unit 6 section c success magnetchcvhvhvhchchchccchc

7/30/2019 Unit 6 Section C Success Magnetchcvhvhvhchchchccchc

1/9

AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623456 Fax : 47623472

(589)

Brain Storming

Comprehensions UNIT6Linked Comprehension Type

C1. 1. Answer (4)

A

B C

P

Q

R

6030

30

60

By Snells law, angles made by BCand QRwith narmal are 30, as shown. Thus, it can be seen from

the figure that angle between BCand QRis 30 + 30 = 60

2. Answer (1)

A

B

C

W

X

Y Z

Grazingemergence

60

90

30

A'

P'

Q'

B'

From geometry

(a) BCfalls normally on interface XW and suffers no deviation.

(b) For AB to be parallel to PQAB should have grazing emergence. Thus by snells law, for refraction

at XZ,

. sin 60 = 1.sin 90

2

3

3. Answer (1)

For no emergence at surfaceXZ,

60 > Qc

sin60 sin cQ

1sin60

cosec60

2

3


2/9


(590)

Success Magnet (Solutions) Brain Storming Comprehensions

C2. 1. Answer (3)

Fringe width = F=d

D

dv

dtdFR

And a = kt

ktdt

dv

2

2ktv

22

2 tRd

kt

dt

dF

R

2. Answer (1)

xcomponent of velocity = velocity of screen =2

2kt

Also, y=d

Dn

ycomponent of velocity =dt

dy

=d

ktn

2

2

j

d

ni

ktv

2

2

3. Answer (4)

Position vector of nth maxima relative to (n 1)th maxima = jd

D

jd

ktj

dt

Dd

da

2

2

C3. 1. Answer (1)

xm

tvtx

2

)1000.1)(10672.1(2

)500.0)(.10054.1(1027

34

mkg

ssjx

x 1.58 102 m


3/9


(591)

Brain Storming Comprehensions Success Magnet (Solutions)

2. Answer (4)

Uncertainity in position

x = 0.53 1010 m

25109.92

x

p kg.m/s

3. Answer (3)

Uncertainity in energy

E = h

Ast

1

thE

E . t h

E . t 2

C4. 1. Answer (1)

As only electrostatic (conservative) force acts, by conservation of energy, final kinetic energy = Initial

kinetic energy. As initial and final interaction potential energies are zero due to large separation

2. Answer(1)

Initial angular momentum about gold nucleus = mvr

= xkm.2

Torque of electrostatic force about nucleus is zero

Angular momentum about nucleus is constant = xkm.2

3. Answer(1)

F

r

x

Gold nucleus(Fixed)

x2

p

In the diagram here, = angle between initial and final velocity vectors of the -particle


4/9


(592)


Change in momentum of the particle )( p

is

pfinal

pinitial

p

As pfinal

=pinitial

=p

2sin2pp

Also, 2 + =

2

By the diagram at any general instant force acting on the particle is F. Its component in the direction

of p

is F cos . Thus

2

2

)(

cos2

sin2 dtFpp

Now as L = constant,

mr2 = mvxdt

dmr

2

2r

vx

dt

d

dvx

rdt

2

2

2

)(

2

.cos2

sin2 dbx

rFpp

Asp = mvand F = 2r

kqQ

2

2

)(

2

cos2

sin2

dkqQ

xmv

=

2

cos2

kqEx

qxmv .24

2cos

2


5/9


(593)


C5. 1. Answer(1)

x

P

dF dF

l

By

xO

C60

Pand Q are two symmetrically located points and magnetic force is shown on them. x- components

of forces on Pand Q cancel out. So, net force is along +ve y-direction.

Magnetic field at point P, B =)sin(2

xl

i

dF= Bidx

=)sin(2

2

xl

dxi

)sin(2

coscos

2

xl

dxidFFd

on integrating fromx= 0 tox= l

2

31ln

.32

2iF

Total Force =

2

31ln3

22iF

2. Answer(1)

Initial angular momentum of conductor ABC is zero about Q.

Due to symmetry, net torque on conductor ABC about point Q is zero.

Angular momentum about Q remains zero.

3. Answer(3)

Centre of mass (CoM) of conductor ABC lies on line BQ. CoM accelerates along +ve y direction.

Total angular momentum = Angular momentum about CoM + Angular momentum of CoM about P.

First term is zero.

So, L = MVcm

.r ; where r= PQ

Vcm

increases, So angular momentum (L) increases.

C6. 1. Answer(2)

For Rotational equilibrium,

mgsin R= MB sin

(M= iN. R2)

RiNmg

iNRmgRB

2


6/9


(594)


2. Answer(3)

f= mgsin

mg cos = mgsin

tan =

3. Answer(2)

T= MBsin ; B =2

B

=2

sinmgR

ImgR

fR2

sin----- (i)

mgsin f= ma ---- (ii)

a = R ---- (iii)

Solving (i), (ii), (iii) we get

a = sin14

5g

C7. 1. Answer(1)

For minimum energy,j= 0

2

hL . Use

I

Lk

2

2

2. Answer(1)IfKT< (KE)R, there will be no rotation.

3. Answer(2)

For nitrogen, minimum KE of rotation is very low.

C8. 1. Answer(1)

N

A B

4I I

N

For the disc A

NRt= 4I1

NRt= I(+ 2)

41

= + 2

------- (i)

21

12

uu

vve

21

21 2202

1

R

RR ------- (ii)


7/9


(595)


4 1 2 =

2 1 + 2 2 =

10

31

52

If the next collision take place ofn1

revolutions of the first disc and n2

revolution of the second disc then

n1 2 = n2 1

2

3

2

1

n

n

2. Answer(2)

So the time after which next collison will take place

= 32

1

or 2

2

2

= 3103

2

=

20

and T=

So = 20 T

3. Answer(2)

IfI1

= I2

and e = 1 then

I 1 = I ( + 2)

10

21

1 2 =

1 + 2 =

1 = and 2 = 0

So next collision will take place after

T22


8/9


(596)


C9. 1. Answer(1)

(V V)5 = V 1 5

4

V

V

2. Answer(2)

Acceleration = drag force / mass = 6r

m

VV

20

0

3. Answer(4)

Velocity increases exponentially from zero to V0.

C10. 1. Answer (3)

dV= Edx

2/3

2/

2

1

a

a

V

V

EdxdV

V2

V1

=

222

000

aEaE

4

021

aEVV

2. Answer (1)

For the gaussian cylinder

E0

A

x a=

E0/20

00

2

AA

EAE

2

3 00E

3. Answer (3)

Consider a gaussian cylinder if thickness dxand area

A. for this surface,

0

)()(

AdxAdEEEA

E

x a= 3

E dE+

000 45tan dx

dE

C11. 1 Answer (3)

For a horizontal axis through C,

222 l

mmRdmRI


9/9


(597)


2. Answer (1)

For the position of centre of mass, we have

)cos1(

sin

2

2

2

20 R

d

dR

x

Also,

l

R

3. Answer (3)

Mgd

IT 2

Here d= Rx0, I= I

cm+ md2 = (MR2 mx

02) + md2

C12. 1. Answer (2)

For equilibrium

Mg N

m l g

E

l

2

mgE

2

2. Answer (2)

The rope starts slipping when Mg= N

i.e. m l g= (lE)

mgE

Now, Edecays exponentially from mg2

Time taken to become half is

2ln

3. Answer (2)

The acceleration becomes non-zero for the first time when

mgE . Also, finally the acceleration

becomes g, as the rope comes out of the electric field.

unit 6 section c success magnetchcvhvhvhchchchccchc

Documents