Question

Find the magnitude of average acceleration of
minute hand of clock (length=20cm) rotate an
angle of a
(A) 5.5 m/s2
(B) 5.6 m/s
(C) 6.5 m/s2
(D) 7.5 m/s2​

Answer 1

Explanation:

dV = $\sqrt{V^{2} + V^{2} + 2VVcos180 }$

            = $V\sqrt{2}$

Time taken

15 x 60 = 900s

Average = $\frac{dV}{dT}$

= $\frac{V\sqrt{2} }{900}$

= $\frac{(W\sqrt{2})(0.2) }{900}$

= $\frac{(\frac{180}{3600})(0.2)(\sqrt{2} ) }{900}$

= 5.48 = 5.5

Answer 2

Answer:

Hence, value of magnitude of average acceleration will be $5.5\:m/s^2$

Explanation:

As per the data given in the question,

We have to determine the magnitude of average accleration of minute hand.

It is given that,

Length of minute hand = 20 cm

We know that,

$dV = \sqrt{V^{2} + V^{2} + 2VVcos180^{\circ}}\\dV = \sqrt{V^{2} + V^{2} + 0}\\ dV={\sqrt{2V^2}}\\dV=V\sqrt{2}$

So, the time taken by minute hand will be: $15 \times 60 = 900\:s$

Hence, the average accleration will be $= \frac{dV}{dT}$

So, computing the value we will get,

$= \frac{V\sqrt{2} }{900}\\\\= \frac{(W\sqrt{2})(0.2) }{900}\\\\= \frac{(\frac{180}{3600})(0.2)(\sqrt{2} ) }{900}\\\\= 5.48 = 5.5\:m\s^2$