Question

[Ans. 0.03636
density 1500 kg m Density of material of body 7800 kg m-3
quid
material of stone is 5200 kg m3, g = 9.8 m s-2.
) ?
(Ans. 5 x 10-5 m3 (1) 0.3685 kgf (iii) 0.3635 kgf]
(1) water (ii) in a liquid of density 800 kg m-3 ?
(Ans. 0.02333
(Ans. 12
apparent weight of body.
. A body weight 5.2 kg in air, 4.6 kg in water and 4.1 kg in a liquid. Calculate the specific gravity of the liquid
A metal block of specific gravity 6.80 floats in mercury of specific gravity 13.6 with 25 x 10 m3 of its vol
above the surface. Calculate the mass of the block.
A substance of mass 0.45 kg has a density of 2500 kg m-3. If three fourth of the substance is
immersed in a liquid of density 1150 kg m-3, calculate the density of liquid.
[Ans. 0.295
A coil having uniform area of cross-section 2 x 10-5 m2 and relative density 8.4 is found to weigh 1200 kgf
completely immersed in water. Find the length of coil.
[Ang
log of wood floats in water with th of its volume above the surface of water. What fraction of its volum
[Ans. 0.340
compl
1
6
e below the surface when floating in a liquid of density 1500 kg m-3 ?
An
e density of
11 vey, when weighed in air. has a weight of 0.045 kgf. What will be ils WURDT
3. A stone is suspended from a spring balance with a fine thread. The readings of the spring balance when the so
is in air and in liquid are 0.065 kg and 0.052 kg, respectively. Calculate the density of liquid if the density
[Ans. 1040 kg ml
4. An iron piece of density 7870 kg m weighs 0.3935 kg in air. Calculate its volume. Find the apparent weight of
the piece when (1) half its volume is immersed in water (ii) fully immersed in a liquid of density 600 kg m 3
5. A metallic piece weighs 0.1 kgf and has a density of 7100 kg m-3. How much will it weight when immersed
[Ans. (i) 0.0859 kgf (ii) 0.08872 k
[ks
​

Answer 1

Answer:

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Explanation:

aotszjc-&)$, akgzjfajvxvmsudft

Answer 2

Explanation:

ANSWER

P

1

=1 atmP

2

=1.5 atm

V

1

=2.5 LP

2

=2.0 L

T

1

=0°C=273KT

2

=?

According to Boyle's law,

T

1

P

1

V

1

=

T

2

P

2

V

2

∴

273K

1×25

=

T

2

1.5×2

∴ T

2

=

1×2.5

1.5×2×273

=327.78K=54.63°C

hope it helpss