JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 25)
A wire of length L and mass per unit length
6.0 × 10–3 kgm–1 is put under tension of
540 N. Two consecutive frequencies that it
resonates at are : 420 Hz and 490 Hz. Then L
in meters is :
5.1 m
2.1 m
1.1 m
8.1 m
Explanation
Fundamental frequency = 70 Hz.
70 = $${1 \over {2l}}\sqrt {{T \over \mu }} $$
$$ \Rightarrow $$ $$l$$ = 2.14 m
70 = $${1 \over {2l}}\sqrt {{T \over \mu }} $$
$$ \Rightarrow $$ $$l$$ = 2.14 m
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