Answer:
a
b

Explanation:
From the question we are told that
The pressure of the manometer when there is no gas flow is 
The level of mercury is 
The drop in the mercury level at the visible arm is 
Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

Here
is the density of mercury with value 
and
is the difference in the level of gas in arm one and two
So


Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

=> 
Generally from manometry principle we have that
![P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0](https://tex.z-dn.net/?f=P_G%20%2B%20%5Crho%20%2A%20g%20%20%2A%20d%20%20%20-%20%20%5Crho%20%2A%20%20g%20%20%2A%20%5Bh%20-%20%28h_m%20%2B%20d%29%5D%20%3D%200)
Here
is the pressure of the gas
![P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0](https://tex.z-dn.net/?f=P_G%20%2B13.6%20%2A10%5E%7B3%7D%20%2A%209.8%20%20%2A%200.039%20%20%20%20-%20%2013.6%20%2A10%5E%7B3%7D%20%20%2A%20%209.8%20%20%2A%20%5B0.950%20-%20%280.148%20%2B%200.039%29%5D%20%3D%200)

converting to psig
Answer:
b
Explanation:
I think so I'm not 100% sure sorry
it's how much it weighs and how much force is pushing on it like a egg if i drop it the weigh can cause it to break and how much force the gravity is pushing on it.