Chapter 8, Problem 1FC

To determine

The correct option.

Answer to Problem 1FC

The correct option is (d) Dime.

Explanation of Solution

Given data:

$\text{Arc}{\text{length}}_{\left(\text{Moon}\right)}$ is $3.48\times {10}^6\text{m}$.

${\text{Radius}}_{\left(\text{Moon}\right)}$ is $3.85\times {10}^8\text{m}$

$\text{Arc}{\text{length}}_{\left(\text{Pea}\right)}$ is $0.50\text{cm}$.

${\text{Radius}}_{\left(\text{Pea}\right)}$ is $71\text{m}$

$\text{Arc}{\text{length}}_{\left(\text{Dime}\right)}$ is $1.8\text{cm}$.

${\text{Radius}}_{\left(\text{Dime}\right)}$ is ${\text{Radius}}_{\left(\text{Pea}\right)}$ that is $71\text{m}$

Formula used:

Consider the expression for angle $\left(\theta \right)$.

$\theta =\frac{\text{Arc}\text{length}}{\text{Radius}}$ (1)

Explanation:

Modify equation (1) for Moon.

${\theta }_{\text{Moon}}=\frac{\text{Arc}{\text{length}}_{\left(\text{Moon}\right)}}{{\text{Radius}}_{\left(\text{Moon}\right)}}$

Substitute $3.48\times {10}^6\text{m}$ for $\text{Arc}{\text{length}}_{\left(\text{Moon}\right)}$ and $3.85\times {10}^8\text{m}$ for ${\text{Radius}}_{\left(\text{Moon}\right)}$,

$\begin{array}{c}{\theta }_{\text{Moon}}=\frac{3.48\times {10}^6\text{m}}{3.85\times {10}^8\text{m}}\\ =9.0\times {10}^{-3}\text{rad}\end{array}$

Modify equation (1) for Pea.

${\theta }_{\text{Pea}}=\frac{\text{Arc}{\text{length}}_{\left(\text{Pea}\right)}}{{\text{Radius}}_{\left(\text{Pea}\right)}}$

Substitute $0.50\text{cm}$ for $\text{Arc}{\text{length}}_{\left(\text{Pea}\right)}$ and $71\text{cm}$ for ${\text{Radius}}_{\left(\text{Pea}\right)}$,

$\begin{array}{c}{\theta }_{\text{Pea}}=\frac{0.50\text{cm}}{71\text{cm}}\\ =7.0\times {10}^{-3}\text{rad}\end{array}$

Modify equation (1) for Dime.

${\theta }_{\text{Dime}}=\frac{\text{Arc}{\text{length}}_{\left(\text{Dime}\right)}}{{\text{Radius}}_{\left(\text{Dime}\right)}}$

Substitute $1.8\text{cm}$ for $\text{Arc}{\text{length}}_{\left(\text{Dime}\right)}$ and $71\text{cm}$ for ${\text{Radius}}_{\left(\text{Dime}\right)}$,

$\begin{array}{c}{\theta }_{\text{Dime}}=\frac{1.8\text{cm}}{71\text{cm}}\\ =25.35\times {10}^{-3}\text{rad}\end{array}$

By comparing the angles obtained from Moon, Pea and Dime, Since ${\theta }_{\text{Dime}}>{\theta }_{\text{Moon}}>{\theta }_{\text{Pea}}$, therefore Dime covers the overview of Moon.

Hence the option (d) is correct and remaining options (a), (b) and (c) are incorrect.

Conclusion:

Thus, the correct option is (d) Dime.