Numerical Questions-2 - Chapter 1 Nature of Physical World and Measurement 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Questions
Question 1
Determine the number of light years in one metre.
Answer:
$
\begin{aligned}
& 1 \mathrm{ly}=9.46 \times 10^{15} \mathrm{~m} \\
& 1 \mathrm{~m}=\frac{1}{9.46 \times 10^{15}}=1.057 \times 10^{-16} \mathrm{ly}
\end{aligned}
$

Question 2
The mass of a box measured by a grocer's balance is $2.3 \mathrm{~kg}$. Two gold pieces $20.15 \mathrm{~g}$ and $20.17 \mathrm{~g}$ are added to the box.
(i) What is the total mass of the box?
(ii) The difference in masses of the pieces to correct significant figures.
Answer:
(i) Mass of box $=2.3 \mathrm{~kg}$
Mass of gold pieces $=20.15+20.17=40.32 \mathrm{~g}=0.04032 \mathrm{~kg}$.
Total mass $=2.3+0.04032=2.34032 \mathrm{~kg}$
In correct significant figure mass $=2.3 \mathrm{~kg}$ (as least decimal)
(ii) Difference in mass of gold pieces $=0.02 \mathrm{~g}$
In correct significant figure ( 2 significant fig. minimum decimal) will be $0.02 \mathrm{~g}$.
Question 3 .
$5.74 \mathrm{~g}$ of a substance occupies $1.2 \mathrm{~cm}^3$. Express its density to correct significant figures. Answer:
Density $=\frac{\text { Mass }}{\text { Volume }}=\frac{5.74}{1.2}=4.783 \mathrm{~g} / \mathrm{cm}^3$
Here least significant figure is 2, so density $=4.8 \mathrm{~g} / \mathrm{cm}^3$.

Question 4
If displacement of a body $\mathrm{s}=(200 \pm 5) \mathrm{m}$ and time taken by it $\mathrm{t}=(20+0.2) \mathrm{s}$, then find the percentage error in the calculation of velocity.

Answer:
Percentage error in measurement of displacement $=\frac{5}{200} \times 100$
Percentage error in measurement of time $=\frac{0.2}{20} \times 100$
$\therefore$ Maximum permissible error $=2.5+1=3.5 \%$
Question 5
If the error in measurement of mass of a body be $3 \%$ and in the measurement of velocity be $2 \%$. What will be maximum possible error in calculation of kinetic energy.
Answer:
$
\begin{aligned}
& \text { K.E. }=\frac{1}{2} m v^2 \\
& \therefore \frac{\Delta k}{k}=\frac{\Delta m}{m}+\frac{2 \Delta v}{v} \Rightarrow \frac{\Delta k}{k} \times 100=\frac{\Delta m}{m} \times 100+2\left(\frac{\Delta v}{v}\right) \times 100 \\
& \therefore \text { Percentage error in K.E. }=3 \%+2 \times 2 \%=7 \%
\end{aligned}
$
Question 6
The length of a rod as measured in an experiment was found to be $2.48 \mathrm{~m}, 2.46 \mathrm{~m}, 2.49 \mathrm{~m}$, $2.50 \mathrm{~m}$ and $2.48 \mathrm{~m}$. Find the average length, absolute error and percentage error. Express the result with error limit.

Answer:
$
\begin{aligned}
& \text { Average length }=\frac{2.48+2.46+2.49+2.50+2.48}{5}=\frac{12.41}{5}=2.48 \mathrm{~m} \\
& \text { Mean absolute error }=\frac{0.00+0.02+0.01+0.02+0.00}{5}=\frac{0.05}{5}=0.01 \mathrm{~m} \\
& \text { Percentage error }=\frac{0.01}{2.48} \times 100 \%=0.04 \times 100 \%=0.40 \% \\
& \text { Correct length }=(2.48 \pm 0.01) \mathrm{m} \\
& \text { Correct length }=(2.48 \mathrm{~m} \pm 0.40 \%)
\end{aligned}
$
Question 28.
A physical quantity is measured as $\mathrm{Q}=(2.1 \pm 0.5)$ units. Calculate the percentage error in (1) $\mathrm{Q}^2(2) 2 Q$.
Answer:
$
\begin{gathered}
\mathrm{P}=\mathrm{Q}^2 \\
\frac{\Delta p}{p}=\frac{2 \Delta \mathrm{Q}}{\mathrm{Q}}=2\left(\frac{0.5}{2.1}\right)=\frac{1.0}{2.1}=0.476 \\
\frac{\Delta p}{p} \times 100 \%=47.6 \%=48 \% \\
\mathrm{R}=2 \mathrm{Q} \\
\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{Q}}{\mathrm{Q}} \Rightarrow \frac{\Delta \mathrm{R}}{\mathrm{R}} \frac{0.5}{2.1}=0.238 \\
\frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100 \%=24 \%
\end{gathered}
$

Question 29.
When the planet Jupiter is at a distance of 824.7 million $\mathrm{km}$ from the earth, its angular diameter is measured to be $35.72^{\prime \prime}$ of arc. Calculate diameter of Jupiter.
Answer:
$
\begin{aligned}
& \theta=35.72^{\prime \prime} \\
& l^{\prime \prime}=4.85 \times 10^{-6} \mathrm{radian} \Rightarrow 35.72^{\prime \prime}=35.72 \times 4.85 \times 10^{-6} \mathrm{rad} \\
& d=\mathrm{D} \theta=824.7 \times 35.72 \times 4.85 \times 10^{-6}=1.4287 \times 10^5 \mathrm{~km}
\end{aligned}
$
Question 30.
A laser light beamed at the moon takes $2.56 \mathrm{~s}$ and to return after reflection at the moon's surface. What will be the radius of lunar orbit?
Answer:
$
\mathrm{t}=2.54 \mathrm{~s}
$
$\therefore t=$ time taken by laser beam to go to the moon $=\frac{t}{2}$
Distance between earth and moon $=d=c \times \frac{t}{2}=3 \times 10^8 \times \frac{2.56}{2}=3.84 \times 10^8 \mathrm{~m}$
Question 31.
Convert

$\text { (i) } 3 \mathrm{~m}^{-2} \mathrm{~s}^{-2} \mathrm{~km} \mathrm{~h}^{-2}$
(ii) $\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2}$ to $\mathrm{cm}^3 \mathrm{~g}^{-1} \mathrm{~s}^{-2}$
Answer:
(i)
$
3 \mathrm{~ms}^{-2}=\left(\frac{3}{1000} \mathrm{~km}\right)\left(\frac{1}{60 \times 60} \mathrm{hr}\right)^{-2}
$

$
\begin{aligned}
& =\frac{3 \times(60 \times 60)^2}{1000}=3.8880 \times 10^4 \mathrm{~km} \mathrm{~h}^{-2}=3.9 \times 10^4 \mathrm{~km} \mathrm{~h}^{-2} \\
& \text { (ii) } \mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2} \\
& =6.67 \times 10^{-11}\left(\mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-2}\right)\left(\mathrm{m}^2 \mathrm{~kg}^{-2}\right) \\
& =6.67 \times 10^{-11} \mathrm{~kg}^{-1} \mathrm{~m}^3 \mathrm{~s}^{-2} \\
& =6.67 \times 10^{-11}(1000 \mathrm{~g})^{-1}(100 \mathrm{~cm})^3\left(\mathrm{~s}^{-2}\right) \\
& =6.67 \times 10^{-11} \times \frac{1}{1000} \times 100 \times 100 \times 100 \mathrm{~g}^{-1} \mathrm{~cm}^3 \mathrm{~s}^{-2} \\
& =6.67 \times 10^{-8} \mathrm{~g}^{-1} \mathrm{~cm}^3 \mathrm{~s}^{-2} \text {. } \\
&
\end{aligned}
$
Question 32 .
A calorie is a unit of heat or energy and it equals $4.2 \mathrm{~J}$ where $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}$. Suppose we employ a system of units in which unit of mass is $\alpha \mathrm{kg}$, unit of length is $\beta \mathrm{m}$, unit of time $\gamma \mathrm{s}$. What will be magnitude of calorie in terms of this new system.
Answer:
$
\begin{aligned}
& n_2=n_1\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]^a\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^b\left[\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right]^c=4.2\left(\frac{\mathrm{kg}}{\alpha \mathrm{kg}}\right)^1\left(\frac{m}{\beta m}\right)^2\left(\frac{s}{\gamma s}\right)^{-2} \\
& n_2=4.2 \alpha^{-1} \beta^{-2} \gamma^2
\end{aligned}
$

Question 33.
The escape velocity $\mathrm{v}$ of a body depends on-
(i) the acceleration due to gravity ' $\mathrm{g}$ ' of the planet,
(ii) the radius $\mathrm{R}$ of the planet. Establish dimensionally the relation for the escape velocity.
Solution:
$v \propto g^a \mathrm{R}^b \Rightarrow v=k g^a \mathrm{R}^b, \mathrm{~K} \rightarrow$ dimensionless proportionality constant
$\begin{aligned} {[v] } & =[g]^a[\mathrm{R}]^b \\ \Rightarrow \quad\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-1}\right] & =\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-2}\right]^a\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^0\right]^b\end{aligned}$
equating powers
$
\begin{aligned}
1 & =a+b \\
-1 & =-2 a \Rightarrow a=\frac{1}{2} \\
b & =1-a=1-\frac{1}{2}=\frac{1}{2} \\
\therefore \quad v & =k \sqrt{g R}
\end{aligned}
$
Question 34.
The frequency of vibration of a string depends of on,
(i) tension in the string
(ii) mass per unit length of string,
(iii) vibrating length of the string. Establish dimensionally the relation for frequency.
Answer:

$
\begin{aligned}
& n \propto \mathrm{I}^a \mathrm{~T}^b m^c, \quad[\mathrm{I}]=\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^0\right] \\
& {[\mathrm{T}] }=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2}\right](\text { force) } \\
& {[\mathrm{M}] }=\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^0\right] \\
& {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right] }=\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^0\right]^a\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2}\right]^b\left[\mathrm{M}^0 \mathrm{~L}^{-1} \mathrm{~T}^0\right]^c \\
& b+c=0 \\
& a+b-c=0 \\
&-2 b=-1 \Rightarrow \mathrm{b}=\frac{1}{2} \\
& c=-\frac{1}{2} a=1 \\
& n \propto \frac{1}{l} \sqrt{\frac{\mathrm{T}}{m}}
\end{aligned}
$
Question 35.
One mole of an ideal gas at STP occupies $22.4 \mathrm{~L}$. What is the ratio of molar volume to atomic volume of a mole of hydrogen? Why is the ratio so large? Take radius of hydrogen molecule to be $1^{\circ} \mathrm{A}$.
Answer:
$
\mathrm{A}^0=10^{-10} \mathrm{~m}
$
Atomic volume of 1 mole of hydrogen
$=$ Avagadro's number $\times$ volume of hydrogen molecule
$
\begin{aligned}
& =6.023 \times 10^{23} \times \frac{4}{3} \times \pi \times\left(10^{-10} \mathrm{~m}\right)^3 \\
& =25.2 \times 10^{-7} \mathrm{~m}^3
\end{aligned}
$
Molar volume $=22.4 \mathrm{~L}=22.4 \times 10^{-3} \mathrm{~m}^3$
$\frac{\text { Molar volume }}{\text { Atomic volume }}=0.89 \times 10^4 \approx 10^4$
This ratio is large because actual size of gas molecule is negligible in comparison to the inter molecular separation.