Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A small sphere of radius $r$ falls from rest in a viscous liquid. As a result, heat is produced due to the viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to:

$r^3$

$r^2$

$r^5$

$r^4$

Step-by-Step Solution

Terminal Velocity ( $v_t$ ): When a sphere of radius $r$ falls through a viscous fluid, it attains a constant terminal velocity given by the formula $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$ . This implies that $v_t \propto r^2$ .
Viscous Force ( $F$ ): According to Stokes' Law, the viscous drag force acting on the sphere moving with terminal velocity is $F = 6\pi \eta r v_t$ . Substituting the proportionality for $v_t$ , we get $F \propto r(r^2) \propto r^3$ .
Rate of Heat Production (Power): The rate of heat production is equal to the power ( $P$ ) dissipated by the viscous force, which is the product of the force and velocity: $P = F \times v_t$ .
Calculation: Substituting the proportionalities: $P \propto (r^3) \times (r^2)$ $P \propto r^5$ Thus, the rate of heat production is proportional to the fifth power of the radius.

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