Question

A piece of iron has dimensions 3 cm × 1.5 cm × 6 cm. If its mass is 205.2 g calculate:
(i) volume of iron
(ii) density in C.G.S. system
(iii) density in S.I. system
(iv) R.D. of iron​

Answer 1

Explanation:

Volume of body V=10

−4

m

3

Density of iron, ρ

i

=7800kgm

−3

Density of water ρ

w

=1000kgm

−3

Weight in air,W

a

=ρ

i

Vg=7800×10

−4

×10=7.8N

We know that,

Weight in water = weight in air – buoyancy force

W

w

=W

a

−ρ

w

Vg

W

w

=7.8−1000×10

−4

×10

W

w

=6.8N

Hence, the weight of body in water is 6.8 N

Answer 2

Answer:

Given that:-

Dimensions of a piece of iron = 3 cm × 1.5 cm × 6 cm.

Mass of iron = 205.2 g

To find:-

1. Volume of iron
2. Density in C.G.S system
3. Density in S.I. system
4. R.D. of iron

Solution:-

1. Volume of iron = l × b × h

= 3 cm × 1.5 cm × 6 cm

$= 27 \: cm {}^{3}$

2. Density in C.G.S system

= M/V

$\frac{205.2 \: g}{27 \: cm } \\ \\ = 7.6 \: gcm {}^{ - 3}$

3. Density in S.I. system

= Density in C.G.S system × 1000 kgm$^{–3}$

$= 7600 \: kgm {}^{ - 3}$

4. R.D of iron = Density of iron

$= 7.6 \: gcm {}^{ - 3} \\ \\ = 7.6$