Question

A gold ornament weighs 570 gram in air nad 520 gram in water. If the specific gravity of gold is 19, find the difference in the volume of water displaced when the ornament is immersed in water and the actual volume of the gold in the ornament. How do you account for this difference in volume ?

Answer 1

Answer Text
Solution :
Find the weight of gold ornament in air (w1)
(
w
1
)
and water (w2)
(
w
2
)
.
Find the density of gold ornament using formula,
d=w1w1−w2
d
=
w
1
w
1
-
w
2

Find the volume of gold ornament using formula, V=md=w1d
V
=
m
d
=
w
1
d

Is this equal to the apparent loss of weight of the gold ornament ?
Find the volume of 570 grams of pure gold using formula, d=mv
d
=
m
v

IS thereany difference in the volume of pure gold and gold ornament ?
Does the gold ornament contain a cavity ?

Answer 2

Explanation:

$1116161111221 \frac{ \frac{ \frac{ \frac{ \frac{ \frac{11646 \times \frac{2255}{2442} }{113} \times \frac{11216}{2} \times \frac{24}{24224} }{44242} \times \frac{242424}{4242424} }{42242424244} \times \frac{2424111214222}{61221} \times \frac{2212111}{222222} }{24334100} \times \frac{222}{311111} }{1} \times \frac{16444}{241} \times \frac{2221211}{23113} \times \frac{4461}{2222} \times \frac{22}{22} }{22}51 2222545$