solutions to mock iit advanced/test - 3[paper-1]/2013100p.s3. solutions to mock iit advanced/test -

Download Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013100p.s3. Solutions to Mock IIT Advanced/Test -

If you can't read please download the document

Post on 18-Jun-2020

1 views

Category:

Documents

0 download

Embed Size (px)

TRANSCRIPT

  • Vidyamandir Classes

    VMC/2013/Solutions 1 Mock IIT Advanced/Test - 3/Paper-1

    Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013

    [CHEMISTRY]

  • Vidyamandir Classes

    VMC/2013/Solutions 2 Mock IIT Advanced/Test - 3/Paper-1

  • Vidyamandir Classes

    VMC/2013/Solutions 3 Mock IIT Advanced/Test - 3/Paper-1

  • Vidyamandir Classes

    VMC/2013/Solutions 4 Mock IIT Advanced/Test - 3/Paper-1

  • Vidyamandir Classes

    VMC/2013/Solutions 5 Mock IIT Advanced/Test - 3/Paper-1

  • Vidyamandir Classes

    VMC/2013/Solutions 6 Mock IIT Advanced/Test - 3/Paper-1

  • Vidyamandir Classes

    VMC/2013/Solutions 7 Mock IIT Advanced/Test - 3/Paper-1

    Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013

    [PHYSICS]

    24.(A)

    1 1 1

    20 10V − = −

    20 1V m= = −

    Co-ordinate is (20 cm, 0.2 cm).

    25.(C) For inductor di

    V L dt

    = −

    Or L V

    di dt L

    = −∫ ∫

    V = (10 t − 20) where 0 4t ms≤ < ( ) 2

    0

    1 1 10 10 20 20

    2

    t

    L t

    i t dt t L L

      ⇒ = − + = − − 

       ∫

    25 20

    L t t

    i L L

    = − +

    26.(B) Force of buoyancy

    ( )0 0BF A x L x gρ ρ = + −  Weight sW LA gρ=

    For equilibrium BF W=

    ( )0 0SLA g A x L x gρ ρ ρ ⇒ = + −  ( )0s wL x L xρ ρ ρ⇒ = + −

    ( )0 0

    1000 800 2

    1000 300 7

    w s s w w

    w

    x x

    L L

    ρ ρ ρ ρ ρ ρ

    ρ ρ − −

    ⇒ = − + ⇒ = = = − −

    27.(C) x a cos tω= v a sin tω ω= −

    1 1 1

    2 2

    a a cos t cos tω ω= ⇒ =

    1 1 3 3 2 6

    T T t t

    π π ω

    π ⇒ = ⇒ = =

    ×

    ( ) 6

    0

    6

    0

    3

    T /

    T /

    a sin t dt

    a v

    T dt

    ω ω−

    < >= = ∫

    28.(A) Given 2 4y x= −

    When 0 2x y= , = ±

    The separation between open ends = 4.

    Magnetic force on a current carrying element placed in a uniform field B is ( ) ɵ ɵ( )3 4 5 60i L B j k i× = × = �� ��

    ɵ

    29.(B) Here at the instant of sliding limiting friction f Nµ= will act on the two blocks for limiting equilibrium, we use:

    F = T Nµ+ …(i)

    and T = Nµ …(ii)

    and 2T + N = 50 …(iii)

    On solving (ii) and (iii), we get

    50

    25 2 1

    N N µ

    = = +

    Now from (i) and (ii)

    F = 2 Nµ = 2 0 5 25 25. N× × =

  • Vidyamandir Classes

    VMC/2013/Solutions 8 Mock IIT Advanced/Test - 3/Paper-1

    30.(C)

    31.(B) After 2 sec speed of boy will be 2 2 4v m / s= × =

    At this moment centripetal force on boy is 2 30 16

    80 6

    r mv

    F N R

    × = = = .

    Tangential force on boy is 30 2 60tF ma N= = × = .

    Total friction acting on boy is 2 2 100r tF F F N= + = .

    At the time of slipping F mgµ=

    or 100 30 10µ= × × ⇒ 1

    3 µ = .

    32.(B)

    33.(C) By conserving angular momentum and linear momentum and also using concept of coefficient restitution, we get

    1

    21 cmv v

    e v cos

    ω

    θ

    + + = =

    . . . . . .(i)

    1cmmv cos mv mvθ = − . . . . . .(ii)

    2

    1 2 12 2

    m mv cos mvθ ω= −

    ℓ ℓ ℓ . . . . . .(iii)

    Now solving, we get : 2

    5 cmv v cosθ= ,

    12 ω

    5 v cosθ= ℓ

    , 1 3

    5 v v cosθ= −

    34.(B) Change in angular momentum of the rod is 2 12 1

    Δ 12 5 5

    cm m

    L I v cos m v cosω θ θ= = × = ℓ

    ℓ ℓ

    35.(B) 1 3 2

    Δ 5 5

    v v cos v v cos v cos v cosθ θ θ θ= + = − = 36.(AB)

    37.(AB) ( )

    ( ) 11 1

    20 20 1

    R

    F R

    µ

    µ

    − = = ⇒

    − . . . . .(i)

    If equiconvex lens of f = 20

    Then ( )1 21 20 R

    µ= −

    From eq. (1) and (2) we conclude.

    ∴ R = 40 ( )1µ − . . . . .(ii) That option (A) is correct

    After Silvering :

    10Feq cm= − (Combination will behave like a concave mirror)

    (B) 15u = −

    1 1 1 1 1 1

    15 10eqv u F v + = ⇒ − = −

    1 1 1

    30 So Bis correct 15 10

    v cm v = − ⇒ = −

    (C) 20u = − 20v = − So C is wrong

    Alternative :

    For planoconvex ( ) 1 2

    1 1 1 1

    f R R µ

      = − − 

     

    1 2R R R= ∞ = r = 20 ( )1µ −

    For equiconvex ( )1 1 11 20 R R

    µ  = − +   

    ( )40 1R µ= −

  • Vidyamandir Classes

    VMC/2013/Solutions 9 Mock IIT Advanced/Test - 3/Paper-1

    38.(ABD)

    Speed will be maximum when the block passes through the equilibrium position.

    springF mg=

    (B) From conservation of energy,

    ( ) 2 21 1 2 2

    maxmg l x Kx mv+ = + . . . .(i)

    where 4 4

    mg mg l x l

    K mg = = =

    Putting 4

    l x = in equation (1), we get

    3

    2 maxv gL=

    (C) is incorrect as in equilibrium compression is 4l / which is more than 8l / .

    (D) Time taken is 1

    2 4 4 4

    T m L

    K g

    π π= =

    39.(BD) Equivalent circuit :

    Induced emf in a rotating rod with constant angular

    Velocity 2ω

    ω 2

    B a e; = , (here a = Radius)

    Now use Kirchoff’s Law.

    40.(3) 1v

    m u n

    = = −

    ( )0

    1L

    L

    X m

    – X X n = = −

    Lens formula ⇒ ( )0

    1 1 1

    L LX X X f − = − −

    ( ) 0

    2 2 0

    1 1 0L L

    L L

    dXdX dX .

    dt dt dtX X X

      − − − = 

     −

    ( ) ( ) 0

    2 2 2 0 0

    1 1 1L

    L L L

    dXdX

    dt dtX X X X X

       − + =  − − 

    0

    0; LdX dXV V

    dt dt = =

    ⇒ 2 2

    0 1 1

    1V V n n

         − + + = +         

    ⇒ [ ]0 0 4

    1 4 4 3

    V V V V− + = ⇒ = ⇒ 0

    3

    4

    V V =

    41.(1) In first case 10

    30 10 1 1

    K R

    × = − × +

    In second case 5

    10 5 2 2

    K R

    × = − × +

    Solving above two equations we get : 1R =

    42.(1) Work done ΔQ= (heat dissipated) Let e : induced emf

    = 2 2( )e N B v

    t t R R

    × = ×

    ℓ =

    2 2 3 2100 (0 4) 2 5 10 1

    10 000

    . . v

    ,

    −× × × × ×

    2 32 5 10. −= ×ℓ

    4 2

    25 10 5 10 m − −= × = ×ℓ

    25 10

    1 v

    −× =

    ⇒ 86 25 10 16 1W . J Jµ−= × × =

  • Vidyamandir Classes

    VMC/2013/Solutions 10 Mock IIT Advanced/Test - 3/Paper-1

    43.(9) 1

    2 2 3

    2 0 3

    2

    P

    R kQ

    kQ E

    RR = − =      

    ⇒ 12 4

    9 2

    Q Q =

    2

    1

    9 =

    8

    Q

    Q

    2 9 8

    9 8

    Q Cµ ×

    = =

    44.(8) Initially 0100 2x g=

    0 20x cm=

    Finally, equilibrium position

    0100 1x g′ =

    0 10′ =x cm

    1

    2 100 10

    T π

    π 2

    = =

    2 10

    T t

    π = =

    21

    10 50 2 100

    S cm π

    = × × =

    Total distance = 50 + 10 + 20 = 80cm.

    So, x = 8

    45.(4) Paschen series 7 6 2

    1 1 1 1 097 10 10

    9 .

    nλ  

    = × − ≤   

    2

    1 1 1

    9 10 97.n − ≤

    2

    1 1 97

    9 10 97

    .

    .x ≥

    ×

    29 1097

    197 x

    × ≥

    3 1097

    7 197

    n≤ ≈ ⇒ 7 3 6 3 5 3 4 3 4lines− , → , − , − ⇒

    46.(1) 2 2 3

    24 (ω )

    10 10 z L R

    − = = +

    ×

    2 2 2 2 1(ω ) ω

    ω R L R L

    C

      + = + − 

     

    ( ) 1ω ω ω

    L L C

    = − +

    2 6

    1 1 5

    2ω 2 100 100 10 L H

    C π π − = = =

    × × ×

    ( ) ( )2 2 22400 500 Rπ= +

    R = 2 2(2400) (5 100)π− × = 2 210 (24) 25π−

    = 10 326 180× ≈

    I = 1A

  • Vidyamandir Classes

    VMC/2013/Solutions 11 Mock IIT Advanced/Test - 3/Paper-1

    Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013

    [MATHEMATICS]

    47.(B) Let ( )2 2P t , t and 21 2Q , tt  

    −   

Recommended

View more >