The angle of $1°$ (degree) will be equal to:

(Use $360°=2\mathrm{\pi } \mathrm{rad}, 1°=60\text{'} \mathrm{and} 1\text{'}=60\text{'}\text{'}$)

$$\begin{gathered} \left(1\right) 1.034\times {10}^{-3} \mathrm{rad} \\ \left(2\right) 1.745\times {10}^{-2} \mathrm{rad} \\ \left(3\right)1.524\times {10}^{-2} \mathrm{rad} \\ \left(4\right) 1.745\times {10}^{+3} \mathrm{rad} \end{gathered}$$

A man wishes to estimate the distance of a nearby tower from him. He stands at point A in front of tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m, and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO; but he finds the line of sight of C shifted from the original line of sight by an angle θ = $40°$ (θ is known as ‘parallax’), the distance of the tower C from his original position A is:

(1) 119 m

(2) 126 m

(3) 320 m

(4) 219 m

The moon is observed from two diametrically opposite points A and B on Earth. The angle θ subtended at the moon by the two directions of observation is $1°$ 54′. Given the diameter of the Earth to be about 1.276 × ${10}^7$ m, the distance of the moon from the Earth is:

$$\begin{gathered} \left(1\right) 2.91\times {10}^8 \mathrm{m} \\ \left(2\right) 3.11\times {10}^9 \mathrm{m} \\ \left(3\right) 3.84\times {10}^8 \mathrm{m} \\ \left(4\right) 1.91\times {10}^7 \mathrm{m} \end{gathered}$$

The Sun’s angular diameter is measured to be 1920′′. The distance D of the Sun from the Earth is $1.496\times {10}^{11} \mathrm{m}.$ The diameter of the Sun:

1.  $2.39\times {10}^9 \mathrm{m}$

2.  $0.39\times {10}^9 \mathrm{m}$

3.  $1.39\times {10}^9 \mathrm{m}$

4.  $1.39\times {10}^{10} \mathrm{m}$

If the size of a nucleus (in the range of ${10}^{-15} \mathrm{to} {10}^{-14} \mathrm{m}$) is scaled up to the tip of a sharp pin, what roughly is the size of an atom? Assume tip of the pin to be in the range ${10}^{-5} \mathrm{m} \mathrm{to} {10}^{-4} \mathrm{m}.$

1.  $1 \mathrm{m}$

2.  $10 \mathrm{m}$

3.  ${10}^{-10} \mathrm{m}$

4.  ${10}^{-5} \mathrm{m}$

We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. The average absolute error and percentage error, respectively, are:

1.  0.22 s and 4

2.  0.11 s and 4

3.  4 and 0.11 s

4.  5 and 0.22 s