हिंदी

A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate:

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प्रश्न

A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate:

  1. The weight of the piece of stone in air,
  2. The volume of the piece of stone,
  3. The relative density of stone,
  4. The relative density of the liquid.
योग
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उत्तर

1) The mass of stone is 15.1 g. Hence, its weight in air will be Wa = 15.1 gf

2) When stone is immersed in water its weight becomes 9.7 gf. So, the upthrust on the stone is 15.1 − 9.7 = 5.4 gf, Since the density of water is 1 g cm−3, the volume of stone is 5.4 cm3.

3) Weight of stone in liquid is Wl = 10.9 gf

Weight of stone in water is Ww = 9.7 gf

Therefore, the relative density of stone is 

`R.D_("stone") = (W_a)/(W_a - W_w)`

= `(15.1  gf)/(15.1 - 9.7  "gf")`

= `15.1/5.4`

= 2.8

4) Relative density of liquid is 

`R.D_("liquid") = (W_a - W_l)/(W_a - W_w)`

= `(15.1 - 10.9)/(15.1 - 9.7)`

= `4.2/5.4`

∴ R.DStone = 0.7777 ≈ 0.78

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Determination of Relative Density of a Solid Substance by Archimedes’ Principle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Upthrust in Fluids, Archimedes’ Principle and Floatation - Exercise 5 (B) [पृष्ठ ११७]

APPEARS IN

सेलिना Concise Physics [English] Class 9 ICSE
अध्याय 5 Upthrust in Fluids, Archimedes’ Principle and Floatation
Exercise 5 (B) | Q 16 | पृष्ठ ११७

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