Free Essay

Advantage and Disadvantages of Media

In: Social Issues

Submitted By fourier
Words 10145
Pages 41
Note:

For the benefit of the students, specially the aspiring ones, the question of JEE(advanced), 2013 are also given in this booklet. Keeping the interest of students studying in class XI, the questions based on topics from class XI have been marked with ‘*’, which can be attempted as a test. For this test the time allocated in Physics, Chemistry & Mathematics and Physics are 22 minutes, 21 minutes and 25 minutes respectively.

FIITJEE SOLUTIONS TO JEE(ADVANCED)-2013
CODE

PAPER 2

3

Time: 3 Hours Maximum Marks: 180ase read the instructions carefully. You are allotted 5 minutes specifically for this purpose. INSTRUCTIONS A. General: 1. This booklet is your Question Paper. Do not break the seals of this booklet before being instructed to do so by the invigilators. 2. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadgets are NOT allowed inside the examination hall. 3. Write your name and roll number in the space provided on the back cover of this booklet. 4. Answers to the questions and personal details are to be filled on a two-part carbon-less paper, which is provided separately. These parts should only be separated at the end of the examination when instructed by the invigilator. The upper sheet is a machine-gradable Objective Response Sheet (ORS) which will be retained by the invigilator. You will be allowed to take away the bottom sheet at the end of the examination. 5. Using a black ball point pen darken the bubbles on the upper original sheet. Apply sufficient pressure so that the impression is created on the bottom duplicate sheet. B. Question Paper Format 6. The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of three sections. Section 1 contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE are correct 7. Section 2 contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has ONLY ONE correct answer among the four choices (A), (B), (C) and (D). 8. Section 3 contains 4 multiple choice questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has ONLY ONE CORRECT ANSWER among the four choices (A), (B), (C) and (D). C. Marking Scheme 9. For each question in Section 1, you will be awarded 3 marks if you darken all the bubble(s) corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded. 10. For each question Section 2 and 3, you will be awarded 3 marks if you darken the bubble corresponding to only the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-2

PART - I: PHYSICS
SECTION – 1 (One or more options correct Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. *1. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are) (A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
GM L (B) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4 GM . L (C) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2 2GM L (D) The energy of the mass m remains constant.

Sol.

(B)
2GMm 1  mv 2  0 L 2 GM  v2 L Note: The energy of mass ‘m’ means its kinetic energy (KE) only and not the potential energy of interaction between m and the two bodies (of mass M each) – which is the potential energy of the system.

*2.

A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity u 0. When the speed of the particle is 0.5 u0. It collides elastically with a rigid wall. After this collision, (A) the speed of the particle when it returns to its equilibrium position is u0. (B) the time at which the particle passes through the equilibrium position for the first time is t   (C) the time at which the maximum compression of the spring occurs is t  m . k

4 m . 3 k (D) the time at which the particle passes through the equilibrium position for the second time is t 5 m . 3 k

Sol.

(A, D) u0   u 0 cos t1  t1  2 3 2 Now the particle returns to equilibrium position at time t2 = 2t1 i.e. with the same mechanical energy 3 i.e. its speed will u0 . Let t3 is the time at which the particle passes through the equilibrium position for the second time.

v  u 0 sin t (suppose t1 is the time of collision)

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-3
T  2t1 2  2 5     3 3


 t3 

5 m 3 k Energy of particle and spring remains conserved.

3.

A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is (are) (A) In the region 0 < r < R, the magnetic field is non-zero (B) In the region R < r < 2R, the magnetic field is along the common axis. (C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis. (D) In the region r > 2R, the magnetic field is non-zero. (A, D) Due to field of solenoid is non zero in region 0 < r < R and non zero in region r>2R due to conductor. Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f1 . An observer in the other vehicle hears the frequency of the whistle to be f 2 . The speed of sound in still air is V. The correct statement(s) is (are) (A) If the wind blows from the observer to the source, f 2  f1 . (B) If the wind blows from the source to the observer, f 2  f1 . (C) If the wind blows from observer to the source, f 2  f1 . (D) If the wind blows from the source to the observer f 2  f1 .

Sol.

*4.

Sol.

(A, B) If wind blows from source to observer Vwu f 2  f1   Vwu  When wind blows from observer towards source Vwu f 2  f1   Vwu In both cases, f 2  f1 . Using the expression 2d sin = , one calculates the values of d by measuring the corresponding angles  in the range  to 90°. The wavelength  is exactly known and the error in  is constant for all values of . As  increases from 0°, (A) the absolute error in d remains constant. (B) the absolute error in d increases (C) the fractional error in d remains constant. (D) the fractional error in d decreases. (D)
 2sin   ln d  ln    ln sin  2 d

*5.

Sol.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-4
d cos d 0 d sin  d     =  cot   d max Also (d)max  d cot   cot  2sin   cos    2 sin 2 

As  increases cot decreases and 6.

cos  sin 2 

also decreases.

Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities +  and –, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region, (A) the electrostatic field is zero (B) the electrostatic potential is constant (C) the electrostatic field is constant in magnitude (D) the electrostatic field has same direction (C, D) In triangle PC1C2    r2  d  r1 The electrostatic field at point P is  4 3  4   K   R1  r2 K   R 3    r1  2   3    3  E 3 3 R1 R2  4   E  K (r2  r1 ) 3    E d 30

 R1

– R2

Sol.

  r1 –  r2 P C2 C1 R2 R1 d *7.

The figure shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the following statement(s) is (are) correct to a reasonable approximation. (A) the rate at which heat is absorbed in the range 0-100 K varies linearly with temperature T. (B) heat absorbed in increasing the temperature from 0-100 K is less than the heat required for increasing the temperature from 400 – 500 K. (C) there is no change in the rate of heat absorption in range 400 – 500 K. (D) the rate of heat absorption increases in the range 200 – 300 K.

C

100

200

300 400 T (K)

500

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-5
Sol. (A, B, C, D) Option (A) is correct because the graph between (0 – 100 K) appears to be a straight line upto a reasonable approximation. Option (B) is correct because area under the curve in the temperature range (0  100 K) is less than in range (400  500 K.) Option (C) is correct because the graph of C versus T is constant in the temperature range (400  500 K) Option (D) is correct because in the temperature range (200 – 300 K) specific heat capacity increases with temperature. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0 where a0 is the Bohr radius. Its 3h orbital angular momentum is . It is given that h is Planck’s constant and R is Rydberg constant. The 2 possible wavelength(s), when the atom de-excites, is (are) 9 9 9 4 (A) (B) (C) (D) 32R 16R 5R 3R (A, C) Given data
4.5a 0  a 0 n2 Z

8.

Sol.

…(i)

nh 3h  …(ii) 2 2 So n = 3 and z = 2 So possible wavelength are 1 9 1 1   RZ2  2  2   1  1 32R 1 3 
1 1 1 1   RZ 2  2  2    2  2 3R 1 2  1 9 1 1  RZ2  2  2   3  3 5R 2 3 

SECTION – 2 : (Paragraph Type)
This section contains 4 paragraphs each describing theory, experiment, date etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of paragraph has only one correct answer along the four choice (A), (B), (C) and (D). Paragraph for Questions 9 to 10 A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s2).
R 30 P

Q

R

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-6

*9.

The speed of the block when it reaches the point Q is (A) 5 ms1 (B) 10 ms1 (C) 10 3 ms1 (B) Using work energy theorem 1 mg R sin 30 + Wf = mv 2 2 2 v 200  150 = 2 v = 10 m/s The magnitude of the normal reaction that acts on the block at the point Q is (A) 7.5 N (B) 8.6 N (C) 11.5 N (A) N  mg cos 60 =
N  5 mv 2 R

(D) 20 ms1

Sol.

*10.

(D) 22.5 N

Sol.

5  7.5 Newton. 2

Paragraph for Questions 11 to 12 A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers’ end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with the power factor unity. All the currents and voltage mentioned are rms values. 11. If the direct transmission method with a cable of resistance 0.4  km1 is used, the power dissipation (in %) during transmission is (A) 20 (B) 30 (C) 40 (D) 50 (B) For direct transmission P  i 2 R  (150) 2 (0.4  20) = 1.8  105 w fraction(in %)  1.8 105 100  30% 6 105

Sol.

12.

In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is (A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1 (A) 40000  200 200

Sol.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-7
Paragraph for Questions 13 to 14 A point Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity . This can be Q considered as equivalent to a loop carrying a steady current . A uniform magnetic field along the positive z-axis 2 is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant . 13. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change, is BR BR (A) (B) (C) BR (D) 2BR 4 2 (B) E(2R) = R 2
E RB 2 dB dt

Sol.

14.

The change in the magnetic dipole moment associated with the orbit, at the end of time interval of the magnetic field change, is BQR 2 BQR 2 (A) BQR 2 (B)  (C)  (D) BQR 2 2 2 (B) L   dt
R   Q  B  R(1) 2  QR 2 B  , in magnitude 2  = L BQR 2 = – (taking into account the direction) 2 v Sol.

Paragraph for Questions 15 to 16 The mass of nucleus A X is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in Z the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of mass m 1 and m2 only if (m1 + m2) < M. Also two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M only if (m3 + m4) > M. The masses of some neutral atoms are given in the table below:
1 1 6 3

H Li Gd

1.007825 u 6.015123 u 151.919803 u

2 1 7 3

H Li Pb

2.014102 u 7.016004 u 205.974455 u

3 1

H Zn Bi

3.016050 u 69.925325 u 208.980388 u

4 2

He Se Po

4.002603 u 81.916709 u 209.982876 u

70 30

82 34

152 64

206 82

209 83

210 84

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-8

15.

The correct statement is 6 (A) The nucleus 3 Li can emit an alpha particle
210 (B) The nucleus 84 Po can emit a proton. (C) Deuteron and alpha particle can undergo complete fusion. 70 (D) The nuclei 30 Zn and 82 Se can undergo complete fusion. 34

Sol.

(C) 6 4 2 3 Li  2 He 1 H
Q  6.015123  4.002603  2.014102 C2 0  0.001582  0 So no -decay is possible 210 1 209 84 P0 1 H  83 Bi Q  209.9828766  1.007825  208.980388  0.005337  0 C2 So, this reaction is not possible 2 4 6 1 H  2 He 3 Li Q  2.014102  4.002603  6.015123  0.001582  0 C2 So, this reaction is possible 70 82 152 30 Zn  34 Se 64 Gd Q  69.925325  81.916709  151.919803  0.077769  0 C2 So this reaction is not possible

16.

The kinetic energy (in keV) of the alpha particle, when the nucleus (A) 5319 (B) 5422 (C) 5707 (A) 210 4 206 84 Po  2 He  82 Pb Q = (209.982876  4.002603  205.97455)C2 = 5.422 MeV from conservation of momentum 2K1 (4)  2K 2 (206) 4K1 = 206K2 103  K1  K2 2 K1 + K2 = 5.422 2 K1  K1  5.422 103 105  K1  5.422 103  K1 = 5.319 MeV = 5319 KeV

210 84

Po at rest undergoes alpha decay, is (D) 5818

Sol.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-9 SECTION – 3 (Matching List Type)
This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 17. A right angled prism of refractive index 1 is placed in a rectangular block of refractive index 2, which is surrounded by a medium of refractive index 3, as shown in the figure. A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between 1, 2 and 3, it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’, or ‘ei’. f e i 2 3 450 1 h g

Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists: P. Q. R. S. List I ef eg eh ei 1. 2. 3. 4. List II 1 > 2 2 2 > 1 and 2 > 3 1 = 2 2 < 1 < 2 2 and 2 > 3

Codes: (A) (B) (C) (D) Sol. P 2 1 4 2 Q 3 2 1 3 R 1 4 2 4 S 4 3 3 1

*18.

(D) P.  (2) ; Q.  (3); R.  (4); S.  (1) P. 2 > 1… (towards normal) 2 > 3 … (away from normal) Q. 1 = 2… (No change in path) i = 0  r = 0 on the block. R. 1 > 2 … (Away from the normal) 2 > 3 … (Away from the normal) 1 1 1    2 sin r  sin r = . Since sin r < 1  1 < 2 2 2 2 2 1  S. For TIR : 450 > C  sin 450 > sin C   2  1  2 2 2 1 Match List I with List II and select the correct answer using the codes given below the lists: List I P. Boltzmann Constant Q. Coefficient of viscosity R. Plank Constant S. Thermal conductivity Codes: P Q R (A) 3 1 2 (B) 3 2 1 (C) 4 2 1 (D) 4 1 2 List II [ML2T-1] [ML-1T-1] [MLT-3K-1] [ML2T-2K-1]

1. 2. 3. 4. S 4 4 3 3

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-10

Sol.

(C) P.  (4) ; Q.  (2); R.  (1); S.  (3) 3 P. KE = K T  [ML2T-2] = K[K]  K = [ML2T-2K-1] 2 Q. F = 6rv  [MLT-2] = [L][LT-1]   = [ML-1T-1] h R. E = hf  [ML2T-2] =  h = [ML2T-1] [T] S. dQ KA(  [ML2T 2 ] k[L2 ][K ]    dt x [T] [L] K = [MLT-3K-1]

*19.

One mole of mono-atomic ideal gas is taken along two cyclic processes EFGE and EFHE as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

P 32P0 F

P0

E V0

H

G

V

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists. P. Q. R. S. List I GE GH FH FG 1. 2. 3. 4. List II 160 P0V0 ln2 36 P0V0 24 P0V0 31 P0 V0

Codes: (A) (B) (C) (D) Sol. P 4 4 3 1 Q 3 3 1 3 R 2 1 2 2 S 1 2 4 4

(A) P.  (4) ; Q.  (3); R.  (2); S.  (1) Apply PV1+2/3 = constant for F to H. 5/3 5/3 (32P0) V0 = P0 VH  VH = 8V0 For path FG PV = constant  (32P0)V0 = P0VG  VG = 32V0 Work done in GE = 31 P0V0 Work done in GH = 24 P0V0 P V  PF VF Work done in FH = H H = 36P0V0 ( 2 / f )
V  Work done in FG = RT ln  G   VF  = 160P0V0ln2.

P 32P0 F n =1 f =3

Isothermal

Adiabatic P0 E V0 H 8V0 G 32V0 V

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-11

20.

Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists: List I Alpha decay + decay Fission Proton emission List II 1. 2. 3. 4.
15 15 8 O 7 N  ..... 238 234 92 U  90 Th  ..... 185 184 83 Bi 82 Pb  ..... 239 140 94 Pu  57 La  .....

P. Q. R. S.

Codes: (A) (B) (C) (D) P 4 1 2 4 Q 2 3 1 3 R 1 2 4 2 S 3 4 3 1

Sol.

(C) P.  (2) ; Q.  (1); R.  (4); S.  (3) 15 15 0 8 O  7 N 1  (Beta decay)
238 234 4 92 U 90 Th  2 He 185 184 1 83 Bi 82 Pb 1 H 239 140 99 94 Ph  57 La  37 Rb

(Alpha decay) (Proton emission) (fission)

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-12

PART - II: CHEMISTRY
SECTION –1 (One or more options correct Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. *21. The Ksp of Ag2CrO4 is 1.1  10–12 at 298K. The solubility (in mol/L) of Ag2CrO4 in a 0.1M AgNO3 solution is (A) 1.1  10–11 (B) 1.1  10–10 –12 (C) 1.1  10 (D) 1.1  10–9 (B)
K sp  1.11012   Ag   CrO 2  4    
1.1  10 12   0.1  s 
2

Sol.

2

s  1.1 1010
22. In the following reaction, the product(s) formed is(are) OH

CHCl3  ?  OH 

CH3 OH OHC CHO O OH OH CHO

CH3 P (A) P(major) (C) R(minor) (B, D) OH

H3C Q

CHCl 2

H3C

CHCl 2

CH3 S

R (B) Q(minor) (D) S(major)

Sol.

O

OH CHO

  OH

CHCl3

+

CH3

H3C CHCl 2 (Minor)


CH3 (major)

CHCl3  OH  : CCl 2  H 2O  Cl 

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-13
OH O

+

 OH 

H 2O

CH3

CH3

O

O CCl 2
 : CCl 2  

O CHCl 2
 
OH 

OH CHO

H

CH3
O

CH3
O

CH3
O

CH3 (major)

 

H 2O  

CH3

:CCl 2 H3C

CCl 2

H3C

CHCl 2 (minor)

23.

The major product(s) of the following reaction is (are) OH

 2  ?  aqueous Br 3.0 equivalents

SO3H OH Br Br Br

OH Br

OH

OH Br

Br SO3H P (A) P (C) R Sol. (B) Br Q

Br Br R (B) Q (D) S

Br

Br SO3H S

Br

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-14
OH Br
 2  
Br 3equivalents

OH Br

SO3H 24.

Br (Q)

After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are) O
Br2 1.0 mol  Reaction I : H C  CH3  aqueous/ NaOH 3 1.0 mol 

O Reaction II :
Br2 1.0 mol  H3C CH3  CH3COOH 1.0 mol  

O H3C P CH2Br H3C

O CBr3 Br 3C Q

O CBr3 BrH2C R

O CH2Br S H3C

O ONa T CHBr3 U

(A) (B) (C) (D) Sol.

Reaction I: P and Reaction II : P Reaction I: U, acetone and Reaction II:Q, acetone Reaction I : T, U, acetone and Reaction II : P Reaction I : R, acetone and Reaction II : S, acetone

(C) Solve as per law of limiting reagent. OH O C H3C CH3
  H   H3C

OH CH3
H    


C

C H2C CH3

O
Br   BrCH2

C

Br2 CH3 

25.

The correct statement(s) about O3 is(are) (A) O–O bond lengths are equal. (C) O3 is diamagnetic in nature. (A, C, D)
1.21Å 1170

(B) Thermal decomposition of O3 is endothermic. (D) O3 has a bent structure.

Sol.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-15
*26. In the nuclear transmutation
Be  X  4 Be  Y  (X, Y) is (are) (A) (, n) (C) (n, D)
9 4 8

(B) (p, D) (D) (, p)

Sol.

(A, B) 8 9 1  4 Be    4 Be  0 n
9 4 2 Be 1 P  4 Be 1 H  1 8

Hence (A) and (B) are correct 27. The carbon–based reduction method is NOT used for the extraction of (A) tin from SnO2 (B) iron from Fe2O3 (C) aluminium from Al2O3 (D) magnesium from MgCO3.CaCO3 (C, D) Fe2O3 and SnO2 undergoes C reduction. Hence (C) and (D) are correct. The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions.  CaCO3  s   CaO  s   CO 2  g  For this equilibrium, the correct statement(s) is(are) (A) H is dependent on T (B) K is independent of the initial amount of CaCO3 (C) K is dependent on the pressure of CO2 at a given T (D) H is independent of the catalyst, if any Sol. (A, B, D)  For the equilibrium CaCO3  s   CaO  S  CO 2  g  . The equilibrium constant (K) is independent of initial amount of CaCO3 where as at a given temperature is independent of pressure of CO2. ∆H is independent of catalyst and it depends on temperature. Hence (A), (B) and (D) are correct.

Sol.

*28.

SECTION-2 (Paragraph Type)
This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).

Paragraph for Question Nos. 29 and 30
An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H2S in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H2O2 in an aqueous NaOH medium. 29. The precipitate P contains (A) Pb2+ (C) Ag+ (A) (B) Hg 2  2 (D) Hg2+

Sol.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-16
30. The coloured solution S contains (A) Fe2(SO4)3 (C) ZnSO4

(B) CuSO4 (D) Na2CrO4

Sol. (D) Solution for the Q. No. 29 to 30. Hot Pb 2  2HCl  PbCl 2  soluble.   Water
H2 S Cr 3  Cr  OH 3   ammoniacal

NaOH Cr  OH 3  Na 2 CrO 4  H 2 O2

Yellow solution

Paragraph for Question Nos. 31 to 32
P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2/H2O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U. COOH H H OH OH COOH S *31. H HO COOH OH H COOH T HO H COOH H OH COOH U

Compounds formed from P and Q are, respectively (A) Optically active S and optically active pair (T, U) (B) Optically inactive S and optically inactive pair (T, U) (C) Optically active pair (T, U) and optically active S (D) Optically inactive pair (T, U) and optically inactive S (B)
H C C H P COOH COOH
4   OH

Sol.

COOH
KMnO

H H

OH OH COOH

H C C HOOC Q

COOH
KMnO4  OH

S  optically inactive COOH

COOH OH H H OH COOH U

H HO

OH H COOH T

H

optically inactive pair

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-17
*32. In the following reaction sequences V and W are, respectively H2 / Ni Q  V  
  3  V   W  1. Zn HHg / HCl  2. 3 PO 4
AlCl anhydrous

(A)

O O and

(B)

CH2OH and CH2OH V W

V (C)
O

O

W

O (D)
HOH2C and CH2 OH V W CH2OH

O and

V

O

W

Sol.

(A)

O
H 2 / Ni Q   

O

O O
AlCl3

O C

+

 O 

Zn  Hg, HCl  

C O HO O OH

C O H3PO4

O

Paragraph for Question Nos. 33 to 34
A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-18

K

L

Pressure

N Volume

M

*33.

The succeeding operations that enable this transformation of states are (A) Heating, cooling, heating, cooling (B) Cooling, heating, cooling, heating (C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling (C) The pair of isochoric processes among the transformation of states is (A) K to L and L to M (B) L to M and N to K (C) L to M and M to N (D) M to N and N to K

Sol. *34.

Sol. (B) Solution for the Q. No. 33 to 34.
K P N L

M

V K – L heating, isobaric L – M cooling, isochoric M – N cooling, isobaric N – K heating, isochoric

Paragraph for Question Nos. 35 to 36
The reactions of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T. 35. P and Q, respectively, are the sodium salts of (A) hypochlorus and chloric acids (C) chloric and perchloric acids (A) R, S and T, respectively, are (A) SO2Cl2, PCl5 and H3PO4 (C) SOCl2, PCl3 and H3PO2 (A)

(B) hypochlorus and chlorus acids (D) chloric and hypochlorus acids

Sol. 36.

(B) SO2Cl2, PCl3 and H3PO3 (D) SOCl2, PCl5 and H3PO4

Sol.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-19
Solution for the Q. No. 35 to 36 2NaOH  Cl 2  NaCl  NaClO  H 2O 
 Cold  dil 
P

6NaOH  3Cl2  5NaCl  NaClO3  3H 2 O 
 Hot  conc.
Q
Charcoal

SO 2  Cl2  SO 2Cl 2 
R

SO 2Cl 2  P4  PCl5  SO 2 
S

PCl5  H 2O  H 3 PO 4  HCl 
T

SECTION – 3: (Matching List Type) This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 37. The unbalanced chemical reactions given in List – I show missing reagent or condition (?) which are provided in List – II. Match List – I with List – II and select the correct answer using the code given below the lists: List – I List - II ? (P) (1) NO PbO  H SO  PbSO  O  other product 
2 2 4 4 2

(Q) (R) (S) Codes: (A) (B) (C) (D) Sol. (D)

Na 2S2 O3  H 2 O  NaHSO 4  other product  N 2 H 4  N 2  other product  XeF2  Xe  other product 
? ?

?

(2) (3) (4)

I2 Warm Cl2

P 4 3 1 3

Q 2 2 4 4

R 3 1 2 2

S 1 4 3 1

1  (P) PbO 2  H 2SO4  PbSO 4  H 2 O  O 2  2 (Q) 2Na 2S2 O3  Cl 2  2H 2 O  2NaCl  2NaHSO4  2S 
(R) N 2 H 4  2I 2  N 2  4HI  (S) XeF2  2NO  Xe  2NOF  *38. Match the chemical conversions in List – I with appropriate reagents in List – II and select the correct answer using the code given below the lists: List – I List - II (P) (1) (i) Hg(OAc)2; (ii) NaBH4 Cl  

(Q) ONa   OEt

(2)

NaOEt

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-20
(R)
 

OH

(3)

Et-Br

(S)
 

(4)

(i) BH3; (ii) H2O2/NaOH

OH Codes: (A) (B) (C) (D) Sol. (A) (P)
NaOEt Cl 

P 2 3 2 3

Q 3 2 3 2

R 1 1 4 4

S 4 4 1 1

(Q)

ONa

EtBr  

OEt

(R)

2   ii NaBH 4

(i) HgOAc

OH

(S)

3 2 2    

i BH (ii) H O / NaOH

OH

39.

An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List – I. The variation in conductivity of these reactions in List – II. Match List – I with List – II and select the correct answer using the code given below the lists: List – I List - II (P) (1) Conductivity decreases and then increases C2 H5  N CH3COOH
3 X Y

(Q) (R) (S) Codes: (A) (B) (C) (D) (A)

KI 0.1M  AgNO3 0.01M 
X Y

(2) (3) (4)

CH3 COOH KOH
X Y

NaOH HI
X Y

Conductivity decreases and then does not change much Conductivity increases and then does not change much Conductivity does not change much and then increases

P 3 4 2 1

Q 4 3 3 4

R 2 2 4 3

S 1 1 1 2

Sol.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-21
  C2 H 5 3 N CH3COOH   C2 H 5 3 NH CH 3COO 
X Y

(P)

Initially conductivity increases due to ion formation after that it becomes practically constant because X alone can not form ions. Hence (3) is the correct match.

 (Q) KI  0.1 M  AgNO3  0.01M   AgI   KNO3
X Y

Number of ions in the solution remains constant until all the AgNO3 precipitated as AgI. Thereafter conductance increases due to increases in number of ions. Hence (4) is the correct match. (R) Initially conductance decreases due to the decrease in the number of OH ions thereafter it slowly increases due to the increases in number of H+ ions. Hence (2) is the correct match. (S) Initially it decreases due to decrease in H+ ions and then increases due to the increases in OH ions. Hence (1) is the correct match. 40. The standard reduction potential data at 25C is given below:

  E Fe 2 , Fe  0.44V E Cu 2 , Cu   0.34V; E Cu , Cu   0.52V

E Fe3 , Fe 2  0.77V;

E  O2 g   4H  4e  2H 2 O  1.23V;     E  O2 g   2H 2 O  4e  4OH   0.40V  

  E Cr 2 , Cr  0.91V  

E Cr3 , Cr  0.74V;

Match E0 of the redox pair in List – I with the values given in List – II and select the correct answer using the code given below the lists: (P) (1) 0.18 V E Fe3 , Fe (Q) (R) (S) Codes: (A) (B) (C) (D) Sol. P 4 2 1 3 Q 1 3 2 4 R 2 4 3 1 S 3 1 4 2
 E 4H 2 O  4H  4OH 





(2) (3) (4)

0.4 V 0.04 V 0.83 V

 E Cr3 , Cr 2 

E Cu 2  Cu  2Cu  



(D) o (P) G o 3 / Fe  G o 3 /Fe2  G Fe2  / Fe Fe Fe
 3  FE oFe 3 /Fe  1 FE oFe3 / Fe2  2  FE o 2 / Fe Fe













 E o 3 / Fe  0.04 V Fe

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-22


(Q) O 2  g   2H 2 O  4e   4O H 
2H 2 O  O 2  g   4H  4e 
 So 4H 2 O  4H   4 O H


E o  0.40 V o ... (i) .. (ii) ... (iii)





E  1.23 V

E for III reduction = 0.40 – 1.23 = - 0.83 V. (R) G oCu 2 /Cu  G oCu2 /Cu   G oCu  /Cu

o

rd













2  FE

o Cu 2 /Cu

 1 FE

o Cu 2 /Cu 

o  1 F  E Cu /Cu





E

o Cu 2 /Cu

 0.18 V . o o  G Cr 3 /Cr  G Cr /Cr2

(S) G

o Cr 3 /Cr 2

o o 1 F  E o 3 /Cr 2  3  F  E Cr3 / Cr  2  F  E Cr /Cr 2 Cr





E

o Cr 3 /Cr 2

 0.4 V .

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-23

PART - III: MATHEMATICS
SECTION  1 : (One or more option correct Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

41.

For a  R (the set of all real numbers), a  1, lim Then a = (A) 5 15 (C) 2

1
 n  1 a 1

a

 2a  ...  n a



n 

 na  1   na  2   ...   na  n    



1 . 60

(B) 7 17 (D) 2

Sol.

(B, D)
1

 x dx
Required limit =
0 1

a



  a  x  dx
0

2 2   2a  1 (a  1) 120

17  a = 7 or  . 2

*42.

Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length 2 7 on y-axis is (are) (A) x2 + y2  6x + 8y + 9 = 0 (B) x2 + y2  6x + 7y + 9 = 0 2 2 (C) x + y  6x  8y + 9 = 0 (D) x2 + y2  6x  7y + 9 = 0 (A), (C) Equation of circle can be written as (x  3)2 + y2 +  (y) = 0  x2 + y2  6x + y + 9 = 0. Now, (radius)2 = 7 + 9 = 16

Sol.

2  9  16 4  2 = 64   =  8.  Equation is x2 + y2  6x  8y + 9 = 0.
 9 43. Two lines L1 : x = 5, (A) 1 (C) 3 Sol. (A, D) x 5 y 0 z 0   0 3  2 y z y z  and L2 : x = ,  are coplanar. Then  can take value(s) 3   2 1 2   (B) 2 (D) 4

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-24 x y z   0 1 2   will be coplanar if shortest distance is zero 5 0 0

 0 0
2

3   2 1 2

0

(5 – )( – 5 + 4) = 0,  = 1, 4, 5 so  = 1, 4 Alternate Solution: As x = 5 and x =  are parallel planes so the remaining two planes must be coplanar. 3  2 So,   2 – 5 + 4 = 0   = 1, 4. 1 2 *44.
1 . Further the incircle of the triangle touches the sides 3 PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are) (A) 16 (B) 18 (C) 24 (D) 22

In a triangle PQR, P is the largest angle and cos P =

Sol.

(B), (D) Let s  a = 2k  2, s  b = 2k, s  c = 2k + 2, k  I, k > 1 Adding we get, s = 6k So, a = 4k + 2, b = 4k, c = 4k  2 1 Now, cos P = 3 

P sa N sc M

Q

sb

L

R

b2  c2  a 2 1   3 [(4k)2 + (4k  2)2  (4k + 2)2] = 2  4k (4k  2) 2bc 3  3 [16k2  4 (4k)  2] = 8k (4k  2)  48k2  96k = 32k2  16k  16k2 = 80k  k = 5 So, sides are 22, 20, 18

*45.

3 i and P = {wn : n = 1, 2, 3, ….. }. Further H1 = 2 1    z  C : Re z   , where C is the set of all complex numbers. If z1  2  represents the origin, then z1 Oz2 =   (A) (B) 2 6 2 5 (C) (D) 3 6

Let w =

1   z  C : Re z   and H2 = 2 

P  H1, z2  P  H2 and O

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-25
Sol. (C), (D) i  3 i = e 6 , so w n  e  6  2 n 1 n 1 Now, for z1, cos  and for z2, cos  6 2 6 2

i

 n 

B1 B2 B3 /6 O

A3 A2 A1

w=

x = 1/2 x = 1/2 Possible position of z1 are A1, A2, A3 whereas of z2 are B1, B2, B3 (as shown in the figure) 2 5 So, possible value of z1Oz2 according to the given options is or . 3 6

*46.

If 3x = 4x1, then x = 2 log3 2 (A) 2 log3 2  1 (C)
1 1  log 4 3

(B) (D)

2 2  log 2 3 2 log 2 3 2 log 2 3  1

Sol.

(A, B, C) log23x = (x  1) log24 = 2 (x  1)  x log23 = 2x – 2 2 x= 2  log 2 3 Rearranging, we get 2 log 3 2 2 x=  1 2 log 3 2  1 2 log 3 2 Rearranging again, 1 log3 4 log 4 3 1  x= = . 1 log3 4  1 1  log 4 3 1 log 4 3 Let  be a complex cube root of unity with   1 and P = [pij] be a n  n matrix with pij = i+j. Then P2  0, when n = (A) 57 (B) 55 (C) 58 (D) 56 (B, C, D) 2 3 4  3 4 5  P    n  2 n  3 ..... 
 4  6 .....  5    7  9 ..... 2 P    n  4  n 6 ..... 

47.

Sol.

..... n  2   ..... n  3      ..... 2n  4   5  7  9 ..... .....  .....   ..... .....    .....  ..... 2n  4  2n  6 .....  .....

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-26
P2 = Null matrix if n is a multiple of 3 48. The function f(x) = 2 |x| + |x + 2|  ||x + 2|  2 |x|| has a local minimum or a local maximum at x = 2 (A) 2 (B) 3 2 (C) 2 (D) 3 (A), (B) f x  gx  f x  gx As, = Min (f (x), g (x)) 2 2 x  x2  x 2 2 x  = Min (|2x|, |x + 2|) 2

Sol.

2x + 4  2x  4 2x + 4 2 2/3 8/3 4x 2 4x

According to the figure shown, points of local minima/maxima are x = 2,

2 , 0. 3

SECTION  2 : (Paragraph Type)
This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Paragraph for Questions 49 and 50 Let f : [0, 1]  R (the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) = f(1) = 0 and satisfies f (x)  2f (x) + f(x)  ex, x  [0, 1]. 49. Which of the following is true for 0 < x < 1 ? (A) 0 < f(x) < 
1 (C)  < f(x) < 1 4

(B) 

1 1 < f(x) < 2 2

(D)   < f(x) < 0

Sol.

(D) Let g (x) = e–x f (x) and g (x) > 1 > 0 So, g (x) is concave upward and g (0) = g (1) = 0 Hence, g (x) < 0  x  (0, 1)  e–x f (x) < 0 f (x) < 0  x  (0, 1) Alternate Solution f (x) – 2f(x) + f (x)  ex

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-27

 x2   f  x  ex   2

  0 

Let g (x) = f (x) e–x – g (0) = 0, g (1) = 
1 2

x2 2

Since g is concave up so it will always lie below the chord joining the extremities which is y =   f  x  e x   f (x) <

x 2

x2 x  2 2
2  0  x  (0, 1)

x2  x  ex

50.

If the function ex f(x) assumes its minimum in the interval [0, 1] at x =
1 3  x 4 4 1 (C) f   x   f  x  , 0  x  4 C Let, g (x) = e–x f (x) As g (x) > 0 so g (x) is increasing. So, for x < 1/4, g (x) < g (1/4) = 0  (f (x) – f (x))e–x < 0  f (x) < f (x) in (0, 1/4).

(A) f   x   f  x  ,

1 , which of the following is true ? 4 1 (B) f   x   f  x  , 0  x  4 3 (D) f   x   f  x  ,  x  1 4

Sol.

Paragraph for Questions 51 and 52 Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a > 0. *51. Length of chord PQ is (A) 7a (C) 2a (B)

(B) 5a (D) 3a

Sol.

2a  a Let P  at 2 , 2at  , Q  2 ,   as PQ is focal chord t  t Point of intersection of tangents at P and Q   1  a, a  t     t   as point of intersection lies on y = 2x + a  1  a  t    2a  a  t
1  1 t   1   t    5 t t 
 1 length of focal chord = a  t   = 5a t 
2 2

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-28

*52.

If chord PQ subtends an angle  at the vertex of y2 = 4ax, then tan = 2 2 (A) 7 (B) 7 3 3 2 2 (C) 5 (D) 5 3 3 (D) Angle made by chord PQ at vertex (0, 0) is given by 1  2  2  t   t  2t   t  = 2 5 tan     =  1 4  3 3 Paragraph for Questions 53 and 54

Sol.

Let S = S1  S2  S3, where
   z 1  3 i    S1 = {z  C : |z| < 4}, S2 =  z  C : Im    0  and S3 = {z  C : Re Z > 0}.    1 3 i   

*53.

Area of S = 10 (A) 3 16 (C) 3 (B) Area of region   42 42     4 6 1 1 = 42   4 6 20  = 3

20 3 32 (D) 3

(B)

Sol.

y  3x  0

S1

 S2

 S3

=

shaded area

 

60º 60º x2 + y2 < 16

*54.

min 1  3i  z  z S

2 3 2 3 3 (C) 2

(A)

2 3 2 3 3 (D) 2

(B)

Sol.

(C) Distance of (1, –3) from y  3x  0 >
3  3  1 2



3 3 2

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-29

Paragraph for Questions 55 and 56 A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls. 55. If 1 ball is drawn from each of the boxes B1, B2 and B3, the probability that all 3 drawn balls are of the same colour is 82 90 (A) (B) 648 648 558 566 (C) (D) 648 648 (A) P (required) = P (all are white) + P (all are red) + P (all are black) 1 2 3 3 3 4 2 4 5 =         6 9 12 6 9 12 6 9 12 6 36 40 82 =    . 648 648 648 648 If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2 is 116 126 (A) (B) 181 181 65 55 (C) (D) 181 181 (D) Let A : one ball is white and other is red E1 : both balls are from box B1 E2 : both balls are from box B2 E3 : both balls are from box B3 E  Here, P (required) = P  2  A 
 A  P   P E2   E2  = A A  A P    P  E1   P    P  E2   P    P  E3  E1  E2     E3 
2

Sol.

56.

Sol.

1 55 6 = 1 =  . 3 2 3 3 4 1 1 2 181 C1  C1 1 C1  C1 1 C1  C1 1     9   12  6 3 3 3 5 6 11 C2 C2 C2
9

C1  3 C1 C2



1 3

SECTION  3 : (Matching list Type)
This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct. *57. Match List I with List II and select the correct answer using the code given below the lists :

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-30
List  I P.
  cos tan 1 y  y sin tan 1 y    1    y 4  takes value  y 2  cot sin 1 y  tan sin 1 y         If cosx + cosy + cosz = 0 = sinx + siny + sinz then x y possible value of cos is 2   If cos   x  cos2x + sinx sin2x secx = cosx sin2x secx 4     + cos   x  cos2x then possible value of secx is 4 

List  II





 





1/2

1.





1 5 2 3

Q.

2.

2

R.

3.

1 2

S.

If cot sin 1 1  x 2  sin tan 1 x 6 possible value of x is









 ,

x  0, then

4.

1

Codes : (A) (B) (C) (D) (B) P P 4 4 3 3 Q 3 3 4 4 R 1 2 2 1 S 2 1 1 2

Sol.

cos  tan 1 y   ysin  tan 1 y  cot sin 1 y   tan  sin 1 y 
1  y2 1  y2

=

1  y2

1  y2 y  y 1  y2

=

1  y2  y 1  y4 1 y 1  y2
2

1 1 1  cos  tan y   ysin  tan y    2   y4  cot  sin 1 y   tan  sin 1 y   y   1 2 4 4 = 2 y 1  y 4   y 4 = 1  y  y  1 y Q  cos x + cos y + cos z = 0 sin x + sin y + sin z = 0 cos x + cos y = – cos z sin x + sin y = – sin z (1)2 + (2)2 1 + 1 + 2(cos x cos y + sin x sin y) = 1 2 + 2 cos (x – y) = 1 2 cos (x – y) = –1 1 cos (x – y) =  2 1 xy 2 cos 2   1    2  2 xy 1 2 cos 2    2  2





..... (1) ..... (2)

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-31
xy 1 cos 2    2  4 xy 1 cos    2  2   R  cos   x  cos 2x  sin x sin 2x sec x 4    = cos x sin 2x sec x  cos   x  cos 2x 4          cos  4  x   cos  4  x   cos 2x = (cos x sin 2x – sin x sin 2x) sec x      2 sin x cos 2x   cos x  sin x  sin 2x sec x 2 2 sin x cos 2x   cos x  sin x  2sin x

1   x 4 2 cos x  sin x  sec x  sec  2 4 1 
S  cot  sin 1 1  x 2  x cot   1 x2 tan 1  x 6   

sin  

x 6 6x 2  1
2

1 x 6x 2  1 2 6x + 1 = 6 – 6x 12x2 = 5
2



x



x 6

x

5 1 5 = 12 2 3

*58.

A line L : y = mx + 3 meets y-axis at E(0, 3) and the arc of the parabola y2 = 16x, 0  y  6 at the point F(x0, y0). The tangent to the parabola at F(x0, y0) intersects the y-axis at G(0, y1). The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum. Match List I with List II and select the correct answer using the code given below the lists : List  I P. m= 1. 2. 3. 4. S 3 2 4 2
1 2 4 2 1

List  II

Q. Maximum area of EFG is R. y0 = S. y1 = Codes : P Q R (A) 4 1 2 (B) 3 4 1 (C) 1 3 2 (D) 1 3 4

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-32

Sol.

(A) A(t) = 2t2(3  4t) For max. A(t), t = m=1
1 sq. units 2 y0 = 4 and y1 = 2 1 2

E G

 4t 2 , 8t 
F

 A t 

max .



59.

Match List I with List II and select the correct answer using the code given below the lists : P. List  I   Volume of parallelepiped determined by vectors a, b  and c is 2. Then the volume of the parallelepiped       determined by vectors 2 a  b , 3 b  c and  c  a  is   Volume of parallelepiped determined by vectors a, b  and c is 5. Then the volume of the parallelepiped       determined by vectors 3 a  b , b  c and 2  c  a  is List  II 1. 100



  



Q.

2.

30





R.

Area of a triangle with adjacent sides determined by   vectors a and b is 20. Then the area of the triangle   with adjacent sides determined by vectors 2a  3b   and a  b is

3.

24









S.

Area of a parallelogram with adjacent sides determined   by vectors a and b is 30. Then the area of the parallelogram with adjacent sides determined by vectors    a  b and a is

4.

60





Codes : (A) (B) (C) (D) Sol. (C) P 4 2 3 1 Q 2 3 4 4 R 3 1 1 3 S 1 4 2 2

 P  a b c  2           2a  b 3b  c c  a  = 6 a     Q  a b c  5        6 a  b b  c c  a  = 12 a    1   R  a  b  20 2 1      2a  3b  a  b 2

 2 b c   6  4  24 
 b c   60 

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

JEE(ADVANCED)2013-Paper 2-PCM-33
1     2  a  b   3  a  b  2 5  40  100 2   S  a  b  30        a  b   a = b  a  30

60.

x 1 y z  3 x4 y 3 z 3   , L2 :   and the planes P1 : 7x + y + 2z = 3, P2 2 1 1 1 1 2 : 3x + 5y  6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L1 and L2, and perpendicular to planes P1 and P2. Match List I with List II and select the correct answer using the code given below the lists :

Consider the lines L1 :

List  I P. Q. R. S. Codes : (A) (B) (C) (D) Sol. P 3 1 3 2 Q 2 3 2 4 R 4 4 1 1 S 1 2 4 3 a= b= c= d= 1. 2. 3. 4. 13 3 1 2

List  II

(A) Plane perpendicular to P1 and P2 has Direction Ratios of normal ˆ ˆ k i j ˆ ˆ ˆ ˆ 7 1 2 =  16i  48j  32k … (1)
3 5 6

For point of intersection of lines (21 + 1,  1, 1  3)  (2 + 4, 2  3, 22  3)  21 + 1 = 2 + 4 or 21  2 = 3  1 = 2  3 or 1 + 2 = 3  1 = 2, 2 = 1  Point is (5,  2,  1) … (2) From (1) and (2), required plane is  1 (x  5) + 3 (y + 2) + 2 (z + 1) = 0 or  x + 3y + 2z =  13 x  3y  2z = 13  a = 1, b =  3, c =  2, d = 13.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.…...

Similar Documents

Free Essay

What Are the Advantages and Disadvantages of Social Media?

... Brandy Thompson Hum/176 - Week 4 Option #1: What are the advantages and disadvantages of social media? Social media has many advantages as well as disadvantages. On a positive aspect, social media can be used to build a community on either a personal level or even a business level. Different social media sites offer different features to their users. There are sites that can be used for advertising videos as well as photos. Some are a good source mainly to reach a worldwide connectivity. There is no faster way to reach people all over the world within minutes, like you can on social media sites. Having the ability to connect to others whom share personal interest and the ability to have real-time information sharing, also known as instant messaging is a feature that has become very convenient between users. There is no limit to how much advertising you can do; it all depends on how much time you are willing to put into it. The best part of it is that’s its free. Of course there is a downfall to social media sites. One that concerns a lot of people is the face-to-face connections. Not only is it very easy to create a persona or an image of who you want people to see, but that becomes a big problem with the younger generation today. Another major concern that falls in......

Words: 513 - Pages: 3

Premium Essay

He Advantages and Disadvantages of Mass Media

...In the modern twenty-first century, media has played an irreplaceable role of people’s daily life. Among many kinds of mass media like TV, newspaper, books… the Internet stands out to be one of the most important and essential kinds of media, even considered as the most crucial one. The Internet has greatly benefit human in the modern life since the first day of its existence, despite having been existing for only one century, much less than other kind of media, such as the radio and the television, the Internet has proved itself to be very effective, convenient and entertaining to its users. The first to mention is that Information has been made available easily because of this wonderful technology. Search engines, websites dedicated to different subjects and large amount of articles and papers are available for everyone in a matter of few seconds. Whether this information is about the latest news happenings in the world or information about your favorite celebrity, everything is available at your fingertips. With this housing tool of information, people cannot only easily improve their knowledge, but also save much time of reading hundreds of books at the library or doing some exhaustive researches. This enormous amount of information is particularly quite valuable to students who need it for their school subjects and further study in their favorite subjects. Furthermore, now many books are being converted into the digital editions, and e-education have become very......

Words: 268 - Pages: 2

Premium Essay

The Advantages and Disadvantages of One Mass Media

...Topic: The advantages and disadvantages of one mass media Writing In the modern twenty-first century, media has played an irreplaceable role of people’s daily life. Among many kinds of mass media like TV, newspaper, books… the Internet stands out to be one of the most important and essential kinds of media, even considered as the most crucial one. The Internet has greatly benefit human in the modern life since the first day of its existence, despite having been existing for only one century, much less than other kind of media, such as the radio and the television, the Internet has proved itself to be very effective, convenient and entertaining to its users. The first to mention is that Information has been made available easily because of this wonderful technology. Search engines, websites dedicated to different subjects and large amount of articles and papers are available for everyone in a matter of few seconds. Whether this information is about the latest news happenings in the world or information about your favorite celebrity, everything is available at your fingertips. With this housing tool of information, people cannot only easily improve their knowledge, but also save much time of reading hundreds of books at the library or doing some exhaustive researches. This enormous amount of information is particularly quite valuable to students who need it for their school subjects and further study in their favorite subjects. Furthermore, now many books are being......

Words: 328 - Pages: 2

Premium Essay

Advantages and Disadvantages of Social Media

...Advantages and Disadvantages of social media Tabitha HUM/186 May 7, 2012 University of Phoenix Advantages and Disadvantages of social media In today’s world, many people depend on the internet for news and entertainment, yet other people worry about how reliable it can really be. Now days you can sign on the internet and find any information that you want, from the weather, to the news, to your favorite video. Many people have started to rely on the internet for research for school, but with anyone being able to post anything, how do you know what you find is creditable? There are many advantages of easily obtainable information on the internet, such as services, searching things, and being able to download things. The internet offers many companies ways to help service customers today. Now instead of mailing in your bills, you are able to sign onto the web and make a payment with one click, or even do some shopping without even leaving your home. This leads to searching things, ‘thousands of groups and organizations have moved online, many of them aggressively promoting their presence through the use of search engines, email lists, and banner advertisements.” (Wright 2005) Also you are able to download videos, games, favorite television shows and movies for free to watch at your leisure. While the advantages seem great, there are also some disadvantages that come with easily obtainable information. Some of those disadvantages are theft of personal information,...

Words: 1032 - Pages: 5

Premium Essay

Mass Media Advantage and Disadvantages

...15 MASS MEDIA A) Types of mass media (advantages, disadvantages) B) Press media C) Radio, television A) Types of mass media (advantages, disadvantages) Mass media – an important role in the society, too powerful and influential Importance: source of information (easily available, complex, cheap) We learn about: world around us, inventions, new technologies, achievements B) Press media: newspapers, magazines and journals - Different sorts of articles: • on economics, politics, social life, sports • Editorials (personal opinion of the editor about something in the news) • Features (special reports about subject, place, person) Newspapers - published every day/ even at weekends - Serious papers : objective + serious news - Tabloids : entertaining stories, gossips about famous people/celebrities Magazines – published 1x a week/month - National and international news - various articles, stories - different according to: hobby/age/gender Journals – serious magazines written by experts C) Radio, television Radio - If we don’t like reading/ it’s expensive to buy a newspaper every day - We listen to the radio news read by a presenter - Music - Is very popular, variety of channels - Advantage: listen throughout the day, no matter what we’re doing - Variety of radio programmes to choose from: o news + weather forecast o chat /quiz/ breakfast/morning shows o educational programmes o phone- in programmes o documentaries Television - The most popular source of information - Much time...

Words: 403 - Pages: 2

Premium Essay

Advantages and Disadvantages of Social Media and Easily Obtainable Information

...Internet Education HUM/186 December 5, 2013 Internet Education In the modern world people operate their businesses and practically live their lives through the Internet, the use of social media, and easily obtainable information. Although easy access to nearly everything and everyone has many advantages, there are plenty disadvantages that accompany social media and easily obtainable information. If the world only knew how becoming aware of the rewards and drawbacks of social media and the ability to access information with ease people may change how they practice and consume it, especially when understanding the need to further question the reliability of their sources. It is important to understand the advantages, disadvantages, and knowledge of use pertaining to easily obtainable information and social media and the integrity of a source. Access to easily obtainable information has many advantages. Most people have a cell phone these days that can access the Internet nearly everywhere allowing them to consume almost whatever media they wish and whenever they want it. Enriquez, Wagner, and Marsh (2012) conclude that Sacramento County’s Environmental Management Department has a new application compatible with the IPhone and Android that allows users to view the health inspection scores of more than 5000 restaurants. People can have cell phone apps that range from locating places to eat and gas stations to counting how many steps a person takes......

Words: 1030 - Pages: 5

Premium Essay

The Advantages and Disadvantages of Different Types of Media

...1. Introduction In this report I will be analysing the advantages and disadvantages of the main types of media used for promotional activities. The main types of media I shall be analysing are: newspapers, television, radio, and billboards. 2. Findings a. Newspapers Promoting a business through a newspaper may seem ‘outdated’ by some as the world of technology continues to grow, develop and capture audiences of all ages. However, there are still some advantages to using a newspaper as a promotional tool. A national newspaper can help you reach large amount of people whereas a regional newspaper has reader loyalty and they will trust the brands they see in this paper a lot than if they were to see an advert online. A regional newspaper can also help a business to gain high coverage within a geographical area so if you were trying to advertise a small business in a local area a regional newspaper would be perfect for this. A newspaper advert is quick; it will often come out the next day. You can also get a lot of room to promote, if you need it. There are also a few disadvantages to using newspapers for promotional activities. Newspapers are known as ‘flat’ advertising. This means that they can’t make noise or grab attention in any other way for the audience. You can’t make a last minute change with a newspaper, because of tight printing schedules. Large space adverts are also very expensive. Lastly, newspapers tend to deliver only to an adult audience, so if you...

Words: 1080 - Pages: 5

Premium Essay

Advantages and Disadvantages of Social Media

...Media plays an important role in our society as a useful platform to get the updated information all over the world (“Role of Media in Society,” 2011). Media has developed gradually from text in papers, to movie in television and finally, the Internet, which then created varieties of social media. Social media helps to connect people through online activities (Jones, 2009). This essay is to discuss the two advantages, and one disadvantage of social media to society. In the first place, the use of social media enables easy access to the obtainable information. In brief, public opinion about certain events, current issues can be simply obtained from the social media. For instance, information including entertainments, sports, and weather reports through channels like Internet forums and social networks can be searched. Also, teenagers can learn lots of new things by joining a community of their passions. Teenagers would usually prefer to learn from others through the social media (Goff, 2009). Similarly, social media is a powerful source of connecting all sorts of people around the world. By accessing social media, people would have the opportunities to meet those who share the same goals, beliefs and interest. Meanwhile, it also helps to stay in contact with friends and families around the world. Still, companies interact and build relationship with their customers through establishing widely used platforms such as Linkedln and Facebook. The two-way communication improves......

Words: 419 - Pages: 2

Premium Essay

Media - Advantages and Disadvantages

...Sampson • "A Real Miracle Happened to Me" By Mrs. John Burger • "Setbacks" BY Ferenc Molnar • "I Was So Worried I Didn't Eat a Bite of Solid Food for Eighteen Days" By Kathryne Holcombe Farmer -----------------------------Sixteen Ways in Which This Book Will Help You 1. Gives you a number of practical, tested formulas for solving worry situations. “How To Stop Worrying And Start Living” By Dale Carnegie 4 2. Shows you how to eliminate fifty per cent of your business worries immediately. 3. Brings you seven ways to cultivate a mental attitude that will bring you peace and happiness. 4. Shows you how to lessen financial worries. 5. Explains a law that will outlaw many of your worries. 6. Tells you how to turn criticism to your advantage. 7. Shows how the housewife can avoid fatigue-and keep looking young. 8. Gives four working habits that will help prevent fatigue and worry. 9. Tells you how to add one hour a day to your working life. 10. Shows you how to avoid emotional upsets. 11. Gives you the stories of scores of everyday men and women, who tell you in their own words how they stopped worrying and started living. 12. Gives you Alfred Adler's prescription for curing melancholia in fourteen days. 13. Gives you the 21 words that enabled the world-famous physician, Sir William Osier, to banish worry. 14. Explains the three magic steps that Willis H. Carrier, founder of the air-conditioning industry, uses to conquer worry. 15. Shows you how to use what William James......

Words: 115134 - Pages: 461

Free Essay

What Are the Advantages and Disadvantages of Social Media?

...Brandy Thompson Hum/176 - Week 4 Option #1: What are the advantages and disadvantages of social media? Social media has many advantages as well as disadvantages. On a positive aspect, social media can be used to build a community on either a personal level or even a business level. Different social media sites offer different features to their users. There are sites that can be used for advertising videos as well as photos. Some are a good source mainly to reach a worldwide connectivity. There is no faster way to reach people all over the world within minutes, like you can on social media sites. Having the ability to connect to others whom share personal interest and the ability to have real-time information sharing, also known as instant messaging is a feature that has become very convenient between users. There is no limit to how much advertising you can do; it all depends on how much time you are willing to put into it. The best part of it is that’s its free. Of course there is a downfall to social media sites. One that concerns a lot of people is the face-to-face connections. Not only is it very easy to create a persona or an image of who you want people to see, but that becomes a big problem with the younger generation today. Another major concern that falls in that category is the lack of interaction between......

Words: 302 - Pages: 2

Premium Essay

Advantages and Disadvantages of Social Media in Searching for a Job

...The Advantages and Disadvantages of Social Media in a Job Search Executive Summary In recent years, the amount of people with access to the internet has grown immensely. This has caused an enormous boom in people using social media. Social media sites like Facebook and Twitter can be used to connect with friends, but also make connections with professionals and other people that give you job opportunities. While Facebook and Twitter are used by a small portion of job recruiters, LinkedIn is by far the top social media site to find a job. There are some disadvantages to using social media to find a job, but the advantages greatly outweigh the bad. The Advantages and Disadvantages of Social Media in a job search. In 2014, 87% of the population of the United States had internet access, up 7% from 2013. That is over 300 million people that will be able to freely surf the internet (Internet Users, 2014). Many of them have some sort of social media account. These accounts are a great way to meet new people, keep acquainted with old friends, and even search for new sources of employment. There are some disadvantages to using social media sites to find employment, but when used correctly, they can be a great tool to find new jobs. Social Media Sites Finding a job using social media can be a daunting task. With hundreds of different sites to choose from, finding the right one is a very important decision. There are many factors that should go into your......

Words: 1028 - Pages: 5

Free Essay

Disadvantage and Advantages of Media

...and Shopian in south and central Kashmir are completely cut off with roads and bridges swept away. The state government has ordered all schools to remain closed till September 12. Many people are now living in relief camps and temporary shelters. There has been no fresh rainfall today, but with virtually all phone lines down in the main city of Srinagar and roads cut off, the exact scale of the disaster is still unclear. No rain is expected in Jammu today, but light showers are likely in Srinagar. There has been no electricity in many areas for hours now all over the state including Srinagar. Phone lines are down and people have reported that they have been unable to contact stranded families since last night. Helplines and even social media sites are flooded with SOS messages giving details of families awaiting rescue. The state was battered by heavy rain for six days which caused major rivers and streams to overflow. As rain let up yesterday, rescue operations intensified. An Indian Air Force's Flight Lieutenant KS Lamba said the high terrain has made rescue operations tougher. "We have experienced pilots to complete this task. We have the immediate relief materials with us which is required as of now. We have medicines, blankets, water and utensils needed etc," he said. The Army has deployed 184 columns (75-100 personnel each) while IAF has deployed 29 planes and helicopters. Over 13,000 people have been rescued from various areas by the Army and the IAF. The National......

Words: 730 - Pages: 3

Premium Essay

Advantages and Disadvantages of Media

...Social media can actually do more harm than good Today, social media is everywhere. Playing a big role in people’s life, social media has made interactions among people easier. Although social media is so easy to use, it brings us many problems as well. In other words, social media has been an essential means of communication, but it can be harmful intentionally or unintentionally. Since we now live in this modern internet society, social media users should think twice before publicizing their posts on the internet. The website called CyberBullyHotline indicates that 4500 kids commit suicide each year due to cyber bullying. An example of unintentional harm with social media is seen in Sue Scheff’s recent writing about “accidental” cyberbullying for the Huffington Post. She describes incidents where teens say things to others, usually online, that are not intended to be hurtful, but are experienced as such. Even though it is not our objective, our words can be taken out of context by others when they are read and regurgitated, amplifying our digital footprint. This can happen offline as well, of course, but technology certainly makes it easier to obscure actual intent. Many of us know from experience that social media often leads to more frequent misunderstandings as communication occurs without important facial expressions, vocal intonations, or other interpretive behavioral cues that provide color and context to what is being conveyed. I agree that it is common for......

Words: 469 - Pages: 2

Premium Essay

Advantages & Disadvantages of Social Media

...study in June 2013 from the University of Michigan found that it was the students with narcissistic tendencies who posted most often, using social media in different ways to boost their egos and control other people’s perceptions of them – something that could have a dangerous impact on young girls at the very time when they have tremendous developmental changes taking place. Another study, from Flinders University in Australia, found that the amount of time teenage girls spend online was associated with low self-esteem and dissatisfaction with their body image (although cause and effect cannot be determined). In interviewing more than 1,000 high-school girls, they found that social media intensified conversations about appearance. And as these conversations involved their peers, they were also more influential in the girls’ lives. Even though 80% of the girls in the survey were of normal weight, almost half of them (46%) reported being dissatisfied with their body size. This evidence may suggest that, for many youngsters, their worries around body image aren’t actually connected with being overweight, but about body shape or concerns with not measuring up to unrealistic, celebrity-driven ‘norms’ that are, in truth, anything but normal. Research from Stanford University suggests that too much ‘screen time’ and constant social media juggling and multitasking are linked with negative feelings and experiences: feeling less socially adequate, for example, and having friends......

Words: 997 - Pages: 4

Premium Essay

Mass Media-Disadvantages and Advantages

...impose health and clean-up costs on the whole society, whereas the neighbors of an individual who chooses to fire-proof his home may benefit from a reduced risk of a fire spreading to their own houses. If external costs exist, such as pollution, the producer may choose to produce more of the product than would be produced if the producer were required to pay all associated environmental costs. Because responsibility or consequence for self-directed action lies partly outside the self, an element of externalization is involved. If there are external benefits, such as in public safety, less of the good may be produced than would be the case if the producer were to receive payment for the external benefits to others. Sources and causes Play media Air pollution produced by ships may alter clouds, affecting global temperatures. Air pollution comes from both natural and human-made (anthropogenic) sources. However, globally human-made pollutants from combustion, construction, mining, agriculture and warfare are increasingly significant in the air pollution equation.[20] Motor vehicle emissions are one of the leading causes of air pollution.[21][22][23] China, United States, Russia, India[24] Mexico, and Japan are the world leaders in air pollution emissions. Principal stationary pollution sources include chemical plants, coal-fired power plants, oil refineries,[25] petrochemical plants, nuclear waste disposal activity, incinerators, large livestock farms (dairy cows, pigs,......

Words: 5408 - Pages: 22