x – y + 1 = 0
The roots of the equation x^{4}  2x^{3} + x = 380 are:
Given equation is x^{4} – 2x^{3} + x – 380 = 0
⇒ (x^{2 }x  20)(x^{2} x +19)= 0
⇒ (x  5)(x + 4)(x^{2} x + 19)= 0 Hence, the required roots of the equation are
Roots of the equation x^{2} + bx  c = 0(b,c > 0) are
Since b, c > 0 Therefore a + b = b < 0 and ab = c < 0 Since product of the roots is –ve therefore roots must be of opposite sign.
In how many ways can 12 gentlemen sit around a round table so that three specified gentlemen are always together?
It is obvious by fundamental property of circular permutations.
The number of ways in which first, second and third prizes can be given to 5 competitors is
First prize can be given in 5 ways. Then second prize can be given in 4 ways and the third prize in 3 ways (Since a competitior cannot get two prizes) and hence the no. of ways. = 5 × 4 × 3 = 60 ways
The coefficient of x^{3} in the expansion of is:
Given, and the (r + 1)^{th} term in the expansion of (x + a)^{n} is T_{(r + 1)} = ^{n}C_{r}(x)^{n – r}
α^{r}
∴ (r + 1)^{th} term in expansion of
= 7C^{r}(x)^{7 – 2r} (– 1)^{r}
Since x^{3} occurs in T_{r + 1}
∴ 7 – 2r = 3 ⇒ r = 2
thus the coefficient of x^{3} = ^{7}C_{2} ( – 1)^{2}
= 7 x 6/2 x 1 = 21.
Q a, b, c are in G.P.
⇒ a^{2}, b^{2}, c^{2 }are in G.P.
The locus of the point of intersection of the lines represent (t being a parameter)
To eliminate the parameter t, square and add the equations, we have
Which is the equation of a circle.
The equation of the circle which passes through the point (4, 5) and has its centre at (2, 2) is
As the circle is passing through the point (4, 5) and its centre is (2, 2) so its radius is
∴ The required equation is:
(x – 2)^{2} + (y – 2)^{2} = 13
Eccentricity of ellipse x^{2} + α^{2} + y^{2}/b^{2} = 1 if it passes through point (9, 5) and (12, 4) is
We have ....(1)
......(2)
From eq. (2) – eq. (1):
Consider the equation of a parabola y^{2} + 4ax = 0, where a > 0 which of the following is/are correct?
Equation of parabola is y^{2} = – 4ax. Its focus is at(– a, 0).
(By using sum of n natural number 1+ 2 + 3 + .... + n = n(n + 1)/2)
Take n^{2} common from Nr and Dr.
The probability of getting 10 in a single throw of three fair dice is :
Exhaustive no. of cases = 6^{3}
10 can appear on three dice either as distinct number as following (1, 3, 6) ; (1, 4, 5); (2, 3, 5) and each can occur in 3! ways. Or 10 can appear on three dice as repeated digits as following (2, 2, 6),
(2, 4, 4), (3, 3, 4) and each can occur in 3!/2!
ways.
∴ No. of favourable cases = 3 x 3! + 3 x 3!/2! = 27
Number of solutions of the equation tan^{1} (1 + x) + tan^{1} (1 x) π/2 are
tan^{1} (1 + x) + tan^{1} (1 x) π/2
⇒ tan^{1} (1 + x) = π/2  tan^{1} (1  x)
⇒ tan^{1} (1 + x)= cot^{1}(1  x)
⇒ tan^{1} (1 + x) = tan^{1}(1/1x)
⇒ 1 + x = 1/1x
⇒ 1  x^{2} = 1 ⇒ x = 0
As A is an orthogonal matrix, AA^{T }= I
⇒ a + 4 + 2b = 0, 2a + 2 – 2b = 0 and a^{2} + 4 + b^{2} = 9
⇒ a + 2b + 4 = 0 , a – b + 1 = 0 and a^{2} + b^{2} = 5
⇒ a = – 2, b = – 1
The points represented by the complex numbers 1 + i, 2 + 3i, 5/3 i on the argand plane are
Let z_{1} = 1 + i, z_{2} = – 2 + 3i and z_{3} = 0 + 5/3 i
If matrix and A^{1} = 1/k adj(A) , then k is
A^{1} = 1/k adj(A) .....(i)
Also, we know .....(ii)
∴ By comparing (i) and (ii)  A  = k
= 3 (2 + 1) + 2 (1 + 0) + 4 (1 – 0) = 9 + 2 + 4 = 15
If x, y, z are complex numbers, and then Δ is
We have
[Interchanging rows and columns]
[Taking –1 common from each row]
∴ + Δ = 0 ⇒ 2Re(Δ) = 0
∴ Δ is purely imaginary.
The function can be contin uous only at those points for which
sin x = cos x ⇒ x = nπ + π/4
Hence f (x) is continuous at x = 3π/4
The value of c in (0, 2) satisfying the mean value theorem for the function f(x) = x(x – 1)^{2}, x ∈ [0, 2] is equal to
f(x) = x (x – 1)2 ; x ∈ [0, 2]
f '(x) = 3x^{2} – 4x + 1 ⇒ f '(c) = 3c^{2} – 4c + 1
Given expression can be written as
A ball is dropped from a platform 19.6m high. Its position function is –
We have, a = d^{2}x/dt^{2} = 9.8
The initial conditions are x (0) = 19.6 and v (0) = 0
So, v = dx/dt = 9.8t v(0) 9.8t
∴ x = – 4.9t^{2} + x (0) = – 4.9t^{2 }+ 19.6
Now, the domain of the function is restricted since the ball hits the ground after a certain time. To find this time we set x = 0 and solve for t. 0 = – 4.9t^{2} + 19.6 ⇒ t = 2
Put x^{2} = t ⇒ 2 x dx = dt
If f (2a  x )dx = m and f (x)dx= n, then f (x)dx is equal to
Put x = 2a – t
so that dx = – dt
when x = a, t = a and when x = 2a, t = 0
An integrating factor of the differential equation sin x dx/dy + 2 y cos x = 1 is
Given differential equation is
The expression satisfying the differ ential equation (x^{2}  1) dy/dx + 2xy = 1 is
Rewrite the given differential equation as follows :
which is a linear form The integrating factor I.F.
Thus multiplying the given equation by (x^{2}  1),
we get (x^{2}  1) dy/dx + 2xy = 1
⇒ d/dx [y(x^{2 } 1)] = 1
On integrating we get y(x^{2}  1) = x + c.
Hence is independent of x and y both.
If are the position vectors of the vertices of a triangle ABC taken in order, then ∠A is equal to
The projection of line joining (3, 4, 5) and (4, 6, 3) on the line joining (–1, 2, 4) and (1, 0, 5) is
Let A = (3, 4, 5), P = (–1, 2, 4) B = (4, 6, 3) and Q = (1, 0, 5)
∴ Dr's of line AB are (4 – 3), (6 – 4), (3 – 5) = 1, 2, –2
and Dr's of line PQ are (1 + 1), (0 – 2), (5 – 4) = 2, –2, 1
∴ Dr's of line PQ =
∴ Projection of line segment AB on the line PQ is
Which of the following statements is correct?
If the constraints in a linear programming problem are changed then
In a binomial distribution, the mean is 4 and variance is 3. Then its mode is :
In Bin omi al di str ibuti on , Mea n = np, Variance = npq and the mode is r if for x = r, the probability function p(x) is maximum.
Given np = 4 and npq = 3
⇒ The distribution will have unique mode (unimodal) & the mode = 4
The given series is
If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately
We know that,
Mode = 3 Median – 2 Mean = 3(22) –2(21)
= 66 – 42 = 24
3 videos15 docs70 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
3 videos15 docs70 tests









