Question

A 0.50N metal sinker appears (as measured using a spring scale) to have a weight of 0.45 N when submerged in water. The relative density of the metal is

Answer 1

its wt = 0.50 = V d g    (d is density of metal, v is volume of metal )

bouyant force = upthrust applied by water = 0.50 - 0.45 = 0.05 N

this upthrust = V w g ( V is volume, w is density of water)

therefore vdg=0.50

   and       vwg=0.05

divide them to get d/w= 10  (relative density of metal with respect to water)

Answer 2

Answer: D. 10

Explanation:

⇒apparent weight=weight-Fb

0.45=0.50-Fb

Fb=0.05 N

⇒Fb= density×gravity×volume

0.05= 1 ×9.8×volume

• volume= 1/196 cm³

⇒weight=mass×gravity

0.50=mass×9.8

• mass=5/98g

⇒density= mass÷volume

density= 5/98 ÷ 1/196

=10