JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 27)
The end correction of a resonance column is 1 cm. If the shortest length resonating with the tunning fork is 10 cm, the next resonating length should be :
28 cm
32 cm
36 cm
40 c
Vysvětlení
Given, End correction (e) = 1 cm
For first resonance,
$${\lambda \over 4} = {l_1} + e$$ = 10 + 1 = 11 cm
For second resonance,
$${3\lambda \over 4} = {l_2} + e$$
$$ \Rightarrow $$ $${l_2}$$ = 3 $$ \times $$ 11 - 1 = 32 cm
For first resonance,
$${\lambda \over 4} = {l_1} + e$$ = 10 + 1 = 11 cm
For second resonance,
$${3\lambda \over 4} = {l_2} + e$$
$$ \Rightarrow $$ $${l_2}$$ = 3 $$ \times $$ 11 - 1 = 32 cm