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The velocity and acceleration vectors of a particle undergoing circular motion are and respectively at an in stant of time. The radius of the circle is –
It can be observed th at component of acceleration perpendicular to velocity is a = 4 m/s^{2}
Wave pulse can travel along a tense string like a violin spring. A series of experiments showed that the wave velocity V of a pulse depends on the following quantities, the tension T of the string, the crosssection area A of the string and then as per unit volume ρ of the string. Obtain an expression for V in terms of the T, A and r using dimensional analysis.
Let V = kT^{a }A^{b}ρ^{c},
k = dimensional constant
Writing dimension on both we side
[LT^{–1}] = [MLT^{–2}]^{a} [L^{2}]^{b} [ML^{–3}]^{c} = [M^{a+c}L^{a+2b–3c}T^{–2a}]Comparing power on both sides we have a + c = 0, a + 2b – 3c = 1, –2a = –1
A man runs at a speed of 4 m/s to overtake a standing bus. When he is 6 m behind the door at t = 0, the bus moves forward and continuous with a constant acceleration of 1.2 m/s^{2}. The man reaches the door in time t. Then,
Let us draw the figure for given situation,
⇒ 4 t = 6 + 0.6 t^{2}
A body is projected, making an acute angle with the horizontal. If angle between velocity and acceleration is θ, then
Here velocity is acting upwards wh en projectile is going upwards and acceleration is downwards. The angle θ between and ar is more than 0º and less than 180º.
The minimum velocity (in ms^{–1}) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is
The condition to avoid skidding,
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 –
A block of mass m is placed on a smooth inclined wedge ABC of inclination q as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and θ for the block to remain stationary on the wedge is
Let the mass of block is m. It will remains stationary if forces acting on it are in equilibrium. i.e., ma cos θ = mg sin θ
⇒ a = g tan θ
Here ma = Pseudo force on block, mg = weight.
A 3.628 kg freight car moving along a horizontal rail road spur track at 7.2 km/hour strikes a bumper whose coil springs experiences a maximum compression of 30 cm in stopping the car. The elastic potential energy of the springs at the instant when they are compressed 15 cm is
A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/s^{2}, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.
Given m = 0.36 kg, M = 0.72 kg.
The figure shows the forces on m and M.
When the system is released, let the acceleration be a. Then
T – mg = ma
Mg – T = Ma
∴
and T = 4 mg/3
For block m :
u = 0, a = g/3, t = 1, s = ?
∴ Work done by the string on m is
Two rings of radius R and nR made of same material have the ratio of moment of inertia about an axis passing through centre is 1 : 8. The value of n is
Ratio of moment of inertia of the rings
(λ = linear density of wire = constant)
∴ n^{3} = 8 ⇒ n = 2
A particle of mass ‘m’ is projected with a velocity v making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height ‘h’ is
Angular momentum L_{0} = pr ⊥
(∴ linear momentum p = mv cos θ and r_{⊥} = H)
⇒ L_{0} = mv cos θH
There is a shell of mass M and density of the shell is uniform. The work done to take a point mass from point A to B is [AB = r]
Gravitaional field inside the shell is zero, so no work required.
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc –
As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure
v_{net} (for lowest point) = v – Rω = v – v = 0 and acceleration
(since linear speed is constant)
A cube is subjected to a uniform volume compression. If the side of the cube decreases by 2% the bulk strain is
A ball whose density is 0.4 × 10^{3} kg/m^{3} falls into water from a height of 9 cm. To what depth does the ball sink ?
Figure shows a copper rod joined to a steel rod.
The rods have equal length and equal crosssectional area. The free end of the copper rod is kept at 0ºC and that of steel rod is kept at 100ºC.
Find the temperature of the junction of the rod.
Conductivity of copper = 390 W/mºC.
Conductivity of steel = 46 W/m ºC
Heat current in first rod (copper)
Here q is ltemperature of the junction and A & l are area and length of copper rod.
Heat current in second rod (steel)
In series clombination, heat current remains same. So,
– 390 θ = = 46 θ – 4600 436 θ = 4600 ⇒ θ = 10.6ºC
If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q ?
Stefan’s law for black body radiation
Q = σe AT^{4}
Here e = 1
A = 4πR^{2}
A thermodynamical system is changed from state (P_{1}, V_{1}) to (P_{2}, V_{2}) by two different process, the quantity which will remain same will be
For all process
ΔU = ΔQ – ΔW
does not change as it depends on initial final states.
A Car not’s heat engine works between the temperatures 427°C and 27°C. What amount of heat should it consume per second to deliver mechanical work at the rate of 1.0 kW?
The efficiency of the heat engine is
But
Thus, the engine would require 417 cal of heat per second, to deliver the requisite amount of work.
A vessel containing 1 more of O_{2 }gas (molar mass 32) at temperature T. The pressure of the gas is p.An identical vessel containing one mole of he gas (molar mass 4) at temperature 2T has a pressure of
Applying gas equation, pV = nRT We can write, p_{1}V = n_{1}RT_{1} and p_{2}V = n_{2}RT_{2}
The temperature of an ideal gas is increased from 27°C to 127°C, then percentage increase in v_{rms} is
We know,
⇒ % increase in
Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)
We know that
P_{A}V_{A} = n_{A} RT, P_{B}V_{B} = n_{B}RT and P_{f} (V_{A} + V_{B}) = (n_{A} + n_{B}) R_{T}
P_{f} (V_{A} + V_{B}) = P_{A}V_{A} + P_{B}V_{B}
An instan tan eous displacemen t of a simple harmonic oscillator is x = A cos (ωt + p/4). Its speed will be maximum at time
Velocity,
Velocity will be maximum, when wt + ω/4 = π/2 or ωt = π/2 – π/4 = π/4
or t = π/4ω
Two waves of wavelengths 99 cm and 100 cm both travelling with velocity 396 m/s are made to interfere. The number of beats produced by them per second is
Velocity of wave v = nλ
where n = frequency of wave
no. of beats = n_{1} – n_{2} = 4
If equation of transverse wave is y = x_{0} cos Maximum velocity of particle istwice of wave velocity, if l is
Three equal charges (q) are placed at corners of an equilateral triangle of side a. The force on any charge is
Two identical capacitors, have the same capacitance C. One of them is charged to potential V_{1} and the other to V_{2}. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is –
Initial energy of combined system
Final common potential,
Final energy of system,
Hence loss of energy = U_{1} – U_{2}
What should be the characteristic of fuse wire?
Fuse wire should be such that it melts immediatley when strong current flows through the circuit. The same is possible if its melting point is low and resistivity is high.
In the circuit shown in figure potential difference between points A and B is 16 V. the current passing through 2Ω resistance will be
∴ 4i_{1} + 2(i_{1} + i_{2}) – 3 + 4i_{1} = 16V ...(i) Using Kirchhoff’s second law in the closed loop we have
9 – i_{2} – 2(i_{1} + i_{2}) = 0 …(ii)
Solving equations (i) and (ii), we get
i_{1} = 1.5 A and i_{2} = 2 A
∴ current through 2W resistor = 2 + 1.5 = 3.5 A.
Two parallel conductors carry current in opposite directions as shown in figure. One conductor carries a current of 10.0 A. Point C is a distance d/2 to the right of the 10.0 A current. If d = 18 cmand I is adjusted so that the magnetic field at C is zero, the value of the current I is
The magnetic field at C due to first conductor is (since, point C is separated by from 1st conductor). The direction of field is perpendicular to the plane of paper and directed outwards.
The magnetic field at C due to second conductor is (since, point C is separated by d/2 from 2nd conductor)
The direction of field is perpendicular to the plane of paper and directed inwards.
Since, direction of B_{1} and B_{2} at point C is in opposite direction and the magnetic field at C is zero, therefore,
B_{1} = B_{2}
On solving I = 30.0 A
A uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron
and are in same direction so th at magnetic force on electron becomes zero, only electric force acts. But force on electron due to electric field is opposite to the direction of velocity.
Two coaxial solenoids are made by winding thin insulated wire over a pipe of crosssectional area A = 10 cm^{2} and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is
(μ_{0} = 4p × 10^{ –7} Tm A^{–1})
The ratio of secondary and primary turns of stepup transformer is 4 : 1. If a current of 4 A is applied to the primary, the induced current in secondary will be
Which of the following electromagnetic radiations has the smallest wavelength?
When light is refracted, which of the following does not change ?
Frequency does not change on refraction.
The given len s is broken into four parts and rearranged as shown. If the initial focal length is f then after rearrangement the equivalent focal length is –
Cutting a lens in tra nsver se direction doubles their focal length i.e. 2f.
Using the formula of equivalent focal length,
We get equivalent focal length as f/2.
In Young's double slit experiment 10th order maximum is obtained at the point of observation in the interference pattern for λ = 7000 Å. If the source is replaced by another one of wavelength 5000 Å then the order of maximum at the same point will be –
n_{1} λ_{1} = n_{2} λ_{2}
10 × 7000 = n_{2} × 5000 ⇒ n_{2} = 14
Transfer characteristics [output voltage (V_{0}) vs input voltage (V_{1})] for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used
I → ON
II → OFF
In II^{nd} state it is used as a amplifier it is active region.
Y = A + (.B) = (A + ).(A+ B) = A + B
⇒ OR gate
How many grams of con centrated nitric acid solution should be used to prepare 250 mL of 2.0M HNO_{3}? The concentrated acid is 70% HNO_{3}.
Molarity (M) =
wt. of 70% acid =
The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) orbit is (in Å)
Radius of hydrogen atom = 0.530 Å, Number of excited state (n) = 2 and atomic number of hydrogen atom (Z) = 1. We know that the Bohr radius.
= 4 × 0.530 = 2.12 Å
The screening effect of inner electron of the nucleus causes the decrease in ionization potential, therefore the order of the screening effect is
f < d < p < s
Hence, the screening effect of delectron is less than pelectron.
When the first ionisation energies are plotted against atomic number, the peaks are occupied by
Rare gases; as the e^{–} is to be removed from stable configuration.
The ions O^{2–}, F^{–}, Na^{+}, Mg ^{2+} a nd Al^{3+} are isoelectronic. Their ionic radii show :
Amongst isoelectronic species, ionic radii of anion is more than that of cations. Further size of anion increase with increase in –ve charge and size of cation decrease with increase in + ve charge. Hence ionic radii decreases from O^{–} to Al3^{+}.
Using MOT, which of the following pairs denote paramagnetic species?
B_{2 }and O_{2} are paramagnetic due to presence of unpaired electron. MO electronic configuration of B_{2} is :
MO electorn i.e. configuration of O_{2} is :
Increasing order of rms velocities of H_{2}, O_{2}, N_{2} and HBr is
RMS velocity of molecules depends on mass. If mol. wt. increases, rms velocity of melocules decreases.
The order of increasing m. wt. is
H _{2} < N _{2} < O _{2} < HBr
Order of V_{rms} of molecules.
HBr < O_{ 2} < N_{ 2} < H_{ 2} .
For the dissociation reaction,
H_{2} (g) → 2H (g) ΔH = 162 Kcal,
heat of atomisation of H is
ΔH = ΔH_{(product)} – DH_{(reactant)}
162 = 2 × ΔH_{H} DH_{H2}
ΔH_{H} = 81 Kcal
The enthalpy of combustion of 2 moles ofbenzene at 27°C differs from the value determinedin bomb calorimeter by
By bomb calorimeter we get DE . 2C_{6} H _{6} (l) + 15 O_{2} (g) → 12CO_{2}(g) + 6H_{2}O (l)
ΔH – ΔE = ΔnRT
= (12 – 15) × 8.314 × 300 = – 7.483 kJ
If 1.0 mole of I_{2} is introduced into 1.0 litre flask at 1000 K, at equilibrium (K_{c} = 10^{–6}), which one is correct?
It shows that (1 – x) < 2x
For the reaction CO (g) + (1/2) O_{2} (g) → CO_{2} (g), K_{p} / K_{c} is
K_{p} = K_{c}(RT)^{Δn};
The oxidation state of sulphur in Na_{2}S_{4}O_{6} is
Na_{2}S_{4}O_{6}
2 + 4x – 12 = 0
4 x – 10 = 0
Oxidation state of S is =
When same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide solution, the ratio of volumes of hydrogen evolved is:
Zn + H_{2}SO_{4} → ZnSO_{4} + H_{2}
Zn + 2NaOH → Na_{2}ZnO_{2} + H_{2}
[ Ratio of volumes of H2 evolved is 1 : 1
The alkali metals form saltlike hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders ?
The s tability of alkali metal h ydr ides decreases from Li to Cs. It is due to the fact that M–H bonds becomes weaker with increase in size of alkali metals as we move down the group from Li to Cs. Thus the order of stability of hydrides is LiH > NaH > KH > RbH > CsH
The most stable carbanion among the following is :
NO_{2} group, being strong electron withdrawing, disperses the ve charge, hence stabilizes the concerned carbanion.
Among the following four structures I to IV,
it is true that
A chiral object or structure has four different groups attached to the carbon.
The number of enan tiomers of the compound CH_{3}CHBrCHBrCOOH is
The structure
has two different chiral carbon atoms, hence number of enantiomers (optically active forms) is 2^{n} = 2^{2} = 4
Which one of the following reactions is expected to readily give a hydrocarbon product in good yields ?
Electrolysis of a concentrated aqueous solution of either sodium or potassium salts of saturated carboxylic acids yields higher alkane at anode.
What will be the main product when acetylene reacts with hypochlorous acid?
Green house gases such as CO_{2}, ozon e, methane, the chlorofluoro carbon compounds and water vapour form a thick cover around the earth which prevents the IR rays emitted by the earth to escape. It gradually leads to increase in temperature of atmosphere.
Due to Frenkel defect, the density of ionic solids
No change in density.
Equal weights of NaCl and KCl are dissolved separately in equal volumes of solutions. Molarity of the two solutions will be :
When equal weights of different solutes are present in equal volumes of solution the molarity is inversely related to molecular mass of the solute. Mol. mass of NaCl is less than KCl. Hence, molarity of NaCl solution will be more.
A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is
The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be
Velocity constant of a reaction at 290 K was found to be 3.2×10^{–3}. At 300 K it will be
Th e velocity con stant doubles for every 10°C rise in temperature.
At high pressure, the entire surface gets covered by a monomolecular layer of the gas follows
At high pressure the extent of adsorption follows zero order kinetics.
Which of the following is incorrect with respect to property indicated ?
F_{2} is expected to have highest bond en ergy but the correct decreasing order is Cl_{2} > Br_{2} > F_{2} because of fluorine atom has very small size due to which there is a high inter electronic repulsion between two fluorine atoms so the bond between two fluorine gets weaker and need less energy.
Strong reducing behaviour of H_{3}PO_{2} is due to
The acids which contain P–H bond have strong reducing properties. Thus H_{3}PO_{2} acid is good reducing agent as it contains two P–H bonds. For example, it reduces AgNO_{3} to metallic silver.
4 AgNO_{3} + 2H_{2}O + H_{3}PO_{2} → 4Ag + 4HNO_{3} + H_{3}PO_{4}
The pair in which both species have same magnetic moment (spin only value) is :
[Cr(H_{2}O)_{6}]^{2+} Cr is in Cr^{2+} form
In [Fe(H_{2}O)]^{2+} Fe^{2+} form. Both will have 4 unpaired electrons.
Which of the following is less acidic among the given halogen compounds?
Due to stronger–Ieffect of F than that of Cl, CHF3 should be more acidic than CHCl3. But actually reverse is true.
This is due to : CCl_{3}^{} left after the removal of a proton from CHCl_{3} is stabilised due to presence of dorbitals in Cl than: CF_{3}^{–} left after the removal of a proton from CHF_{3} which is not stabilised due to the absence of dorbitals on F.
In a S_{N}2 substitution reaction of the type
which one of the following has the highest relative rate ?
S_{N}2 mech an ism is followed in case of primary and secondary halides i.e., S_{N}2 reaction is favoured by small groups on the carbon atom attached to halogens so
Which of th e following is not the product of dehydration of
What will be the correct structural formula of product for the following reaction?
Nucleophilic addition reaction will be most favoured in
Aldehydes are more reactive than ketones due to +I effect of –CH_{3} group. There are two – CH_{3} group in acetone which reduces +ve charge density on carbon atom of carbonyl group. More hindered carbonyl group becomes less reactive. So in the given case CH_{3}CHO is the right choice.
When NH3 is passed over heated metal A, its amide is formed. The metal is
When Potassium is treated with ammonia, then potassium amide is obtained.
Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories ?
Insulin is a biochemically active peptide harmone secreted by pancreas.
Which statement is incorrect about peptide bond?
Due to resonance,
bond acquires some double bond character, hence shorter in length.
A mixture of chlorides of copper, cadmium, chromium, iron and aluminium was dissolved in water acidified with HCl and hydrogen sulphide gas was passed for sufficient time. It was filtered, boiled and a few drops of nitric acid were added while boiling. To this solution ammonium chloride and sodium hydroxide were added and filtered. The filterate shall give test for.
CuS and CdS are precipitated by H_{2}S. Hydroxide of A1 will pass into the solution in the form ofNaAlO_{2}, being amphoteric in nature. Hence filtrate will give test for sodium and aluminium.
Volume of 3% solution of sodium car bonate necessary to neutralise a litre of 0.1 N sulphuric acid
Normality of 3% Na_{2}CO_{3}.
For H_{2}SO_{4} sol. N1 = 0.1, V_{1} = 100 mL For Na_{2}CO_{3} sol. N_{2} = 0.566.
Now apply N_{1}V_{1} = N_{2}V_{2}
DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the word.
Q. SAGACIOUS
Sagacious means ‘judicious’, so ‘wise’ is correct answer.
DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the word.
Q. REMEDIAL
Remedial means ‘reformative’, so ‘corrective’ is correct answer.
DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the word.
Q. RETICENT
Reticent mean s ‘quiet’ so ‘secretive’ is correct answer
DIRECTIONS: Choose the word opposite is meaning to the given word.
Q. FIDELITY
Fidelity means ‘faithfulness in relation s’, so ‘treachery’ is correct antonym.
DIRECTIONS Choose the word opposite is meaning to the given word.
Infrangible means ‘strong’, so ‘breakable’ is correct antonym.
DIRECTIONS: Choose the word opposite is meaning to the given word.
Q. PROGENY
Progeny means ‘child’, so ‘parent’ is correct antonym.
DIRECTIONS: A part of sentence is underlined. Below are given alternatives to the underlined part (a), (b), (c) and (d) which may improve the sentence. Choose the correct alternative.
Q. It was not possible to drag any conclusion so he left the case.
Use of ‘draw’ is more suitable for using before word ‘conclusion’, so option (c) is correct.
DIRECTIONS: A part of sentence is underlined. Below are given alternatives to the underlined part (a), (b), (c) and (d) which may improve the sentence. Choose the correct alternative.
Q. I am looking after my pen which is missing.
Use of ‘looking for’ is proper because look for means ‘to search for something’ which suits here.
DIRECTIONS: A part of sentence is underlined. Below are given alternatives to the underlined part (a), (b), (c) and (d) which may improve the sentence. Choose the correct alternative.
Q. “Mind your language !” he shouted.
‘Mind your language’ is proper to use here because it gives proper sense of sentence.
DIRECTIONS: Fill in the blanks.
Q. I ..... to go there when I was student.
‘Use to’ is used when any h abit is to be shown, so use of option (b) is proper.
DIRECTIONS: Fill in the blanks.
Q. She was angry ...... me
‘Angry’ agrees with preposition ‘with’, so use of option (c) is correct here.
DIRECTIONS: Fill in the blanks.
Q. You should not laugh ...... the poor.
Laugh agrees with preposition ‘at’, so use of option (b) is correct here.
DIRECTIONS: In the questions below, each passage consists of six sentences. The first and sixth sentence are given in the beginning. The middle four sentences in each have been removed and jumbled up. These are labeled as P, Q, R and S. Find out people order for the four sentences.
Q. 1. He is a famous doctor.
P. Once I had to consult with him.
Q. I never believed him.
R. He suggested me a proper remedy.
S. I become completely fine. 6. Now I also admit this fact.
DIRECTIONS: In the questions below, each passage consists of six sentences. The first and sixth sentence are given in the beginning. The middle four sentences in each have been removed and jumbled up. These are labeled as P, Q, R and S. Find out people order for the four sentences.
Q. 1. We don’t know the plan of Ram.
P. He cares for his friends.
Q. He is a complete person.
R. We want some help and advice.
S. As we are in a trouble. 6. We hope he will do his best for us.
DIRECTIONS: In the questions below, each passage consists of six sentences. The first and sixth sentence are given in the beginning. The middle four sentences in each have been removed and jumbled up. These are labeled as P, Q, R and S. Find out people order for the four sentences.
Q. 1. It is not my problem.
P. All residents of this society are careless.
Q. I am unable to convince anyone.
R. They don’t want to do some good.
S. Every one seems to be unwise here. 6. We all have to suffer one day.
In a certain code language ‘DOME’ is written as ‘8943’ and ‘MEAL’ is written as ‘4321’. What group of letters can be formed for the code ‘38249’?
Find the missing number from the given response.
From fig. 1, 93 – (27 + 63) = 93 – 90 = 3 From fig. 2, 79 – (38 + 37) = 79 – 75 = 4 From fig. 3, 67 – (16 +42) = 67 – 58 = 9 Hence, option (d) is correct.
Which of the following correctly represents the relationship among illiterates, poor people and unemployed?
Hence, option (b) is correct.
Sushma walks 20m towards north. Then she turns right and walks 30m. Now, she turns right and walks 35m. Now turning left, she walks 15m.Again, she turns left and moves 15m. Finally, she turns left and walks 15m. In which direction and how far is she from her original position.
In a classroom, there are 5 rows and 5 children A, B, C, D and E are seated one behind the other in 5 separate rows as follows.
From the information given in the question the arrangement of students is
Which of the following will fill the series?
2, 9, 28, ? , 126
The given series follows the pattern
1^{3} + 1 = 2
2^{3} + 1 = 8 + 1 = 9
3^{3} + 1 = 27 + 1 = 28
4^{3} + 1 = 64 + 1 = 65
5^{3} + 1 = 125 + 1 = 126
Two signs in the equations have been interchanged, find out the two signs to make equation correct.
3 ÷ 5 x 8 + 2  10 = 13
Interchanging symbols ÷ and + as given in option (d), we get 3 + 5 × 8 ÷ 2 – 10
Assertion: [A] India is a democratic country.
Reason: [R] India has a constitution of its own.Choose the correct alternative from the given options.
Both Assertion and Reason ar e cor rect but India is a democratic country because the government is elected by its citizens and not because India has its own constitution.
Which one of the following figures completes the original figure?
How many squares are there in the following figure?
A class has 175 students. The following data shows the number of students obtaining one or more subjects. Mathematics 100, Physics 70, Chemistry 40; Mathematics and Physics 30, Mathematics and Chemistry 28, Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered Mathematics alone?
n (M alone)
= n(M) – n(M ∩ C) – n(M ∩ P) + n(M ∩ C ∩ P) = 100 – 28 – 30 + 18 = 60
If x sin^{3} θ + y cos^{3} θ = sin θ cos θ and x sin θ = y cos θ, then x^{2} + y^{2} =
x sin ^{3 }θ + y cos^{ 3} θ = sin θ cos θ ....(i) and x sin θ = y cos θ ...(ii) Equation (i) may be written as x sin q.sin^{ 2 }θ + y cos ^{3} θ = sin θ cos θ
⇒ y cos θ sin^{ 2} θ + y cos^{ 3} θ = sin θ cos θ
⇒ y cos θ(sin^{ 2} θ + cos^{ 2} θ) = sin θ cos θ
⇒ y cos θ = sin q cos θ
∴ y = sin θ ...(iii) Putting the value of y from (iii) in (ii), we get x sin θ = sin θ. cos θ
⇒ x = cos θ ...(iv) Squaring (iii) and (iv) and adding , we get
x^{2} + y^{2} = cos^{2} θ + sin^{2} θ = 1
If cos 7θ = cos q – sin 4θ, then the general value of θ is
cos 7θ = cos θ  sin4θ
⇒ sin 4θ = cos θ  cos 7θ
⇒ sin 4θ = 2 sin 4θ sin 3θ
⇒ sin 4θ (1  2 sin 3θ)= 0
∴ sin 4θ = 0 or
If the real part of is 4, z ≠ 1, then the locus of the point representing z in the complex plane is
Real part of is given by
Comparing with the equation
= 2. Thus, the locus of z given by the equation
(i) is a circle with centre 3/2 and radius = 1/2
If α and β are the roots of x^{2 }– x + 1 = 0, then the equation whose roots are α100 and β100 are
We have x^{3} + 1 ≡ ( x + 1) ( x ^{2}  x + 1) .
Therefore, α and β are the complex cube roots of –1 so that we may take a = –ω and b =–ω^{2}, where w ≠ 1 is a cube root of unity..
Thus α^{100} = ( ω)^{100} = ω and β^{100} = (–ω2)100 = ω2, so that the required equation is x^{2} + x + 1 = 0.
The set of all real x satisfying the inequality
Given,
⇒ 3 x  < 0 and 4 x  < 0
or 3 x  > 0 and 4 x  < 0
⇒  x  > 3 and x >4
or  x  < 3 and x <4
⇒  x> 4 orx < 3
⇒ x∈(∞,4) ∪[3, 3]∪ (4, ∞)
If x satisfies  3 x – 2  +  3x – 4  +  3x – 6  > 12, then
Dividing R at analyse 4 cases.
When x the inequality becomes
2 – 3x + 4 – 3x + 6 – 3x > 12.
implying – 9x > 0 ⇒ x < 0.
when x > 2 the ineqality becomes
3x – 2 + 3x – 4 + 3x – 6 > 12,
Implying 9x > 24 Þ x > 8/3 The inequality in invalid in the other two sections.
∴ either x < 0 or x > 8/3
In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together ?
Leaving one seat vacant between two boys, 5 boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls any sit in 5! ways.
Hence the required number = 4! × 5!
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?
Atleast one black ball can be drawn in the following ways (i) one black and two other colour balls = ^{3}C_{1} × ^{6}C_{2} = 3 × 15 = 45
(ii) two black and one other colour balls = ^{3}C_{2} × ^{6}C_{1} = 3 × 6= 18
(iii) All the three are black = ^{3}C_{3} × ^{6}C_{0} = 1
∴ Req. no. of ways = 45 + 18 + 1 = 64
The coefficient of th e middle ter m in the expansion of (2 + 3x)^{4} is :
When exponent is n then total number of terms are n + 1. So, total number of terms in (2 + 3x)^{4} = 5
Middle term is 3rd. ⇒ T_{3} = ^{4}C_{2}(2)^{2} . (3x)^{2}
If C_{0}, C_{1}, C_{2}, .............. C_{n} denote the binomial coefficients in the expansion of (1 + x)^{n}, then the value of
C_{0} + (C_{0} + C_{1}) + (C_{0}+ C_{1} + C_{2}) + ....+ (C_{0}+ C_{1} + .....+ C_{n–1})
C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + ................ + (C_{0} + C_{1} + ............ C_{n–1}) = nC_{0} + (n  1)C_{1} + (n  2)C_{2}+ .......C_{n 1} = C_{1} + 2C_{2} + 3C_{3} + 4C_{4} .......nC_{n}= n.2^{n1}
The sum of the series 1 + 2 .2 + 3 .2^{2} + 4 .2^{3} + .... + 100 .2^{99} is
Let S = 1 + 2. 2 + 3. 2^{2} +4. 2^{2} + .... + 100 .299 ....(i) It is an arithmetico  geometric series.
On multiplying Eq. (i) by 2 and then subtracting it from Eq. (i), we get
The quadratic equation whose roots are the x and y intercepts of the line passing through (1, 1) and making a triangle of area A with the coordinate axes is
Equation of the line making intercepts a and b on the axes is
Since, it passes through (1, 1)
Also the area of the triangle formed by the line and the axes is A.
From eqs. (i) and (ii), we get, a + b = 2A Hence, a and b are the roots of the eq.
x^{2} (a +b)x +ab = 0 ⇒ x^{2} 2Ax +2A = 0
If 4a^{2} + b^{2} + 2c^{2} + 4ab – 6ac – 3bc = 0, the family of lines ax + by + c = 0 is concurrent at one or the other of the two points
4a^{2} + b^{2} + 2c^{2} + 4ab – 6ac – 3bc ≡ (2a + b)^{2} – 3(2a + b) c + 2c^{2} = 0
⇒ (2a + b – 2c) (2a + b – c) = 0 ⇒ c = 2a + b or c =
The equation of the family of lines is
a(x + 2) + b(y + 1) = 0 or a(x + 1) + b giving the point of concurrence (–2, –1) or
A pair of tangents are drawn from the origin to the circle x^{2} + y^{2}+ 20 (x + y) + 20 = 0, then the equation of the pair of tangent are
Equation of pair of tangents is given by SS_{1} = T^{2},
or S = x^{2} + y^{2} + 20 (x + y ) + 20, S_{1} = 20,
T = 10 (x + y) + 20 = 0
∴ SS_{1} = T^{2}
⇒ 20 (x^{2} + y^{2} + 20 (x + y ) + 20) = 10^{2} (x + y + 2)^{2}
⇒ 4x^{2} + 4y^{2} + 10xy = 0 ⇒ 2x^{2} + 2y^{2} + 5xy = 0
An ellipse has OB as semi minor axis, F and F ' its focii and the angle FBF ' is a right angle.
Then the eccentricity of the ellipse is
∵ ∠FBF ' = 90° ⇒ FB^{ 2} + F'B^{ 2}= FF'^{2}
Also, e^{2} = 1 b ^{2} / a^{ 2} =1e^{2}
(By using equation (i))
If the lin e 2x – 3y = k touches th e parabola y2 = 6x, then find the value of k.
⇒ y^{2} = 3 (3y + k) ⇒ y^{2} – 9y – 3k = 0 ........(3)
If line (1) touches parabola (2) then roots of quadratic equation (3) is equal
∴ (–9)^{2} = 4 × 1 × (– 3k) ⇒ k = – 27/4
S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is
Let eq. of ellipse be
S is (–ae, 0), T is (ae, 0) and B is (0, b).
Also SB^{2} = ST^{2} ⇒ 4a^{ 2}e^{ 2} = a^{ 2} e^{ 2} + b^{ 2}
Let f(x) = (x^{5 }– 1)(x^{3} + 1), g(x) = (x^{2} – 1)(x^{2} – x + 1) and let h(x) be such that f(x) = g(x)h(x). Then
Given f (x) = g(x) h(x)
In the truth table for the statement (p ∧ q) ® (q ν ~ p), the last column has the truth value in the following order is
T T T T
If the value of mode and mean is 60 and 66 respectively, then the value of median is
Mode = 3 Median – 2 Mean
∴
Variance
= 1.32
Let R be the relation on the set R of all real numbers, defined by aRb If a – b < 1. Then, R is
Since,  a  a  = 0 < 1, so aRa,
∴ R is reflexive.
Now, aRb ⇒  a  b  < 1⇒  b  a  < 1⇒ bRa
∴ R is symmetric.
But R is not transitive as
1R2, 2R3 but 1 3
[∵1 – 3 = 2 > 1]
The greatest and least values of (sin^{–1} x)^{2} + (cos^{–1} x)^{2} are respectively
We have, (sin^{–1} x)^{2} + (cos^{–1} x)^{2} = (sin^{–1} x + cos ^{–1} x)^{2} – 2 sin^{–1} x. cos^{–1} x
Thus, the least value is 2
and the greatest value is
The given trigonometric ratio
The given determinant vanishes, i.e.,
Expanding along C_{1}, we get (x – 4)(x – 5)^{2} – (x – 5)(x – 4)^{2} – {(x – 3)(x –5)^{2} – (x – 5)(x – 3)^{2}} + (x – 3)(x – 4)^{2} – (x – 4)(x – 3)2 = 0 Þ (x – 4)(x – 5)(x – 5 – x + 4) – (x – 3)(x – 5)(x – 5 – x + 3) +(x – 3)(x – 4) (x – 4 –x + 3) = 0
⇒ – (x – 4)(x – 5) + 2(x – 3)(x – 5) – (x – 3) (x – 4) = 0
⇒ – x2 + 9x – 20 + 2x2 – 16x + 30 – x2 + 7x – 12 = 0
⇒ – 32 + 30 = 0 Þ –2 = 0
Which is not possible, hence no value of x satisfies the given condition.
If the lines lx + my + n = 0, mx +ny + l = 0 and nx + ly + m = 0 are concurrent then
Since the lines are concurrent, so
⇒ (l + m + n) (l^{2} + m^{2} + n^{2} – lm – mn – nl) = 0
⇒ l + m + n = 0 [∵ l^{2} + m^{2} + n^{2} > lm + mn + n]
is continuous at x = –5, then the value of ‘a’ will be
The equation of all lines having slope 2 which are tangent to the curve
The equation of the given curve is
The slope of the tangent to the given curve at any point (x, y) is given by For tangent having slope 2, we must have
which is not possible as square of a real number cannot be negative.
Hence, there is no tangent to the given curve having slope 2.
The function f (x) = (x(x–2))^{2} is increasing in the set
Here, f(x) = (x(x – 2))2
⇒ f '(x) = 4x (x – 2) (x – 1) For f(x) as increasing, f '(x) > 0 So, 4x (x – 1) (x – 2) > 0 ⇒ x(x – 1) (x – 2) > 0
From the above figure required interval is, (0, 1) ∪ (2, ∞)
If a^{2} x^{4} + b^{2} y^{4} = c^{4}, then the maximum value of xy is
If the sum of two positive quantities is a constant, then their product is maximum, when they are equal. ∴
a^{2} x^{4}. b^{2}y^{2} is maximum when
∴ maximum value of
Maximum value of xy =
Let
Let cos x = t and – sin x dx = dt.
Now, x = 0 ⇒ t = cos 0 = 1 and
Area intercepted by the curves y = cos x, x ∈ [0, π] and y = cos 2x, x ∈ [0, π], is
The general solution of the differential equation
The equation is,
The solution to the differential equation
where f (x) is a given function is
are three unit vectors such that where is null vector, then
If vectors 2i – j + k, i + 2j – 3k and 3i + aj + 5k are coplanar, then the value of a is
If given vectors are coplanar, then there exists two scalar quantities x and y such that
Comparing coefficient of and on both sides of (1) we get x + 3y = 2 , 2x + ay = –1 , –3x + 5y = 1 ...(2)
Solving first and third equations, we get x = 1/2, y = 1/2 Since the vectors are coplanar, therefore these values of x and y will satisfy the equation 2x + ay = –1
∴ 2 (1/2) + a (1/2) = – 1 ⇒ a = –4
The coordinates of the poin t wh ere the lin e through the points A (3, 4, 1) and B (5, 1, 6) crosses the XYplane are
Equation of the line through the given points is
Any point on this line can be taken as (3 + 2λ, 4 – 3λ, 1 + 5λ) If this point lies on XYplane then the zcoordinate is zero
Thus the required coordinates of the point are
Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7.
The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are
Therefore,
Hence,
For k = 1, 2, 3 the box Bk contains k red balls and (k + 1) white balls. Let and A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box B_{2}, is
In a box, B_{1} = 1R, 2W; B_{2} = 2R, 3W and B_{3} = 3R, 4W
Also, given that, and
The probability of India winning a test match against West Indies is 1/2. Assuming independence from match to match, the probability that in a 5 match series India’s second win occurs at the third test, is –
The sample space is [LWW, WLW]
∴ P (LWW) + P (WLW)
= Probability that in 5 match series, it is India’s second win = P (L) P (W) P (W) + P (W) P (L) P (W)
An object is obseved from the points A, B and C lying in a horizontal straight line which passes directly underneath the object. The angular elevation at B is twice that at A and at C three times that at A. It AB = a, BC = b, then the height of the object is
Let ED = h, ∠EAB = a ∠EBD = 2α, ∠ECD = 2α Now, ∠DBE = ∠EAB + ∠BEA
⇒ 2α = α + ∠BEA
⇒∠BEA = α = ÐEAB
⇒ AB = EB = α Similarly, ∠EBC = α
A shopkeeper wants to purchase two articles A and B of cost price ` 4 and ` 3 respectively. He thought that he may earn 30 paise by selling article A and 10 paise by selling article B. He has not to purchase total articles worth more than ` 24. If he purchases the number of articles of A and B, x and y respectively, then linear constraints are
x, y > 0 and 4x + 3y < 24
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