2009 cbse pmt 2

Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42- 1 -

CBSE - 2009 (Pre)PHYSICS & CHEMISTRY SET-A

1. If the dimensions of a physical quantity are given by Ma Lb Tc, then the physical quantity will be(1) Force if a = 0, b = –1, c = –2 (2) Pressure if a = 1, b = –1, c = –2(3) Velocity if a = 1, b = 0, c = –1 (4) Acceleration if a = 1, b = 1, c = –2

Sol: Ans [2]

Pressure = ForceArea

Dimension of pressure = [M1 L–1 T–2]

i.e., a = 1, b = –1, c = –2

2. A particle starts its motion from rest under the action of a constant force. If the distance covered in first10 seconds is S1 and that covered in the first 20 seconds is S2, then :(1) S2 = S1 (2) S2 = 2S1 (3) S2 = 3S1 (4) S2 = 4S1

Sol: Ans [4]

Here u = 0, acceleration = aDistance travelled in first 10 seconds

1110 10 10 0 50 502

S u a a a

Distance travelled in first 20 seconds

2120 20 20 0 200 2002

S u a a a

So, S2 = 4S1

3. A bus is moving with a speed of 10 ms–1 on a straight road. A scooterist wishes to overtake the busin 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooteristchase the bus?

(1) 10 ms–1 (2) 20 ms–1 (3) 40 ms–1 (4) 25 ms–1

Sol: Ans [2]Let V be the speed of scooterist

S = V × 100 = 100 + 10 × 100V = 20 m/sec

4. The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its accelerationis(1) 14 ms–2 upwards (2) 30 ms–2 downwards (3) 4 ms–2 upwards (4) 4 ms–2 downwards

Sol: Ans [3]

R – mg = ma

228000 2000 9.8 4.2 m /sec upwards2000

R mgam

4 m/sec2 upwards


5. An explosion blows a rock into three parts. Two parts go off at right angles to each other. These twoare, 1 kg first part moving with a velocity of 12 ms–1 and 2 kg second part moving with a velocity of8 ms–1 and 2 kg second part moving with a velocity of 8 ms–1. If the third part flies off with a velocityof 4 ms–1, its mass would be

(1) 3 kg (2) 5 kg (3) 7 kg (4) 17 kg

Sol: Ans [2]

4 m cos = 12 ....(i)

4 m sin = 2 × 8 = 16 ....(ii)

(4m)2 × (cos2 + sin2) = 144 + 256

16 m2 = 400

m

8 m/sec

12 m/sec1 kg2 kg

4

m2 = 25

m = 5 kg

6. A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceilingand has force constant value k. The mass is released from rest with the spring initially unstretched. Themaximum extension produced in the length of the spring will be

(1) Mg/2k (2) Mg/k (3) 2Mg/k (4) 4Mg/k

Sol: Ans [3]

21 0 02

mgx kx

2mgxk

7. Two bodies of mass 1 kg and 3 kg have position vectors ˆ ˆˆ ˆ ˆ ˆ2 and 3 2 ,i j k i j k respectively..

The centre of mass of this system has a position vector

(1) ˆˆ ˆ–i j k (2) ˆˆ2 2i k (3) ˆˆ ˆ2i j k (4) ˆˆ ˆ2 2i j k

Sol: Ans [3]

m1 = 1 kg; m2 = 3 kg

1 2ˆ ˆˆ ˆ ˆ ˆ2 ; 3 2r i j k r i j k

1 1 2 2

1 2

m r m rrm m

ˆ ˆˆ ˆ ˆ ˆ2 3( 3 2 ) ˆˆ ˆ21 3

i j k i j kr i j k


8. Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of thisframe about an axis through the centre of the square and perpendicular to its plane is

(1) 213

Ml (2) 243

Ml (3) 223

Ml (4) 2133

Ml

Sol: Ans [2]22

412 2

ML LI M

m

mm

m

2

243

I ML

9. A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to itsplane with a constant angular velocity . If two objects each of mass m be attached gently to theopposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

(1)M

M m

(2)( 2 )

2M m

M m

(3)

2M

M m

(4)( 2 )M m

M

Sol: Ans [3]I11 = I22

(Mr2) = (M + 2m) r2

= 2

MM m

10. A body, under the action of a force ˆˆ ˆ6 8 10 ,F i j k

acquires an acceleration of 1 m/s2. The mass ofthis body must be

(1) 10 2 kg (2) 2 10 kg (3) 10 kg (4) 20 kg

Sol: Ans [1]

| | 36 64 100 200 10 2F N

| | 10 2 10 2 kg1

Fma

11. If F

is the force acting on a particle having position vector r

and

be the torque of this force aboutthe origin, then

(1) . 0 and . 0r F

(2) . 0 and . 0r F

(3) . 0 and . 0r F

(4) . 0 and . 0r F

Sol: Ans [4]

r F

So, . 0r and . 0F


12. The figure shows elliptical orbit of a planet m about the sun S. The shadedarea SCD is twice the shaded area SAB. If t1 is the time for the planet tomove from C to D and t2 is the time to move from A to B then(1) t1 = t2 (2) t1 > t2

(3) t1 = 4t2 (4) t1 = 2t2

Sol: Ans [4]

1 2

1 2

dA dAdt dt

1 2

2x xt t

t1 = 2t213. An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is

the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

(1) 2 212

m v (2) 312

mv (3) mv3 (4) 212

mv

Sol: Ans [2]

ddt

(KE) = ddt

(21

2mv

= 21 .2

2dm dvv m vdt dt

ddt

(KE) = 21 . 02

mvv 0dvdt

Here m = a , dmdt

= a v = mv .

= 312

mv

14. A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest afterattaining a height of 18 m. How much energy is lost due to air friction? (g = 10 m/s2)(1) 10 J (2) 20 J (3) 30 J (4) 40 J

Sol: Ans [2]

K.E. at ground = 1 1 20 20 200 J2

P.E. at height = 1 × 10 × 18 = 180 JLoss in energy due to friction = 200 – 180 = 20 J.

15. The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T1

and T2(T1 > T2). The rate of heat transfer, ,dQdt

through the rod in a steady state is given by

(1) 1 2( )kA T TdQdt L

(2) 1 2( )kL T TdQ

dt A

(3) 1 2( )k T TdQdt LA

(4) 1 2( )dQ kLA T T

dt

Sol: Ans [1]

Self explanatory


16. In thermodynamic processes which of the following statements is not true?(1) In an adiabatic process PV = constant(2) In an adiabatic process the system is insulated from the surroundings(3) In an isochoric process pressure remains constant(4) In an isothermal process the temperature remains constant

Sol: Ans [3]In isochoric process, volume remains constant.

17. A black body at 227°C radiates heat at the rate of 7 Cals/cm2s. At a temperature of 727°C, the rate ofheat radiated in the same units will be(1) 80 (2) 60 (3) 50 (4) 112

Sol: Ans [4]E = T4

4 4

4 42

7 (227 273) (500)(727 273) (1000)E

E2 = 11218. The internal energy change in a system that has absorbed 2 Kcals of heat and done 500 J of work is

(1) 7900 J (2) 8900 J (3) 6400 J (4) 5400 J

Sol: Ans [1]

dQ = 2 kcal = 2 × 4.2 × 1000 J = 8400 J

dW = 500 J

dU = dQ – dW = 8400 – 500 = 7900 J

19. The driver of a car travelling with speed 30 m/sec towards a hill sounds a horn of frequency 600 Hz. Ifthe velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is(1) 500 Hz (2) 550 Hz (3) 555.5 Hz (4) 720 Hz

Sol: Ans [4]

1330600

330 30f

2 1330 30 600 330 360 720 Hz

330 300 330f f

20. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time periodT. The speed of the pendulum at x = a/2 will be

(1)3a

T

(2)3

2aT

(3)

aT

(4)23 a

T

Sol: Ans [1]

22 2 2

4av a y a

2 3 3.2

a avT T


21. Which one of the following equations of motion represents simple harmonic motion?(1) acceleration = kx (2) acceleration = –k0x + k1x2

(3) acceleration = –k(x + a) (4) acceleration = k(x + a)Where k, k0, k1 and a are all positive.

Sol: Ans [3]

In S.H.M., acceleration displacement

22. The electric field part of an electromagnetic wave in a medium is represented by Ex = 0

6 22.5 cos 2 10 10 ;yN rad radE t xC m s

Ez = 0. The wave is

(1) moving along –x direction with frequency 106 Hz and wavelength 200 m.(2) moving along y direction with frequency 2 × 106 Hz and wavelength 200 m.(3) moving along x direction with frequency 106 Hz and wavelength 100 m.(4) moving along x direction with frequency 106 Hz and wavelength 200 m.

Sol: Ans [4]E = 2.5 cos [(2 × 106)t – ( × 10–2)x]Compare with

2cos ( )E a vt x

62 2 10vt t

... (i)

22 10x x

... (ii)

So, = 200 m; v = 200 × 106 m/sec.

610 Hzvn

23. A wave in a string has an amplitude of 2 cm. The wave travels in the +ve direction of x axis with aspeed of 128 m/sec, and it is noted that 5 complete waves fit in 4 m length of the string. The equationdescribing the wave is

(1) y = (0.02)m sin (7.85x – 1005t) (2) y = (0.02)m sin (7.85x – 1005t)

(3) y = (0.02)m sin (15.7x – 2010t) (4) y = (0.02)m sin (15.7x – 2010t)

Sol: Ans [1]a = 0.02 mv = 128 m/sec.

4 m5

2 2siny a x vt


2 20.02sin 5 5 284 4

y x t

y = (0.02)m sin (7.85 x – 1005 t)

24. Each of the two string of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass perunit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneouslythe number of beats is

(1) 3 (2) 5 (3) 7 (4) 8

Sol: Ans [3]

1 = 100 20

2 51.6 .001 = 13.82

2 = 100 20

2 49.1 .001 = 143.78

number of beats = 1 –2 = 143.78 – 136.82 = 6.96 7

25. Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitanceand breakdown voltage of the combination will be

(1) 3C, 3V (2) ,3 3C V

(3) 3 ,3VC (4) , 3

3C V

Sol: Ans [4]

C = 3C

C C C

V = 3V

26. A wire of resistance 12 ohms per meter is bent to form a complete circle of radius 10 cm. Theresistance between its two diametrically opposite points, A and B as show in the figure, is

(1) 6 (2) 0.6

(3) 3 (4) 6

Sol: Ans [2]

Circumference circle = 10 22100 10 5

Resistance of wire = 1212

5 5

Resistance of each section = 1210 A B

Eq. resistance =

12 122 610 10 0.612 12 1010 10


27. A bar magnet having a magnetic moment of 2 × 104 JT–1 is free to rotate in a horizontal plane. Ahorizontal magnetic field B = 6 × 10–4 T exists in the space. The work done in taking the magnet slowlyfrom a direction parallel to the field to a direction 60° from the field is

(1) 2J (2) 0.6 J (3) 12 J (4) 6 J

Sol: Ans [4]

M = 2 × 104 J/T ; B = 6 × 10–4 T ; Q1 = 0° ; Q2 = 60°

Work = 2 × 104 × 6 × 10–4 (cos 0 – cos 60) = 6 J.

28. The magnetic force acting on a charged particle of charge –2µc in a magnetic field of 2T acting in ydirection, when the particle velocity is

6 1ˆ ˆ(2 3 ) 10 ms ,i j is

(1) 8 N in z direction (2) 8 N in –z direction

(3) 4 N in z direction (4) 8 N in y direction

Sol: Ans [2]

6 6ˆ ˆ ˆ( ) 2 10 (2 3 ) (2 )10F q v B i j j

ˆ ˆ2(4 0) 8k k newton.

29. A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular tothe magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop whenthe radius is 2 cm is

(1) 1.6 µV (2) 3.2 µV (3) 4.8 µV (4) 0.8 µV

Sol: Ans [2]

2. 2d d dre B r B rdt dt dt

–62 20.04 3.2 10 V = 3.2 V.100 1000

e

30. The electric potential at a point (x, y, z) is given by

V – x2y – xz3 + 4

The electric field E

at that point is

(1) 3 2 2ˆˆ ˆ(2 ) 3E i xy z j xy k z x

(2) 3 2 2ˆˆ ˆ(2 ) 3E i xy z j x k xz

(3) 2 2 2ˆˆ ˆ2 ( ) (3 )E i xy j x y k xz y

(4) 2 2ˆˆ ˆE iz jxyz k z

Sol: Ans [2]

ˆˆ ˆdV dV dVE i j kdx dy dz

3 2 2ˆˆ ˆ[(2 ) ( ) ( 3 )]xy z i j x k z x

3 2 2ˆˆ ˆ(2 ) 3 .i xy z j x k xz


31. See the electrical circuit shown in this figure. Which of the following equations is a correct equationfor it?

(1) 1 – (i1 + i2) R + i1 r1 = 0

(2) 1 – (i1 + i2) R – i1 r1 = 0

(3) 2 – i2 r2 – 1 – i1 r1 = 0

(4) –1 – (i1 + i2) R + i2 r2 = 0

Sol: Ans [2]

1 = (i1 + i2)R + i1r1

1 = (i + i2)R – i1r1 = 0.

32. A galvanometer having a coil resistance of 60 shows full scale deflection when a current of 1.0 amppasses through it. It can be converted into an ammeter to read currents upto 5.0 amp by

(1) putting in parallel a resistance of 15 (2) putting in parallel a resistance of 240

(3) putting in series a resistance of 15 (4) putting in series a resistance of 240

Sol: Ans [4]

1 60 60 155 1 4

g g

g

I RR

I I

and to convert a galvanometer into ammeter put the resistance in parallel.

33. Under the influence of a uniform magnetic field, a charged particle moves with constant speed V in acircle of radius R. The time period of rotation of the particle

(1) depends on both V and R (2) depends on V and not on R

(3) depends on R and not on V (4) is independent of both V and R

Sol: Ans [4]

2 .mtBq

34. Power dissipated in an LCR series circuit connected to an a.c. source of emf is

(1)2

2 2 1/R R LwCw

(2)2

2 2 1/R R LwCw

(3)2

2 2 1 /R Lw RCw

(4)

22 2 1R Lw

CwR

Sol: Ans [2]

2cos EIR EERP EIz z

2

22

.1

E RPR L

C


35. Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities ,– and respectively. if VA, VB and VC denote the potentials of the three shells, then, for c = a + b, wehave(1) VC = VB = VA (2) VC = VA VB (3) VC = VB VA (4) VC VB VA

Sol: Ans [2]

2 2 2

0 0 0

1 ( 4 ) 1 4 1 4. .4 4 4A

a b cVa b c

0 0

( ) (2 )a b c a

2 2 2

0 0 0

1 4 1 4 1 4. .4 4 4B

a b cVb b c

2 2

0 0

a ab c ab b

2 2 2

0 0 0

1 4 1 4 1 4. . .4 4 4C

a b cVc c c

2 2

0 0

2 .a b c ac c

36. A student measures the terminal potential difference (V) of a cell (of emf and internal resistance r)as a function of the current (I) flowing through it. The slope, and intercept, of the graph between V andI, then, respectively, equal(1) – and r (2) and –r (3) –r and (4) r and –

Sol: Ans [3]

V = – IrV = –Ir + Slope is –r and intercept is .

37. A rectangular, a square, a circular and an elliptical loop, all in the (x – y) plane, are moving out of a

uniform magnetic field with a constant velocity, ˆ.V vi The magnetic field is directed along the

negative z axis direction. The induced emf, during the passage of these loops, out of the field region,will not remain constant for(1) any of the four loops (2) the rectangular, circular and elliptical loops(3) the circular and the elliptical loops (4) only the elliptical loop

Sol: Ans [3]

Area coming out per second from the field is not constant.38. If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is

(1) attracted by both the poles(2) repelled by both the poles(3) repelled by the north pole and attracted by the south pole(4) attracted by the north pole and repelled by the south pole


Sol: Ans [2]

Diamagnetic material is repelled by magnetic field.

39. The number of photo electrons emitted for light of a frequency v (higher than the threshold frequencyv0) is proportional to(1) Frequency of light (v) (2) v – v0

(3) Threshold frequency (v0) (4) Intensity of light

Sol: Ans [4]

Number of photo electrons emitted is directly proportional to the intensity of light.

40. Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is

9 mW. The number of photons arriving per sec. on the average at a target irradiated by this beam is

(1) 3 × 1019 (2) 9 × 1017 (3) 3 × 1016 (4) 9 × 1015

Sol: Ans [3]

= 667 nm = 667 × 10–9 m

E = 9 mW sec = 9 × 10–3 J

3 916

34 8

9 10 667 10 3 10 .6.6 10 3 10

n

41. The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three

different radiations. Which one of the following is a correct statement?

(1) curves (b) and (c) represent incident radiations of same frequency having same intensity.

(2) curves (a) and (b) represent incident radiations of different frequencies and different intensities.

(3) curves (a) and (b) represent incident radiations of same frequency but of different intensities.

(4) curves (b) and (c) represent incident radiations of different frequencies and different intensity.

Sol: Ans [3]

Stopping potential depends upon frequency and photo current depends upon intensities.

42. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles

emitted by it. The resulting daughter is an:

(1) isotope of parent (2) isobar of parent

(3) isomer of parent (4) isotone of parentSol: Ans [1]

Due to alpha emission atomic number decreases by 2 and due to beta emission atomic number increasesby 1.


43. The ionization energy of the electron in hydrogen atom in its ground state is 13.6 eV. atoms are excited

to higher energy levels to a radiations of 6 wavelengths. Maximum wavelength of emitted radiation

corresponding the transition between

(1) n = 4 to n = 3 states (2) n = 3 to n = 2 states

(3) n = 3 to n = 1 states (4) n = 2 to n = 1 states

Sol: Ans [1]

( 1) 62

n n

n = 4

4

3

2

1

44. In a Rutherford scattering experiment when projectile of charge z1 and mass M1 approach a target

nucleus of charge z2 and mass M2 distance of closest approach is r0. The energy of the projectile is

(1) directly proportional to mass M1 (2) directly proportional to M1 × M2

(3) directly proportional to z1 z2 (4) inversely proportional to z1

Sol: Ans [3]

k = potential energy = 1 2

0 0

1 .4

q qr

So, k q1q2

45. In the nuclear decay given below:

A A A 4 A 4Z Z 1 Z 1 Z 1X Y B B,*

the particles emitted in the sequence are:

(1) (2) (3) (4)

Sol: Ans [2]

4 41 1 1

A AA AZ Z ZZ X Y B B

,

46. The mean free path of electrons in metal is 4 × 10–8 m. The electric field which can give on an average

2 eV energy to an electron in the metal will be in units of V/m

(1) 5 × 107 (2) 8 × 107 (3) 5 × 10–11 (4) 8 × 10–11

Sol: Ans [1]

2eV = eE × 4 × 10–8

78

2 5 10 V / m4 10

E


47. Sodium has body centred packing. Distance between two nearest atoms is 3.7 Å. The lattic parameter

is

(1) 8.6 Å (2) 6.8 Å (3) 4.3 Å (4) 3.0 Å

Sol: Ans [4]

3 1.7 3.7 3Å2 2

a

48. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal

of wavelength:

(1) 4000 Å (2) 6000 Å (3) 4000 nm (4) 6000 nm

Sol: Ans [1]

34 87

19

6.6 10 3 10 4.95 102.5 1.6 10

hcE

= 4950 Å

4000 Å

49. The symbolic representation of four logic gates are given below

(i)

(ii)

(iii)

(iv)

The logic symbols for OR, NOT and NAND gates are respectively :(1) (i), (iii), (iv) (2) (iii), (iv), (ii) (3) (iv), (i), (iii) (4) (iv), (ii), (i)

Sol: Ans [4]

These are standard conventions.

50. A transistor is operated in common-emitter configuration at Vc = 2V such that a change in the basecurrent from 100 µA to 200 µA produces a change in the collector current from 5 mA to 10 mA. Thecurrent gain is(1) 50 (2) 75 (3) 100 (4) 150

Sol: Ans [3]

Ib = 200 – 100 = 100 A = 100 × 10–6 A

IC = 10 – 5 = 5 mA = 5 × 10–3 A

= 3

6

5 10 50100 10

c

b

II


51. 10 g of hydrogen and 64g of oxygen were filled in a steel vessel and exploded. Amount of waterproduced in this reaction will be

(1) 1 mol (2) 2 mol (3) 3 mol (4) 4 mol

Sol.: Ans. [4]

2H2 + O2 2H2O

Oxygen is limiting reagent.

64g O2 = 2 mole of oxygen

Hence 4 mole of water will be produced.

52. Oxidation numbers of P in 34PO of S in 2

4SO and that of Cr in 272OCr are respectively

(1) – 3 , +6 and +6 (2) +5, +6 and +6 (3) +3, +6 and +5 (4) +5, +3 and +6

Sol.: Ans. [2]

5OP 34

6OS 24

6OrC 272

53. Maximum number of electrons in a subshell of an atom is determined by the following

(1) 2n2 (2) 4l + 2 (3) 2l +1 (4) 4l–2

Sol.: Ans. [2]

Total number of orbitals in subsheel = 2l +1

Hence total number of e = 4l + 2

54. Which of the following is not permissible arrangement of electrons in an atom?

(1) n = 3, l = 2, m = – 2, s = –½ (2) n = 4, l = 0, m = 0, s = – ½

(3) n = 5, l = 3, m = 0, s = + ½ (4) n = 3, l =2, m = –3, s = –½

Sol.: Ans. [4]

Value of m cannot be greater then l.

55. From the following bond energies

(1) H–H bond energy: 431.37 kJ mol–1 (2) C = C bond energy: 606.10 kJ mol–1

(3) C – C bond energy: 336.49 kJ mol–1 (4) C – H bond energy: 410.50 kJ mol–1

Enthalpy for the reaction;

HHHH||||

HCCHHHCC||||HHHH

will be

(1) 553.0 kJ mol–1 (2) 1523.6 kJ mol–1 (3) –243.6 kJ mol–1 (4) –120.0 kJ mol–1

Sol.: Ans. [4]

gH {Bond energy of the reactants – Bond energy of product}


ΔH(C C) ΔH 4(C H ΔH(H H)] [ C C) ΔH 6(C H)]

= – 120.0 kJ mol–1

56. The ionization constant of ammonium hydroxide is 1.77 ×10–5 at 298 K. Hydrolysis constant of ammo-nium chloride is

(1) 5.65 × 10–12 (2) 5.65 × 10–10 (3) 6.50 × 10–12 (4) 5.65 ×10–13

Sol.: Ans. [2]

bH K

kwK

1095

14

1065.510565.01077.1

10

57. Given

(i) 2 oCu 2e Cu, E 0.337V

(ii) 2 oCu e Cu , E 0.153V Electrode potential, E° for the reaction

CueCu will be;(1) 0.38 V (2) 0.52 V (3) 0.90 V (4) 0.30 V

Sol.: Ans. [2]

2 01 1Cu 2e Cu E 0.337V; ΔG nFE 2 F 0.337

2 02 2Cu e Cu E 0.1537V; ΔG nFE 1 F 0.153

V0.52ΔGΔGΔG?E.CueCu 213

58. What is the [OH–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of0.10 M Ba(OH)2?

(1) 0.12 M (2) 0.10 M (3) 0.40 M (4) 0.0050 M

Sol.: Ans. [2]

No. of milli moles of HCl = M×V = 0.050 ×10 = 1No. of millimoles of Ba(OH)2 = M×V = 0.10 × 30 = 32HCl + Ba(OH)2 BaCl2 + 2H2O1×10–3 mole 0.5×10–3 mole 0.5×10–3 mole 1×10–3 mole2.5×10–3 moles of Ba(OH)2 will left in resulting solution hence moles of OH– = 2×2.5×10–3 = 5×10–3

mole 50 ml solution contains 5×10–3 mole of [OH]

1000 ml solution contains M1.0M1050

1000105 13

59. The energy absorbed by each molecule (A2) of a substance is 4.4×10–19 J and bond energy per mol-ecule is 4.0 ×10–19 J. The kinetic energy of the molecule per atom will be

(1) 4.0×10–20 J (2) 2.0 ×10–20 J (3) 2.2×10–19 J (4) 2.0×10–19 J

Sol.: Ans. [2]


60. For the reaction 322 NH2H3N , if 1143 sLmol102dt

]d[NH , the value of dt

]d[H2 would be

(1) 114 sLmol101 (2) 114 sLmol103 (3) 114 sLmol104 (4) 114 sLmol106

Sol.: Ans. [2]

dt]d[NH

21

dt]d[H

31

dt]d[N 322

4432 10310223

dt]d[NH

23

dt]d[H

61. For the reaction A+ B products, it is observed that

(a) on doubling the initial concentration of A only, the rate of reaction is also doubled and

(b) on doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rateof the reaction

The rate of this reaction is given by:

(1) rate = k[A] [B] (2) rate = k[A]2 [B] (3) rate = k[A] [B]2 (4) rate = k [A]2 [B]

Sol.: Ans. [3]

Rate = k[A] [B]2

Because with respect to A; it is 1st order and with respect to B; it is 2nd order.

62. The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mhos cm2 and at infinitedilution is 400 mhos cm2. The dissociation constant of this acid is

(1) 1.25 ×10–4 (2) 1.25 × 10–5 (3) 1.25 ×10–6 (4) 6.25 ×10–4

Sol.: Ans. [2]

542 1025.1104321;02.0

4000.8 CK

63. A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at –0.00732°C. Numberof moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = –1.86°C/m)

(1) 1 (2) 2 (3) 3 (4) 4

Sol.: Ans. [2]

Calculated freezing point depression = Kf.m = 1.86 ×0.0020

Observed freezing point depression = 0.00732

Van’t Hoff factor pointfreezingCalculatedpointfrezzingObserved

i

20020.086.1

00732.0


64. In the reaction 3 2 2BrO (aq) 5Br (aq) 6H 3Br ( ) 3H O( )l l . The rate of appearance of bromine(Br2) is related to rate of disappearance of bromide ions a following:

(1)dt

)d(Br53

dt)d(Br2

(2)dt

)d(Br53

dt)d(Br2

(3)dt

)d(Br35

dt)d(Br2

(4)dt

)d(Br35

dt)d(Br2

Sol.: Ans. [2]

+3 (aq) 2 2BRO (aq) + 5 Br + 6 H 3Br (P) + 3H O(P)

dtO]d[H

31

dt]d[Br

31

dt]d[Br

51

dt]d[BrO 223

dtd[Br]

53

dt]d[Br2

65. Lithium metal crystallises in a body centred cube crystal. If the length of the side of the unit cell oflithium is 351pm, the atomic radius of the lithium will be

(1) 300.5 pm (2) 240.8 pm (3) 151.8 pm (4) 75.5 pm

Sol.: Ans. [3]

Radius for BCC 8.1514

351732.043

ar

66. The dissociation constants for acetic acid and HCN at 25°C are 1.5×10–5 and 4.5×10–10 respectively.The equilibrium constant for the equilibrium

COOHCHCN 3 HCN + CH3COO– would be

(1) 3.0 × 104 (2) 3.0 × 105 (3) 3.0 × 10–5 (4) 3.0 × 10–4

Sol.: Ans. [1]

CH3COOH CH3COO– + H+; K1 = 1.5×10–5 ...(i)

HCN H+ + CN– ; K2 = 4.5 ×10–10 ...(ii)

CN– + CH3COOH HCN + CH3COO–; K3 = ?

410

5

2

13 103.0

104.5101.5

KKK

67. The values of H and S for the reaction C(graphite) +CO2(g) 2CO(g) are 170 kJ a 170 JK–1 respec-tively. This reaction will spontaneous at:

(1) 510 K (2) 710 K (3) 910 K (4) 1110 K

Sol.: Ans. [4]

Using formula G = H – TS


68. Half life period of a first order reaction is 1386 seconds. The specific rate constant of reaction is

(1) 5.0×10–2 s–1 (2) 5.0×10–3 s–1 (3) 0.5×10–2 s–1 (4) 0.5×10–3 s–1

Sol.: Ans. [4]

13

2/1sec105.0693.0

t

69. In which of the following molecules /ions BF3, NO2–, NH2

– and H2O the central atom is sp2 hybridized?

(1) BF3 and NO2– (2) NO2

– and NH2– (3) NH2

– and H2O (4) NO2– and H2O

Sol.: Ans. [1]

70. Among the following which is the strongest oxidising agent?

(1) Cl2 (2) F2 (c) Br2 (4) I2

Sol.: Ans. [2]

71. According to MO theory which of the following lists ranks the nitrogen species in terms of increasingbond order?

(1) 2222 NNN (2) 2

222 NNN (3) 2222 NNN (4) 2

222 NNN

Sol.: Ans. [3]

Bond order of 2N2 , Bond order of 5.2N2 , Bond orders of N2 = 3

72. In the case of alkali metals, the covalent character decreases in the order

(1) MI > MBr > MCl > MF (2) MCl > MI > MBr > MF

(3) MF > MCl > MBr > MI (4) MF > MCl > MI > MBr

Sol.: Ans. [1]

As difference in electronegativity increases, covalent character decreases.

73. Which of the following oxides is not expected to react with sodium hydroxide?

(1) BeO (2) B2O3 (3) CaO (4) SiO2

Sol.: Ans. [3]

CaO because metal oxide does not react with metal oxide.

74. Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0×104 amperes of current ispassed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% cur-rent efficiency, At mass of Al = 27 g mol–1)

(1) 1.3 × 104 g (2) 9.0 ×103 g (3) 8.1 ×104g (4) 2.4 ×105 g

Sol.: Ans. [3]

q = i.t. = 4.0 ×104 × 6 × 60 × 60 C

3 × 96500 C deposits 27 g of Al

4.0 × 104 × 6 × 60 × 60 will deposit 4

427 4.0 10 6 60 60 8.1 103 96500

g

75. The stability of +1 oxidation state increases in the sequence

(1) Ga < ln < Al < Tl (2) Al < Ga < ln < Tl (3) Tl < ln < Ga < Al (4) ln < Tl < Ga < Al


Sol.: Ans. [2]

Down the gp; stability of lower oxidation state increases.

76. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radiusof copper atom in pm?(1) 108 (2) 128 (3) 157 (4) 181

Sol. Ans. [2]

Radius for FCC cube: 128414.12

36122

a

pm

77. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OHto a gas?(1) London dispersion force (2) Hydrogen bonding(3) Dipole-dipole interaction (4) Covalent bonds

Sol: [2]

Due to presence of intermolecular H-bonding; CH3OH is liquid at room temperature

78. Which of the following complex ions is expected to absorb visible light?(1) [Zn(NH3)6]2+ (2) [Sc(H2O)3(NH3)3]3+

(3) [Ti(en)2(NH3)2]4+ (4) [Cr(NH3)6]3+

(Atomic number Zn = 30, Sc = 21, Ti = 22, Cr = 24)

Sol: [4]

due to unpaired electrons, it shows d–d transition by absorbing visible light.

79. Out of 2 26 6TiF ,COF , Cu2Cl2 and 2

4NiCl (Z of Ti = 22, CO = 27, Cu = 29, Ni = 28) the colourlessspecies are:

(1) 3 26 4COF and NiCl (2) 2 3

6 6TiF and COF

(3) 22 2 5Cu Cl and NiCl (4) 2

6 2 2TiF and Cu Cl

Sol: [4]In 2

6TiF , Ti+4 exists, electronic configuration is 1s22s22p63s23p6

In Cu2Cl2 Cu+ exists. electronic configuration is 1s22s22p63s23p63d10

There are no unpaired electrons in both the cases.

80. Which of the following does not show optical isomerism?

(1) [CO(en)3]3+ (2) [CO(en)2Cl2]+

(3) [CO(NH3)3Cl3]0 (4) [CO(en)Cl2(NH3)2]+

(en = ethylenediamine)

Sol: [3]

81. Which one of the elements with the following outer orbital configurations may exhibit the largest numberof oxidation states?

(1) 3d24s2 (2) 3d34s2 (3) 3d54s1 (4) 3d54s2


Sol: [4] This is the outer orbital configuration of Mn; it shows variable oxidation states from +2 to +7.

82. Which of the following molecules acts as a Lewis acid?(1) (CH3)3N (2) (CH3)3B (3) (CH3)2O (4) (CH3)3P

Sol: [2]

Central atom Boron is electron deficient.

83. Amongst the elements with following electronic configurations, which one of them may have the highestionization energy?(1) Ne[3s23p1] (2) Ne[3s23p3] (3) Ne[3s23p2] (4) Ar[3d104s24p3]

Sol: [2]

Ne[3s23p3]

84. The straight chain polymer is formed by:(1) hydrolysis of (CH3)2 SiCl2 followed by condensation polymerisation(2) hydrolysis of (CH3)3 SiCl followed by condensation polymerisation(3) hydrolysis of CH3 SiCl3 followed by condensation polymerisation(4) hydrolysis of (CH3)4Si by addition polymerisation

Sol: [1]

hydrolysisSi Cl

CH3

CH3

Cl Si OH

CH3

CH3

OH Condensation Si O

CH3

CH3 n

85. The IUPAC name of the compound having the formula 2CH C CH CH is:

(1) 1-butene-3-yne (2) 3-butene-1-yne (3) 1-butyn-3-ene (4) but-1-yne-3-ene

Sol: [1]

86. Which of the following compounds will exhibit cis-trans (geometrical) isomerism?(1) 2 Butenol (2) 2-Butene (3) Butanol (4) 2-Butyne

Sol: [2]

2-Butene; 2 butenol may show geometrical isomerism but it is misprinted actually it should be2-butanol.

87. H2COH.CH2OH on heating with periodic acid gives:

(1) C O

H

H

2 (2) 2 CO2 (3) 2 HCOOH (4)CHO

CHO

Sol: [1]

4HIO2 32HCHO H O HIO

CH2

CH2

OH

OH


88. Consider the following reaction,

3 2 4

2

PBr i) H SO room temperaturealc.KOHii)H O, heatethanol X Y

the product Z is:(1) CH3CH2OH (2) CH2 = CH2

(3) CH3CH2 – O – CH2 – CH3 (4) CH3 – CH2 – O – SO3HSol: [1]

3 2 4

2

PBr i) H SO room temperatureKOH alc.3 2 3 2 2 2 3 2ii) H O;

(Y)(X) (Z)CH CH OH CH CH Br CH CH CH CH OH

89. Benzene reacts with CH3Cl in the presence anhydrous AlCl3 to form:(1) Xylene (2) Toluene (3) Chlorobenzene (4) Benzylchloride

Sol: [2]according to NCERT book; However xylene will also be formed because –CH3 group is weaklyactivating towards electrophilic substitution hence further electrophilic substitution occurs

3

3

CH Clanhy. AlCl

CH3

3

3

CH Clanhy. AlCl

CH3

CH3

CH3

CH3

90. Nitrobenzene can be prepared from benzene using a mixture of conc. HNO3 and conc. H2S. In themixture, nitric acid acts as a/an:

(1) catalyst (2) reducing agent (3) acid (4) base

Sol: [4]

HNO3 acts as a base because it accepts H+ from H2SO4

2H O2 4 2 2 2H SO HO NO H O NO NO

|H

91. Which of the following reactions is an example of nucleophilic substitution reaction?

(1) RX Mg RMgX (2) RX KOH ROH KX

(3) 2RX 2Na R R 2NaX (4) 2RX H RH HX

Sol: [2]

X– is substituted by OH– (nucleophile)

92. Which one of the following is employed a tranquilizer?

(1) Chloropheninamine (2) Equanil (3) Naproxen (4) Tetracycline

Sol: [2]

Equanil


93. Structures of some common polymers are give which one is not correctly presented?(1) Nylon 66

2 6 2 4NH(CH ) NHCO(CH ) CO 2

(2) Teflon

2 2CF CF n

(3) Neoprene

CH2 C CH CH2 CH2

Cl n

(4) Terylene

nOC C

O

O CH2 CH2 O

Sol: [1]

Nylon 66 is polymer of adipic acid and hexamethylene diamine it should be

2 6 2 4N H (C H ) N H C O (C H ) C O n

94. Predict the productNH

CH32NaNO HCl Product

(1)N

CH3

OH

(2)N

N

CH3

O

(3)N

NO2

CH3

(4)

NHCH3

NO

NHCH3

NOSol: [2]

2° amine with NaNO2 + HCl give Nitrosamine

95. Propionic acid with Br2 | P yields a dibromo product. Its structure would be:

(1) CH3 C

Br

Br

COOH (2) CH2Br – CHBr – COOH

(3) H C

Br

Br

CH2

COOH(4) CH2Br – CH2 – COBr


Sol: [1]

2Br(HVZ reaction)

CH3 CH2

C O

OH

C COOH

Br

Br

CH3

96. Trichloroacetaldehyde, CCl3CHO reacts with chlorobenzene in presence of sulphuric acid and produces:

(1) Cl CH

Cl

Cl

Cl

Cl

(2)Cl C

Cl

CH2Cl

Cl

(3)Cl C

H

Cl

Cl

(4) Cl C

OH

Cl

Cl

Sol: [1]

CH O

CCl

Cl

Cl

H Cl

H Cl

2 4dil. H SO

Cl

Cl

Cl

Cl

Cl97. Consider the following reaction:

3 4

3

CH Cl Alka line KMnOZn dust Anhydrous AlClPhenol X Y Z

the product Z is:(1) Benzene (2) Toluene (3) Benzaldehyde (4) Benzoic acid

Sol: [4]

ZnZnO

OH

3

3

CH Clanhy.AlCl(Friedel Craftreaction)

CH3

4alkaline KMnO

COOH


98. The state of hybridization of C2, C3, C5 and C6 of the hydrocarbon,

C CH

CH3

CH3

CH3 CH CH

CH3

C CH1234567

is in the following sequence:(1) sp, sp2, sp3 and sp2 (2) sp, sp3, sp2 and sp3 (3) sp3, sp2, sp2 and sp (4) sp, sp2, sp2 and sp3

Sol: [2] sp, sp3, sp2, sp3

99. The segment of DNA which acts as the instrumental manual for the synthesis of the protein is:(1) nucleoside (2) nucleotide (3) ribose (4) gene

Sol: [4] gene

100. Which of the following hormones contains iodine?(1) thyroxine (2) insulin (3) testosterone (4) adrenaline

Sol: [1] Thyroxine

2009 cbse pmt 2

Documents

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centre of mass

block of mass

3s14 s2

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s12 s2

competitive examinations

circular ring of mass