JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 9)
A clock has a continuously moving second's hand of 0.1 m length. The average acceleration of the
tip of the hand (in units of ms–2) is of the order of :
10-3
10-1
10-2
10-4
Explanation
R = 0.1 m
$$\omega $$ = $${{2\pi } \over T}$$ = $${{2\pi } \over {60}}$$ = 0.105 rad/sec
a = $${\omega ^2}R$$
= (0.105)2(0.1)
= 0.0011
= 1.1 $$ \times $$ 10-3
Average acceleration is of the order of 10–3.
$$\omega $$ = $${{2\pi } \over T}$$ = $${{2\pi } \over {60}}$$ = 0.105 rad/sec
a = $${\omega ^2}R$$
= (0.105)2(0.1)
= 0.0011
= 1.1 $$ \times $$ 10-3
Average acceleration is of the order of 10–3.
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