1.A 20N block is suspended by a spring balance.A beaker containing some water is placed on a weighing machine which reads 40N.The the spring balance is lowered inside the water such that the block is fully immersed.The spring balance now reads 16N,What will the weighing machine read now? Ans:44N
2.A piece of wood is floating in water kept in a bottle.the bottle is connected to an air pump.When more air is pushed into the bottle from the pump ,the piece of wood would float with
A) larger part in water
B)lesser part in water
C)same part in water as before
Ans :c
1not really sure, but here goes...
given that the reading of the spring balance is 16N. So it implies that the force applied by the water on the block is 4N i.e. the apparent weight of the block in the water is 4N. On the weighing machine, the total weight is recorded, so 40 + 4=44N.
Waiting for the actual answer, now.
1yeah 44,what abt the second one
what i thought is that more air means more pressure,which will tend to push the block down slightly but the answer is that it will remain as it is
1i dunno, water gets compressed, density increases nd more thrust on the block so same?
33The amount of block inside the water is independent of atm pressure.
Force on block is only mg and v'Ïg
Vdg=v'Ïg
v'=Vd/Ï
nowhere is atm pressure
It pushes both block and water.
106Q1. From FBD of block inside water,
T = 16N so, U = mg-T = 20-16 = 4N
Now, As water has exerted U force on block so it will get back U force from the block .. as the FBD of water in the 3rd diag...
So, mg + U = R
==> R = 40+4 = 44 N
Q2. 
h = height of wood immersed in water. Initially... forces on block were. P downwards, P+dgh upwards (d=density) and mg downward.. the resultant of P and P+dgh = dgh upwards.. = mg
Finally if the pressure due to air became P' ... the resultant of P' and P'+dgx = dgx ... now it needs to balance mg ... = dgh
So, x=h hence the height of immersed water wont change...