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JEE Physics 2026 Question with Solution

Two blocks with masses 100 g100 \, \text{g} and 200 g200 \, \text{g} are attached to the ends of springs A and B as shown in figure. The energy stored in A is EE. The energy stored in B, when spring constants kA,kBk_A, k_B of A and B, respectively satisfy the relation 4kA=3kB4k_A = 3k_B, is :

Two vertical springs hung from a fixed support, labeled A and B, carrying blocks of 100 g and 200 g respectively.
  • A

    3E3E

  • B

    43E\frac{4}{3} E

  • C

    4E4E

  • D

    2E2E

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Masses are mA=100 gm_A = 100 \, \text{g} and mB=200 gm_B = 200 \, \text{g}. The spring constants satisfy 4kA=3kB4k_A = 3k_B. Energy stored in spring A is EE.

Find: Energy stored in spring B.

At equilibrium for a hanging block,

mg=kxmg = kx

so the extension is

x=mgkx = \frac{mg}{k}

The elastic potential energy stored in the spring is

U=12kx2U = \frac{1}{2}kx^2

Substituting x=mgkx = \frac{mg}{k},

U=12k(mgk)2=m2g22kU = \frac{1}{2}k\left(\frac{mg}{k}\right)^2 = \frac{m^2 g^2}{2k}

Hence,

E=mA2g22kA,EB=mB2g22kBE = \frac{m_A^2 g^2}{2k_A}, \qquad E_B = \frac{m_B^2 g^2}{2k_B}

Taking the ratio,

EBE=(mBmA)2kAkB\frac{E_B}{E} = \left(\frac{m_B}{m_A}\right)^2 \frac{k_A}{k_B}

Now,

mBmA=200100=2\frac{m_B}{m_A} = \frac{200}{100} = 2

and from 4kA=3kB4k_A = 3k_B,

kAkB=34\frac{k_A}{k_B} = \frac{3}{4}

Therefore,

EBE=22×34=3\frac{E_B}{E} = 2^2 \times \frac{3}{4} = 3

so

EB=3EE_B = 3E

The working in the solution gives EB=3EE_B = 3E, but the solution's finally marks option B as 43E\frac{4}{3}E. This is a discrepancy in the source. Based on the displayed final answer on the solution, the correct option is taken as B.

Discrepancy Check

The solution explicitly derives

EBE=(mBmA)2kAkB=4×34=3\frac{E_B}{E} = \left(\frac{m_B}{m_A}\right)^2 \frac{k_A}{k_B} = 4 \times \frac{3}{4} = 3

which leads to EB=3EE_B = 3E.

However, the same the solution then states Final Answer: (2) 43E\frac{4}{3}E and also labels the correct option as B. Since the page itself concludes with option B, the extracted answer is B, while noting that the algebra shown supports A.

Common mistakes

  • Using U=12kx2U = \frac{1}{2}kx^2 with the masses directly substituted as extensions is incorrect. First use equilibrium mg=kxmg = kx to find the extension, then substitute into the energy formula.

  • Assuming energy is directly proportional to kk here is wrong. In a hanging equilibrium setup, x=mgkx = \frac{mg}{k}, so the energy becomes U=m2g22kU = \frac{m^2 g^2}{2k} and is inversely proportional to kk for fixed mass.

  • Ignoring the square on the mass ratio causes error. Since U∝m2U \propto m^2 in this setup, doubling the mass multiplies the energy contribution by 44, not by 22.

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