Question

A block of weight 5kgf and dimensions 5cm * 2cm *1 (* = multiplication sign) rests on a table in three different positions with its base as (i) 5cm *2 cm (ii) 2cm * 1cm (iii) 1 cm *5 cm . Calculate the maximum and minimum pressure in S.I units exerted by a block on its base (take g = 10 ms-2)

Answer 1

Answer:

maximum pressure = 2,50,000 Pa

minimum pressure = 50,000 Pa

Explanation:

mass of block = 5kg

Pressue = Thrust/contact area

When it is kept by 5cm * 2cm as its base :

Thrust = mg

= 5 * 10

50 N

Contact area = 5*2 = 10 cm^2 = 0.001 m^2

Pressure = 50/0.001 = 50000 Pascal

When it is kept by 2cm*1 cm as base :

Thrust = mg =50 N

Contact area = 2*1 =2 cm^2 = 0.0002 cm^2

Pressure = 50/0.0002 = 2,50,000 Pascal

When it is kept by 1cm*5cm as base :

Thrust = mg = 50 N

Contact area = 5*1 = 5cm^2 = 0.0005 m^2

Pressure = 50/0.0005 = 1,00,000 Pascal

Answer 2

Answer:

The maximum pressure is 2,50,000 Pa and minimum pressure is 50,000 Pa in S.I units exerted by a block on its base (take g = 10 ms-2)

Explanation:

it is given in the question that mass of block is 5kg

Again, we know that Pressure is,  $\frac{Force}{Unit Area}$

Case-1

When the block is kept by $5cm \times 2cm$ as it's base :

area = mg

$A= 5\times 10\\A=50N$

Contact area = $5cm\times 2cm$

Contact area = $10 cm^2$

Contact area = $0.001 m^2$

Now,

$Pressure = \frac{50N}{0.001m^2} \\Pressure = 50000 Pascal$

Case-2

When the block is kept by $2cm\times1 cm$ as it's base :

area = mg

$A= 5\times 10\\A=50N$

Contact area = $2cm\times 1cm$

Contact area = $2 cm^2$

Contact area = $0.0002 m^2$

Now,

$Pressure = \frac{50N}{0.0002m^2} \\Pressure = 2,50,000 Pascal$

Case-3

When the block is kept by $1cm\times5cm$ as it's base :

area = mg

$A= 5\times 10\\A=50N$

Contact area = $5cm\times 1cm$

Contact area = $5 cm^2$

Contact area = $0.0005 m^2$

Now,

$Pressure = \frac{50N}{0.0005m^2} \\Pressure = 1,00,000 Pascal$

Hence, The maximum pressure is 2,50,000 Pa and minimum pressure is 50,000 Pa in S.I units exerted by a block on its base (take g = 10 ms-2).

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