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प्रश्न
A solid sphere moves at a terminal velocity of 20 m s−1 in air at a place where g = 9.8 m s−2. The sphere is taken in a gravity-free hall having air at the same pressure and pushed down at a speed of 20 m s−1.
(a) Its initial acceleration will be 9.8 m s−2 downward.
(b) It initial acceleration will be 9.8 m s−2 upward.
(c) The magnitude of acceleration will decrease as the time passes.
(d) It will eventually stop
उत्तर
(b) There is no gravitational force acting downwards. However, when the starting velocity is 20 m/s, the viscous force, which is directly proportional to velocity, becomes maximum and tends to accelerate the ball upwards.
\[\text{ When the ball falls under gravity, }\]
\[\text{ neglecting the density of air: } \]
\[\text{ Mass of the sphere = m }\]
\[\text{ Radius = r }\]
\[\text{ Viscous drag coeff . }= \eta\]
\[\text{Terminal velocity is given by}: \]
\[\text{ mg }= 6\pi\eta r v_T \]
\[ \Rightarrow \frac{6\pi\eta r v_T}{m} = g . . . (1)\]
\[\text{ Now, at terminal velocity, the acceleration of the ball due to the viscous force is given by: } \]
\[a = \frac{6\pi\eta r v_T}{m}\]
\[\text{ Comparing equations (1) and (2), we find that : } \]
\[ \text{ a = g }\]
Thus, we see that the initial acceleration of the ball will be 9.8 ms - 2 .
(c) The velocity of the ball will decrease with time because of the upward viscous drag. As the force of viscosity is directly proportional to the velocity of the ball, the acceleration due to the viscous force will also decrease.
(d) When all the kinetic energy of the ball is radiated as heat due to the viscous force, the ball comes to rest.
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