The angle of \(1^\circ\) (degree) will be equal to:
(Use \(360^\circ=2\pi\) rad)
1. \(1.034\times10^{-3}\) rad
2. \(1.745\times10^{-2}\) rad
3. \(1.524\times10^{-2}\) rad
4. \(1.745\times10^{3}\) rad

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 \(\theta=40^\circ\) (\(\theta\) is known as ‘parallax’), the distance of the tower C from his original position A is: (Given \(\tan40^\circ=0.8391\))
1. 119 m
2. 126 m
3. 320 m
4. 219 m

The Sun’s angular diameter is measured to be 1920′′. The distance D of the Sun from the Earth is $1.496\times {10}^{11}<mspace\; width="1em"></mspace>\mathrm{m}.$ The diameter of the Sun:

1.  $2.39\times {10}^9<mspace\; width="1em"></mspace>\mathrm{m}$

2.  $0.39\times {10}^9<mspace\; width="1em"></mspace>\mathrm{m}$

3.  $1.39\times {10}^9<mspace\; width="1em"></mspace>\mathrm{m}$

4.  $1.39\times {10}^{10}<mspace\; width="1em"></mspace>\mathrm{m}$

If the size of a nucleus (in the range of \(10^{-15}\) to \(10^{-14}\) 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 \(10^{-5}\) m to \(10^{-4}\) m.
1. \(1\) m
2. \(10\) m
3. \(10^{-10}\) m
4. \(10^{-5}\) m

We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be \(2.63~\text s, 2.56~\text s, 2.42~\text s, 2.71~\text s,\) and \(2.80~\text s.\) The average absolute error and percentage error, respectively, are:
1. \(0.22~\text s\) and \(4\%\)
2. \(0.11~\text s\) and \(4\%\)
3. \(4~\text s\) and \(0.11\%\)
4. \(5~\text s\) and \(0.22\%\)