Question

In a sonometer wire the fundamental frequency of the wire is 260 Hz. In sonometer the tension is maintained by suspending a 50.7 kg mass. When the suspended mass is completely submerged in water the fundamental frequency becomes 240 Hz. The suspended mass has a volume

Answer 1

Solution:

Tension in Wire , Tₐ = mg = 50.7 x 10 = 507 N

Tension in Wire after mass is completely submerged, $T_{w}$= Tₐ -loss in weight

                                                                                          = 507 - loss in weight

Now,

$\frac{n_{w} }{n_{a} } = \sqrt{\frac{T_{w} }{T_{a} } }$

Squaring both sides

$(\frac{n_{w} }{n_{a} })^{2} = {\frac{T_{w} }{T_{a} } }$

$(\frac{240}{260})^{2} = {\frac{507 - loss\ in\ weight }{507 } }$

$(\frac{12}{13})^{2} = {\frac{507 - loss\ in\ weight }{507 } }$

$\frac{144}{169} = {\frac{507 - loss\ in\ weight }{507 } }$

$\frac{144 * 507}{169} = {507 - loss\ in\ weight }$

$\frac{144 * 39}{13} = {507 - loss\ in\ weight }$

$144 * 3 = {507 - loss\ in\ weight }$

$432 = {507 - loss\ in\ weight }$

$- 75 = - loss \ in \ weight$

Loss in weight will tell us about the volume of suspended mass = m * 10³ * g

[We used 10³ as we have to take out volume]

$m * 10^{3} * g = 75$

$m = \frac{75}{10^{3} * 10 }$

$m = 0.0075\ m^{3}$

The suspended mass has a volume of 0.0075 m³