Laws of Motion

  • Question 545
    CBSEENPH11020751

    A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are F1, F2 and F3 respectively and are in the plane of the paper and along the directions shown, the force on the segment of QP is

    • F3 - F1-F2

    • square root of left parenthesis straight F subscript 3 minus straight F subscript 2 right parenthesis squared space plus straight F subscript 2 squared end root
    • square root of left parenthesis straight F subscript 3 minus straight F subscript 1 right parenthesis squared minus straight F subscript 2 squared end root
    • F3 - F2 + F1

    Solution

    B.

    square root of left parenthesis straight F subscript 3 minus straight F subscript 2 right parenthesis squared space plus straight F subscript 2 squared end root
    F4 sin θ = F2
    F4 cos θ = (F3-F1)
    straight F subscript 4 space equals space square root of left parenthesis straight F subscript 3 minus straight F subscript 1 right parenthesis squared space plus straight F subscript 2 squared end root
    Question 546
    CBSEENPH11020752

    Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed is

    • 0.5 N

    • 1.5 N

    • fraction numerator square root of 3 over denominator 4 end fraction space straight N
    • square root of 3 space straight N

    Solution

    B.

    1.5 N

    Minimum additional force needed
    F = - (Fresultant)x
    straight F subscript resultant space equals space left square bracket left parenthesis 4 minus 2 right parenthesis left parenthesis cos space 30 space bold j with hat on top space minus space sin space 30 space bold i with hat on top right parenthesis space plus space left parenthesis cos space 60 space bold i with hat on top space plus sin space 60 space bold j with hat on top right parenthesis
space equals space open square brackets 2 open parentheses fraction numerator square root of 3 over denominator 2 end fraction bold j with hat on top minus 1 half bold i with hat on top close parentheses space plus open parentheses 1 half bold i with hat on top space plus fraction numerator square root of 3 over denominator 2 end fraction bold j with hat on top close parentheses close square brackets
equals open square brackets open parentheses square root of 3 space plus fraction numerator square root of 3 over denominator 2 end fraction close parentheses bold j with hat on top space plus open parentheses negative bold i with bold hat on top space plus fraction numerator bold i with bold hat on top over denominator 2 end fraction close parentheses close square brackets
equals open square brackets negative 1 half bold i with bold hat on top space plus fraction numerator 3 square root of 3 over denominator 2 end fraction bold j with hat on top close square brackets
equals fraction numerator bold i with hat on top over denominator 2 end fraction plus fraction numerator 3 square root of 3 over denominator 2 end fraction space bold j with hat on top
therefore comma
straight F space equals space minus space open parentheses fraction numerator bold i with hat on top over denominator 2 end fraction close parentheses space equals space 1 half bold i with bold hat on top
Hence comma space space vertical line straight F vertical line space equals space 0.5 space straight N

    Question 547
    CBSEENPH11020753

    A shell of mass 200 g is ejected from a gun of mass 4 g by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is

    • 100 ms-1

    • 80 ms-1

    • 40 ms-1

    • 120 ms-1

    Solution

    A.

    100 ms-1

    In the given problem conservation of linear momentum and energy hold good.
    Conservation of momentum yields.
    m1v1 + m2v2 = 0
    4v1 + 0.2 v2  = 0
    Conservation of energy yields

    1 half straight m subscript 1 straight v subscript 1 squared space plus 1 half space straight m subscript 2 straight v subscript 2 squared space equals space 1050 space space space... left parenthesis straight i right parenthesis
or
1 half space straight x 4 straight v subscript 1 squared space plus space 1 half 0.1 straight v subscript 2 squared space equals space 1050 space space space.... space left parenthesis ii right parenthesis

solving space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space have
straight v subscript 1 space equals space 100 space straight m divided by straight s

    Question 548
    CBSEENPH11020759

    Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be

    • Mv newton

    • 2 Mv newton

    • Mv/2 Newton

    • zero

    Solution

    A.

    Mv newton

    Force required,
    straight F space equals space fraction numerator straight d left parenthesis mv right parenthesis over denominator dt end fraction
equals space straight v open parentheses dm over dt close parentheses space equals space vM

    as velocity v is constant, hence
    F= Mv newton

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