Hindi, asked by pulastnandlal36, 5 months ago

सिद्धेश्वरी ने देखा कि उसका बेटा किसान धीरे-धीरे घर की तरफ आ रहा है कि सलमा को बताता है कि उसे अच्छी नौकरी मिल गई आगे


की कहानी आप पूरी कर लिखिए​

Answers

Answered by freefirepros1
12

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.

Answered by manshi999
1

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