JEE  >  JEE Main & Advanced Mock Test Series  >  Part Test - 5 (JEE Advanced) Download as PDF

Part Test - 5 (JEE Advanced)


Test Description

66 Questions MCQ Test JEE Main & Advanced Mock Test Series | Part Test - 5 (JEE Advanced)

Part Test - 5 (JEE Advanced) for JEE 2022 is part of JEE Main & Advanced Mock Test Series preparation. The Part Test - 5 (JEE Advanced) questions and answers have been prepared according to the JEE exam syllabus.The Part Test - 5 (JEE Advanced) MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Part Test - 5 (JEE Advanced) below.
Solutions of Part Test - 5 (JEE Advanced) questions in English are available as part of our JEE Main & Advanced Mock Test Series for JEE & Part Test - 5 (JEE Advanced) solutions in Hindi for JEE Main & Advanced Mock Test Series course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Part Test - 5 (JEE Advanced) | 66 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study JEE Main & Advanced Mock Test Series for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
Part Test - 5 (JEE Advanced) - Question 1

4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 1

Heat released by steam inconversion to water at 100°C is Q1 = mL = 4 × 540 = 2160 cal.
Heat required to raise temperature of water from 46°C t 100°C is Q2 = mS Δθ = 20 × 1 × 54 = 1080 J 
Hence all steam is not converted to water only half steam shall be converted to water
So Final mass of water = 20 + 2 = 22 gm

Part Test - 5 (JEE Advanced) - Question 2

A diatomic ideal gas undergoes a thermodynamic change according to the P–V diagram shown in the figure. The total heat given to the gas is nearly (use ln2 = 0.7) :     

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 2

Part Test - 5 (JEE Advanced) - Question 3

Two rods are joined between fixed supports as shown in the figure. Condition for no change in the lengths of individual rods with the increase of temperature will be ( α1, α2 = linear expansion co-efficient A1, A2 = Area of rods Y1, Y2 = Young modulus )    

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 3

Part Test - 5 (JEE Advanced) - Question 4

For a gas sample with N0 number of molecules, function N(V) is given by : N(V) =   =   V  V0 and N(V) = 0 for V > V0. Where dN is number of molecules in speed range V to V+ dV. The rms speed of the molecules is :

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 4

Part Test - 5 (JEE Advanced) - Question 5

The co-efficient of thermal expansion of a rod is temperature dependent and is given by the formula α = a T, where a is a positive constant and T in ºC. If the length of the rod is l at temperature 0 ºC, then the temperature at which the length will be 2 l is:

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 5

Part Test - 5 (JEE Advanced) - Question 6

 A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength
at which the radiation has maximum intensity is λ0. If at another temperature T' the power radiated is
P' and wavelength at maximum intensity is then

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 6

Part Test - 5 (JEE Advanced) - Question 7

Thermal coefficient of volume expansion at constant pressure for an ideal gas sample of n moles having pressure P0, volume V0 and temperature T0 is    

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 7

Part Test - 5 (JEE Advanced) - Question 8

There are two thin spheres A and B of the same material and same thickness. They emit like black bodies. Radius of A is double that of B. A and B have same temperature T. When  A and B are kept in a room of temperature T0 (< T), the ratio of their rates of cooling (rate of fall of temperature) is: [ assume negligible heat exchange between A and B ]

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 8

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 9

During an experiment, an ideal gas is found to obzey a condition The
gas is initially at temperature T, pressure P and density ρ. The gas expands such that density changes to .

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 9

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 10

When the temperature of a copper coin is raised by 80 oC, its diameter increases by 0.2%, 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 10

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 11

Graph shows a hypothetical speed distribution for a sample of N gas particles (for V > V0 ;   =0)

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 11

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 12

Pick the correct statement(s) :    

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 12

Since PV = nRT therefore P and V both can change simultaneously keeping temperature constant. 

Part Test - 5 (JEE Advanced) - Question 13

STATEMENT–1 : A gas is kept in an insulated cylinder with a movable piston, in compressed state. As the piston is released, temperature of the gas decreases. 

STATEMENT–2 : According to the kinetic theory of gas, a molecule colliding with the piston must rebound with less speed than it had before the collision. Hence average speed of the molecules is reduced.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 13

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Part Test - 5 (JEE Advanced) - Question 14

Statement-1 : As the temperature of the blackbody increases, the wavelength at which the spectral intensity (Eλ) is maximum decreases.

Statement-2 : The wavelength at which the spectral intensity will be maximum for a black body is proportional to the fourth power of its absolute temperature.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 14

From Wein's law λmT = constant i.e., peak emission wavelength 
. Hence as T increase λm decreases.

Part Test - 5 (JEE Advanced) - Question 15

The atmospheric lapse rate
For small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.

The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.

The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.

Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantity TP(1 – λ)/λ has a uniform value through the layers of troposphere.

(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)

Q.    If behaviour of the mixing of parcels of air is approximately assumed to be adiabatic then lapse rate can be expressed as :

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 15

Part Test - 5 (JEE Advanced) - Question 16

The atmospheric lapse rate
For small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.

The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.

The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.

Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantity TP(1 – λ)/λ has a uniform value through the layers of troposphere.

(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)

Q.    Mechanical equilibrium of the atmosphere requires that the pressure decreases with altitude according to  . Assuming free fall acceleration to be uniform, then lapse rate is given by

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 16

Part Test - 5 (JEE Advanced) - Question 17

The atmospheric lapse rate
For small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.

The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.

The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.

Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantity TP(1 – λ)/λ has a uniform value through the layers of troposphere.

(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)

Q.    The value of theoretical lapse rate on the earth is (use g = 9.8 m/s2 ;  R = 8.3 J/mol-k and M = 29 g/mol)

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 17

Part Test - 5 (JEE Advanced) - Question 18

The figures given below show different processes (relating pressure P and volume V) for a given amount
of an ideal gas. ΔW is work done by the gas and ΔQ is heat absorbed by the gas. 

Column-I                                                                      Column-II
(A) In Figure (i)                                                                (p) ΔQ > 0.
(B) In Figure (ii)                                                               (q) ΔW < 0.
(C) In Figure (iii)                                                              (r) ΔQ < 0.
(D) In Figure (iv) (for complete cycle)                                (s) ΔW > 0.
                                                                                     (t) ΔU < 0.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 18

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 19

The thermal conductance (reciprocal of the thermal resistance) for axial flow of a truncated cone of length 
the radius of the two ends are r1 and r2 assuming that thermal conductivity of the material is K is 
. Find the value of n2.


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 19

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 20

Consider a vertical tube open at both ends. The tube consists of two parts, each of different crosssections
and each part having a piston which can move smoothly in respective tubes. The two pistons are joined together by an inextensible wire. The combined mass of the two piston is 5 kg and area of cross-section of the upper piston is 10 cm2 greater than that of the lower piston. Amount of gas enclosed by the pistons is one mole. When the gas is heated slowly, pistons move by 50 cm. Find rise in the temperature of the gas, in the form (X/R) K where R is universal gas constant. Use g = 10 m/sand outside pressure = 105N/m2). Fill value of X in the answer sheet.  


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 20

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 21

One mole of monoatomic ideal gas undergoes a process ABC as shown in figure. The maximum temperature of the gas during the process ABC is in the form  . Find X.


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 21

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 22

One mole of an ideal gas is kept enclosed under a light piston (area=10-2m2) connected by a compressed
spring (spring constant 200 N/m). The volume of gas is 0.83 m3 and its temperature is 100K. The gas
is heated so that it compresses the spring further by 0.1 m. Find the work done by the gas in the
process in joules ? (Take R = 8.3 J/K-mole and suppose there is no atmosphere). 

  


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 22

Part Test - 5 (JEE Advanced) - Question 23

Which of the following statements is correct ?

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 23

(C) The hydrogen atoms are in a vertical plane with the axial fluorine atoms.

Hydrogen atoms lie in the CSF2 axial plane. We know that the π bond involving a p-orbital on the carbon atom must lie in the equatorial plane of the molecule.

Part Test - 5 (JEE Advanced) - Question 24

The hybridisation and geometry of the anion of ICl3 in liquid phase is :

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 24

Part Test - 5 (JEE Advanced) - Question 25

Select the incorrect statement about the XeF6.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 25

In tetramer, four square pyramidal XeF5+ ions are joined to two similar ions by means of two bridging F- ions. The
Xe-F distances are 1.84 Å on the square pyramidal units and 2.23 Å and 2.60 Å in the bridging groups. The solid has various crystalline forms, of which three are tetrameric and a fourth has both hexamers and tetramers.

Part Test - 5 (JEE Advanced) - Question 26

Based on the Valence Shell Electron Pair Repulsion (VSEPR) model, what is the geometry around the sulphur, carbon and nitrogen in the thiourea – S , S – dioxide, O2SC(NH2)2 ? (consider the Lewis structure with zero formal charges on all atoms).

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 26

Part Test - 5 (JEE Advanced) - Question 27

The following flow diagram represents the manufacturing of sodium carbonate

Which of the following options describes the reagents, products and reaction conditions (given in parentheses)?

 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 27

Part Test - 5 (JEE Advanced) - Question 28

When boron is fused with potassium hydroxide which pair of species are formed ?

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 28

Part Test - 5 (JEE Advanced) - Question 29

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 29

Part Test - 5 (JEE Advanced) - Question 30

Orthosilicate ions,(SiO44-) undergo polycondensation to forms pyrosilicate, [O3Si – O – SiO3]6-, in presence
of :

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 30

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 31

Select the correct statement(s) with respect to the pπ–dπ dative bond.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 31

(A) Steric repulsions of bulkier groups and pπ–dπ dative bonding favour for a linear Si–O–Si group.
(B) Due to stabillization of the conjugate base anion by O(pπ) → Si(dπ) bonding motion.
(C) It is pyramidal because pπ–dπ bonding is not effective due to the bigger size of phosphorus atom.
(D) Most effective pπ–dπ overlapping due to small size of chlorine.

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 32

Which of the following statements is/are correct ?      

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 32

(A) Nitrogen is more electronegative than phosphorus.

So, dipole moment of trimethylamine is greater than trimethy phosphine.

In trisilyl ether the lone pair of electron on oxygen atom is less easily available for donation because of pπ-dπ
delocalisation due to presence of the vacant d-orbital with Si. This however is not possible with carbon in CH3–O–CH3 due to the absence of d-orbital making it more basic.
(C) Bond order of C2
and O2 are same i.e., 2. In C2 molecules both bonds are π-bonds whereas, there is one  σ and one π-bond in O2 molecule
C2 = 131 pm ; O2 = 121 pm.

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 33

Highly pure dilute solution of sodium in liquid ammonia :

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 33

Blue colour of the solution is due to ammoniated electrons and
good conductor of electricity because of both ammoniated cations and ammoniated electrons.

(change of oxidation states from 0 to – 1).

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 34

Select the correct statement (s)

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 34

Part Test - 5 (JEE Advanced) - Question 35

Statement-1 : The melting points of the various fluorides of xenon decrease in the order XeF2 > XeF4 > XeF6.

Statement-2 : Xe – F bond length in XeF2, XeF4 and XeF6 decreases in the order XeF2 > XeF4 > XeF6.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 35

S-1 : The melting point depends on the packing of the molecules. Symmetrical molecules are more closely packed as compared to unsymmetrical molecules. Hence symmetrical molecules will have higher melting points as compared to unsymmetrical molecules.

S-2 : This may be owing to the increase in charge on xenon atom with increasing number of F atoms.

 XeF2 XeF4 XeF6   
  2.00   1.95  1.89

Part Test - 5 (JEE Advanced) - Question 36

Statement-1 : Ethers behave as bases in the presence of mineral acids.

Statement-2 : It is due to the presence of lone pair of electrons on the oxygen atom.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 36

Part Test - 5 (JEE Advanced) - Question 37

The elements of group 1 describe, more clearly than any other group of elements, the effects of increasing the size of atoms or ions on the physical and chemical properties. The chemical and physical properties of the elements are closely related to their electronic structures and sizes. These metals are highly electropositive and thus form very strong bases, and have quite stable oxo-salts. In the manufacturing of sodium hydroxide, chlorine and sodium carbonate, the sodium chloride is used as starting material.

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 37

Basic strength of the oxides increase in the order Li2O < Na2O < K2O < Rb2O < Cs2O. The increase in basic strength is due to the decrease in I.E. and increase in electropositive character.The melting points of the halides decrease in the order NaF > NaCl > NaBr > NaI, as the size of the halide ion increases. The decrease in melting point is due to increase the covalent character with increase in the size of anion according to Fajan's rule.

Part Test - 5 (JEE Advanced) - Question 38

The elements of group 1 describe, more clearly than any other group of elements, the effects of increasing the size of atoms or ions on the physical and chemical properties. The chemical and physical properties of the elements are closely related to their electronic structures and sizes. These metals are highly electropositive and thus form very strong bases, and have quite stable oxo-salts. In the manufacturing of sodium hydroxide, chlorine and sodium carbonate, the sodium chloride is used as starting material.

Q. Which of the following acts as an oxidising as well as reducing agent ?            

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 38

NaNO2 (sodium nitrite) acts both as oxidising agent and reducing agent because the nitrogen atom in it is in +3 oxidation state (+3 is intermediate oxidation state for nitrogen)
Oxidising property : 2NaNO2 + 2KI + 2H2SO→ NaSO4 + K2SO4 + 2NO + 2H2O + I2
Reducing property : NaNO2 + H2O2 → NaNO3 + H2O.

Part Test - 5 (JEE Advanced) - Question 39

The elements of group 1 describe, more clearly than any other group of elements, the effects of increasing the size of atoms or ions on the physical and chemical properties. The chemical and physical properties of the elements are closely related to their electronic structures and sizes. These metals are highly electropositive and thus form very strong bases, and have quite stable oxo-salts. In the manufacturing of sodium hydroxide, chlorine and sodium carbonate, the sodium chloride is used as starting material.

Q. Sodium hydroxide is manufacture by the electrolysis of brine solution. The reaction by-products are :

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 39

Part Test - 5 (JEE Advanced) - Question 40

Match the entries in Column I with the correctly related characteritic(s) in Column II.
        Column – I                                           Column – II    
    (A) Cyclic silicates                                   (p) Thortveitite
    (B) Single chain silicates                          (q) Beryl
    (C) Pyro silicates                                     (r) General formula is (SiO3)n2n – 
    (D) Sheet silicates (two dimensional)         (s) Two oxygen atoms per tetrahedron are shared
                                                                  (t) Mica (muscovite)

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 40

(A) Two oxygen atoms per tetrahedron are shared forming rings. They have general formula (SiO3)n2n-.
Be3Al2[Si6O18], beryl.
(B) Two oxygen atoms per tetrahedron are shared forming a chain of tetrahedron. They have general
formula (SiO3)n2n-
(C) One oxygen atom per tetrahedron is shared. They have general formula Si2O72- . Thortveitite.
(D) Three oxygen atoms per tetrahedron are shared. They have general formula (Si2O5)n2- .
Mica (muscovite).

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 41

Give the atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference in the number of total p- and s-electrons.


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 41

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 42

How many maximum atoms lie in the same plane in the sructure of methylenesulphurtetrafluoride (CH2SF4)?


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 42

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 43

There will be three different flourine-flourine distances in molecule CF2(C)2CF2. Assuming ideal bond angles for a particular hybridisation (assume no distortion due to double bonds) find out the smaller flourine-flourine distance and fill the result (in pm) in the increasing order in your answer sheet. Round off your answer to the nearest whole number.

(Given that C–F bond length = 134 pm, C=C bond length = 134 pm √3 = 1.73)


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 43

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 44

Answer the following questions with respect to the compound NO[BF4].
    (P) Bond order of the part underlined.
    (Q) Total number of σ bonds in the compound.
    (R) Total number of π bonds in the compound.
    (S) Number of hybrid orbitals involved in the hybridisation of boron.    


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 44

NO+[BF4]-
(P) Molecular orbital electronic configuration ;
NO+ is derivative of O2, so MOT configuration is :
σ1s2 σ*1s2 σ2s2 σ*2s2 σ2px2py2 = π2pz2
 (x is taken as molecular axis)
Bond order = = 3
One σ bond and two π bonds.
(Q) Number of σ bonds in NO+ is one and in BF4- are four. So total number of σ bonds are five. 

(R) Number of σ bonds in NO+ are two.

(S) 

Steric number of central atom boron is 4 + 0 = 4 ; so its hybridisation is sp3 and thus the number of hybrid orbitals
involved in sp3 hybridisation is four.

Part Test - 5 (JEE Advanced) - Question 45

From a point P(1, 2) pair of tangent’s are drawn to a hyperbola ‘H’ in which one tangent to each arm of hyperbola. Equation of asymptotes of hyperbola H are √3x – y + 5 = 0 & √3x + y – 1 = 0 then eccentricity of ‘H’ is    

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 45

Part Test - 5 (JEE Advanced) - Question 46

Length of latus rectum of parabola which touches the line 3x + 4y + 7 = 0 having focus at (2, 3) and tangent at vertex passes through a point (1, 1), is    

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 46

Part Test - 5 (JEE Advanced) - Question 47

An ellipse whose focii are (2, 4) & (14, 9) and touches x-axis then its eccentricity is 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 47

Part Test - 5 (JEE Advanced) - Question 48

If tangent and normal at P to ellipse  intersect major axis at T and N in such a way that ratio of area of PTN and PSS` is 91/60, then area of PSS` is (S and S` are focii)

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 48

Part Test - 5 (JEE Advanced) - Question 49

An ellipse E has its centre at (1, 2), focus at (6, 2) and passes through the point P(4, 6). A hyperbola H is concentric with the ellipse and its transverse axis coincides with the major axis of the ellipse. If E and H intersect each other at a right angle at P, then the equation of the hyperbola is

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 49

Part Test - 5 (JEE Advanced) - Question 50

Let ‘P’ be a point which does not lie outside the triangle ABC, A = (3, 2), B = (0, 0), C = (0, 4) which satisfy d (P, A) < maximum {d (P, B), d (P, C)} then maximum distance of P from side BC, where d (P, A) gives the distance between P &  A, is 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 50

Part Test - 5 (JEE Advanced) - Question 51

Let P, Q, R and S be the feet of the perpendiculars drawn from a point (1, 1) upon the lines x + 4y = 12, 4y – x = 4 and their angle bisectors respectively, then equation of the circle which passes through Q, R, S is  

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 51

Part Test - 5 (JEE Advanced) - Question 52

Equation of the straight line meeting the circle with centre at origin and radius equal to 5 in two points at equal distances of 3 units from the point A(3, 4), is

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 52

Part Test - 5 (JEE Advanced) - Question 53

Let R(x1, y1) and S(x2, y2) be the end points of latus rectum of parabola y2 = 4x. The equation of ellipse with latus rectum RS and eccentricity 1/2 are (a > b) 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 53

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 54

If equation of tangents at P, Q and vertex A of a parabola are 3x + 4y – 7 = 0, 2x + 3y – 10 = 0 and x – y = 0 respectively, then

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 54

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 55

If H = (3, 4) and C = (1, 2) are orthocentre and circumcentre of ΔPQR and equation of side PQ is x – y + 7 = 0, then 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 55

*Multiple options can be correct
Part Test - 5 (JEE Advanced) - Question 56

If a variable tangent of circle x2 + y2 = 1 intersects the ellipse x2 + 2y2 = 4 at points P and Q then the locus of the point of intersection of tangents at P and Q is :

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 56

Part Test - 5 (JEE Advanced) - Question 57

Statement-1 :     Let L1 = 0 & L2 = 0 are two non perpendicular intersecting straight lines in two dimensional plane and if for λ = 1,  L1 + λL2 = 0 gives line passes through acute angular region between the lines then for λ = – 4,  L1 + λL2 = 0 gives line passes through obtuse angular region.

Statement-2 :    Points (x1, y1) and (x2, y2) are on the same / opposite sides of a line ax + by + c = 0 if and only if ax1 + by1 + c and  ax2 + by2 + c are of same/opposite signs respectively.
 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 57

Part Test - 5 (JEE Advanced) - Question 58

Statement-1 : Normal drawn at a fixed point P(t1), t1  0, on the parabola y2 = 4ax again intersects the
parabola at point t2 for all non-zero real values of t2.
Statement-2 : Normal drawn at a point P(t1), t1  0, on the parabola y2 = 4ax again intersects the parabola
at the point t2, where t2 = – t1 – .

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 58

Part Test - 5 (JEE Advanced) - Question 59

An ellipse has semi-major axes of lengths 2 and semi-minor axis of length 1. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both x-axis and y-axis.

  

Q.    The locus of centre of the ellipse is 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 59

The coordinate axes are tangent to the ellipse at any position. So,
origin lies on the director circle of the ellipse.
Let C be (α, β). The equation of director circle depends only on the location of centre of ellipse and its semi-major and minor axes, not on their orientation.
Hence, director circle is (x – α)2 + (y – β)2 = a2 + b2 = 22 + 12 = 5
Since it passes through (0, 0)
α2 + β2 = 5
so locus is x2 + y2 = 5 

Part Test - 5 (JEE Advanced) - Question 60

An ellipse has semi-major axes of lengths 2 and semi-minor axis of length 1. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both x-axis and y-axis.

Q. The locus of foci of the ellipse is 

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 60

Part Test - 5 (JEE Advanced) - Question 61

An ellipse has semi-major axes of lengths 2 and semi-minor axis of length 1. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both x-axis and y-axis.

Q. When the ordinate of the centre of ellipse is 1, a light ray passing through origin and centre of ellipse strikes the tangent at the vertex of the parabola y2 + 4x – 10y – 15 = 0. The reflected ray cuts y-axis at the point

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 61

You may use the fact that the product of perpendicular distances of any tangent on ellipse from its foci is
always equal to square of semi-minor axis.
Since the path of centre is x2 + y2 = 5 when y = 1, x = 2.
The incident ray passes through origin and (2, 1) ⇒ its equation is y = 1/2x
Given parabola is (y – 5)2 = – 4 (x – 10)
whose vertex (10, 5) lies on y = 1/2x.
The tangent at the vertex is the line x = 10.
Figure shows that OP = 2 (OR) = 10
so reflected ray cuts y-axis at (0, 10) 

Part Test - 5 (JEE Advanced) - Question 62

Detailed Solution for Part Test - 5 (JEE Advanced) - Question 62

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 63

Given two circle  let the radius of the third circle, which
is tangent to the two given circles and to their common diameter, be  , then find the value of p.


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 63

Let O1 and O2 be the centres of the circles of radius R = 5 and r = 3 respectively and O3 be the centre of the third circle. Let x be the radius of the third circle. and P be the point of tangency of the circle and the diameter O1 O2.

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 64

If (α, β) is the centroid of a triangle PQR where P, Q and R three points lying on the parabola
(y – 4)2 = 15 (x + 3) and normals at P and Q meet at R. Find the largest value of m satisfying
α + β > m for all possible values of α and β.


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 64

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 65

Consider a family of circles cutting the circles x2 + y2 + 2x – 4y – 4 = 0 and x2 + y2 – 4x + 4y + 4 = 0
orthogonally. Show that the chords in which the circle 2x2 + 2y2 – 5x – 6y – 2 = 0 cuts the member of
the family are concurrent at a point. If the coordinates of this point are (a, b), then find the value of
50(a + b).


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 65

*Answer can only contain numeric values
Part Test - 5 (JEE Advanced) - Question 66

An isosceles ΔABC is inscribed in the circle x2 + y2 = 16. From points A, B and C three ordinates are drawn
which cut the ellipse  at the points P, Q and R respectively such that A and P are on the
opposite sides, Q and B are on the same side and C and R are on the same side of the major axis. If ΔPQR
is right angled at P and θ is the smallest angle of the ΔABC, then find the value of 16tan2θ + 8 tan θ + 9.


Detailed Solution for Part Test - 5 (JEE Advanced) - Question 66

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about Part Test - 5 (JEE Advanced) Page
In this test you can find the Exam questions for Part Test - 5 (JEE Advanced) solved & explained in the simplest way possible. Besides giving Questions and answers for Part Test - 5 (JEE Advanced), EduRev gives you an ample number of Online tests for practice