A brick of weight 20N and dimensions 20 cm x10cm x 5cm is placed once on its largest surface area and then on its smallest surface area on the ground. Find out the ratio of the pressure exerted by the brick.
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Answers
Answer:
Surface Area Of Cuboid = 2LW+2LH+2HW
Step-by-step explanation:
=2(20cm x 10cm)+2(10cm x 5cm)+2(5cm x 20cm)
=400cm + 100cm+ 200 cm
=700cm²
Hope It Helps You
Answer:
Let P
1
,P
2
and P
3
be the pressures exerted by the brick while resting on different faces.
The dimensions of the given brick are 20cm×10cm×5cm
Case (i) : When the block is resting on 20cm×10cmface.
Thrust acting= Weight of the brick
T=500gwt
Area of constant (A)=20cm×10cm
Pressure exerted (P
1
)
=
Area
Thrust
=
20×10
500
∴P
1
=2.5gwtcm
−2
Case (ii) : When the block is resting on 20cm×5cmface
Thrust= Weight of the brick
=500gwt
Area of constant (A)=20cm×5cm
Pressure exerted
(P
2
)=
Area
Thrust
=
20×5
500
∴P
2
=5gwtcm
−2
Case (iii) : When the block on 10cm×5cmface
Thrust= Weight of the brick =500g.wt.
Area of contact =10cm×5cm
Pressure=
Area
Thrust
=
10×5
500
P
3
=10gwtcm
−2