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Laws of Motion question

2025 · 28 Jan · Shift 2 · Q57
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  5. /2025 · 28 Jan · Shift 2 · Q57

Laws of Motion question

2025 · 28 Jan · Shift 2 · Q57

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A balloon and its content having mass M is moving up with an acceleration ‘a’. The mass that must be released from the content so that the balloon starts moving up with an acceleration ‘3a’ will be (Take ‘g’ as acceleration due to gravity)
  1. A
    3Ma2a+g\frac{3Ma}{2a + g}2a+g3Ma​
  2. B
    2Ma3a+g\frac{2Ma}{3a + g}3a+g2Ma​
  3. C
    3Ma2a−g\frac{3Ma}{2a - g}2a−g3Ma​
  4. D
    2Ma3a−g\frac{2Ma}{3a - g}3a−g2Ma​
View written solutionFree

Correct answer: B

  1. Let the upward buoyant/lifting force be constant and equal to FFF.

    Initially, the balloon of mass MMM moves upward with acceleration aaa.

    Taking upward as positive, by Newton’s second law: F−Mg=MaF - Mg = MaF−Mg=Ma F=M(g+a)F = M(g+a)F=M(g+a)

  2. After releasing some mass

    Let the released mass be mmm.

    Then the new mass of the balloon system becomes: M−mM-mM−m

    The balloon now moves upward with acceleration 3a3a3a.

    Again applying Newton’s second law: F−(M−m)g=(M−m)3aF - (M-m)g = (M-m)3aF−(M−m)g=(M−m)3a

  3. Substitute F=M(g+a)F = M(g+a)F=M(g+a)

    M(g+a)−(M−m)g=3a(M−m)M(g+a) - (M-m)g = 3a(M-m)M(g+a)−(M−m)g=3a(M−m)

    Expand the left side: Mg+Ma−Mg+mg=3aM−3amMg + Ma - Mg + mg = 3aM - 3amMg+Ma−Mg+mg=3aM−3am Ma+mg=3aM−3amMa + mg = 3aM - 3amMa+mg=3aM−3am

  4. Collect terms containing mmm

    mg+3am=3aM−Mamg + 3am = 3aM - Mamg+3am=3aM−Ma m(g+3a)=2Mam(g+3a) = 2Mam(g+3a)=2Ma

    Therefore, m = rac{2Ma}{g+3a}

    Rearranging: m = rac{2Ma}{3a+g}

  5. Match with the options

    This corresponds to: 2Ma3a+g\boxed{\frac{2Ma}{3a+g}}3a+g2Ma​​

    So the correct option is B.

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