If the pressure in a gas container that is connected to an open-end u-tube manometer is 116 kpa and the pressure of the atmosphere at the open end of the tube is 752 mm hg, the level of mercury in the tube will a. be 870 mm higher in the arm open to the atmosphere. b. be 870 mm higher in the arm connected to the gas cylinder. c. be 118 mm higher in the arm connected to the gas cylinder. d. be 118 mm higher in the arm open to the atmosphere.

P = absolute pressure in gas container connected to open-end U-tube = 116 kPa = 116000 Pa

we know that, 1 atm = 101325 Pa   and  1 atm = 760 mm of Hg , hence

P = 116000 Pa  (1 atm/101325 Pa) (760 mm of Hg/1 atm)

P = (116000 x 760/101325)  mm of Hg

P = 870.07 mm of Hg

P₀ = atmospheric pressure = 752 mm Hg

Level of mercury is given as

P' = P - P₀

inserting the values

P' = 870.07 - 752

P' = 118 mm

so correct choice is

d.be 118 mm higher in the arm open to the atmosphere.

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards