### JEE Main On 08 April 2018 Question 15

##### Question: A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $ 2.7\times 10^{3}kg/m^{3} $ and its Youngs modulus is $ 9\text{.27}\times 1{0^{10}}\text{ Pa} $ . What will be the fundamental frequency of the longitudinal vibrations? [JEE Main Online 08-04-2018]

#### Options:

A) $ 10kHz $

B) $ 7.5kHz $

C) $ 5kHz $

D) $ 2.5kHz $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

$ \ell =60cm $ $ \rho =2.7\times 10^{3}kg/m^{3} $

$ y=9.27\times 10^{10}Pa $

For fundamental frequency $ \frac{\lambda }{2}=\ell \Rightarrow \lambda =2\ell $

$ f=\frac{v}{\lambda }\Rightarrow f=\frac{1}{2\ell }\sqrt{\frac{Y}{\rho }} $

$ f=\frac{1}{2\times 06}\sqrt{\frac{9.27\times 10^{10}}{2.7\times 10^{3}}} $

$ f\approx 5kHz $