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Question

Length of a minute hand of a clock is $$4.5\ cm$$. Find average velocity of time of the minute hand between $$6\ AM$$. to $$6.30\ A.M$$ and $$6\ A.M.$$ to $$6.30\ P.M$$.


Solution

 Given that,

Time $$t = 30\ min = 1800\ sec$$

Now, the length of the minute hand is almost as long as the radius of the clock

The displacement is

  $$ D=2\times 4.5 $$

 $$ D=9\,cm $$

 $$ D=0.09\,m $$

Now, the velocity is

  $$ v=\dfrac{D}{t} $$

 $$ v=\dfrac{0.09}{1800} $$

 $$ v=5\times {{10}^{-5}}\,m/s $$

Hence, the average velocity is $$5\times {{10}^{-5}}\,m/s$$


Displacement during $$6:00 \ AM$$ to $$6:30\ PM$$ will be:

$$D=4.5\times 2=9cm$$

$$D=0.09 \ m$$

Time taken:

$$t=12.5$$ Hours.

$$t=12.5\times 3600$$ sec

Therefore,

Velocity will be given by:

$$v=\dfrac{0.09}{12.5\times 3600}=2\times 10^{-6}\ m/s$$

  


Physics

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