Numerical Problems-2 - Chapter 1 - Nature of Physical World and Measurement - 11th Science Guide Samacheer Kalvi Solutions

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Numerical Problems
Question 1.
In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 sec. If the speed of sound in water is $1460 \mathrm{~ms}^{-1}$. What is the distance of enemy submarine?
Answer:
Given:
Speed of sound in water $=1460 \mathrm{~ms}^{-1}$
Time delay $=80$ s
Distance of enemy ship $=$ ?
Solution:
Total distance covered $=$ speed $\times$ time
$
=1460 \mathrm{~ms}^{-1} \times 80 \mathrm{~s}=116800 \mathrm{~m}
$
Time taken is for forward and backward path of sound waves.
$
\begin{aligned}
& \text { Distance of enemy ship }=\frac{\text { total distance covered }}{2}=\frac{116800}{2} m \\
& =58400 \mathrm{~m} \text { (or) } 58.4 \mathrm{~km}
\end{aligned}
$
Question 2.
The radius of the circle is $3.12 \mathrm{~m}$. Calculate the area of the circle with regard to significant figures.
Answer:
Given: radius : $3.12 \mathrm{~m}$ (Three significant figures)
Solution:
Area of the circle $=\pi r^2=3.14 \times(3.12 \mathrm{~m})^2=30.566$
If the result is rounded off into three significant figure, area of the circle $=30.6 \mathrm{~m}^2$

Question 3.
Assuming that the frequency $\mathrm{v}$ of a vibrating string may depend upon
(i) applied force (F)
(ii) length (1)
that $v \propto \frac{I}{l} \sqrt{\frac{F}{m}}$ using dimensional analysis. [Related to
(iii) mass per unit length (m), prove that JIPMER 2001]
Answer:
Given: The frequency $v$ of a vibrating string depends
(i) applied force (F)
(ii) length (1)
(iii) mass per unit length (m)
Solution:
$v \propto \mathrm{F}^x p^y m^z \propto v=\mathrm{K} \mathrm{F}^x l^y m^z$
Substitute the dimensional formulae of the above quantities
$
\begin{aligned}
{\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right] } & =\left[\mathrm{MLT}^{-2}\right]^x[\mathrm{~L}]\left[\mathrm{ML}^{-1}\right]^z \\
{\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right] } & =\left[\mathrm{M}^{x+z} \mathrm{~L}^{x+y-z} \mathrm{~T}^{-2 x}\right]
\end{aligned}
$
Comparing the powers of $M, L, T$ on both sides, $x+z=0, x+y-z=0,-2 x=-1$
Solving for $\mathrm{x}, \mathrm{y}, \mathrm{z}$, we get
$
x=\frac{1}{2} \quad y=-1 \quad z=-\frac{1}{2}
$
Substitute $\mathrm{x}, \mathrm{y}, \mathrm{z}$ values in equ(1)

$v \propto \mathrm{F}^{1 / 2} l^{-1} m^{-1 / 2} \quad \therefore \quad v \propto \frac{1}{l} \sqrt{\frac{\mathrm{F}}{m}}$
Question 4.
Jupiter is at a distance of 824.7 million $\mathrm{km}$ from the Earth. Its angular diameter is measured to be $35.72^{\prime \prime}$. Calculate the diameter of Jupiter.

Given, 
Given Distance of Jupiter = 824.7 × 10

6 km = 8.247 × 1011 m 

angular diameter = 35.72 × 4.85 × 10

-6rad = 173.242 × 10-6 rad 

= 1.73 × 10

-4 rad 
∴ Diameter of Jupiter D = D × d = 1.73 × 10

-4 rad × 8.247 × 1011 m 

= 14.267 × 1o

7 m = 1.427 × 108 m (or) 1.427 × 105</sup km

Question 5. 
The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy.
The time for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the
percentage of accuracy in the determination of acceleration due to gravity ‘g’ from the above
measurement. 
Answer: 
Given, 
Length of simple pendulum (l) = 20 cm 
absolute error in length (∆l) = 2 mm = 0.2 cm 
Time taken for 50 oscillation (t) = 40 s 
error in time ∆T = 1 s 
Solution: Time period for one oscillation (T)