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AIEEE-2012-3 Ltd , FIITJEE House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph: 46106000, 26569493, Fax: 26513942 website: www fiitjee com (4) Statement 1 is true, statement 2 is false 4 2 Sol Statement 1 has 20 terms whose sum is 8000 And statement 2 is true and supporting statement 1

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RESONANCE Page # 7 9 A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure The mass is undergoing circular motion is the x-y plane with centre at O and constant angular speed

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RESONANCEPage # 2

PHYSICS

PART -  : PHYSICS

Section I : Single Correct Answer Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)

out of which ONLY ONE is correct.

1.Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of

potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45° to

the vertical JUST after release. Then X is nearly (A) 1 × 10-5 V(B) 1 × 10-7 V(C) 1 × 10-9 V(D) 1 × 10-10 V

Ans.(C)

Sol. mg = qE 1.67

× 10-27 × 10 = 1.6 × 10-19 × 01.0X

X = 9106.167.1l V

X = 1

× 10-9 V

2 .A mixture of 2 moles of helium gas (atomic mass = 4 amu), and 1 mole of argon gas (atomic mass = 40 amu)

is kept at 300 K in a container. The ratio of the rms speeds )) 99
8 )on(argv)helium(vrmsrms is : (A) 0.32(B) 0.45(C) 2.24(D) 3.16

Ans.(D)

Sol. ArHe

RmsRmsmRT3mRT3

v v ArHe  = 10440 mm HeAr  e 3.16

RESONANCEPage # 3

3.A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other end of the

spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2

m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency s/rad3.

Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 45º as shown in the

figure. Point P is at a horizontal distance of 10 cm from O. If the pebble hits the block at t = 1s, the value of v is

(take g = 10 m/s2) (A) s/m50(B) s/m51(C) s/m52(D) s/m53

Sol.Time of flight for projectile

T = gsinu2  = 1 sec. g45sinu2 = 1 sec. u = 2g u = 50 m/s

4.Three very large plates of same area are kept parallel and close to each other. They are considered as ideal

black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures

2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is

(A) T265 41)(
:(B) T497 41)(
:(C)T297 41)(
:(D)T9741

Ans.(C)

Sol. In steady state energy absorbed by middle plate is equal to energy released by middle plate. A(3T)4 - A(T")4 = A(T")4 - A(2T)4 (3T)4 - (T")4 = (T")4 - (2T)4 (2T")4 = (16 + 81) T4

T" = T297

4/1

PHYSICS

RESONANCEPage # 4

5.A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed , as shown in the

figure. At time, t = 0, a small insect starts from O and moves with constant speed v with respect to the rod

towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains

 throughout. The magnitude of the torque (||) on the system about O, as a function of time is best represented

by which plot ? (A) (B) (C) (D)

Ans.(B)

Sol.

L = [m(vt)2]

L = mv2t2

So = dtdL = 2mv2t

 k t

Bstraight line passing through (0, 0)

6.In the determination of Young"s modulus )(

2ldMLg4Y by using Searle"s method, a wire of length L = 2 m and

diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension  = 0.25 mm in the length of the wire is

observed. Quantities d and  are measured using a screw gauge and a micrometer, respectively. They have the

same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum

probable error of the Y measurement

PHYSICS

RESONANCEPage # 5

(A) due to the errors in the measurements of d and  are the same.

(B) due to the error in the measurement of d is twice that due to the error in the measurement of .

(C) due to the error in the measurement of  is twice that due to the error in the measurement of d.

(D) due to the error in the measurement of d is four time that due to the error in the measurement of .

Ans.(A)

Sol. d =  = 1005.0 mm y =

2dMLg4

99
8 maxyy + dd2 error due to  measurement mm25.0mm100/5.0 error due to d measurement 25.0100/5.0 mm5.01005.02 dd2 l So error in y due to  measurement = error in y due to d measurement

7.Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface charge

density. The variation of the magnitude of the electric field )r(E and the electric potential V(r) with the distance r from the centre, is best represented by which graph? (A) (B) (C) (D)

Ans.(D)

PHYSICS

RESONANCEPage # 6

Sol.

8.A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first

lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R =

14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be

(A) -280.0 cm(B) 40.0 cm(C) 21.5 cm(D) 13.3 cm

Ans.(B)

Sol. '65

7

211R1

R1)1(f1

'65 7 1

141)15.1(f1

1 145.0
f11 '65 7 142.0
f12 21f1
f1 f1 = 142.0

145.0

147.0
f1 401
1407
v1 = 401

2014012

v1 v = 40 cm

PHYSICS

RESONANCEPage # 7

9.A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass

is undergoing circular motion is the x-y plane with centre at O and constant angular speed . If the angular

momentum of the system, calculated about O and P are denoted by OL and PL respectively, then

(A) OL and PL do not vary with time. (B) OL varies with time while PL remains constant. (C) OL remains constant while PL varies with time. (D)

OL and PL both vary with time.

Ans.(C)

10.

Young"s double slit experiment is carried out by using green, red and blue light, one color at time. The fringe

widths recorded are G, R and B, respectively. Then (A)

G > B > R(B) B > G > R(C) R > B > G(D) R > G > B

Ans.(D)

Sol.  = dD

VIBGYOR  increase

R > G > BSoR > G > B

Section II : Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out

of which ONE or MORE are correct.

11.A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down the pipe.

When this pulse reaches the other end of the pipe. (A) a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is open. (B) a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is open. (C) a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed. (D) a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed.

Ans.(B), (D)

Sol.

At open end phase of pressure wave charge by  so compression returns as rarefraction. While at closed

end phase of pressure wave does not change so compression return as compression.

PHYSICS

RESONANCEPage # 8

12.A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle  with the horizontal. A

horizontal force of 1 N on the block through its center of mass as shown in the figure. The block remains

stationary if (take g = 10 m/s2) (A)  = 45° (B)  > 45° and a frictional force acts on the block towards P. (C)  > 45° and a frictional force acts on the block towards Q. (D)  < 45° and a frictional force acts on the block towards Q.

Ans.(A), (C)

Sol. f = 0, If sin = cosB = 45° f towards Q, sin > cosB > 45° f towards P, sin < cosB < 45°

13.A cubical region of side a has its centre at the origin. It encloses three fixed point charges, -q at (0, -a/4, 0), +

3q at (0, 0, 0) and -q at (0, +a/4, 0). Choose the correct option(s).

(A) The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane

x = -a/2.

(B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing the plane

y = -a/2. (C) The net electric flux crossing the entire region is 0 q

(D) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the plane

x = +a/2.

Ans.(A), (C), (D)

PHYSICS

RESONANCEPage # 9

Sol.Position of all the charges are symmetric about the planes x = 2a  and x = 2a . So net electric flux through them will be same.

Similarly flux through y =

2a  is equal to flux through y = 2a 000in qqqq3q

By symmetry flux through z = 2a

 is equal to flux through x = 2a

14.For the resistance network shown in the figure, choose the correct option(s).

(A) The current through PQ is zero. (B) I

1 = 3 A.

(C) The potential at S is less than that at Q. (D) I

2 = 2 A.

Ans.(A), (B), (C), (D)

Sol. Due to input and output symmetry P and Q and S and T have same potential.

PHYSICS

RESONANCEPage # 10

Req = 18126

l = 4 1 = 412 = 3A 2 = 312612l)( 2 = 2 A V

A - VS = 2 × 4 = 8V

VA - VT = 1 × 8 = 8V

VP = VQ B Current through PQ = 0(A)

V

P = VQ B VQ > VS(C)

I

1 = 3A(B)

I

2 = 2A(D)

15.Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and

magnetic fields jˆEE0 and jˆBB0. At time t = 0, this charge has velocity v in the x-y plane, making an

angle  with x-axis. Which of the following option(s) is(are) correct for time t > 0 ? (A) If  = 0°, the charge moves in a circular path in the x-z plane. (B) If  = 0°, the charge undergoes helical motion with constant pitch along the y-axis.

(C) If  = 10°, the charge undergoes helical motion with its pitch increasing with time, along the y-axis.

(D) If  = 90°, the charge undergoes linear but accelerated motion along the y-axis.

Ans.(C), (D)

Sol.

If  = 0° then due to magnetic force path is circular but due to force qE0 (s) q will have accelerated motion along

y-axis. So combined path of q will be a helical path with variable pitch so (A) and (B) are wrong.

If  = 10° then due to vcos, path is circular and due to qE0 and vsin, q has accelerated motion along y-axis so

combined path is a helical path with variable pitch (C) is correct. If  = 90° then FB = 0 and due to qE0 motion is accelerated along y-axis. (D)

PHYSICS

RESONANCEPage # 11

Section III : Integer Answer Type

This section contains 5 question. The answer to each question is a single digit integer, ranging from 0

to 9 (both inclusive).

16.A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge.

It makes a closest approach of 10 fm to the nucleus. The de Brogle wavelength (in units of fm) of the proton at

its start is : (take the proton mass, m p = (5/3) × 10-27 kg, h/e = 4.2 × 10-15 J.s/C ;

041 = 9 × 109 m/F ; 1 fm = 10-15 m)

Ans.7 Sol.

1591010ee120109l

l= m2p 2 ph 22 2 hp :l= 22m2h (120) (3)10-27+15+92 = (4.2)2 × 10-30

330210360102.42.4

l ll= 29103604242ll = 72 ×10-30 = 7×10-15 m = 7 fm

17.A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius

2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is IO and

IP, respectively. Both these axes are perpendicular to the plane of the lamina. The ratio

OPII to the nearest integer

is : Ans.3

PHYSICS

RESONANCEPage # 12

Sol.

I0 = 2)R2()m4(

2 - 23 mR2 = mR2 [8 - 23]

213mR2

IP = 23 (4m) (2R)2 - 

66
5

7]R)R2[(m2mR222

= 24 mR2 - 2mR211 = 2mR237

213237

I I OP  = 31337e

Ans. 3

PHYSICS

RESONANCEPage # 13

18.An infinitely long solid cylinder of radius R has a uniform volume charge density . It has a spherical cavity of

radius R/2 with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field

at the point P, which is at a distance 2R from the axis of the cylinder, is given by the expression

0k16R23

. The value of k is Ans.6

E1 = R2.R.02

E2 = 23

0)R2(8R.34.

.41

E1 - E2 = 424.R.

4R00l

'65

7

24114R0

096R23

0K16R23

BK = 6

19.A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a shown in the figure. Both the cylinder

and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the

magnetic field at the point P is given by 12N0 aJ, then the value of N is : Ans5

PHYSICS

RESONANCEPage # 14

Sol12J

2JB aoao1 12aJ5 2a J 65
6112a
J

0oo)(

: = 12N0aJ

N = 5

20.A circular wire loop of radius R is placed in the x-y plane centered at

the origin O. A square loop os side a (a << R) having two turns is placed with its center at a = 3R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45° with respect to the z-axis. If the mutual inductance between the loops is given by R2 a 2/p2 0 , then the value of p is Ans.7 Sol.

B = 2/3222

0)XR(2iR

B = 2/3222

0)R3R(2iR

 = 2/322

0)R4(2iR

 = R16i

R.2.2iR

0 32

0

 = NBA cos45° = 21aR16i2

20

R28ia 2

0

M = i

R2aM2/72

0  = R2a2/P2 0 P = 7

PHYSICS

RESONANCEPage # 15

PART - II : CHEMISTRY

SECTION - I : Single Correct Answer Type

This section contains 10 multiple choice questions, Each question has four choices, (A), (B), (C) and (D)

out of which ONLY ONE is correct.

21.Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

(A) HNO3, NO, NH4Cl, N2(B) HNO3, NO, N2, NH4Cl (C) HNO3, NH4Cl, NO, N2(D) NO, HNO3, NH4Cl, N2

Ans.(B)

Sol.

HNO3 = + 5

NO = +2

NH

4Cl = -3

N2 = 0

So correct order will be HNO

3, NO, N2, NH4Cl.

22.The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius] :

(A) 2022ma4h(B) 2022ma16h (C) 2022ma32h(D) 2022ma64h

Ans.(C)

Sol. mv (4a0) =  h so, v =

0am4h

soKE = 21 mv2 = 21 m. 20222am16h = 2022am32h

23.The number of aldol reaction (s) that occurs in the given transformation is :

CH3CHO + 4HCHO bbbbbbbrbNaOH.aq.conc

(A) 1(B) 2(C) 3(D) 4

Ans.(C)

CHEMISTRY

RESONANCEPage # 16

Sol.CH3-CHO + HCHO

oncondensatialdolst1OH bbrb oncondensatialdolnd2HCHO/OHquotesdbs_dbs19.pdfusesText_25