Question

1) i) Radius of an iron sphere is 0.21 cm. If density of iron is 7.8 g/cm³, calculate it's mass.

ii) A pressure of 1000 Pa, acts on a surface of area 15 cm² by a block of mass 'm'. Calculate 'm'. Calculate the new pressure exerted by the same block if the area of contact with the surface becomes 10 cm².​

Answer 1

Answer:

Explanation:

1)

volume of iron sphere = 4/3πr³

= 4/3× 22/7 × (0.21)³ cm³

=4× 22× 0.1 ×0.0441 cm³

=0.38808 cm ³

density = mass/volume

mass = density × volume

= 7.8 × 0.38808 g

=3.02 g

2)

pressure = force / area

force = pressure ×area

force = 1000×15×10-4 = 1.5 N

pressure exerted by the area 10 cm2 is given by:

P =1.5/10×10-4

= 1500 pascal

Hope it helps

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\mathcal\purple{Solutions:-}$

⏩ i) D = Mass/Volume

=> M = D × V

=> D = 7.8 g/cm³

Now,

Volume of the sphere, V= 4/3 πr³

= 4/3 × 22/7 × 21/100 × 21/100 × 21/100

$= \frac{4 \times 22 \times 21 \times 21}{100 \times 100 \times 100} cm^{3}$

M $= \frac{7.8 \times 4 \times 22 \times 21 \times 21}{100 \times 100 \times 100}$

$= \frac{302702.4}{100 \times 100 \times 100}$

$= 0.3027024 \: g$

$= 0.31 \: g$

ii) P = F/A = mg/A

=) m= PA/g

$= \frac{1000 \times 15}{10 \times 100 \times 100}$

$= 0.15 \: kg$

Now, P= mg/A

$= \frac{0.15 \times 10}{10 \times 10^{( - 4)} }$

$= 1</em><em>,</em><em>500 \: </em><em>Pa</em><em>$

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Hope it helps...:-)

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