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JEE-MAIN 2017

Offline Paper Set-A

Questions with Solutions PART A - PHYSICS

ALL THE GRAPHS / DIAGRAMS GIVEN ARE SCHEMATIC AND NOT DRAWN TO SCALE. 1. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density

remains same, the stress in the leg will change by a factor by: (1) 9 (2) 1/9 (3) 81 (4) 1/81 Ans.: (1)

Sol.: Initial stress mgA

= d b h g dhgb

× × × ×= =

×l

l

Where d = density, l= Length, b=width, h=height Final stress = d 9h g× × = 9 (Initial stress).

2. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time

?

(1) (2) (3) (4) Ans.: (3) Sol.: For upward motion, velocity will be positive but decreases till maximum height and reaches zero. Then it will

increase in negative direction. 3. A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is

v0 = 10ms–1. If, after 10 s, its energy is (1/8)mv02, the value of k will be :

(1) 10–3 kg m–1 (2) 10–3 kg s–1 (3) 10–4 kg m–1 (4) 10–1 kg m–1 s–1 Ans.: (3)

Sol.: 2F kv= − K.E. at 10 s

2dvm kvdt

= − 0vv

2=

v 10

2v 0

dv k dtmv

= −∫ ∫ 0 0

0

mv vm 10kv 2

=+

1 1 k 10v v m

− + = − × 2m = m + 10 kv0

= +0

1 1 10kv v m

m = 10 kv0

0

0

mvv

m 10k v=

+

2410k 10 kg / s

10 10

−−= =

×

4. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by

the force during the first 1 sec. will be :

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(1) 4.5 J (2) 22 J (3) 9 J (4) 18 J Ans.: (1) Sol.: F = 6t

dvm 6tdt

=

v

1

00

dv 6 t d t=∫ ∫

1v 6 32

= × =

w K.E.= ∆ = 21 1 32

× × = 4.5 J.

5. The moment of inertia of a uniform cylinder of length I and radius R about its perpendicular bisector is I. What is

the ratio l / R such that the moment of inertia is minimum?

(1) 32

(2) 3

2 (3) 1 (4)

32

Ans.: (2)

Sol.: 2 2 2 2ML MR L RI M

12 4 12 4

= + = +

2 2

2 L Rd R L12 4

= π +

( )2

2 2d R L L 3R12

π= +

4 2

2

d R L L 312 R

π= +

5 2

2

d R L L. 312 R R

π= +

Let L xR

=

So, ( )5

2d RI .x x 312π

= +

As x increases, I increases for all x O≥

Lowest option is 3

2.

6. A slender uniform rod of mass M and length I is pivoted at one end so that it can rotate in a vertical plane (see

figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle 6 with the vertical is:

(1) 3 sin2gl

θ (2) 2 sin3gl

θ (3) 3 cos2gl

θ (4) 2 cos3gl

θ





Ans.: (1) Sol.: 0 0Iτ = α

2mmg sin .

2 3θ = α

l l

3gsin

2θ

α =l

7. The variation of acceleration due to gravity g with distance d from centre of

the earth is best represented by (R = Earth's radius) :

(1) (2) (3) (4) Ans.: (4)

Sol.: 3

3

GMd for d RRg

GMd for d RR

≤= ≥

8. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled

with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by: (Given: room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C) (1) 800 °C (2) 885 °C (3) 1250 °C (4) 825 °C

Ans.: (2) Sol.: Heat lost by copper ball = Heat gained by copper calorimetric + Heat gained by water. 100 0.01 (T 75) 100 0.1 (75 30) 170 1 (75 30)× × − = × × − + × × −

1 (T 75) 45 756φ − = φ + φ

T = 810 + 75 = 885°C 9. An external pressure P is applied on a cube at 0°C: so that it is equally compressed from all sides. K is the bulk

modulus of the material of the cube and a is coefficient of linear expansion. Suppose We Want to bring the cube to its original fize by heating. The temperature should be raised

(1) 3

PKα

(2) PKα

(3) 3PKα

(4) 3PKα

Ans.: (1)

Sol.: Pkv / v

= −∆

Or PvVk

∆ = −

For expression, PvVk

∆ =

Pv3 V Tk

α ∆ =

⇒ PT3 k

∆ =α

Z

O X

mg

l

θ

θ2 sin





10. Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that

Cp – Cv = a for hydrogen gas Cp – Cv = b for nitrogen gas The correct relation between a and b is :

(1) 1

14a b= (2) a = b (3) a = 14 b (4) a = 28 b

Ans.: (3) Sol.: For molar heat capacities p vC C R− =

If pC and pC are specific heats, then

p pRC CM

− =

Where M ⇒ molar mass

So, 2

2

N

H

Ma 28 14b M 2

= = =

A = 14 b. 11. The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The

atmospheric pressure in the room remains 1 × 105 Pa. If nf and ni are the number of molecules in the room before and after heating, then nf – ni will be: (1) –1.61 ×1023 (2) 1.38 ×1023 (3) 2.5 ×1025 (4) –2.5 ×1025

Ans.: (4)

Sol.: f iif

PV PVn nRT RT

− = −

510 30 3 1 1

25 300 290× × = −

4 1 19 4 1010 30 29

− = × × ×

312 10 6 10

29 10× ×

= − ×

2572 1029

= − × 252.5 10= − × .

12. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of

equilibrium. The kinetic energy - time graph of the particle will look like :

(1) (2)

(3) (4) Ans.: (4) Sol.: At t = 0 i.e.at equilibrium position, K.E. is maximum.

At Tt4

= i.e. at extreme position, K.E. zero

At Tt2

= , K.E. is again maximum and so on.





13. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at

frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light= 3 × 108 ms–1) (1) 10.1 GHz (2) 12.1 GHz (3) 17.3 GHz (4) 15.3 GHz Ans.: (3) Sol.: For Doppler effect in light

2s2

0s

V1

Cf fV

1C

−=

−

Where, f = frequency of observation 0f = frequency of source

sV = speed of source

C = speed of light

So, f = 10 GHz.

2

2

C21CC21C

−

−

= 10 GHz.

1 314 2101 112 2

−= ×

−

10 3= =17.32 GHz.

14. An electric dipole has a fixed dipole moment pr

, which makes angle 8 with respect to x-axis. When subjected to

an electric field 1ˆE Ei=

uur, it experiences a torque 1

ˆT k= τuur

. When subjected to another electric field 2 1ˆE E j= −

uur; it

experiences a torque 2 1T T= −uur uur

. The angle θ is:

(1) 30° (2) 45° (3) 60° (4) 90° Ans.: (3) Sol.: 1T pEsin= θ

2 2T pE cos= θ

3 p Ecos= θ

1 2T T=

pEsin 3 pEcosθ = θ

tan 3θ = 60θ = ° .

p

E1=Eiθ

p

90°−

θ





15. A capacitance of 2 µF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 µF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :

(1) 2 (2) 16 (3) 24 (4) 32 Ans.: (4) Sol.: Desired configuration is

So, minimum number of capacitor required = 4 8 32× = 16. In the given circuit diagram when the current readies steady state in the circuit the charge on the capacitor of

capacitance C will be:

(1) CE (2)

1

2 1( )rCE

r r+ (3)

2

2( )rCE

r r+ (4)

1

1 1( )rCE

r r+ Ans.: (3) Sol.: At steady state, current through capacitor will be zero.

So, current through 2r p.d.= across capacitor, 22

2

ErV I r

r r= =

+

Change on capacitor = 2

2

CE rCV

r r=

+

17.

In the above circuit the current in each resistance is: (1) 1 A (2) 0.25 A (3) 0.5 A (4) 0 A Ans.: (4) Sol.: Taking potential of right side bottom end as zero and using concept of relative potential

p.d. across each resistor is zero, so current in each resistance is zero.

2v

6v

6v

2v

2v

2v

2v

2v

2v

2v

4v

4v

O

O

1Ω 1Ω 1Ω





18. A magnetic needle of magnetic moment 6.7×10–2 Am2 and moment of inertia 7.5×10–6 kg m2 is performing

simple harmonic oscillation in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is: (1) 6.65 s (2) 8.89 s (3) 6.98 s (4) 8.76 s Ans.: (1)

Sol.: IT 2M B

= π

6

2

7.5 102 3.146.7 10 0.01

−

−

×= × ×

× ×

= 6.66 s. 19. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 to, it shows full scale

deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 –10 V is :

(1) 1.985 × 103 Ω (2) 2.045 × 103 Ω (3) 2.535 × 103 Ω (4) 4.005 × 103 Ω Ans.: (1)

Sol.: s g 3g

V 10R R 15 2000 15I 5 10−= − = − = −

×

1985= Ω

31.985 10= × Ω 20. In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the

figure. The magnitude of change in flux through the coil is:

(1) 200 Wb (2) 225 Wb (3) 250 Wb (4) 275 Wb Ans.: (2) Sol.: q R∆φ = ∆ ×

= Area under i-t curve R×

= 250 wb 21. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It

produces continuous as well as characteristic X-rays. If is the smallest possible wavelength of X-ray in the spectrum, the variation of log X min with log V is correctly represented in:

(1) (2)

(3) (4)





Ans.: (1)

Sol.: minhcev

λ =

Lag minhclag lagVe

λ = −

So, graph is as shown

minlag λ

22. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of

magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is: (1) real and at a distance of 40 cm from convergent lens (2) virtual and at a distance of 40 cm from convergent lens (3) real and at a distance of 40 cm from the divergent lens (4) real and at a distance of 6 cm from the convergent lens Ans.: (1) Sol.: f 25cm= −

For lens -1; u . f 25= ∞ = −

So, v 25= − For lens – 2; 4 = - 40, f = 20 So, v 40 cm=

23. In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A

beam of light consisting of two wavelengths, 650 nm and 520 ran, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :

(1) 1.56 mm (2) 7.8 mm (3) 9.75 mm (4) 15.6 mm Ans.: (2)

Sol.: λβ =

Dd

For light of 650 nm

9

1 4

650 10 1.5 1.95 mm5 10

−

−

× ×β = =

×

For light of 520 nm,

9

2 4

520 10 1.5 1.56 mm5 10

−

−

× ×β = =

×

Least distance from the common central maximum = L.C.M. of 1β and 2β = 7.8 mm

log λmin

log V

15cm

f= -25cm f=20cm





24. A particle A of mass m and initial velocity v collides with a particle B of mass (m/2) which is at rest. The collision

is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is:

(1) A

B

13

λ=

λ (2) A

B

2λ=

λ (3) A

B

23

λ=

λ (4) A

B

12

λ=

λ

Ans.: (2)

Sol.: Let 1v and 2v be velocities after collision

By momentum conservation,

1 2mmv v mv2

+ = Solving (1) & (2), we get

Or 21

vv v

2+ = 1

vv3

=

Or 1 22v v 2v+ = ……..(1) & 24 vv3

=

Using equation of e, A B B

B A A

p php h P

λ= =

λ

2 1v v v− = ……..(2) =

m 4v. 22 3v 1m.3

= =

25. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ1 / λ2 is given by :

(1) r = 4/3 (2) r = 2/3 (3) r = ¾ (4) r = 1/3 Ans.: (4)

Sol.: hcE∆ =λ

Or hcE

λ =∆

So, 1 2

2 1

EE 13rE E 3

λ ∆= = = =

λ ∆

26. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t,

the ratio of the number of B to that of A is 0.3. Then, t is given by :

(1) T log 2t2 log1.3

= (2) log1.3t Tlog 2

= (3) t = T log (1.3) (4) Tt

log (1.3)=

vm

m 2

rest

v1m

v2

m 2





Ans.: (2)

Sol.: A Bλ→ t = 0 N0 0

t x 0.3x Now, 0x 0.3x N+ =

Or 0Nx

1.3=

Since, tN N.e−λ=

⇒ 00

tNN e

1.3−λ= ⇒ te 1.3−λ = ⇒ ln 1.3t =

λ ⇒

ln (1.3)t .Tln 2

=

⇒ lag (1.3)t T

lag 2=

27. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the

output voltages will be : (1) 45º (2) 90 º (3) 135 º (4) 180 º Ans.: (4) Sol.: Based on theory 28. In amplitude modulation, sinusoidal carrier frequency used is denoted by ω and the signal frequency is denoted

by ωm. The bandwidth (∆ωm) of the signal is such that ∆ωm << ωc. Which of the following frequencies is not contained in the modulated wave ?

(1) ωm (2) ωc (3) ωm + ωc (4) ωc – ωm Ans.: (1) Sol.: Frequency contained in modulated wave c m c m(w w ) w w w− ≤ ≤ +

So, mw is not contained in the modulated wave.

29. Which of the following statements is false ?

(1) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude. (2) In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed. (3) A rheostat can be used as a potential divider. (4) Kirchhoff's second law represents energy conservation.

Ans.: (2) Sol.: Based on theory. 30. The following observations were taken for determining surface tension T of water by capillary method:

diameter of capillary, D = 1.25×10–2 m; rise of water, h = 1.45×l0–2 m.

Using g = 9.80 m/s2 and the simplified relation 3rhgT 10 N / m2

= × ,

the possible error in surface tension is closest (1) 0.15% (2) 1.5% (3) 2.4% (4) 10% Ans.: (2)





Sol.: − −∆ × × × ×

= =2 2hg 1.25 10 1.45 10 9.8T

4 4

T D hT D h

∆ ∆ ∆= +

4 4

2 2

T 10 10T 1.25 10 1.45 10

− −

− −

∆= +

× ×

T 1 1100 100T 125 145

∆ × = + ×

270 100125 145

= ××

=1.5%.

PART B - CHEMISTRY 31. Given

(graphite) 2 2C O (g) CO (g)+ → ; ∆rH° = – 393.5 kJ mol–1

2 2 21H (g) O (g) H O( )2

l+ → ; ∆rH° = – 285.8 kJ mol–1

2 2 4 2CO (g) 2H O( ) CH (g) 2O (g)l+ → + ; ∆rH° = + 890.3 kJ mol–1

Based on the above thermochemical equations, the value of ∆rH° at 298K for the reaction

(graphite) 2 4C 2H (g) CH (g)+ →

(1) – 74.8 kJ mol–1 (2) – 144.0 kJ mol–1 (3) + 74.8 kJ mol–1 (4) + 144.0 kJ mol–1. Ans.: (1)

Sol.: 2 2C O CO+ → H 393.5∆ = − kJ mol-1

2 2 21H O H O2

+ → H 285.8∆ = − kJ mol-1

2 2 4 2CO 2H O CH 2O+ → + H 890.3∆ = kJ mol-1

2 4C 2H CH M ?+ → ∆ =

eq (I) + 2× eq (II) + eq (III) 32. 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of

M2CO3 in g mol–1 is : (1) 118.6 (2) 11.86 (3) 1186 (4) 84.3 Ans.: (4) Sol.: 2 3 2 2 2M CO 2HCl MCl CO H O+ → + +

1 gm .01186 mole means .52184 gm ∵ .52184 gm - 1 gm

∴ 44 gm CO2 - 44 84.3gm.52184

=

33. ∆U is equal to :

(1) Adiabatic work (2) Isothermal work (3) Isochoric work (4) Isobaric work. Ans.: (1) Adiabatic work Sol.: For Adiabatic u w∆ =





34. The Tyndall effect is observed only when following conditions are satisfied : (a) The diameter of the dispersed particles is much smaller than the wavelength of the light used. (b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude. (d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. (1) (a) and (c) (2) (b)and(c) (3) (a)and(d) (4) (b) and (d)

Ans.: (4) Sol.: both b & d correct 35. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach

between two atoms in metallic crystal will be :

(1) 2 a (2) a2

(3) 2a (4) 2 2 a

Ans.: (2)

Sol.: For fcc ad2

=

36. Given

2

oCl /Cl

E 1.36 V=− , 3oCr /Cr

E 0.74 V+ = − , 2 32 7

oCr O /Cr

E 1.33 V+ =− , 24

oMnO /Mn

E 1.51 V+ =−

Among the following, the strongest reducing agent is : (1) Cr3+ (2) Cl– (3) Cr (4) Mn2+

Ans.: (3) Sol.: The strongest reducing agent is Cr 37. The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If

acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (Kf for benzene = 5.12 K kg mol–1) (1) 74.6% (2) 94.6% (3) 64.6% (4) 80.4%

Ans.: (2) Sol.: fT∆ = i Kfm

.45 = .2 1000i 5.1260 20

× ××

×

i = .0527

for association i 11 1n

−∝=

− .0527 1

1 12

−=

− = .945=94.5%

38. The radius of the second Bohr orbit for hydrogen atom is :

(Planck's Const h = 6.6262×10–34 Js; mass of electron = 9.1091×10–31 kg; charge of electron e = 1.60210×10–19 C; permittivity of vacuum ε0 = 8.854185×l0–12 kg–1 m–3 A2) (1) 0.529 Å (2) 2.12 Å (3) 1.65 Å (4) 4.76 Å

Ans.: (2)

Sol.: 2 2

2 2n hr

4 mze=

π

2 2.529 x .529(2)r Aº 2.12Aº

z 1= = =





39. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1, exceeds that of R2 by 10 kj mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to : (R = 8.314Jmol–lK–l) (1) 6 (2) 4 (3) 8 (4) 12

Ans.: (2)

Sol.: logk = logA EaRT

− it is given E1 = 10 + E2 (10 × 103)

log k1 = log A 210 ERT+

−

log k1 = log A 2E10RT RT

− − ......... (I)

log k2 = log A 2ERT

− ................ (II)

2 22 1

E E10logk logK log A log ART RT RT

− = − − + +

3

2

1

K 10 10 10log 4K RT 8.314 300

×= = =

×

40. pKa of a weak acid (HA) and pKb, of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB)

solution is : (1) 7.0 (2) 1.0 (3) 7.2 (4) 6.9 Ans.: (4) Sol.: For weak acid & weak base

a b1 1 1pH pkw pk pk2 2 2

= + − = + −1 1 1(14) (3.2) (3.4)2 2 2

pH = 6.9 41. Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the

one which is incorrect, is : (1) both form nitrides (2) nitrates of both Li and Mg yield NO2 and O2 on heating (3) both form basic carbonates (4) both form soluble bicarbonates

Ans.: (3) both form basic carbonates Sol.: Li & mg not form basic carbonates 42. Which of the following species is not paramagnetic ?

(1) O2 (2) B2 (3) NO (4) CO Ans.: (4) CO Sol.: Total No. of el. in CO are 14 & all are paired. So it is diamagnetic. 43. Which of the following reactions is an example of a redox reaction ?

(1) XeF6 + H2O → XeOF4 + 2HF (2) XeF6 + 2H2O → XeO2F2 + 4HF (3) XeF4 + O2F2 → XeF6 + O2 (4) XeF2 + PF5 → [XeF]+ PF6–.

Ans.: (3) Sol.: XeF4 + O2F2 → XeF6 + O2





44. A water sample has ppm level concentration of following anions

F– = 10; SO42– = 100; NO3

– =50,

The anion/anions that make/makes the water sample unsuitable for drinking is/are : (1) only F (2) only SO4 (3) only NO3

– (4) both SO2– and NO3

– Ans.: (1) only F Sol.: 1.5 ppm of F- causes tooth decay and more concentration causes poison effect. 45. The group having isoelectronic species is:

(1) O2–, F–, Na, Mg2+ (2) O–, F–, Na+, Mg2+ (3) O2–, F–, Na+, Mg2+ (4) O–, F–, Na, Mg+ Ans.: (3) O2–, F–, Na+, Mg2+

Sol.: 2 2O ,F ,Na &mg− − + + all have 10 electronic

46. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are: (1) Cl– and ClO– (2) Cl– and ClO2

– (3) ClO– and ClO3– (4) ClO2

– and ClO3–.

Ans.: (1) Cl– and ClO– Sol.: 2 2NaOH Cl NaCl NaClO H O+ → + +

hence Cl− & −ClO . 47. In the following reactions, respectively acting as a/an :

(a) ZnO + Na2O → Na2ZnO2 (b) ZnO + CO2 → ZnCO3 (1) acid and acid (2) acid and base (3) base and acid (4) base and base

Ans.: (2) acid and base Sol.: ZnO is amphoteric oxide. 48. Sodium salt of an organic acid 'X' produces effervescence with conc. H2SO4. 'X' reacts with the acidified aqueous

CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. 'X' is: (1) CH3COONa (2) Na2C2O4 (3) C6H5COONa (4) HCOONa

Ans.: (2) Na2C2O4

Sol.:

COONa

COONa

H SO2 4CO + Na SO + H O2 2 4 2

CaCl2

COO

CaCOOH

COOH

KMnO4

H+ CO + H O2 2

COO 49. The most abundant elements, by mass in the body of a healthy human adult are :

Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is: (1) 7.5 kg (2) 10 kg (3) 15 kg (4) 37.5 kg

Ans.: (1) 7.5 kg Sol.: for 1H 10% Total mass of body 75 kg 1H = 7.5 kg. H∆ = 7.5 kg 2H = 15 kg.





50. On treatment of 100 mL of 0.1 M solution of CoCl3.6H2O with excess AgNO3; 1.2×1022 ions are precipitated. The complex is : (1) [Co(H2O)6]Cl3 (2) [Co(H2O)5Cl]Cl2.H2O (3) [Co(H2O)4Cl2]C1.2H2O (4) [Co(H2O)3Cl3].3H2O

Ans.: (2)

Sol.: mole 1000Mvol. in ml

×=

mole 1000.1100

×=

mole = .01 mole of CoCl3 6 H2O for AgNO3 ∴ 6.023 × 1023 ions - 1 mole

∴ 1.2 × 1022 ions - 22

231.2 10 .02

6.023 10×

=×

mole

∵ .01 mole - .02 mole

∴ 1 mole .02 2.01

= = mole of Cl will ppt

51. Which of the following compounds will form significant amount of meta product during mono-nitration

reaction?

(1) (2) (3) (4) Ans.: (1) Sol.:

NH2 group is O & P directing but in this reaction most of the aniline convert into Anilinium ion (NH3

+) which is meta directing and about 47% m - Nitro compound is formed.

52. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to

decolourize the colour of bromine ?

(1) (2) (3) (4) Ans.: (3) Sol.:

Here is no 'H'β

∴ No double bond formation in presence of base





53. The formation of which of the following polymers involves hydrolysis reaction ? (1) Nylon 6, 6 (2) Terylene (3) Nylon 6 (4) Bakelite

Ans.: (3) Sol.:

CH2CH2

CH2

CN

O

CH2

CH2

HOH HO C

O

(CH )2 5 NH 2

54. Which of the following molecules is least resonance stabilized ?

(1) (2) (3) (4) Ans.: (2) Sol.: Aromaticity is not present in structure 2 55. The increasing order of the reactivity of the following halides for the SNl reaction is:

(1) (I) < (III) < (II) (2) (II) < (III) < (I) (3) (III) < (II) < (I) (4) (II) < (I) < (III)

Ans.: (4) 56. The major product obtained in the following reaction is :

(1) (+) C6H5CH(OtBu)CH2C6H5 (2) (–) C6H5CH(OtBu)CH2C6H5 (3) (±) C6H5CH(OtBu)CH2C6H5 (4) (+) C6H5CH=CHC6H5

Ans.: (4) Sol.:

C H6 5 C CH 2 C H6 5

H

Br

t. BuOK∆→

C H6 5 C OHC H6 5

H

OH H∆→

6 5 6 5C H CH CH C H= −

57. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?

(1) (2)





(3) (4) Ans.: (3) 58. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of

possible stereoisomers for the product is : (1) Two (2) Four (3) Six (4) Zero Ans.: (4) Sol.:

CH3 CH C

CH3

CH2CH3 Or

C C

CH 3

H

CH 3

C H2 52 2

HBrH O→

C CCH 3

H

CH3

H

Br

* *C H2 5

two chiral carbon atoms are present in the product hence 2n = 22 59. The correct sequence of reagents for the following conversion will be :

(1) CH3MgBr, [Ag(NH3)2]+OH–, H+/CH3OH (2) [Ag(NH3)2]+OH–, CH3MgBr, H+/CH3OH (3) [Ag(NH3)2]+OH–, H+/CH3OH, CH3MgBr (4) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH–. Ans.: (3) Sol.:

O

CHO

[ ]3Ag(NH) OH+ −+ →

O

COOH

3H /CH OH+

→

O

COOCH3

32CH HgBr→

CH3HO

HO CH3

CH3





60. The major product obtained in the following reaction is :

(1) (2)

(3) (4) Ans.: (4) Sol.:

OO

COOH

DIBAL H−→

OH

CHO

CHO

DlBAL reduces the ester into Aldehydes and acids into aldehyde

H C3

CH3

H

AlCH 3

CH 3

PART C - MATHEMATICS

61. The function 1 1f : R ,2 2

→ − defined as 2

xf (x)1 x

=+

is:

(1) injective but not surjective. (2) surjective but not injective. (3) neither injective nor surjective. (4) invertible.

Ans.: (2)

Sol.: 1 1f : R ,2 2

→ −

2

xf(x)1 x

=+

x R∀ ∈

2

2 2 2 2' (1 x ).1 x.2x [x 1).(x 1)f (x)

(1 x ) (1 x ) + − − + −

= = + +

So, f(x) is subjective but not injective

12

-1 2

(0, 0)

(1, )

(-1, )

-1





62. If, for a positive integer n, the quadratic equation,

x(x + l) + (x + l)(x + 2) + .... ( )x n 1 (x n) 10n+ + − + =

has two consecutive integral solutions, then n is equal to : (1) 9 (2) 10 (3) 11 (4) 12

Ans.: (3)

Sol.: 2nx [(0 1) (1 2) (2 3).........(n 1) n)x

sum of roots+ + + + + + − +

[1.2 2.3 (n 1)n) 10 n+ + − − =

⇒ 2 [1 2 ........n 1] n+ + − +

2(n 1) n2 n n2

− ×+ =

nα + β = − 2n n

n −

= − ∵

. n(n 1)α β = +∑

n(n 1)(n 1) 10n3

− += −

[ ) 1α − β =

( ) 2( ) 4 1α − β = α + β − α + β =

2

2 n 1 30n 4 13

− −− =

2 23n 4n 124 3− + =

2n 121=

n = 11

63. Let ω be a complex number such that 2ω + l = z where z 3= − .

If 2 2

2 7

1 1 1

1 1 3k

1

−ω − ω =

ω ω

, then k is equal to:

(1) z (2) – 1 (3) 1 (4) – z Ans.: (4)

Sol.: 2 1 Z−ω + = Z 3 31− = = °

2 2

2 7

1 1 11 1 3k1

−ω − ω =ω ω

9 7 4 7 2 2 21( ) 1( 1) 3k−ω − ω − ω − ω − ω + ω + ω + =

2 2 21 1 3k− − ω − ω − ω + ω + + ω + ω =

23 3 3k− ω + ω =

w 1 k− − − ω =

( )Z k−− =

k 7= − .





64. If A = 2 34 1

− −

, then adj (3A2 + 12A) is equal to

(1) 51 6384 72

(2) 51 8463 72

(3) 72 6384 51

− −

(4) 72 8463 51

− −

Ans.: (3)

Sol.: 22 3 2 3 2 3A , A

4 1 4 1 4 1− − −

= = − − − ⇒ 2 16 9

A12 13

− = −

2 48 27 24 363A 12A

36 39 48 12− −

+ = + − −

72 6384 51

− = −

adj ( )23A 12A+51 6384 72

=

65. If S is the set of distinct values of 'b' for which the following system of linear equations

x + y + z = l x + ay + z = 1 ax + by + z = 0 has no solution, then S is : (1) an infinite set (2) a finite set containing two or more elements (3) a singleton (4) an empty set

Ans.: (3)

Sol.: 1 1 11 a 1 0a b 1

∆ = =

21 (a b) 1(1 a) 1.(b a ) 0− − − − + + =

a 1=

1 2 3 0∆ = ∆ = ∆ =

But a = 1 and b = 1

Then x y 2 1x y 2 0

+ + = + + =

there is no solution.

66. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is : (1) 468 (2) 469 (3) 484 (4) 485

Ans.: (4)

Sol.: X Y L M L M

3 0 0 3 2 1 1 2 1 2 2 1 0 3 3 0





4 3 3 43 0 0 3c c c c 16× × × =

+ 4 3 3 4

2 1 1 2c c c c 324× × × =

+ 4 3 3 4

1 2 2 1c c c c 144× × × =

+ 4 3 3 4

0 3 3 0c c c c 1× × × =

Ans. 485 67. The value of

(21C3 – 10C1) + (21C4 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4) + …. + (21C10 – 10C10) is: (1) 221 – 210 (2) 220 – 29 (3) 220 – 210 (4) 221 – 211 Ans.: (3)

Sol.: ( ) ( ) ( )21 10 21 10 21 101 1 2 2 10 10C C C C ....... C C− + − + −

( )10 10 10 101 2 3 10C C C ....... C− + + + +

⇒ 21 21 21 21 212 2 3 4 10C C ....... C C ......... C− + −

⇒ ( )20 102 1 2 1− − −

⇒ 20 102 1 2 1− − +

⇒ 20 102 2− . 68. For any three positive real numbers a, b and c,

9(25a2 + b2) + 25(c2 - 3ac) = 15b(3a + c). Then : (1) b, c and a are in AP (2) a, b and c are in AP (3) a, b and c are in GP (4) b, c and a are in GP.

Ans.: (1)

Sol.: ( ) ( ) ( )2 2 215a 13b 5c (15 a)( 5c)+ + − (15a)(3b) 3b 5c 0− − × =

2 2 21 (15a 3b) (13b 5c) (5c 15a) 02

− + − + − =

cb5

= ca3

=

a + b = 2c b, c, a an in A.P. 69. Let a, b, c ∈ R. If f(x) = a2 + bx + c is such that a + b + c=3 and

f(x + y) = f(x) + f(y) + xy, ∀ x, y ∈ R, then 10

n 1f (n)

=∑ equal to :

(1) 165 (2) 190 (3) 255 (4) 330 Ans.: (4)

Sol.: 2f(x) ax bx c= + + , a b c 3+ + =

and f(x y) f(x) f(y) x.y x, y R+ = + + ∀ ∈

2 2 2a(x y) b(x y) ax bx c ay by c+ + + = + + + + +





Comparing coefficient of 2x ,x and find 1a2

=5b2

= , C = 0

⇒ 2x 5f(x) x

2 2= +

⇒ 10

n 1f(n) ?

=

=∑

⇒ 210

n 1

n 5 .n2 2=

+ ∑ ⇒

2 2 21 5 2 5 10 5.1 .2 ....... .102 2 2 2 2 2

+ + + + + +

⇒ 2 2 2 21 5.(1 2 3 ...... 10 ) (1 2 3 4 .... 10)2 2

+ + + + + + + + + + ⇒ 330.

70. ( )3

x2

cot x cos xlim2xπ

→

−

π − is equal to:

(1) 1/16 (2) 1/8 (3) 1/4 (4) 1/24 Ans.: (1)

Sol.: 3x

2

cot x cosxlim( 2x)π

→

−π −

R.H.L. approach

3h 0

2cot h cos h2lim

2 h2

→

π + − + π

π π − × +

3h 0

tanh sinhlim8h→

+−

−

3h 0

(tanh sinh)lim 18h→

−−

+

3h 0

(1 cosh sinhlim8 h→

− ×−

×

2

3h 0

h h hsin sin cos2 2 2lim

28 h→

×−

×

3

3h 0

hsin1 12lim

2 8 16h2

→

=×

71. If for 1x 0,4

∈

the derivative of 13

6x xtan1 9x

− −

is x .g(x) , then g(x) is:

(1) 33x x

1 9x− (2) 3

3x1 9x−

(3) 33

1 9x+ (4) 3

91 9x+

Ans.: (4)

Sol.: 13

6x xy tan1 9x

− = −

1x 0,4

∈





332

1 1 232

2.(3 x )tan 2 tan 3.x1 3x

− −

= = −

As 32 33.x 0,

8 ∈

12

3

dy 1 32x 3 xdx 21 9x

= × × ×+ 3

9 . x1 9x

=+

⇒ 3

9g(x)1 9x

=+

.

72. The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes

through the point:

(1) 1 1,2 2

(2) 1 1,2 3

−

(3) 1 1,2 3

(4) 1 1,2 2

− −

Ans.: (1) Sol.: Normal to the curve at point “P” intersect the y-axis x = 0 y(0 2)(0 3) 0 6− − = +

y = 1 point (0, 1)

2

P(0,1)

dy (x 5x 6) y(2x 5) 1dx

− + + − =

P(0,1)

dy (6) 5) 1dx

− =

p

dy 1dx

=

Equation of normal at point (0,1)

p

1(y 1) (x 0)dydx

−− = −

(y 1) 1x− = − Ans. (1)

Satisfy the point in equation of normal

1x2

= and 1y2

=

1 12 2− −

= .

73. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the

maximum area (in sq. m) of the flower-bed, is : (1) 10 (2) 25 (3) 30 (4) 12.5

Ans.: (2) Sol.: Total length = 20 r r r 20+ + θ =

20 2rr−

θ =





Area = 2 21 1 20 2rr r2 2 r

− θ =

210r r∆ = −

d 0dr∆

= 10 2r 0− =

r 5= 50 25 25∆ = − = .

74. Let nnI tan x dx, (n 1)= >∫ . If I4 + I6 = a tan5 x + bx5 + C, where C is a constant of integration, then the ordered

pair (a, b) is equal to :

(1) 1 ,05

(2) 1 , 15

−

(3) 1 ,05

−

(4) 1 ,15

−

Ans.: (1)

Sol.: In = 4tan xdx, (n 1)>∫

I4 + I6 = 4 6tan x dx tan x.dx+∫ ∫

= ( )4 2tan x dx 1 tan x dx+ +∫

= 2 4sec x .tan x dx∫

Put 2tanx tsec x.dx dt= =

I4 + I6 =5

4 tt dt c5

− = +∫

I4 + I6 = 5tan x c

5+

1a , b 05

= = .

75. The integral

34

4

dx1 cos x

π

π +∫ is equal to

(1) 2 (2) 4 (3) – 1 (4) – 2 Ans.: (1)

Sol.:

34

4

dxI1 cos x

π

π

=+∫ …….(1)

3x x4 4π π → + −

34

4

dxI1 cos( x)

π

π

=+ π −∫

34

4

dxI1 cos x

π

π

=−∫ …….(2)

θ r





(1,0)

(1,0)

(1,2)

x+y=3

y=1+Öx

x = 4y2

(2,1)

(2,0)

34

4

dx dx2I1 cos x 1 cos x

π

π

= ++ −∫

34

2

4

22 I dxsin x

π

π

= ∫

34

2

4

2 I x cosec x dx

π

π

= ∫

[ ]34

4

I cot xπ

π= −

3I cot cot4 4π π = − −

I 2= + .

76. The area (in sq. units) of the region (x, y) : x ≥ 0, x + y ≤ 3, x2 ≤ 4y and y ≤ 1 + x is: (1) 3/2 (2) 7/3 (3) 5/2 (4) 59/12

Ans.: (3)

Sol.: Area = 1 2 2 2

0 1 0

x(1 x )dx (3 x)dx dx7

+ + − −∫ ∫ ∫

Area = 52

77. If ( ) dy2 sin x (y 1)cos x 0dx

+ + + = and y(0) = 1, then y2π

is equal to:

(1) – 2/3 (2) – 1/3 (3) 4/3 (4) 1/3 Ans.: (4)

Sol.: dy(2 sin x) (y 1)cosx 0dx

+ + + =

dy(2 sin x) (y 1)cos xdx

+ = − +

dy cosx dx

(y 1) 2 sinx= −

+ +

[ )log(y 1) log 2 sinx c+ = − + +

Find c y(0) 1= ⇒ n = y

log(1 1) log(2) c+ = − +

log(4) c=

Again y find " y"2π =

log(y 1) log(3) log4+ = − + ⇒ 4y 13

+ = ⇒ 1y3

= .





78. Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocenter of this triangle is at the point:

(1) 31,4

(2) 31,4

−

(3) 12,2

(4) 12,2

−

Ans.: (3)

Sol.: k 3k 1

1 5 k 1 282

k 2 1

−=

−

25k 13k 46 0+ − =

k = 2 and 2 k5

−=

K is integer k = 2

Orthocentre 12,2

79. The radius of a circle, having minimum area, which touches the curve y = 4 – x2

and the lines, y = |x| is :

(1) 2( 2 – 1) (2) 4( 2 – 1) (3) 4( 2 + 1) (4) 2( 2 + 1) Ans.: (2) Sol.: Given curves are

2y 4 x= − and y | x |=

There are two circle satisfy the ratio of circle is “r” Length = (0, 4 - r)

Circle touch the line y = x in first quadrant (4 r) r

2−

=

r 4( 2 1)= − .

80. The eccentricity of an ellipse whose centre is at the origin is (1/2). If one of its directrices is x = – 4, then the

equation of the normal to it at (1) 4x – 2y = l (2) 4x + 2y = 7 (3) x + 2y = 4 (4) 2y – x = 2

Ans.: (1) Sol.: Directrix of ellipse

a4 xe

−− = = ⇒ a e 4= × ⇒ 1e

2= ⇒ a = 2

2 2 2 2a e a b= − ⇒ 214 4 b4

× = − ⇒ 21 4 b− = − ⇒ 2b 3=

2 2

2 2

x y 1a b

+ = ⇒ 2x 2y dy 04 3 dx

+ × = ⇒ 1

1

xdy 3dx 4 y

= − ×

P

dy 1dx 2

= − ×

m = 2 2y 3 4x 4− = −

⇒ 2y 4x 1− −

A (2,-6)

D

H k,

C (-2,2) (-5,2)B





Equation of normal at 31,2

⇒ 3y 2(x 1)2

− = −

⇒ 4x 2y 1− =

81. A hyperbola passes through the point P( 2, 3 ) has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through, the point:

(1) (2 2,3 3 ) (2) ( 3, 2 ) (3) ( 2, 3 )− − (4) (3 2,2 3 )

Ans.: (1) Sol.: Equation of hyperbola

2 2

2 2

x y 1a b

− =

2 2

2 3 1a b

− =

2 2

2 3 1a 4 a

− =−

2 2 2 48 2a 3a 4a a− − = −

4 2a 9a 8 0− + =

4 2 2a 8a a 8 0− − + =

Equation of tangent at “P” ( 2, 3)

1 12 2

x x y y1

a b− =

2 2

x 2 y 3 1a b

− =

And has foci ( 2, 0)±

Find a & b

1

22

be 1a

= +

2 2 2(ae) a b= +

2 24 a b= + ……………(1)

Put 2a 8= 24 8 b= +

2 2b a 1× = 24 1 b− =

2b 3=

x 2 y 3 11 3

− =

Equation of tangent yx 2 13

− =

At point (2 2, 3 3) satisfying the equation of tangent





82. The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal

perpendicular to both the lines x 1 y 2 z 41 2 3− + −

= =−

and x 2 y 1 z 72 1 1− + +

= =− −

(1) 1083

(2) 583

(3) 1074

(4) 2074

Ans.: (1) Sol.: a(x 1) b(y 1) c(z 1) 0− + + + + =

a 2b 3c 0− + = 2a b c 0− − =

a b c2 5 1 b 1 4

−= =

+ − − − +

a b c7 7 3

= = = λ

c 3= λ a 7= λ b 7= λ &

2 1 2

7(1 1) 7(3 1) 3( 7 1)PQ7 7 3

− + + + − +=

+ +

10PQ83

=

83. If the image of the point P(l, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line x y z1 4 5

= = is

Q then PQ is equal to :

(1) 2 42 (2) 42 (3) 6 5 (4) 3 5 Ans.: (1) Sol.: Equation of PQ line

x 1 y 2 z 3 11 4 5− + −

= = = −

R point line in the plane

$x y z and b i 4 j 5k1 4 5

= = = + +r $ $

2 ( 1) 3(4 2) 4(5 3) 22 0× λ + + λ − − λ + + =

1λ = R(2,2,8)

distance between P & R is 2 2 2(2 1) (4) (5)− + +

42= PQ = 2 PR×

PQ = 2 42×

84. Let $a 2i j 2k= + −r $ $ and b i j= +

r $ $ . Let cr

be a vector such that c a 3− =r r

, ( )a b c 3× × =r r r

and the angle between cr

and a b×r r

be 30°. Then a .cr r

is equal to :

(1) 2 (2) 5 (3) 1/8 (4) 25/8 Ans.: (1)

Sol.: $a 2i j 2k= + −r $ $ and b i j= +

r $ $

| c a | 3,|(a b) c) 3− = × × =r r r r r

$| a | . |c |.sin .n 3b× θ =r rr

Q

P (1,3,-7)

7(x 1) 7(y 1) 3(z 1) 0- + + + + =

Q

P (1,-2,4)

2x+3y-4z+22=0





$i j ka b 2 1 2

1 1 0× = −

$ $r r

$a b 2i 2j k× = − +r r $ $

a b 9 3× = =r r

$| a | . |c |.sin .n 3b× θ =r rr

$| a | . |c |.sin30 .n 3b× ° =r rr

13 | c | 32

× × =r

| c | 2=r

| c a | 3− =r r

2| c a | 9− =r r

(c a).(c a). 9− − =r r r r

c.c a.c c .a a.a 9− − + =r r r r r r r r

2 2| c | 2a . c | a | 9− + =r r r r

4 9 9 2a.c+ = +r r

a . c 2=r r

85. A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement,

then the variance of the number of green balls drawn is : (1) 6 (2) 4 (3) 6/25 (4) 12/5

Ans.: (4)

Sol.: 15 G10 Y

10 balls one by one

Variance = ( )22i ix .P xi pi−∑ ∑

2

2 15 151 10 10 125 25

= × × − × ×

2

2

15 151025 25

= −

⇒ 15 1510 125 25

× − 12

5= .

86. For three events A, B and C,

P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = (1/4) and P(All the three events occur simultaneously) = (1/16) Then the probability that at least one of the events occurs, is :

(1) 7/16 (2) 7/64 (3) 3/16 (4) 7/32





Ans.: (1)

Sol.: P(exactly one of A or B occurs) = 1P(A) P(B) 2P(A B)4

+ − ∩ =

P(exactly one of B or C occurs) = 1P(B) P(C) 2P(B C)4

+ − ∩ =

P(exactly one of C or A occurs) = 1P(C) P(A) 2P(C A)4

+ − ∩ =

Add these all 32 P(A) 2 P(A B)4

− ∩ =∑ ∑

3(P(A)) P(A B)8

− ∩ =∑ ∑

Now 1P(A B C)16

∩ ∩ = (Given)

P(A B C) P(A) (P(A B) P(A B C)∪ ∪ = − ∩ + ∩ ∩∑ ∑

3 1 78 16 16

= + = .

87. If two different numbers are taken from the set 0, 1, 2, 3, ......, 10); then the probability that their sum as well

as absolute difference are both multiple of 4, is: (1) 12/55 (2) 14/45 (3) 7/55 (4) 6/55

Ans.: (4)

Sol.: A 0, 1, 2, 3, 4......10=

112n(s) C when'S'denote sample space=

F (0,4), (0,8),(2,6),(2,10),(4,8),(6,10)=

6P(E)55

= .

88. If 5(tan2 x - cos2 x) = 2cos 2x + 9, then the value of cos 4x is :

(1) 1/3 (2) 2/9 (3) – (7/9) (4) – (3/5) Ans.: (3)

Sol.: 2 2 25(tan x cos x) 2cos x 14 5− = + −

2 2 25(tan x cos x) 2cos x 14− + = + 2 2 25(sec x cos x) 2cos x 14− = +

22(2cos x 1) 14= − + 2 2 25sec x 5cos 4cos x 12− = + 2 25sec x 9cos 12− =

2

2

5.2 1 cos x9 1221 cos x

+− = +

2 2 220 9(1 cos x) 24(1 cos x)− + = + 2 2 220 9(1 cos 2x 2cos x) 24 24cos x− + + = +

2 2 211 9cos 2x 18cos x 24 24cos x− − = + 2 29cos 2x 42cos x 13 0+ = =





2 29cos 2x 3cos2x 39cos x 13+ + + 2(3cos x 1) (3cos2x 13) 0+ + =

2 1cos x3

= − 2 2cos2x 2cos x 1= −

12 19

= × −7

9−

= .

89. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the

ground such that AP = 2AB. If ∠BPC = β, then tan β is equal to: (1) 1/4 (2) 2/9 (3) 4/9 (4) 6/7

Ans.: (2) Sol.: PCA∆

x12tan

2x 4α = =

PBA∆

tan( ) tan( )tan( )1 tan .tan

α + βα + β =

− α β

11 tan tan tan1 412 1 tan4

− α + β + β=

− β

11 .tan4 .tan

1 92 4

− ββ

=

2tan9

β =

90. The following statement (p q) [(~ p q) q]→ → → → is:

(1) equivalent to ~ p q→ (2) equivalent to p ~ q→

(3) a fallacy (4) a tautology. Ans.: (4) Sol.: ⇒

hence Tautology



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