Q 1 :    

Mercury is filled in a tube of radius 2 cm up to a height of 30 cm. The force exerted by mercury on the bottom of the tube is ________ N.

(Given, atmospheric pressure=105 Nm-2,density of mercury 1.36×104kg m-3,g=10 ms-2,π=227)                       [2024]



(177)       P=P0+ρgh

               F=P0A+ρghA=227×4×10-4(105+1.36×104×10×30100)

              F=887×10-4(105+0.408×105)

               =887×10×1.408=177

 



Q 2 :    

A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of 10 N is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is _________ N.          [2024]



(1000)

F1A1=F2A2

F1π(7)2=10π(0.7)2F1=1000 N

 



Q 3 :    

Two cylindrical vessels of equal cross-sectional area of 2 m2 contain water upto height 10 m and 6 m, respectively. If the vessels are connected at their bottom, then the work done by the force of gravity is: (Density of water is 103 kg/m3 and g=10 m/s2)          [2025]

  • 1×105 J

     

  • 4×104 J

     

  • 6×104 J

     

  • 8×104 J

     

(4)

Uinitial=(ρA×10)g×5+(ρA6)g×3

Uinitial=ρAg(50+18)=68ρAg

Ufinal=(ρA×16)g×4=(ρAg)×64

W=U=68ρAg64ρAg=4ρAg

W=4×1000×2×10=8×104 J



Q 4 :    

In a hydraulic lift, the surface area of the input piston is 6 cm2 and that of the output piston is 1500 cm2. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is ______ kJ.          [2025]



5

F1A1=F2A2, 1006=F1500, F=503×1500

F=50×500=25×103 N

W=F.S=25×103×20100=5×103=5 kJ



Q 5 :    

A vessel wih square cross-section and height of 6 m is vertically partitioned. A small window of 100 cm2 with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density 1.5×103 kg/m3. What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity = 10 m/s2)          [2025]



150

in equilibrium

Fext+Fw=Fl

 Fext=FlFw

                     =(P0ρlgh)A(P0ρwgh)A

                      =(ρlρw)ghA

                      =(15001000)×10×3×(100×104)

                      = 150 N.