8. A brick of dimensions 20 cm x 10 cm x 5 cm applies a force of 5 N when placed on a surface. Calculate the pressure exerted by it when it rests on different faces.
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
∴ From the above three cases, it is clear that, as the area of contact decreases, the pressure exerted increases and is greater when the brick rests on its 10cm×5cm face.
Answers & Comments
Answer:
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Explanation:
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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
∴ From the above three cases, it is clear that, as the area of contact decreases, the pressure exerted increases and is greater when the brick rests on its 10cm×5cm face.