A body falls freely under the action of gravity from a height \(\mathrm{h}\) above the ground. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) P.E. }=2 \text { (K.E.) } & \text { (P) constant at every point } \\ \hline \text { (b) P.E. }=\text { K.E. } & \text { (Q) at height }(\mathrm{h} / 3) \\ \hline \text { (c) P.E. }=(1 / 2) \text { (K.E.) } & \text { (R) at height }(2 \mathrm{~h} / 3) \\ \hline \text { (d) P.E.+ K.E. } & \text { (S) at height }(\mathrm{h} / 2) \\ \hline \end{array} $$ (A) \(a-P, b-Q, c-R, d-S\) (B) $\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{S}, \mathrm{d}-\mathrm{R}$ (C) $\mathrm{a}-\mathrm{S}, \mathrm{b}-\mathrm{R}, \mathrm{c}-\mathrm{Q}, \mathrm{d}-\mathrm{P}$ (D) $\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{S}, \mathrm{c}-\mathrm{Q}, \mathrm{d}-\mathrm{P}$

Step by step solution

01

Conservation of mechanical energy

As the body falls freely under the action of gravity, we can use the conservation of mechanical energy principle. The total mechanical energy of the body (the sum of its potential energy [P.E.] and kinetic energy [K.E.]) remains constant throughout the fall. Mathematically, this can be expressed as: \[P.E. + K.E. = constant\] We will use this equation to analyze the given expressions in the table.

02

Determine Potential Energy and Kinetic Energy expressions at different heights

We will determine the expressions for potential energy (P.E.) and kinetic energy (K.E.) at different heights and then compare them with the expressions given in Column - i of the table. At height h, the body has no initial velocity, so its potential and kinetic energies are: \(P.E. = mgh\) \(K.E. = 0\) At height \(\frac{h}{3}\), we have: \(P.E. = mg\frac{h}{3}\) Using the conservation of mechanical energy: \(P.E. + K.E. = mgh\) \(mg\frac{h}{3} + K.E. = mgh\) Solving for K.E., we get: \(K.E. = \frac{2}{3}mgh\) At height \(\frac{2h}{3}\), we have: \(P.E. = mg\frac{2h}{3}\) Using the conservation of mechanical energy: \(P.E. + K.E. = mgh\) \(mg\frac{2h}{3} + K.E. = mgh\) Solving for K.E., we get: \(K.E. = \frac{1}{3}mgh\)

03

Compare expressions in Column - i with the derived potential energy and kinetic energy expressions

Now, we will compare the expressions we derived in Step 2 with the expressions given in Column - i of the table. (a) P.E. = 2(K.E.) This expression is true for height \(\frac{h}{3}\): \(mg\frac{h}{3} = 2(\frac{2}{3}mgh)\) (b) P.E. = K.E. This expression is true for height \(\frac{2h}{3}\): \(mg\frac{2h}{3} = (\frac{1}{3}mgh)\) (c) P.E. = (1/2)(K.E.) This expression is not true at any of the heights we calculated. (d) P.E. + K.E. This expression is true at all heights, as it represents the conservation of mechanical energy: \(P.E. + K.E. = constant\)

04

Match the expressions with corresponding heights

Now we will match the expressions in Column - i with the corresponding heights in Column - ii. (a) P.E. = 2(K.E.): At height \(\frac{h}{3}\) (P) (b) P.E. = K.E.: At height \(\frac{2h}{3}\) (R) (c) P.E. = (1/2)(K.E.): This expression does not match any of the heights we considered. (d) P.E. + K.E.: This is constant at every point (S) Therefore, the correct answer is: (A) a-P, b-R, d-S

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