Q69 / 90 · JEE Main 2024 (05 Apr Shift 2) · Physics

Match List-I with List-II : $\begin{array}{|c|c|c|c|} \hline & \text { List-I } & & \text { List-II } \\ \hline \text { (A) } & \text { A force that restores an elastic body of unit area to its original state } & \text { (I) } & \text { Bulk modulus } \\ \hline \text { (B) } & \text { Two equal and opposite forces parallel to opposite faces } & \text { (II) } & \text { Young's modulus } \\ \hline \text { (C) } & \begin{array}{l} \text { Forces perpendicular everywhere to the surface per unit area } \\ \text { same everywhere } \end{array} & \text { (III) } & \text { Stress } \\ \hline \text { (D) } & \text { Two equal and opposite forces perpendicular to opposite faces } & \text { (IV) } & \text { Shear modulus } \\ \hline \end{array}$ Choose the correct answer from the options given below :

  • A
    (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  • Correct answer
    B
    (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  • C
    (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  • D
    (A)-(III), (B)-(I), (C)-(II), (D)-(IV)

Solution

(1) stress $=\frac{F_{\text {restoring }}}{A}$ If $\mathrm{A}=1$ Stress $=F_{\text {restoring }}$ (A)-(III) (2)Figure
(B)-(IV) (3)Figure
(C)-(I) (4)Figure
(D)-(II)

Hence, the correct answer is option: 'B'.

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