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Question: A brick with dimensions of \(20cm\times 10cm\times5 cm\) has a weight \[500gwt\]. Calculate the pres...

A brick with dimensions of 20cm×10cm×5cm20cm\times 10cm\times5 cm has a weight 500gwt500gwt. Calculate the pressure exerted by it when it rests on different faces.

& \text{A}\text{. Case i) }{{\text{P}}_{1}}=1.25g/c{{m}^{2}},Case\text{ ii)}{{\text{P}}_{2}}=5g/c{{m}^{2}},\text{ Case iii)}{{\text{P}}_{3}}=5g/c{{m}^{2}} \\\ & \text{B}\text{. Case i) }{{\text{P}}_{1}}=2.5g/c{{m}^{2}},Case\text{ ii)}{{\text{P}}_{2}}=5g/c{{m}^{2}},\text{ Case iii)}{{\text{P}}_{3}}=10g/c{{m}^{2}} \\\ & \text{C}\text{. Case i) }{{\text{P}}_{1}}=5g/c{{m}^{2}},Case\text{ ii)}{{\text{P}}_{2}}=25g/c{{m}^{2}},\text{ Case iii)}{{\text{P}}_{3}}=15g/c{{m}^{2}} \\\ & \text{D}\text{. Case i) }{{\text{P}}_{1}}=2.5g/c{{m}^{2}},Case\text{ ii)}{{\text{P}}_{2}}=2.5g/c{{m}^{2}},\text{ Case iii)}{{\text{P}}_{3}}=10g/c{{m}^{2}} \\\ \end{aligned}$$
Explanation

Solution

We know that pressure is the thrust applied per unit area of the surface. Here we have the dimensions of the brick which in turn will give the surfaces of the brick. Also the weight of the brick is given hence we can calculate the force.

Formula used:
P=TAP=\dfrac{T}{A}

Complete step by step answer:
We know that a brick is a cuboid. We also know that a cuboid has 66 faces or 33 pairs of rectangular faces. The dimensions of the cuboid are given as 20cm×10cm×5cm20cm\times 10cm\times5 cm and is as shown below

Also the mass of the brick is given as m=500gm=500g, which is also the thrust TT due to the brick.
Since the faces are rectangular, we know that their area is given as l×bl\times b where ll is the length and bb is the breadth of the rectangle.
Let us assume, P1P_{1} to be the pressure on the surface 11 whose dimension is given by 20cm×10cm20cm\times 10cm
Then the area of the surface is given as A1=20cm×10cm=200cm2A_{1}=20cm\times 10cm=200cm^{2}
Then, P1=500200=2.5g/cm2P_{1}=\dfrac{500}{200}=2.5g/cm^{2}
Let us assume P2P_{2}, to be the pressure on the surface 22 whose dimensions is given by 20cm×5cm20cm\times 5cm
Then the area of the surface is given as A2=20cm×5cm=100cm2A_{2}=20cm\times 5cm=100cm^{2}
Then P2=500100=5g/cm2P_{2}=\dfrac{500}{100}=5g/cm^{2}

Let us assume P3P_{3}, to be the pressure on the surface 33 whose dimension is given by 10cm×5cm10cm\times 5cm
Then the area of the surface is given as A3=10cm×5cm=50cm2A_{3}=10cm\times 5cm=50cm^{2}
Then P3=50050=10g/cm2P_{3}=\dfrac{500}{50}=10g/cm^{2}
Hence the answer is B. Case i) P1=2.5g/cm2,Case ii)P2=5g/cm2, Case iii)P3=10g/cm2\text{B}\text{. Case i) }{{\text{P}}_{1}}=2.5g/c{{m}^{2}},Case\text{ ii)}{{\text{P}}_{2}}=5g/c{{m}^{2}},\text{ Case iii)}{{\text{P}}_{3}}=10g/c{{m}^{2}}

Note:
One may assume surface 33 as surface 22 or surface 11 also. It doesn’t matter. The values of the pressure may get interchanged with respect to the order in which the dimensions are taken. Hence don’t expect the values to be in the same order as given in the options.