Numerical Problems-1 - Chapter 7 - Properties of Matter - 11th Science Guide Samacheer Kalvi Solutions

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Numerical Problems
Question 1.
A capillary of diameter dmm is dipped in water such that the water rises to a height of $30 \mathrm{~mm}$. If the radius of the capillary is made of its previous value, then compute the height up to which water will rise in the new capillary?
Answer:
We know the height of the capillary rise $h=\frac{2 \mathrm{~T} \cos \theta}{r d g}$
where $\mathrm{T}=$ surface tension, $r=$ radius, $d=$ density, $g=$ acceleration due to gravity
Now,
Here
$
\begin{aligned}
h_1 r_1 & =h_2 r_2 \Rightarrow \frac{h_2}{h_1}=\frac{r_1}{r_2} \\
r_2 & =\frac{2}{3} r_1 \text { and } h_1=30 \mathrm{~mm} \\
h_2 & =\frac{r_1}{\left(\frac{2}{3}\right) r_1} \times 30=\frac{3}{2} \times 30 \\
h_2 & =45 \mathrm{~mm}
\end{aligned}
$
Question 2.
A cylinder of length $1.5 \mathrm{~m}$ and diameter $4 \mathrm{~cm}$ is fixed at one end. A tangential force of $4 \times 10^5 \mathrm{~N}$ is applied at the other end. If the rigidity modulus of the cylinder is $6 \times 10^{10} \mathrm{Nm}^{-2}$ then, calculate the twist produced in the cylinder.

Answer:
$
\begin{aligned}
& \tau=\frac{\pi \eta r^4 \phi}{2 l} \\
& \text { Twist produced in the cylinder } \phi=\frac{\tau \times(2 l)}{\pi \eta r^4}=\frac{(\mathrm{F} \times l) \times(2 l)}{\pi \eta r^4} \\
& =\frac{4 \times 10^5 \times 2 \times(1.5)^2}{3.14 \times 6 \times 10^{10} \times\left(2 \times 10^{-2}\right)^4}=\frac{18 \times 10^5}{3.01 \times 10^{-6} \times 10^{10}} \\
& \phi=59.80 \\
&
\end{aligned}
$
Torsion of a cylinder
Question 3.
A spherical soap bubble A of radius $2 \mathrm{~cm}$ is formed inside another bubble B of radius $4 \mathrm{~cm}$. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than radius of both soap bubbles A and B.
Answer:
From the excess pressure inside a soap bubble
$
\Delta \mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}}
$
Here the two bubbles having the same pressure and temperature. So the radius of the combined bubbles,
$
\begin{aligned}
& \frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2} \\
& \frac{1}{\mathrm{R}}=\frac{1}{2}+\frac{1}{4}=\frac{2+1}{4}=\frac{3}{4} \\
& \mathrm{R}=\frac{4}{3}=1.33 \\
& \therefore \quad \mathrm{R}=1.33 \mathrm{~cm}
\end{aligned}
$
Question 4.
A block of $\mathrm{Ag}$ of mass $\mathrm{x} \mathrm{kg}$ hanging from a string is immersed in a liquid of relative density 0.72 . If the relative density of $\mathrm{Ag}$ is 10 and tension in the string is $37.12 \mathrm{~N}$ then compute the mass of $\mathrm{Ag}$ block.
Answer:

From the terminal velocity condition, $\mathrm{F}_{\mathrm{G}}-\mathrm{U}=\mathrm{FF}=\mathrm{T}, \mathrm{m}=\mathrm{x}$
$
\begin{aligned}
& m g-m g\left(\frac{\rho_w}{\rho}\right)=\mathrm{T} \\
& m g\left[1-\frac{0.72}{10}\right]=37.12 \\
& 9.8 x[1-0.072)=37.12 \\
& 9.0944 x=37.12 \\
& x=4 \mathrm{~kg}
\end{aligned}
$
Question 5.
The reading of pressure meter attached with a closed pipe is $5 \times 10^5 \mathrm{Nm}^{-2}$. On opening the valve of the pipe, the reading of the pressure meter is $4.5 \times 10^5 \mathrm{Nm}^{-2}$. Calculate the speed of the water flowing in the pipe.
Answer:
Using Bernoulli's equation
$
\begin{aligned}
\mathrm{P}_1+\frac{1}{2} \rho v_1^2 & =\mathrm{P}_2+\frac{1}{2} \rho v_2^2 \\
\frac{1}{2} \rho\left(v_2^2-v_1^2\right) & =\mathrm{P}_1-\mathrm{P}_2
\end{aligned}
$
Here initial velocity $V_1=0$ and density of water $p=1000 \mathrm{~kg} \mathrm{~m}^{-3}$
$
\begin{aligned}
& \frac{1}{2} \times 10^3 \times v_2^2=(5-4.5) \times 10^5 \\
& v_2^2=\frac{0.5 \times 10^5 \times 2}{10^3}=1 \times 10^2 ; v_2^2=100 \\
& v_2=10 \mathrm{~ms}^{-1}
\end{aligned}
$