JEE MAIN MODEL - HINTS AND SOLUTIONS
IIT, MEDICAL & ENGINEERING Entrance COACHING Now @ MUKKAM: BHABHA Institute of Sciences 2nd floor, Karayil Tower, Near Bend Pipe Bridge, Orphanage Road,
Mukkam, PIN - 673602
Ph : 8943300201 , 8943300204.
PART : A — PHYSICS Sol : Answer (1)
ALL THE GRAPHS/DIAGRAMS GIVEN ARE
SCHEMATIC AND NOT DRAWN TO SCALE 01. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is (1) 5.112 cm (2) 5.124 cm (3) 5.136 cm (4) 5.148 cm Sol : Answer (2) Reading = M.S.R + No of division of V.S matching the main scale division (1MSD – 1VSD) 2.45 5.10 240.05 50 = 5.124 cm option (2) is correct. 03. At what angle the vector A B and A B must
02. The velocity - displacement graph of a particle act, so that the resultant is A2 B2 moving along a straight line is shown A2 B2 A2 B2 (1) cos-1 (2) cos-1 2 2 2 2 A B A B A2 B2 A2 B2 (3) cos-1 (4) cos-1 2 A2 B2 2 B2 A2 Sol : Answer (4)
The most suitable acceleration-displacement graph will be
04. A uniform cylinder of radius R is spinned about its
axis to the angular velocity 0 and then placed into a corner (see figure). The coefficient of friction (1) (2) between the corner walls and the cylinder is equal to . How many turns does the cylinder accomplish before it stops?
(3) (4)
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
2 R1 2 2 R1 2 (1) 0 (2) 0 8g1 4g1 2 R1 2 2 R1 2 (3) 0 (4) 0 2g1 6g1 (1) (2) Sol : Answer (1)
(3) (4) Sol : Answer (4)
From the given function we can see that F = 0 at x = 0 i.e., slope of U – x graph is zero at , x = 0. Therefore, the most appropriate option is (4)
06. A particle of mass ‘m’ oscillates along the horizontal diameter AB inside a smooth spherical shell of radius R. At any instant K.E. of the particle is K. Then force applied by particle on the shell at this instant is:
K 2K (1) (2) R R 05. A particle which is constrained to move along x-axis, 3K K is subjected to a force in same direction which varies (3) (4) with distance x of the particle from the origin as R 2R F(x) = - kx + ax3. Here k and a are positive constant. Sol : Answer (3) For x 0, the functional form of the potential energy mv2 U(x) of the particle is: N mgsin R By conservation of energy
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
1 08. A metallic cube as a bulk modulus ‘B ’and a density mgRsin mv2 2 ‘ ’ . A pressure of ‘P ’ is applied uniformly from all sides of the cube. The increase in density is B PB (1) (2) B P B B B (3) (4) B P B P Sol : Answer (1) M PV = , B = V V M M 1 = = V V V V 1 V 07. The magnitudes of gravitational field at distances r1 and r2 from the centre of a uniform sphere of radius R 1 and mass m are E and E , respectively. Then, = = g1 g2 P 1 Eg1 r1 B [a] if r1 < R and r2 < R Eg 2 r2 09. A bubble is rising from bottom of a lake to the top 2 slowly such that its diameter gets doubled. If a Eg1 r2 [b] 2 if r1 > R and r2 > R Eg 2 r1 barometer placed on the bank of lake reads ‘h’, then depth of lake is (Given is the relative density of Eg1 r1 [c] if r1 > R and r2 > R mercury in barometer) Eg 2 r2 (1) 5h (2) 4h 2 Eg1 r2 [d] if r1 < R and r2 < R (3) 7h (4) 2h E r 2 g 2 1 Sol : Answer (3) (1) [a] and [b] are true (2) [b] and [c] are true The rising of bubble is isothermal (3) [b] and [d] are true (4) [a] and [d] are true P1V1= P2V2 Sol : Answer (1) 3 (H w g P2 )V Pa 2 V
Pa 8 V
H w g = 7Pa 7P 7 h g H a Hg w g w g 7h
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
10. A vessel having area of cross-section ‘A’contains a
liquid up to a height ‘h’ . At the bottom of the vessel,
there is a small hole having area of cross-section ‘a’.
Then the time taken for the liquid level to fall from
height ‘H1’ to ‘H2’ is given by A 2 (1) 2gH1 H 2 (2) H1 H 2 a g A g (3) H H (4) 2gH 1 2 a 2
Sol : Answer (2) :
12. Two rods, one of aluminium and the other made of
steel, having initial lengths l1 and l2 are connected
together to form a single rod of length l1 + l2 .The coefficients of linear expansion for aluminium and
steel are a and s , respectively . If the length of each rod increases by the same amount when their temperature are raised by t 0C , then find the ratio l 1 , l1 l2 (1) s (2) a a s A 2 a s t . H H . (3) (4) 1 2 s a s a a g Sol : Answer (4)
11. Steam at 1000 C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 25 0C till the temperature of the calorimeter and its contents rises to 80 0C.The mass of the steam condensed (in kg) is (1) 0.130 (2) 0.065 (3) 0.260 (4) 0.135
13. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied Sol : Answer (1) which increases the internal energy of the gas is 2 3 (1) (2) 5 5
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
3 5 constants. Find the period of small oscillations of the (3) (4) 7 7 particle about its equilibrium position in the field. Sol : Answer (4) 3 3 4mk1 8mk1 (1) 4 (2) 4 k2 2 k2
2mk3 2mk3 (3) 4 1 (4) 2 1 k 4 k 4 2 2
Sol : Answer (3)
14. Which of the following graphs correctly represents dV the variation of dP with P for an ideal gas V at constant temperature?
(1) (2)
(3) (4)
Sol : Answer (1)
16. A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. The speed (v) of the wave 15. In a given forced field, the potential energy of pulse varies with height (h) from the lower end as a particle is given as a function of its x – coordinate
k1 k2 as Ux , where k1 and k2 are positive x 2 x
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
Sol : Answer (4)
(1) (2)
(3) (4)
Sol : Answer (2)
17. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field E at the centre O is
18. Three capacitors C1, C2 and C3 are connected as shown in the figure given below to a battery of V
volt. If the capacitor C3 breaks down electrically, the change in total charge on the combination of capacitors is
q ˆ q ˆ (1) 2 2 j (2) 2 2 j 2 0r 4 0r q q (3) ˆj (4) ˆj 4 2 r 2 2 2 r 2 0 0
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
C (1) C C V 1 3 1 2 Sol : Answer (4) C1 C2 C3
C1 C2 (2) C1 C2 V 1 C C C 1 2 3
C3 (3) C1 C2 V 1 C1 C2 C3
C2 (4) C1 C2 V 1 C C C 1 2 3 Sol : Answer (1)
20. A current I flows around a closed path in the horizontal plane of the circle as shown in the figure given below. The path consists of eight arcs with alternating radii r and 2r. Each segment of arc subtends equal angle at the common centre P. The magnetic field produced by current path at point P is
19. For the given circuit,
3 I (1) . 0 ; perpendicular to the plane of the paper 8 r and directed inwards 3 I (2) . 0 ; perpendicular to the plane of the paper 8 r and outwards 1 I If internal resistance of cell is 1.5 , then (3) . 0 ; perpendicular to the plane of the paper (1) V – V = 0 (2) V – V = 4 V 8 r P Q P Q and inwards (3) VP – VQ = - 4 V (4) VP – VQ = - 2.5V
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
1 I (4) . 0 ; perpendicular to the plane of the paper If a student plots graphs of the square of maximum 8 r charge (Q 2) on the capacitor with time (t) for two and outwards max different values L1 and L2 (L1 > L2) of L then which of Sol : Answer (1) the following represents this graph correctly?
(1) (2)
(3) (4)
21. A domain in ferromagnetic iron is in the form of a Sol : Answer (1) cube of side length 1 m. Estimate the number of iron atoms in the domain. The molecular mass of iron is 55 g/mol and its density is 7.9 g/cm3. (1) 8.65 10-10 atoms (2) 8 10-13 atoms (3) 8 105 atoms (4) 8.65 1010atoms Sol : Answer (4)
22. An LCR circuit is equivalent to a damped pendulum.
In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below 23. An infinitesimally small bar magnet of dipole moment M is pointing and moving with a speed v in the x direction. A small closed circular conducting loop of radius ‘a’ and negligible self inductance lies in the x - y plane with its centre at x = 0 and its axis conciding with x axis. If x = 2a , the emf induced in the loop is
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
3 Mv 3 Mv 24. An electromagnetic radiation represented by (1) 0 (2) 0 2 2 E a1 costcos t falls over lithium surface 16 a 32 a 0 with work function of 2.39 eV. Maximum kinetic 1 Mv 1 Mv 0 0 energy of emitted photoelectrons will be (3) 2 (4) 2 8 a 16 a (Given, a = 4 N/C, 61014 rads -1 and
3.61015 rads-1) Sol : Answer (2) 0 (1) 0 (2) 0.37 eV
(3) 0.70 eV (4) No photoemission occurs
Sol : Answer (2)
25. In the situation as shown in the figure, the focal length of the thin plane concave lens is 20 cm. A point object P is at distance of 20 cm from the lens. Find the velocity of image formed by plane mirror at the instant as shown in the figure.
ˆ ˆ -1 ˆ -1 (1) 2i 3j cms (2) 3j cms (3) 2ˆi 3ˆj cms-1 (4) 3ˆj cms-1
Sol : Answer (2)
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
(1) wavelength is doubled and the frequency remains uncharged (2) wavelength is doubled and frequency becomes half (3) wavelength is halved and frequency remains uncharged (4) wavelength and frequency both remains uncharged
Sol : Answer (3)
26. In an interference pattern of Young’s double still experiment, we observe the 12th order maxima for wavelength 600 nm at a point on the screen. What order will be visible at the same point, if the source is replaced by light of wavelength 480 nm? 28. A diatomic molecules is made of two masses m1 and th th m2 which are separated by a distance r. If we (1) 4 (2) 10 calculate its rotational energy by applying Bohr’s (3) 15th (4) 20th rule of angular momentum quantisation, its energy will be given by (n is an integer) 2 2 2 2 2 Sol : Answer (3) m1 m2 n h n h (1) 2 2 2 (2) 2 2m1 m2 r 2m1 m2 r 2n2h2 m m n2h2 (3) (4) 1 2 m m r 2 2m m r 2 1 2 1 2
Sol : Answer (4)
27. An electromagnetic wave of frequency = 3.0 MHz passes from vacuum into a dielectric medium with permittivity = 4.0. Then
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
30. For the given logic circuit
Which of these is correct?
(1) y = 0 for x1 = x3 = 0 and x2 = 1
(2) y = 0 for x1 = x2 = x3 = 0
(3) y = 1 for x1 = x2 = x3 = 1
(4) y = 1 for x1 = x2 = 1 and x3 = 0 Sol : Answer (1)
when x1 = x3 = 0 and x2 = 1, inputs to AND gate are 1 and 1 so, y = 0 due to inversion of NOT gate
29. Half-life of a neutron is about 693 s and mass of PART : B — CHEMISTRY neutron is 1.6 10-27 kg. A beam of 10000 high 31. A sample of a hydrate of barium chloride weighing speed neutrons with kinetic energy 0.08 eV is fired of 61 g was heated until all the water of hydration is in space. Number of neutrons that possibly decay in removed. The dried sample weighed 52g. The travelling through a distance of 4 km will be formula of the hydrated salt is (1) 10 (2) 100 (atomic mass Ba = 137 amu) (3) 1000 (4) 10000 (1) BaCl2. H2O (2) BaCl2.2H2O
(3) BaCl2.3H2O (4) BaCl2.4H2O Sol : Answer (1) Sol : Answer (2)
Weight of hydrated BaCl2 = 61 g
Weight of anhydrous BaCl2 = 52 g Loss in Mass = 61 – 52 = 9g
Mass of H2O removed = 9g 9 Moles of H2O removed = 0.5 18
Molecular mass of BaCl2 = 208 % of H2O in the hydrated BaCl2 9 = 100 14.75% 61 18x 14.75 100 208 18x x = 2
Formula of the hydrated salt is BaCl2. 2H2O 32. At very high pressure, the compressibility
factor of one mole of a gas is given by:
Pb Pb (1) 1 (2) RT RT
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
Pb b At 298 K, G of formation for CH OH( ), (3) 1 (4) 1 3 RT (VRT ) H2O( ) and CO2(g) are -166.2, -237.2 and Sol : Answer (1) -394.4 kJ/Mol respectively. If standard enthalpy of a combustion of Methanol is -726 kJ/Mol. Efficiency For 1 mole of gas (P ) (V – b) = RT V 2 of the fuel cell will be a a (1) 80% (2) 87% At very high pressure, P > so is V 2 V 2 (3) 90% (4) 97% negligible. Sol : Answer (4) P ( V – b) = RT H 726 KJ Mol 1 PV - Pb = RT G (CH OH) 166.2KJ / Mol PV Pb f 3 Z 1 RT RT G f (H 2O) 237.2KJ / Mol
33. The ionisation enthalpy of hydrogen atom is G f (CO2 ) 394.4KJ / Mol 1.312 x 106 J Mol-1. The energy required to excite the electron in the atom from n = 1 to n = 2 is: Gr [394.4 2(237.2)] (166.2) (1) 9.84 x 105J Mol-1 (2) 8.51 x 105J Mol-1 = -702.6 KJ /Mol 5 -1 5 -1. G (3) 6.56 x 10 J Mol (4) 7.56 x 10 J Mol % Efficiency = 100 Sol : Answer (1) H 702.6 KJ / Mol IE = E E = =96.77% 1 726 KJ Mol 1 1.312 x 106 = 0 - E 1 36. The equilibrium constant at 298 K for a reaction 6 1 E1 1.31210 J Mol A + B C + D is 100. If the Initial concentration of all the four species were 1M each, 1.312106 1.312106 E then equilibrium concentration of D will be: 2 22 4 (1) 0.182 (2) 0.818 (Energy of electron in second orbit n = 2) (3) 1.818 (4) 1.182 Energy required when an electron makes Sol : Answer (3) transition from n = 1 to n = 2 A + B C + D 6 1.31210 6 Initial conc 1 1 1 1 E E2 E1 (1.31210 ) 4 Equilibrium conc 1-x 1- x 1+x 1+x = 9.84 x 105 J Mol-1 1 x2 34. The hybridization of orbitals of N atom in Now Kc 1 x2 - + + NO3 , NO2 and NH4 are respectively 2 2 3 3 2 1 x (1) Sp, Sp , Sp (2) Sp, SP , SP 100 2 (3) SP2, Sp, Sp3 (4) Sp2, Sp3, Sp 1 x Sol : Answer (2) 1 x 10 35. In a fuel cell methanol is used as fuel and oxygen gas 1 x is used as oxidizer. The reaction is: 10 – 10x = 1+ x = 9 3 x = 0.81 CH3OH( ) O (g) CO (g) 2H O() 2 2 2 2 i e [D] at equilibrium = 1+0.81 = 1.818 M
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
2+ o 37. The increasing order of the ionic radii of the given 2Fe (aq)+ 2H2O( ) E = 1.67 V
isoelectronic species is: 2 2 3 2 2- - 2+ + 2+ + - 2- [Fe ] (110 ) 7 (1) S , Cl , Ca , K (2) Ca , K , Cl , S Q 4 3 4 10 PO2 [H ] 0.1(10 ) (4) K+, S2-, Ca2+, Cl- (3) Cl-, Ca2+, K+, S2- 0.059 E E o logQ n Sol : Answer (2) 0.059 38. In a face centered cubic lattice, atom A occupies the 7 E 1.67 log10 1.57V corner positions and atom B occupies the face centre 4 points. If one atom of B is missing from one face 41. A first order reaction goes as follows centre point, the formula of the compound is:
(1) A2B5 (2) A2B
(3) AB2 (4) A2B3 Sol : Answer (1) Share of A from corners = 8 x 1/8 = 1 Share of B, with one face vacant = 5/2 -4 -1 A : B = 1: 5/2 = 2:5 K1 = 1.26 x 10 s -5 -1 Formula is A2B5 K2 = 3.8 x 10 s 39. In 0.2 molal aqueous solution of a weak acid HX, the The percentage of B in the mixture of B and C is likely to be degree of ionization is 0.3.Taking Kf of water is 1.85, the freezing point of the solution will be nearest (1) 80% (2) 76.83% to: (3) 92% (4) 68% (1) - 0.480oC (2) -0.360oC Sol : Answer (2) (3) -0.260oC (4) + 0.480oC 1.26104 Sol : Answer (1) % Of B 76.83% 1.26104 3.8105 HX H+ + X- 42. Coagulation of 90 ml of a negative sol requires 10ml 1 Mole 0 0 of 0.5 M NaCl. The coagulation value of NaCl is 1-0.3 0.3 0.3 (1) 25 (2) 50 i = ( 1- 0.3) + 0.3 + 0.3 = 1.3
Tf = i. Kf . M = 1.3 x 1.85 x 0.2 = 0.48 (3) 75 (4) 100 Freezing point = 0oC – 0.48oC = -0.480C Sol : Answer (2) 40. Consider the following cell reaction: 10ml of 0.5 M NaCl = 10 x 0.5 millimoles of NaCl = 5 millimoles of NaCl. 2Fe (s) + O (g) + 4H+(aq) 2Fe2+(aq)+ 2H O( ) 2 2 o 2+ -3 100 ml of total sol requires E = 1.67 V At [Fe ] = 10 M, Po2 = 0.1atm and NaCl = 5 millimoles pH = 3 the cell potential at 25oC is: 1000 ml (ie 1 litre) sol requires (1) 1.47 V (2) 1.77 V NaCl = 50 millimoles (3) 1.87 V (4) 1.57V 43. In the context of Ha ll Heroult process for the Sol : Answer (4) extraction of Al, which of the following statements is 2+ 2Fe Fe n = 4 false? pH=3 [H+] = 1 x 10-3M (1) Na3AlF6 serves as the electrolyte + 2Fe (s) + O2(g) + 4H (aq) (2) CO and CO2 are produced in this process
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
(3) Al2O3 is mixed with CaF2 which lowers the M.P. (1) reducing agent (2) Oxidising agent of the mixture and brings conductivity. (3) Bleaching agent (4) None of these (4) A13+ is reduced at the cathode to form Al Sol : Answer (1) Sol : Answer (1) 44. Which is thermodynamically the most stable allotropic form of phosphorus?
(1) Red (2) White
(3) Black (4) Yellow Sol : Answer (3) 48. The least stable carbonate of alkali metal is: 45. How many moles of KMnO4 are required to oxidise
one mole of ferrous oxalate FeC2O4 in acidic (1) Na2CO3 (2) Li2CO3 medium? (3) K2CO3 (4) Cs2CO3 2 1 (1) (2) Sol : Answer (2) 5 5 Li2CO3 is unstable to heat, Li+ being very small in 5 3 2- (3) (4) size polarises a large CO3 ion leading to the 4 5 formation of more stable Ci2O and CO2 Sol : Answer (4) 49. Silica is soluble in: - + 2+ 3MnO4 + 24 H + 5FeC2O4 3Mn + (1) HCl (2) HNO3 3 12 H2O + 5Fe + 10 CO2 (3) H2SO4 (4) HF
5 moles of FeC2O4 = 3 Moles of KMnO4 Sol : Answer (4)
3 HF is the only acid in which silica (SiO2) is soluble 1 mole of FeC2O4 = moles of KMnO4 5 due to the formation of Silicon tetra fluoride as 46. Low spin complex of d6 cation in an octahedral field SiO2 + 4HF SiF4 + 2H2O will have the following energy in total. 50. A hydrocarbon contains 20% hydrogen and 80% of 12 12 carbon. The empirical formula is: (1) o P (2) o 3P 5 5 (1) CH4 (2) CH3
2 2 (3) CH2 (4) CH (3) o 2P (4) o P 5 5 Sol : Answer (2) Sol : Answer (2) Elemen Percentag Atomi Relativ Simples 6 222 o For low spin complex d is t2g eg . t e c mass e t ratio number Total energy will be o + energy of 3 pair of of electrons atoms
Net o + 3p = -6 x 0.4 + 3p 20 H 20 1 20 3 12 6.66 = 3P 5 o 47. In the reaction 6.66 C 80 12 6.66 1 6.66 Ag2O + H2O2 2Ag + H2O + O2
H2O2 act as Empirical formula = CH3
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
Electron withdrawing effect is maximum at ortho and
H2SO4 / HgSO4 para position than meta position. 51. CaC2 +H2O AB
(1) C2H2 and CH3CHO (2) CH4 and HCOOH
(3) C2H4 and CH3COOH (4) C2H2 and CH3COOH Sol : Answer (1) 55.
CaC2 + 2H2O CH CH + Ca(OH)2
H 2SO4 Tautomerism CH CH H 2O CH 2 CH OH HgSO4
CH 3CHO
aq.KOH KMnO4 / H What is the major product ‘A’ 52. CH 3CH 2 Br A B
NH3 C Br2 D D is: KOH
(1) CH3Br (2) CH3CONH2
(3) CH3NH2 (4) CHBr2 (1) (2) Sol : Answer (3)
aq.KOH KMnO4 / H CH 3CH 2 Br CH 3CH 2OH
NH3 Br2 CH 3COOH CH 3CONH 2 CH 3 NH 2 KOH
53. Phenol Zndust X CH3ClY KMnO4 /OHZ (3) (4) anhy.AlCl 3 Sol : Answer (3) The product ‘z’ is: Ring expansion (1) Toluene (2) Benzaldehyde 56. Which one is most reactive towards nuceloplilic (3) Benzoic acid (4) Benzene addition reaction? Sol : Answer (3)
54. In the following compounds
Sol : Answer (4)
Electron withdrawing groups increase the reactivity
towards nuclephilic addition. (1) iii > iv> i > ii (2) i > iv > iii > ii (3) ii > i > iii > iv (4) iv > iii > i > ii 57. When CH3CONH2 is heated with P2O5, the product Sol : Answer (4) is:
Electron withdrawing group stabilizes the phenoxid (1) CH3COOH (2) CH3COONH4 anion. (3) CH3CN (4) CH3COCl’
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
Sol : Answer (3) = tan1 cot11 P2O5 CH 3 C NH 2 CH 3C N H 2O 4 ll D O 1 HNO2 PCl3 NH3 = tan cot22 58. C2 H5 NH 2 AB C 2 4 recognise the compound ‘C’ from the following 1 (1) Propane nitrile (2) Methyl amine = tan cot 4 (3) Ethyl amine (4) Acetamide = tan11 Sol : Answer (3) CH CH NH HNO2 CH CH OH PCl3 = 3 2 2 3 2 4 CH CH Cl NH3 CH CH NH 3 2 3 2 2 62. The distance of the point (1, -2, 4) from the plane 59. Polysaccharides have which of the following passing through the point (1, 2, 2) and perpendicular linkage? to the planes x – y + 2z = 3 and 2x – 2y + z = 0 is: (1) Glycosidic linkage (2) H-bond (1) 2 (2) 2 1 (3) Peptide linkage (4) No linkage (3) 2 2 (4) Sol : Answer (1) 2 In polysaccharides different monosaccharides units Sol : Answer (3) are linked to one another by glycosidic linkage iˆ ˆj kˆ 60. Salol is Normal vector = 1 1 2 (1) Acetyl salicylic acid (2) Phenyl salicylate 2 2 1 (3) Methyl salicylate (4) None of these = Sol : Answer (2) iˆ1 4 ˆj1 4 kˆ 2 2 Phenyl salicylate is known as salol. It is used as an antiseptic. It is prepared by the reaction of salicylic = 3iˆ 3ˆj 0kˆ acid with phenol. Equation of the plane is 3x 13y 2 0
PART : C — MATHEMATICS 3x 3y 9 0 43 31 3 2 9 61. The principal value of tan-1(cot ) is: Distance = 4 32 32 02 3 3 (1) (2) 3 6 9 4 4 = 18 (3) (4) 12 12 4 4 4 = 2 2 Sol : Answer (4) 3 2 3 2 2
1 43 tan cot ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 4 63. If the vector ai j k , i bj k and i j ck , are coplanar, then the value of 44 = tan1 cot 1 1 1 4 4 1 a 1 b 1 c
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
1 1 211cos 1 0 (1) -1 (2) 2 2 2cos 3 (3) 1/2 (4) 1 2cos 1 1 cos Sol : Answer (4) 2 a 1 1 3 1 b 1 0 c a 2b 3ab 1 1 c 2 2 cˆ aˆ 2 4bˆ2 9aˆ bˆ 4aˆ.bˆ12bˆ.aˆ bˆ 6aˆ.aˆ bˆ
Taking 2 1 = 1 4 911sin 4 0 0 3 2 C2 R2 – R1, 3 C3 R3 – R2 5 9 2 4 a 1 a 0 27 5 2 1 b 1 1 b 0 4 1 0 c 1 27 = 7 a[b 1c 1 0] 1 ac 1 1 b 0 4 28 27 = ab 1c 11 ac 1 1 a1b 0 4 a 1b 1c 1 a 1c 1 a 1b 0 55 = 4 Dividing by 1 a1b1c 55 a 1 1 cˆ 0 2 1 a 1b 1 c 2cˆ 55 a 1 1 1 1 0 1 a 1b 1 c 4 1 65. If A = then the determinant of the matrix 1 1 1 3 1 1 2019 2018 2017 1 a 1 b 1 c A – 2A - A is: (1) 25 (2) -25 64. Let a and b be two unit vector such that (3) 75 (4) -75 a b 3 . If c a 2b 3(ab) then 2 c = Sol : Answer (1) (1) 55 (2) 51 4 1 A 3 1 (3) 43 (4) 37 2019 2018 2017 Sol : Answer (1) A 2A A 2017 2 a b 3 A A 2A I
2 A2017 A2 2A I a b 3 2017 2 2 2 A A 2A I aˆ 2aˆ .bˆ bˆ 3
2017 20 5 2 ˆ 2 1 1 2 aˆ b cos 1 3 15 5
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
125 25 1 a b 4 1 x a b 3 66. The domain of the function f(x) = sin 1 is 1 b 3 2 (1) [-3, 3] (2) [-1,1] b 4 b 4 (3) (,1][1,) (4) (,3](3,)
Sol : Answer (2) x t dt / 2 11 x 68. The value of lt = 0 sin 2 2 x sin(2x ) 2 1 x (1) (2) / 2 0 1 2 (3) / 4 (4) /8 Sol : Answer (3) 0 1 x 2 x 1 x 1 tdt lt 2 0 0 1 x 1 x sin2x 2 1 x 1 x lt L.H.R Domain = 1,1 x cos2x 2 2 x2 x 1 67. If lt ( ax b) 4 , then x x 1 2 (1) a = 1, b = 4 (2) a = 1, b = - 4 cos 0 2 4 (3) a = 2, b = -3 (4) a = 2, b = 3 69. For xR, f (x) log 2 sin x and g(x) = f(f(x)) then Sol : Answer (2) (1) g1(0) = cos (log 2) (2) g1(0) = - cos (log 2) x 2 x 1 lt ax b 4 (3) g is differentiable at x = 0 and g1(0) = -sin (log 2) x x 1 (4) g is not differentiable at x = 0 x2 x 1 ax bx 1 lt 4 Sol : Answer (1) x x 1 gx f f x x2 x 1 ax2 ax bx 1 lt 4 g'x f ' f x f 'x x x 1 f x log 2 sin x 1 ax2 1 a bx lt 4 log 2 sin x x x 1 f 'x cos x log 2 sin x 1 a 0 a 1 log 2 sin 0 1 a bx f '0 cos 0 lt 4 log 2 sin 0 x x 1 1 a b = - 1 4 1 1 log 2 sinlog 2 f log 2 coslog 2 log 2 sinlog 2 1coslog 2 coslog 2
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
g '0 f ' f 0 f '0 2x3 5x9 = dx 2 5 3 f 'log 2 f '0 1 x x coslog 21 coslog 2 [put , 1 x2 x5 t 5 x dt 70. For x(0, ), define f(x) = t sin t dt then f 2x 3 5x 6 2 0 dx has 3 6 2x 5x dx dt ] (1) Local maximum at and 2
(2) Local minimum at and 2 dt t 2 1 (3) Local minimum at and Local maximum at 2 = = c c 3 2 5 2 (4) Local maximum at and Local minimum at 2 t 2 21 x x Sol : Answer (4) 1 1 c c x 2t 2 2 5 2 f x t sin t dt 21 x x 0 1 ' = c f x x sin x 2 1 1 2 1 1 2 5 f ''x x cos x sin x x x 2 x 1 = 2 c f 'x 0 x sin x 0 x5 x3 1 2 5 sin x 0 x 0, x 10 x , 2 sin , x 0, 5 x = 2 c 2 2x5 x3 1 1 f 11 cos sin 0 2
11 1 f 2 2 cos 2 sin 2 2 0 2 log3 xsin x 2 2 72. The value of 2 2 dx log2 sin x sin(log 6 x ) 2x12 5x9 1 1 71. The integral dx (1) log(3/ 2) (2) log(3/ 2) 5 3 3 (x x 1) 2 4 x10 x5 (3) log(3/2) (4) 1/6 log(3/2) (1) c (2) c 5 3 2 5 3 2 2(x x 1) 2(x x 1) x10 x5 (3) c (4) c 5 3 2 5 3 2 2(x x 1) (x x 1) Sol : Answer (2) Sol : Answer (1) 2 log3 x sin x 2x12 5x9 dx dx log2 sin x2 sin log6 x2 5 3 3 x x 1
2x12 5x9 dx 5 2 5 3 x 1 x x = 2x12 5x9 = dx 15 2 5 3 x 1 x x
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
[Put x2 t 1 1 1 ,3, , dt 2 8 2 2x dx dt x dx 2 x log 2 t log 2 x log 3 t log 3 ]
log3 sint dt
log2 sint sinlog6 t 2
1 log3 sin t dt 2 log2 sin t sinlog 6 t 1 log3 log 2 2 2 b f x dx b a a f x f a b x 2 1 3 log 4 2 74. If a curve y = f(x) passes through the point (1, -1) and satisfies the differential equations y (1 + xy)dx = x dy
then f(-1/2) = 3. The area of the region {(x, y): y2 2xand y 4x 1} (1) -4/5 (2) 2/5 is: (3) 4/5 (4) -2/5 (1) 7/32 (2) 5/64
(3) 9/32 (4) 15/64 Sol : Answer (3)
y 1 xy dx xdy Sol : Answer (3) 2 x, y y 2 2x and y 4x 1 ydx xy dx xdy ydx xdy xy 2dx y2 2x (1) ydx xdy xdx y 2 y 4x 1 (2) x Solving (1) and (2) d xdx y 2 4x 1 2x x 2 d xdx 16x 8x 1 2x 0 y 2 16x 10x 1 0 x x2 2 c 16x 8x 2x 1 0 y 2 8x2x 1 2x 1 0 This pass through ( 1 , - 1) 2x 18x 1 0 1 1 x or x 2 8
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
1 3 x = 4 sec2 dx 1 1 2 2 c 4 1 2 3 1 1 x 4 tan c 1 1 2 2 = 2 2 2 1 x x 1 2 y 2 2 4
1 1 1 1 3 x x 4 2 2y 8 2 tan 1 5 2 4 2y 8 3 8 tan tan 2y 8 8 5 2 1 2 1 2 4 y 5 77. The integral 75. The probability of a man hitting a target is 2/5. He sin 2 xcos2 x fires at the target k times (k, a given number). Then dx 5 3 2 3 2 5 2 the minimum k so that the probability of hitting the (sin x cos xsin x sin xcos x cos x) target at least once is more than 7/10 is: 1 1 (1) 3 c (2) 3 c (1) 3 (2) 5 1 cot x 1 cot x (3) 2 (4) 4 1 1 (3) 3 c (4) 3 c Sol : Answer (1) 3(1 tan x) 3(1 tan x) P [hitting at least 1 time ] = 1 –P [ hitting no time] Sol : Answer (4) k 3 sin2 xcos2 x = 1 dx 5 3 2 3 2 5 2 5 (sin x cos xsin x sin xcos x cos x) k sin 2 x cos 2 x 3 7 dx 1 (sin 3 x cos 3 x) 2 5 10 = 6 k Dividing by cos x 3 3 tan 2 xsec2 x Then = dx 5 10 3 (tan x 1) From options k =3 [ Put
tan3 x 1 t 3 / 4 dx 76. The value of 2 2 dt / 4 1 cos x 3tan xsec x ] dx (1) 2 (2) 4 dt (3) -2 (4) 0 3 dx 2 Sol : Answer (1) (t ) 1 3 1 t 1 dx c I 4 3 1 3t 1 cos x 4 1 3 c 3 dx 3(tan x 1) = 4 x 4 2cos 2 2
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
2 2 107 78. If 5(tan x – cos x) = 2 cos 2x + 9, then the value of 101 107 101 2 cos 4x is: 101 214 (1) 1/3 (2) 2/9 (3) -7/9 (4) -3/5 2 Sol : Answer (3) 2 1 1 5tan2 x cos2 x 2cos 2x 9 80. Let R = { (3, 3), (6, 6), (9, 9), (12, 12), (6, 12), sin 2 x 5 cos2 x 2 2cos2 x 1 9 (3, 9), (3, 12), (3, 6) } be a relation on the set 2 cos x A = { 3, 6, 9 12}. The relation is: 2 4 (1) An equivalence relation 1 cos x cos x 2 5 4cos x 7 (2) Reflexive and symmetric only cos 2 x (3) Reflexive and transitive only 55cos2 x 5cos4 x 4cos4 x 7cos2 x (4) Reflexive only 4 2 9cos x 12cos x 5 0 Sol : Answer (3) 12 144 180 81. Let z, w be complex numbers such that z i w 0 cos 2 x 18 and arg (zw) = , then arg (z) = 12 324 5 = (1) (2) 18 4 2 12 18 3 (3) (4) 18 4 4 6 30 or Sol : Answer (3) 18 18 z i w 0 1 5 or z i w 3 3 z i w 1 cos 2 x argz argiw 3 argi argw 2 1 1 cos 2x 2cos x 1 2 1 3 3 argz cos 4x 2cos 2 2x 1 2
2 3 1 2argz 2 1 2 3 3 2 7 argz 1 4 9 9 82. If twice the 11th term of an AP is equal to 7 times of 79. If , C are the distinct roots of the equation its 21st term, then its 25th term is equal to: x2 – x + 1 = 0, then101 107 (1) 24 (2) 120 (3) 0 (4) 48 (1) 1 (2) 2 (3) -1 (4) 0 Sol : Answer (3) Sol : Answer (1) Let a is the first term of AP and d is common difference ∴ 11th term will be a + 10d 2 , 21st term will be a + 20d We need to find 25th term which is a + 24d Here given that 2 times of 11th term is equals to 7 times the 21st term
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
2( a + 10d) = 7( a + 20d) co-ordinate of the point of intersection of the lines 2a + 20d = 7a + 140d 3x + 4y = 9 and y =mx + 1 is also an integer is: 5a + 120d = 0 or (1) 2 (2) 0 a + 24d =0 (which is 25th term) (3) 4 (4) 1 th ∴ 25 term of given AP = 0 Sol : Answer (1) 83. The number of arrangements of the letters of the 5 word ‘BANANA’ in which two N’s do not appear Solving, 3x 4y 9, y mx 1 we get x adjacently is: 3 4m (1) 40 (2) 60 x is an integer if 3 + 4 m = 1, -1, 5, -5 2 4 2 8 (3) 80 (4) 100 m , , , Sol : Answer (1) 4 4 4 4 so m has two integral values. 6! 720 180 87. The number of common tangents to the circles 2!2! 4 x2 + y2 – 4x – 6y–12 = 0 and 2! 5! 120 x2 + y2 + 6x + 18y + 26= 0 is: two N’s come together 60 2! 2! 2 (1) 1 (2) 2 required 180 60 120 (3) 3 (4) 4 4 th 84. If x occurs in the r term in the expression of Sol : Answer (3) (x4 + 1/x3)15 then r = (1) 7 (2) 8 (3) 9 (4) 10 Sol : Answer (3)
r 4 15r 1 tr1 15 C4 x x3 607r 15 C4 x 60 7r 4 56 7 r 8 r 1 81 9th 85. In ABC, if (a + b + c) ( a – b + c) = 3ac, then (1) B 600 (2) B 300 88. The point of intersection of the normals to the parabola y2 = 4x at the ends of its latus rectum is: (3) C 600 (4) A C 900 (1) (0,2) (2) (3, 0) Sol : Answer (1) (3) (0,3) (4) (2,0) a b ca b c 3ac Sol : Answer (2) a c2 b2 3ac End of the latus rectum are a2 2ac c2 b2 3ac 1,1 2 t 2,2t t 1 2 2 2 a c b ac Equation of the normal at t 2 ,2t is 2 2 2 a c b 1 2 y tx 2t t 2ac 2 Equation of the normal for t 1 are 1 cos B 2 y x 3and y x 3 B 600 86. The number of integral values of m, for which the x They intersect at (3, 0) Or
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018
Where x is mean
x2 y 2 89. Let the equation of two ellipses be E1 : 1 3 2 x2 y 2 and E2 : 2 1, if the product of their 16 b eccentricities is 1/2, then the length of the minor axis
of E2 is: (1) 8 (2) 9 (3) 4 (4) 12 Sol : Answer (3) 1 e1 e2 = 2 c c 1 1 2 a1 a2 2
2c1 c2 = a1a2 …………………….(1)
c1 3 2 1 2 c1 16 b
2 16 b2 = 3 4 4(16 - b2) = 48 64 - 48 = 4b2 16 = 4 b2 b = 2 2b = 4 90. If the standard deviation of the number 2, 3 a and 11 is 3.5, then which of the following is true (1) 3a2 – 32 a + 84 = 0 (2) 3a2 – 34a + 91 = 0 (3) 3a2 – 23a + 44 = 0 (4) 3a2 - 26a + 55 = 0 Sol : Answer (1) As we learnt in , Standard Deviation -
If x1, x2...xn are n observations then square root of the arithmetic mean of
Registration Started BHABHA CRASH Course – NEET/JEE/KERALA ENGG. & PHARMACY. Course commencement: Immediate after the Plus two Board examination.
BHABHA INSTITUTE OF SCIENCES - KOZHIKODE, KOTTAKKAL & MANJERI HINTS & SOLUTIONS/ JEE MAIN MODEL/ 02 -01 -2018