(a) Two bodies \(\mathrm{A}\) and \(\mathrm{B}\) fall freely from heights \(\mathrm{h}_{1}\) and \(\mathrm{h}_{2}\) respectively, where, \(\mathrm{h}_{1}>\mathrm{h}_{2} .\) Which body will take more time to reach the ground? Explain. (b) What is the weight of a body on the moon whose weight is \(12 \mathrm{kgwt}\) on the earth?

Assuming both bodies fall in a vacuum and under the influence of the Earth's gravitational force, we can use the formula: \(t = \sqrt{\frac{2h}{g}}\) where t is the time taken, h is the height and g is the acceleration due to gravity (\(9.81 m/s^2\) on Earth). For body A, since it is falling from height \(h_1\), we'll call the time it takes to reach the ground \(t_1\): \(t_1 = \sqrt{\frac{2h_1}{g}}\) For body B, since it is falling from height \(h_2\), we'll call the time it takes to reach the ground \(t_2\): \(t_2 = \sqrt{\frac{2h_2}{g}}\) Now we need to find which of these two times is greater.

We are given that \(h_1 > h_2\). Since both heights are positive and the numerator of each term is multiplied by 2, we can deduce that numerators in both the expressions \(2h_1\) and \(2h_2\) are positive and that \(2h_1 > 2h_2\). Now, since both bodies are falling under Earth's gravity and experiencing the same acceleration due to gravity (g), the denominators in both expressions are the same. Since \(2h_1 > 2h_2\), the fraction \(\frac{2h_1}{g}\) is greater than \(\frac{2h_2}{g}\). When we take the square root of both expressions, since square roots are increasing functions, this inequality remains the same. Therefore, \(t_1 > t_2\), indicating that body A will take more time to reach the ground than body B.

On Earth, the weight of the body is given by \(12 \mathrm{kgwt}\). The weight of the body on the moon is \(\frac{1}{6}\)th of its weight on Earth. To find the weight of the body on the moon, we'll multiply its weight on Earth by the ratio of the moon's gravity to Earth's gravity: Weight on Moon = Weight on Earth * \(\frac{Gravity_{Moon}}{Gravity_{Earth}}\) Weight on Moon = \(12 \mathrm{kgwt} * \frac{1}{6} = 2 \mathrm{kgwt}\) Therefore, the weight of the body on the moon is \(2 \mathrm{kgwt}\).