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MCQMediumJEE 2025Newton's Second Law & Force

JEE Physics 2025 Question with Solution

A balloon and its content having mass MM is moving up with an acceleration aa. The mass that must be released from the content so that the balloon starts moving up with an acceleration 3a3a will be:

  • A

    3Ma2a+g\frac{3Ma}{2a + g}

  • B

    3Ma2a−g\frac{3Ma}{2a - g}

  • C

    2Ma3a+g\frac{2Ma}{3a + g}

  • D

    2Ma3a−g\frac{2Ma}{3a - g}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A balloon of mass MM moves upward with acceleration aa. After releasing mass mm, it moves upward with acceleration 3a3a.

Find: The mass released.

Let the upward buoyant force be FBF_B. For the initial motion, applying Newton's second law:

FB−Mg=MaF_B - Mg = Ma

Therefore,

FB=M(g+a)F_B = M(g+a)

After releasing mass mm, the new mass becomes M−mM-m. The buoyant force remains the same, so:

FB−(M−m)g=(M−m)3aF_B - (M-m)g = (M-m)3a

Substitute FB=M(g+a)F_B = M(g+a):

M(g+a)−(M−m)g=(M−m)3aM(g+a) - (M-m)g = (M-m)3a Mg+Ma−Mg+mg=3a(M−m)Mg + Ma - Mg + mg = 3a(M-m) Ma+mg=3aM−3amMa + mg = 3aM - 3am

Rearranging,

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

Hence,

m=2Ma3a+gm = \frac{2Ma}{3a+g}

Therefore, the mass to be released is 2Ma3a+g\frac{2Ma}{3a+g}. The solution working gives this value, which matches option C, although the solution incorrectly marks the correct option as A.

Using constant buoyant force

Given: Initial mass MM, initial upward acceleration aa, final upward acceleration 3a3a after releasing some mass.

Find: Released mass.

The key idea is that the displaced air volume does not change, so the buoyant force stays constant.

Initial condition:

FB−Mg=MaF_B - Mg = Ma FB=M(g+a)F_B = M(g+a)

Final condition after releasing mass mm:

FB−(M−m)g=(M−m)3aF_B - (M-m)g = (M-m)3a

Putting the value of FBF_B from the initial condition:

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

Expanding both sides:

Mg+Ma−Mg+mg=3aM−3amMg + Ma - Mg + mg = 3aM - 3am Ma+mg=3aM−3amMa + mg = 3aM - 3am

Collecting terms containing mm on one side:

mg+3am=2Mamg + 3am = 2Ma m(g+3a)=2Mam(g+3a) = 2Ma m=2Mag+3am = \frac{2Ma}{g+3a}

Thus, the required released mass is 2Mag+3a\frac{2Ma}{g+3a}, so the correct option is C.

Common mistakes

  • Assuming the buoyant force changes after releasing mass. This is wrong here because the balloon volume is taken to remain the same, so the upthrust stays constant. Use the same upward force in both equations.

  • Using the new mass as M+mM+m instead of M−mM-m. The balloon loses mass, so its mass must decrease after release.

  • Trusting the marked option letter without checking the algebra. The page labels option A, but the derived expression is 2Mag+3a\frac{2Ma}{g+3a}, which corresponds to option C in the given options.

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