Numerical Questions-2 - Chapter 7 Properties of Matter 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Questions
Question 1.
Two steel wires of lengths $1 \mathrm{~m}$ and $2 \mathrm{~m}$ have diameters $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$ respectively. If they are stretched by forces of $40 \mathrm{~N}$ and $80 \mathrm{~N}$ respectively, find the ratio of their elongations.
Solution:
We know that
$
\mathrm{Y}=\frac{\mathrm{Fl}}{\pi r^2 \Delta l} \Rightarrow \Delta l=\frac{\mathrm{F} l}{\pi r^2 \mathrm{Y}}
$
Therefore, for the two wires,
$
(\Delta l)_1=\frac{\mathrm{F}_1 l_1}{\pi r_1^2 \mathrm{Y}} ;(\Delta)_2=\frac{\mathrm{F}_2 l_2}{\pi r_2^2 \mathrm{Y}}
$
The ratio of their elongations
$
\begin{aligned}
\frac{(\Delta l)_1}{(\Delta l)_2} & =\frac{\mathrm{F}_1 l_1 r_2^2}{\mathrm{~F}_2 l_2 r_1^2}=\frac{40}{80} \times \frac{1}{2} \times\left(\frac{2}{1}\right)^2 \\
\frac{(\Delta l)_1}{(\Delta l)_2} & =\frac{1}{1} \\
(\Delta l)_1:(\Delta l)_2 & =1: 1
\end{aligned}
$

Question 2.
A wire of length $2 \mathrm{~m}$ and cross-sectional area $2 \times 10^{-6} \mathrm{~m}^2$ is made of a material of Young's modulus $2 \times$ $10^{11} \mathrm{Nm}^{-2}$. What is the work done in stretching it through $0.1 \mathrm{~mm}$.
Answer:
$
\begin{aligned}
\text { Work done } & =\frac{1}{2} \mathrm{Y} \times(\text { Strain })^2 \times \text { Volume } \\
& =\frac{1}{2} \times 2 \times 10^{11} \times\left(\frac{0.1 \times 10^{-3}}{2}\right)^2 \times\left(2 \times 10^{-6}\right) \\
& =0.1 \times 10^{11} \times 10^{-12}=0.1 \times 10^{-1} \\
\text { Workdone } & =100 \times 10^{-4} \mathrm{~J}
\end{aligned}
$
Question 3.
A wire is stretched by $0.01 \mathrm{~m}$ when it is stretched by a certain force. Another wire of the same material but double the length and double the diameter is stretched by the same force. What is the elongation in metres?
Answer:
$
\begin{aligned}
\Delta l & =\frac{F l}{\pi r^2 \mathrm{Y}} \\
\frac{(\Delta l)_2}{(\Delta l)_1} & =\left(\frac{l_2}{l_1}\right)\left(\frac{r_1}{r_2}\right)^2=2 \times\left(\frac{1}{2}\right)^2=\frac{1}{2} \\
(\Delta l)_2 & =\frac{1}{2} \times 0.01 \\
(\Delta l)_2 & =0.005 \mathrm{~m}
\end{aligned}
$

Question 4 .
Two wires of same material and same diameter have lengths in the ratio $2: 5$. They are stretched by same force. Calculate the ratio of workdone in stretching them.
Answer:
Workdone $\mathrm{W}=\frac{1}{2} \mathrm{~F}(\Delta l)=\frac{1}{2} \mathrm{~F}\left(\frac{\mathrm{F} l}{\pi r^2 \mathrm{Y}}\right)$
Since F, Y and $r$ are the same for both the wires, we have $\frac{\mathrm{W}_1}{\mathrm{~W}_2}=\frac{l_1}{l_2}=\frac{2}{5}$
Question 5.
The approximate depth of an ocean is $2700 \mathrm{~m}$. The compressibility of water is $45.4 \times 10^{-11} \mathrm{~Pa}^{-1}$ and density of water is $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$. What fractional compression of water will be obtained at the bottom of the ocean?
Answer:
$
\begin{aligned}
\text { Bulk modulus } \mathrm{B} & =\frac{\Delta \mathrm{P}}{-\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)} \\
\text { Compressibility } \mathrm{K} & =\frac{1}{\mathrm{~B}}=\frac{-\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)}{\Delta \mathrm{P}} \\
\frac{\Delta \mathrm{V}}{\mathrm{V}} & =-\mathrm{K} \Delta \mathrm{P}=-\mathrm{K} h \rho g=-45.4 \times 10^{-11} \times 2700 \times 10^3 \times 10 \\
\frac{\Delta \mathrm{V}}{\mathrm{V}} & =-1.226 \times 10^{-2} .
\end{aligned}
$
Question 6.
An iceberg of density $900 \mathrm{kgm}^{-3}$ is floating in water of density $1000 \mathrm{kgm}^{-3}$. What is the percentage of volume of iceberg outside the water?
Answer:
Fraction of volume inside water $=$ Relative density of the body

$
\frac{v^{\prime}}{v}=-\frac{900}{1000}=0.9
$
Fraction of volume outside water $=1-0.9=0.1$
Percentage of volume outside water $=0.1 \times 100=10 \%$
Question 7.
A cubical copper block has each side $2.0 \mathrm{~cm}$. It is suspended by a string and submerged in oil of density $820 \mathrm{~kg} \mathrm{~m}^{-3}$. Calculate the tension in the string.
(density of copper $=8920 \mathrm{~kg} \mathrm{~m}^{-3}$ )
Answer:
Tension $=$ true weight - upthrust due to oil
$
\begin{aligned}
& =\mathrm{V} \rho g-\mathrm{V} \sigma g=\mathrm{V}(\rho-\sigma) g \\
& =\left(2 \times 10^{-2}\right)^3 \times(8920-820) \times 10=8 \times 10^{-6} \times 8100 \times 10
\end{aligned}
$
Question 8.
By sucking through a straw, a student can reduce the pressure in his lungs to $750 \mathrm{~mm}$ of $\mathrm{Hg}$ (density = $13.6 \mathrm{~g} \mathrm{~cm}^{-2}$ ). Using the straw, he can drink water from a glass upto a maximum depth of.
Answer:
Pressure difference between atmosphere and lungs
$
\Delta \mathrm{P}=760-750=10 \mathrm{~mm}=1 \mathrm{~cm} \text { of } \mathrm{Hg}
$
Maximum depth upto which the student can suck water from a glass $=1 \times 13.6=13.6 \mathrm{~cm}$
Question 9.
A sphere made of a material of specific gravity 8 has a concentric spherical cavity and just sinks in water. Calculate the ratio of the radius of the cavity to that of the outer radius of the sphere.
Answer:
Let the radius of the sphere be $R$ and that of the cavity be $r$. Then, since the sphere just sinks in water,
$
\begin{aligned}
\frac{4}{3} \pi\left(\mathrm{R}^3-r^3\right) \rho g & =\frac{4}{3} \pi \mathrm{R}^3 \rho_w g \\
1-\left(\frac{r}{R}\right)^3 & =\frac{\rho_w}{\rho}=\frac{1}{8} \\
\frac{r}{R} & =\frac{7^{\frac{1}{3}}}{2}
\end{aligned}
$
Question 10.
A solid floats in liquid A with half its volume immersed and in liquid B with $\frac{2}{3}$ of its volume immersed. The densities of the liquids $\mathrm{A}$ and $\mathrm{B}$ are in the ratio.

Answer:
Weight of liquid $\mathrm{A}$ displaced is $\mathrm{W}_{\mathrm{A}}=\frac{1}{2} \mathrm{~V} \rho_{\mathrm{A}} g$
Weight of liquid $B$ displaced is $W_B=\frac{2}{3} V \rho_B g$
According to Archimede's principle $\mathrm{W}_{\mathrm{A}}=\mathrm{W}_{\mathrm{B}}$
$
\begin{array}{r}
\frac{1}{2} \mathrm{~V} \rho_{\mathrm{A}} g=\frac{2}{3} \mathrm{~V} \rho_{\mathrm{B}} g \\
\frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}}=\frac{4}{3}
\end{array}
$
Question 11.
An ideal liquid is flowing in a cylindrical tube of internal diameter $4 \mathrm{~cm}$ with a velocity of $5 \mathrm{~ms}^{-1}$. If this tube is connected to another tube of internal diameter $2 \mathrm{~cm}$, then the velocity of the liquid in the second tube will be?
Answer:
According to the equation of continuity
$
\begin{aligned}
& a_2 v_2=a_1 v_1 ; d_2^2 v_2=d_1^2 v_1 \\
& v_2=\left(\frac{d_1}{d_2}\right)^2 \times v_1=\left(\frac{4}{2}\right)^2 \times 5 \\
& v_2=20 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 12.
The cylindrical tube of a spray pump has radius $R$, one end of which has ' $n$ ' fine holes, each of radius $r$. If the speed of the liquid in the tube is $\mathrm{V}$, what is the speed of the ejection of the liquid through the holes.
Answer:
Let the required speed be $v^{\prime}$ '.
According to equation of continuity a ' $v$ ' = av
$
v^{\prime}=\frac{a v}{a^{\prime}}=\frac{\pi \mathrm{R}^2 \mathrm{~V}}{n \pi r^2} \Rightarrow v^{\prime}=\frac{\mathrm{R}^2 \mathrm{~V}}{n r^2}
$
Question 13.
What is the pressure on a swimmer $10 \mathrm{~m}$ below the surface of a lake?

Answer:
Here, $h=10 \mathrm{~m}, g=9.8 \mathrm{~ms}^{-2}, \rho_{\mathrm{w}}=1000 \mathrm{~kg} \mathrm{~m}^{-3}$
Pressure on a swimmer $10 \mathrm{~m}$ below the surface of the lake,
$
\begin{aligned}
\mathrm{P} & =\mathrm{P}_a+h \rho g=\left(1 \times 10^5\right)+\left(10 \times 9.8 \times 10^3\right) \\
& =10 \times 10^4+9.8 \times 10^4=19.8 \times 10^4 \\
\mathrm{P} & =1.98 \times 10^5 \approx 2 \mathrm{~atm}
\end{aligned}
$
Question 14.
A square metal plate of $10 \mathrm{~cm}$ side moves parallel to another plate with a velocity of $10 \mathrm{~cm} \mathrm{~s}^{-1}$, both plates immersed in water. If the viscous force is 200 dyne and viscosity of water is 0.01 poise. What is their distance apart?
Answer:
Here, $\mathrm{A}=10 \times 10=100 \mathrm{~cm}^2, d v=10 \mathrm{~cm} \mathrm{~s}^{-1}, \eta=0.01$ poise, $\mathrm{F}=200$ dyne
As,
$
\begin{aligned}
\mathrm{F} & =\eta \mathrm{A} \frac{d v}{d x} \\
d x & =\frac{\eta \mathrm{A} d v}{\mathrm{~F}}=\frac{0.01 \times 100 \times 10}{200} ; \quad d x=0.05 \mathrm{~cm}
\end{aligned}
$
Question 15.
Find the terminal velocity of a steel ball $2 \mathrm{~mm}$ in diameter falling through glycerine. Relative density of steel is 8 , relative density of glycerine is 1.3 and viscosity of glycerine is 8.3 poise.
Answer:
$
\begin{aligned}
& v=\frac{2}{9} \frac{r^2}{\eta}\left(\rho-\rho^{\prime}\right) g=\frac{2}{9} \frac{(0.1)^2}{8.3}(8-1.3) \times 9.8 \\
& v=1.758 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}
$
Question 16.
The terminal velocity of a copper ball of radius $2 \mathrm{~mm}$ falling through a tank of oil at $20^{\circ} \mathrm{C}$ is $6.5 \mathrm{~cm} \mathrm{~s}^{-1}$. Compute the viscosity of the oil at $20^{\circ} \mathrm{C}$. Density of oil $=1.5 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, density of copper is $8.9 \times$ $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
Answer:
$
\begin{aligned}
& \eta=\frac{2}{9} \frac{r^2}{v}\left(\rho-\rho^{\prime}\right) g=\frac{2 \times\left(2 \times 10^{-3}\right)^2 \times\left(8.9 \times 10^3-1.5 \times 10^3\right) \times 9.8}{9 \times 6.5 \times 10^{-2}} \\
& \eta=0.992 \text { decapoise }
\end{aligned}
$

Question 17.
Water is flowing in a pipe of radius $1.5 \mathrm{~cm}$ with an average velocity of $15 \mathrm{~cm} \mathrm{~s}^{-1}$. What is the nature of flow? Given coefficient of viscosity of water is $10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$ and its density

Answer:
$
\begin{aligned}
& \mathrm{D}=2 \times 1.5 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m} \\
& \nu=15 \mathrm{~cm} \mathrm{~s}^{-1}=0.15 \mathrm{~ms}^{-1} \\
& \eta=10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1} \\
& \mathrm{R}_c=\frac{\rho v \mathrm{D}}{\eta}=\frac{10^3 \times 0.15 \times 3 \times 10^{-2}}{10^{-3}} \\
& \mathrm{R}_c=4500 \\
& \text { As } \mathrm{R}_c>3000, \text { so the flow is turbulent. }
\end{aligned}
$
Question 18.
Water is flowing with a speed of $2 \mathrm{~ms}^{-1}$ in a horizontal pipe with cross- sectional area decreasing from 2 $\times 10^{-2} \mathrm{~m}^{-2}$ to $0.01 \mathrm{~m}^2$ at pressure $4 \times 10^4 \mathrm{~Pa}$. What will be the pressure at small cross-section?
Answer:
$
\begin{aligned}
& \text { Here, } v_1=2 \mathrm{~ms}^{-1}, a_1=2 \times 10^{-2} \mathrm{~m}^2, a^2=0.01 \mathrm{~m}^2, \mathrm{P}_1=3 \times 10^4 \mathrm{~Pa} ; \mathrm{P}_2=? ; \rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3} \\
& \text { : } v_2=\frac{a_1 v_1}{a_2}=\frac{2 \times 10^{-2} \times 2}{0.01}=4 \mathrm{~ms}^{-1} \\
& \text { As, } \\
& \mathrm{P}_1+\frac{1}{2} \rho v_1^2=\mathrm{P}_2+\frac{1}{2} \rho v_2^2 \\
& \mathrm{P}_2=\mathrm{P}_1-\frac{1}{2} \rho\left(v_1^2-v_2^2\right) \\
& =4 \times 10^4-\frac{1}{2} \times 10^3\left(2^2-4^2\right) \\
&
\end{aligned}
$

$
\mathrm{P}_2=3.4 \times 10^4 \mathrm{~Pa}
$
Question 19.
Calculate the height to which water will rise in capillary tube of $1.5 \mathrm{~mm}$ diameter. Surface tension of water is $7.4 \times 10^{-3} \mathrm{Nm}^{-1}$.
Solution:
Here,
$
\begin{aligned}
& r=\frac{1.5}{2}=0.75 \mathrm{~mm}=0.75 \times 10^{-3} \mathrm{~m} \\
& \sigma=7.4 \times 10^{-3} \mathrm{Nm}^{-1} \\
& \rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3} \\
& \theta=0^{\circ} \\
& h=\frac{2 \sigma \cos \theta}{r \rho g}=\frac{2 \times 7.4 \times 10^{-3} \times \cos 0^{\circ}}{0.75 \times 10^{-3} \times 10^3 \times 9.8} \\
& h=0.002014 \mathrm{~m}
\end{aligned}
$
$
\text { Angle of contact } \quad \theta=0^{\circ}
$