1) Yes
2) 
Explanation:
1)
To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.
The pressure exerted by a column of fluid of height h is:

where
is the atmospheric pressure
is the fluid density
is the acceleration due to gravity
h is the height of the column of fluid
Here we have:
is the sea water density
h = 5440 m is the depth at which the bathyscaphe is located
Therefore, the pressure on it is

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.
2)
Here we want to find the force exerted on the bathyscaphe.
The relationship between force and pressure on a surface is:

where
p is hte pressure
F is the force
A is the area of the surface
Here we have:
is the pressure exerted
The bathyscaphe has a spherical surface of radius
r = 3 m
So its surface is:

Therefore, we can find the force exerted on it by re-arranging the previous equation:

First, we write the SI prefixed. The SI unit for distance is meters.
Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²
Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga.
1.5 x 10¹¹ / 10⁹
= 1.5 x 10² Gm or 150 Gm
Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.
1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
Answer:
The total work done is 5997.6 J
Solution:
As per the question:
Mass of the bag, m = 60 kg
Vertical distance, h = 9 m
Mass lost, m' = 12 kg
To calculate the amount of work done:
Lost mass is proportional to the square root of the distance covered while lifting:
m' ∝ 
m' = 
where
K = proportionality constant
12 = 3K
K = 4
Mass of the floor containing bag at a height h:

Work done is given by:


![W = g([60h]_{0}^{9} + 4\times \frac{2}{3}[h^{\frac{3}{2}}]_{0}^{9})](https://tex.z-dn.net/?f=W%20%3D%20g%28%5B60h%5D_%7B0%7D%5E%7B9%7D%20%2B%204%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D%5Bh%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D_%7B0%7D%5E%7B9%7D%29)
![W = 9.8\times ([60\times 9 - 0] + \frac{8}{3}[9^{\frac{3}{2}} - 0^{\frac{3}{2}}])](https://tex.z-dn.net/?f=W%20%3D%209.8%5Ctimes%20%28%5B60%5Ctimes%209%20-%200%5D%20%2B%20%5Cfrac%7B8%7D%7B3%7D%5B9%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20-%200%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D%29)
