Question

From the figure, a thin wire is stretched by a block going over a pulley. The wire vibrates in its fifth harmonic in resonance with a particular tuning fork. When a beaker containing a liquid of density $800{\text{kg/m}}^3$ is brought under the block so that the block is completely dipped into the beaker, the wire vibrates in its sixth harmonic in resonance with the same tuning fork. If the density of material of the block is $2.6\times {10}^x{\text{kg/m}}^3$, then what is the value of $x$?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Let $\rho$ be the density of block and $V$ be the volume of block.
So, weight of block $=\rho Vg$
Let the density of the liquid be ${\rho }_1$.


Let $T$ be the tension produced in the wire and $\mu$ be the linear mass density of wire.

So, from FBD of block (I)
$T=\rho Vg--\left(I\right)$
and from FBD of block (II)
$T^{\prime }+{\rho }_{1}Vg=\rho Vg$
$T^{\prime }=\rho Vg-{\rho }_{1}Vg--\left(II\right)$
[density of liquid ${\rho }_1=800{\text{kg /m}}^3]$

So from question, initially
$f_5=\frac{5}{2L}\sqrt{\frac{T}{\mu }}=\frac{5}{2L}\sqrt{\frac{\rho Vg}{\mu }}$ [From eq (I)]
When block is immersed in the liquid,
$f_6=\frac{6}{2L}\sqrt{\frac{T^{\prime }}{\mu }}=\frac{6}{2L}\sqrt{\frac{\rho Vg-{\rho }_{1}Vg}{\mu }}$ [From eq (II)]

As the tunning fork is the same in both cases, so
$f_5=f_6$
$\Rightarrow \frac{5}{2L}\sqrt{\frac{\rho Vg}{\mu }}=\frac{6}{2L}\sqrt{\frac{\rho Vg-{\rho }_{1}Vg}{\mu }}$
$\Rightarrow 5\sqrt{\rho }=6\sqrt{(\rho -{\rho }_1)}$
$\Rightarrow 5\sqrt{\rho }=6\sqrt{\rho -800}[\because {\rho }_1=800{\text{kg/m}}^3]$
$\Rightarrow 25\rho =36\rho -(36\times 800)$
$\Rightarrow \rho =2618.18{\text{kg/m}}^3=2.6\times {10}^3{\text{kg/m}}^3$
So, $x=3$.