Top Banner

of 7

2012 Sample Physics Solution 2

Apr 03, 2018

Download

Documents

Yash Paunikar
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
  • 7/28/2019 2012 Sample Physics Solution 2

    1/7

    I ITJ EE PHYSICS SAMPLE PAPER - ISOLUTIONS

    SECTIONI

    Straight Objective Type

    1. uM

    vM

    Tdt22

    C

    Tdt

    v

    ..... (i)

    gll

    gu2

    2

    MvTdt v

    Tdt

    A

    T1

    M ..... (ii)

    MaMgT1

    Mg

    T1

    B

    a..... (iii)

    al

    v2

    From (i) and (ii) uM

    vM

    22

    3

    3

    uv

    999

    2 g

    l

    gl

    l

    ua .

    (b)

    2. All the equipotential surfaces of the field between the

    sphere and the plate are convex down ward. Hence on

    any straight line parallel to plate the points farther from

    the sphere will have potential lower than those closer to

    sphere. 1 2

    V3V2 V1

    (b)

    3. Torque about O is zero as well angular momentum hence = 0.

    (d)

    4. )(2

    1)(

    2

    1

    2vvMyHMg 21

    2

    1Mv

    wIvM cc2

    1)(

    2

    1 2 .....(i)

    cMvMv12 ..... (ii)

    vvc2 ..... (iii)

    yvv1

    Solving (i), (ii) and (iii) )(4

    3 2

    yHgv g

    H

    tyH

    g

    dt

    dy

    v

    3

    )(3

    4

    (d)

  • 7/28/2019 2012 Sample Physics Solution 2

    2/7

    5. Both upper half and lower half will have same effective area of

    2

    2

    R so charge in flux will be same and induced emf will have

    some value. But since the resistance is different due to which

    current must be different but ring is as a whole is closed circuit

    so electric field will be generated to make the current flow in

    both parts to be same.

    010riRE ..... (i)

    0riRE ..... (ii)

    092 irRE

    R

    ir

    E 2

    9

    r

    bR

    r

    E

    i 1111

    2

    r

    bR

    R

    rE

    112

    9 2= Rb

    22

    9

    10 r

    i10r

    E

    +

    E

    +

    (b)

    6. twwtatwaE 00 coscoscos )]cos()[cos(2

    cos 000 wwwwa

    twa

    Highest possible energy for photon corresponds to frequency 0ww hence.

    Qww

    hKE

    2

    )( 0max

    (c)

    7. ghhrWST2 ,

    2

    2 ghhrWg

    Heat = Q = gST WW =2

    22 ghrand

    gr

    Sh

    2

    g

    SQ

    22

    (a)

    8.svhat

    22

    1 sp vh

    vat2

    21 tt

    1

    2

    2

    s

    p

    s

    p

    v

    v

    hv

    v

    a

    (a)

  • 7/28/2019 2012 Sample Physics Solution 2

    3/7

    SECTION II

    Reasoning Type

    9. PV= nRT

    V

    nR

    T

    P

    )(

    1

    slopeV

    P

    T

    12

    v2 = constant

    v1 = constant

    (b)

    10. Potential atEandKare different due to which current flows betweenEandKmake current

    flow betweenAB and CD also possible.

    (d)

    11. Assertion and Reason correct and correct explanation.

    (a)

    12. Acceleration relative to cart parallel to incline is always zero only the acceleration

    perpendicular to incline will change in different situation due to which change in tension but

    angle will remain same and string always remain perpendicular to the incline.

    (d)

    SECTION III

    Linked Comprehension Type

    Passage-IWhen the temperature of rods in increased there will be increase in their lengths and thereby

    the springs are compressed, let21 , xx and 3x be the compression in the three springs

    respectively. ThenL

    Kx1 2Kx2 2Kx2 3Kx3

    2212

    xxxTL

    TL

    21 2KxKx and 32 32 KxKx

    321 32 xxx

    TLxx

    x2

    3

    32

    111

    TLx

    11

    91

  • 7/28/2019 2012 Sample Physics Solution 2

    4/7

    2

    2

    1111

    9

    2

    1

    2

    1TLKKxE 222 )(

    242

    81TKL

    Similarly

    222

    2484

    81TKLE

    222

    3242

    27TKLE

    13. (b)

    14. (a)

    15. (a)

    Passage-II16. The charge q on the small disc can he calculated by applying Gauss law

    qdsE 0

    Since one side of the small disc is in contact with plate

    xvVd

    rrEq

    2

    0

    2

    0 d

    rx 0

    2

    .

    (b)

    17. Let sv be the steady velocity of the small disc just after its collision with the bottom plate

    then the steady state kinetic energy sk of the disc just above the bottom plate is given by

    2

    2

    1ss mvk

    for each round trip disc gains electrostatic energy by

    U= 2qV

    afterbeforeloss kkk beforeke )1(2

    afterk112

    .

    Since sk is the energy after collision at the bottom plate and )( mgdqVks is the

    energy before the collision at the top plates total energy loss can be written as

    )()1(1

    1 22 mgdqVkkk sstot

    In its steady state Ushould be compensated by totalk

    )()1(11

    2 22

    mgdqVkkqV ss

    ])1()1[(1

    22

    4

    2

    mgdqVks

    mgdqVmvs 2

    2

    2

    22

    112

    1

    )2(1

    2

    1 22

    2

    2gd

    m

    xVvs

  • 7/28/2019 2012 Sample Physics Solution 2

    5/7

    2Vvs .

    (c)18. The disc will lose kinetic energy eventually cease to move when the disc cannot reach the

    top plate. In other words, the threshold voltage cV can be determined from the condition

    that velocity of the disc at the top plate is zero. In order to have velocity zero at top plate

    kinetic energy at top plate must satisfy the relation

    0mgdqVkK css

    sk is steady state kinetic energy at the bottom plate

    011 2

    2

    2

    2

    mgdqVmgde

    eqV

    e

    ecc

    xmgd

    eeVc 2

    2

    11

    (d)

    SECTION IV

    Matrix Match Type

    1.

    Concave Convex

    (A) m < 0 |m| < 1

    Object as well image is real in case of concave mirror object and image is virtual

    incase of convex.

    (B) m < 0 |m| > 1

    Both real in case of concave both virtual incase of convex.

    (C) |m| < 1 m > 0

    Concave object virtual image real.

    Convexobject real image virtual.

    (D) |m| > 1 m > 0

    Concave object real image virtual.

    Convex object virtual, image real.

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

  • 7/28/2019 2012 Sample Physics Solution 2

    6/7

    2.

    q, v, m

    y0O

    F

    B

    2B

    x

    R/2

    MA

    Ry0B

    R

    C

    D

    qB

    mvR ; 1cos 0

    R

    yR

    = 2 ;)2( Bq

    mtAB ;

    )(Bq

    mtBC

    ))1((cos3)3( 1

    qB

    m

    qB

    mttT BCAB

    sinRXA ; sin2RABXX AB

    sin4RBCXX BC

    sin5RCDXX CD

    When velocity be come parallel

    qB

    m

    qB

    m

    Bq

    m

    qB

    mtt AMFA

    2

    33

    22

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

    SECTION V

    Subjective or Numerical Problems

    1. 2Tcos = W+ 2w

    2Ncos (90

    ) = W2Nsin = W

    Taking torque aboutB

    T AB sin 2 = cossin2

    1OBNABw

    T 4 sin 2 = w 2 sin +N rcot

    (AB = 4 m ; OB = rcosec )

    Solving from above r= 3m

    D

    2

    90

    CA

    90

    O

    NN

    w w

    W

    B

  • 7/28/2019 2012 Sample Physics Solution 2

    7/7

    2.a

    xay seclog ; w

    td

    d(Constant)

    a

    x

    dx

    dytan ;

    a

    x

    adx

    yd 22

    2

    sec1

    Radius of curvature =

    2

    2

    2 23

    1

    dx

    yd

    dx

    dy

    =a

    xa sec

    a

    x

    dx

    dytantan

    a

    x; x = a

    awdt

    da

    dt

    dx; 0

    2

    2

    dt

    xd; aw

    a

    x

    dt

    dx

    dx

    dy

    dt

    dytan

    dxaa

    xaw

    dt

    yd 1sec

    2

    2

    2

    a

    xaw 22 sec

    Now resultant acceleration

    2

    2

    22

    2

    2

    dt

    yd

    dt

    xda

    a

    xwa 442 sec0

    22

    22 sec Ra

    w

    a

    xaw

    a

    a

    4sec

    2

    1

    2

    1

    22 22

    = 2m/sec424

    18