Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

An electric fan has blades of length $30\text{ cm}$ as measured from the axis of rotation. If the fan is rotating at $1200\text{ r.p.m}$ , the acceleration of a point on the tip of the blade is about:

$1600\text{ m/s}^2$

$4740\text{ m/s}^2$

$2370\text{ m/s}^2$

$5055\text{ m/s}^2$

Step-by-Step Solution

Identify Given Values: Radius ( $R$ ) = length of blade = $30\text{ cm} = 0.3\text{ m}$ . Frequency ( $n$ ) = $1200\text{ revolutions per minute (r.p.m)}$ .
Convert Units: Convert frequency to revolutions per second: $v = \frac{1200}{60} = 20\text{ rev/s}$ . Calculate angular velocity ( $\omega$ ): $\omega = 2\pi v = 2\pi(20) = 40\pi\text{ rad/s}$ .
Calculate Acceleration: The centripetal acceleration ( $a_c$ ) is given by the formula $a_c = \omega^2 R$ . $a_c = (40\pi)^2 \times 0.3$ $a_c = 1600\pi^2 \times 0.3$ $a_c = 480\pi^2$ Using $\pi^2 \approx 9.87$ : $a_c \approx 480 \times 9.87 = 4737.6\text{ m/s}^2$ . This value is approximately $4740\text{ m/s}^2$ .

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