Question

One end of a horizontal string is attached to vibrating blade and the other end passes over pulley as shown in the figure. A Sphere of mass 2 kg hangs at the end of the string. The string is vibrating in its second harmonic. A container of water is raised under the sphere so that the sphere is completely submerged. In this configuration, the string vibrates in its fifth harmonic. What is the radius of the square?

• A. 5.38 cm
• B. 6.38 cm
• C. 7.38 cm
• D. 4.38 cm

Answer 1

The correct option is C 7.38 cm

$\begin{array}{c}CaseI:T=2g\\ v=\sqrt{\frac{2g}{\mu }}\\ f=2.\frac{v}{2t}=\frac{2.}{2l}\sqrt{\frac{2g}{\mu }}\\ caseII:\\ f=5.\frac{v^{\prime }}{2t}=\frac{5}{2l}\sqrt{\frac{T^{\prime }}{\mu }}\\ \Rightarrow \frac{5}{2l}\sqrt{\frac{T^{\prime }}{\mu }}=\frac{2}{2l}\sqrt{\frac{T}{\mu }}\\ \Rightarrow 5\sqrt{T^{\prime }}=2\sqrt{T}\\ \Rightarrow 25T^{\prime }=4T\\ \Rightarrow T^{\prime }=\frac{4}{25}.2g\\ \Rightarrow v\left[\rho -\sigma \right]=\frac{8}{25}\\ \Rightarrow v\rho -v\sigma =\frac{8}{25}\\ \Rightarrow 2-\frac{4}{3}\pi R^3.\left(1000\right)=\frac{8}{25}\\ \therefore R=7.38cm\end{array}$

$\therefore$ Option $C$ is correct.