Shouldn't the observer on the Earth find the clock on the rocket to show less time due to time dilation and similarly in the opposite case the observer on rocket will find the clock on earth to show less time?
There is a similar question in Arthur Beiser on which there is a discussion from some time ago on Physics Forums.
Please tell if what they concluded in the end is correct or not.
I think the book has some issues.
If $\Delta t$ is 30 minutes in the moving frame where $v=2\times 10^8$, then $\Delta t\prime$ in the stationary frame, where $v=0$, is 40.25 minutes so that watch should say 10:40:15... plus...
Travelling 30 minutes at $2/3c$, the rocket will travel some distance and thus light (i.e. the displayed time on the watch) will take some time to travel from one observer to the other.
Reciprocal does not mean it is the same in all reference frames. The proper time in each frame needs to be calculated in relation to the proper time in the other frame. When 30 minutes have elapsed for the stationary observer, 30 minutes will not yet have elapsed for the moving observer.
The book is clearly and blatantly wrong! There are more than one issues. I would lay down the major conceptual issues and let you work out the details as is ideal in answering the homework questions.
There are two different questions that the book is miserably confusing with each other. One question is
"What time would Ram observe the time to be on Shyam's clock when (according to Ram) his own clock ticks, say, 10:30?"
and the other is
"What time would Ram see (i.e. while looking through a telescope) the time to be on Shyam's clock when (according to Ram) his own clock ticks, say, 10:30?"
The book claims to answer the latter but seems to employ the method to answer the first one (and fails to even do that correctly).
As you can expect, the first question simply means answering what would the actual time be on Shyam's clock (according to Ram) when Ram's clock says that it is 10:30 (according to Ram). Since, according to Ram, Shyam's clock is ticking slowly due to time-dilation, Ram would observe the display on the clock to be $10: X$ where $X$ is simply $\dfrac{30}{\gamma}$ ($<30$ as you expected! The book, sadly, messed up here in calculating $X$ to be $30\gamma$).
But, actually, the book has asked the second question (even though the book itself is simply attempting at answering the first one as I already said). And this "actually looking" business means that we should account for the time it takes for the light signal to travel from the clock of Shyam to Ram. This means, that at 10:30 in his own clock, Ram would observe the signal that left Shyam's clock when Ram's clock was displaying $10: T$ such that $\dfrac{vT}{c}=30$. And these signals would carry the information about what the dial of Shyam's clock was actually displaying (according to Ram) at $10: T$ in Ram's clock. This would be $10: Y$ such that $Y=\dfrac{T}{\gamma}$ by the same logic used in solving the first question above.
Clearly, $Y<X<30$ (and the book is wrong in its answer whether it claims to have answered the first question (in which its answer should match $X$) or the second question (in which case its answer should match $Y$)).
Edit
The symmetry argument used in the book is correct. That is to say that if Ram observes Ram observes that $X<30$ minutes have elapsed in Shyam's clock when his own clock shows that $30$ minutes have elapsed then Shyam will also observe that $X<30$ minutes have elapsed in Ram's clock when his own clock shows that $30$ minutes have elapsed. There is no contradiction here because simultaneity is frame-dependent in relativity. The similar argument goes for $Y$ as well.
Less time or more time depends on how you measure it.
Moving clock measures shorter time interval, that two (at least) synchronized by Einstein's synchronization method clocks of “stationary” reference frame.
Hence, two clocks of stationary reference frame measure greater time interval, than moving one.
That discussion and the book only lead you astray. The book by Arthur Beiser “Concepts of Modern Physics” contains completely wrong statement (Chapter Relativity, p. 11, Formula 1.5)
http://phy240.ahepl.org/Concepts_of_Modern_Physics_by_Beiser.pdf
Beiser speaks about moving observer, but gives Transverse Doppler Shift formula for stationary one. He says:
Observer moving perpendicular to a line between him and the light source measures transverse frequency shift
$$\nu=\nu_0 \sqrt {1-v^2/c^2}$$
It is wrong. Since “moving” observer’s clock run slower, he sees, that clock “at rest” is ticking $\gamma$ times faster and light received at points of closest approach is $\gamma$ times blueshifed.
The correct formula is:
$$\nu= \frac {\nu_0} {\sqrt {1-v^2/c^2}}$$
Please look for transverse Doppler Effect here:
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
Or Einstein’s 1905 paper, where he speaks about moving observer and gives correct formula for a frequency as measured by moving observer (chapter 7)
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf
Observer "at rest" measures that "moving" clock tick slower. "Moving" observer measures, that clock "at rest" ticks faster than his own.
The only problem is to choose one frame an to stay within it, or to decide who is "moving" and who is at "rest".
If each of them is stubborn like sheep and shouts that it is he who is "at rest", then each of them can really think that the other is slower and come up with all kinds of nonsense.