A piece of iron of density $7800 k g / m^{3}$ and volume $100 c m^{3}$ is completely immersed in water. Calculate the weight of the iron piece in air.
[Take, $g = 10 m / s^{2}$ ]

7.8 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8.8 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.8 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9.8 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $7.8 N$
Density = $\frac{m a s s}{v o l u m e}$

$\Rightarrow$ Mass = $V o l u m e \times D e n s i t y$ = $(1 \times 10^{- 4} \times 7800) k g = 0.78 k g$

So weight of Iron piece = mg = $0.78 \times 10 = 7.8 N$

so option (A) is correct

Suggest Corrections

Similar questions

Q. A piece of iron of density

7800 k g / m^{3}

and volume

100 c m^{3}

is completely immersed in water. Calculate the upthrust on the iron piece.

[Take,

g = 10 m / s^{2}

A piece of iron of density $7.8 \times 10^{3} k g m^{- 3}$ and volume 100 $c m^{3}$ is totally immersed in water ( $ρ$ = 1000 kg $m^{- 3})$ . Calculate its apparent weight in water.

A piece of iron has density 7.8/

c m^{3}

& volume

10^{- 4} m^{3}

. It is totally immersed in the water calculate weight of iron in water

Q. A piece of aluminum with mass

1.00 k g

and density

2700 k g / m^{3}

is suspended in air from a string and then completely immersed in a container of water. Calculate the tension in the string

(a)

before and

(b)

after the metal is immersed. Take,

g = 10 m / s^{2}

, density of water

= 10^{3} k g m^{- 3}

A piece of iron weighs 44.5 gf in the air. If the density of iron is $8.9 \times 10^{3} k g m^{- 3}$ , find the weight of iron piece when immersed in water.

Join BYJU'S Learning Program

A piece of iron of density 7800kg/m3 and volume 100cm3 is completely immersed in water. Calculate the weight of the iron piece in air.[Take, g=10m/s2]

The correct option is B 7.8NDensity = massvolume ⇒ Mass = Volume×Density = (1×10−4×7800)kg=0.78kgSo weight of Iron piece = mg =0.78×10=7.8Nso option (A) is correct

A piece of iron of density $7800 k g / m^{3}$ and volume $100 c m^{3}$ is completely immersed in water. Calculate the weight of the iron piece in air.
[Take, $g = 10 m / s^{2}$ ]

The correct option is B $7.8 N$
Density = $\frac{m a s s}{v o l u m e}$

$\Rightarrow$ Mass = $V o l u m e \times D e n s i t y$ = $(1 \times 10^{- 4} \times 7800) k g = 0.78 k g$

So weight of Iron piece = mg = $0.78 \times 10 = 7.8 N$

so option (A) is correct