### JEE Main On 16 April 2018 Question 17

##### Question: The end correction of a resonance column is $ 1,cm $ If the shortest length resonating with the tuning fork is $ 10,cm, $ the next resonating length should be [JEE Main 16-4-2018]

#### Options:

A) 32 cm

B) 40 cm

C) 28 cm

D) 36 cm

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

Given that end correction $ e=1,cm $ and shortest resonating length= $ 10cm $

$ \frac{\lambda }{4}=l _1+e $ $ \frac{\lambda }{4}=10+1=11cm $

$ \lambda =44cm $ Let the next resonating length be $ \frac{3\lambda }{4}=l _2+e $

$ 33=l _2+1 $ $ l _2=32,cm $

Let the shortest resonating length be $ l _1 $ And the next be $ l _2 $

Then $ , $ $ l _1+e=\frac{\lambda }{4} $ ………… (1)

$ l _2+e=\frac{3\lambda }{4} $ ………….(2)

Here, e = end correction On dividing (1) by (2)

$ \frac{l _1+e}{l _2+e}=\frac{1}{3} $ $ \frac{10+1}{l _2+1}=\frac{1}{3} $

$ 33=l _2+1 $ $ l _2=32cm $ the end correction given is 1 cm

so the shortest length resonating will be one fourth of wavelength

so the wavelength comes out to be 44cm the nest resonating length will be three fourth of wavelength

so it will be 33cm and end correction given is 1cm so the next length resonating will be $ 33-1=32,cm $

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so the correct option is A.
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