
A stone weighs 392.4 N in air and 196.2 N IN the water. Compute the volume of stone and its specific gravity.
Answer
233.1k+ views
Hint: Specific gravity is also known as the relative density and is defined as the ratio of the density of the substance to that of a standard substance. It has no unit and hence it is a dimensionless quantity.
Complete step by step solution:
Given data:
Weight of the stone in the air = 392.4 N
Weight of the stone in the water = 196.2 N
The volume of the stone =?
Specific gravity =?
We know that weight = mg
Where m is the mass and g is the acceleration due to gravity.
Also, mass in terms of volume and density is given as, $m = Vd$
Thus, in air, $Vdg = 392.4\_\_\_\_\_\_\_\_\left( 1 \right)$
Thus, again in water, $Vdg = 196.2\_\_\_\_\_\_\left( 2 \right)$
We know that the density of water is 1000 and also g value is nearly equal to 10.
Substituting these values in equation 2, we get,
$V \times 1000 \times 10 = 196.2$
$ \Rightarrow V = \dfrac{{196.2}}{{{{10}^4}}}{m^3} = 1.962 \times {10^{ - 2}}{m^3}$
Substituting the values of V and g values in equation 1, we get
$1.962 \times {10^{ - 2}} \times d \times 10 = 392.4$
$ \Rightarrow d = 2000$
Specific gravity is given by the formula,
$\text{Specific Gravity} = \dfrac{{\text{density of the object}}}{{\text{density of the water}}}$
$ \Rightarrow \text{Specific Gravity} = \dfrac{{2000}}{{1000}} = 2$
Thus the volume of the stone is \[1.962 \times {10^{ - 2}}{m^3}\] and the specific gravity is 2.
Note: 1. Weight is defined as the force exerted on the object by gravity. The amount of matter contained in the object is called mass. The weight is concerned with gravity and mass while the mass is independent of gravity.
2. The density is defined as the amount of matter that is contained in a unit volume. The density is concerned with mass and volume.
3. If the value of the specific gravity is one, then the object will not sink or float.
Complete step by step solution:
Given data:
Weight of the stone in the air = 392.4 N
Weight of the stone in the water = 196.2 N
The volume of the stone =?
Specific gravity =?
We know that weight = mg
Where m is the mass and g is the acceleration due to gravity.
Also, mass in terms of volume and density is given as, $m = Vd$
Thus, in air, $Vdg = 392.4\_\_\_\_\_\_\_\_\left( 1 \right)$
Thus, again in water, $Vdg = 196.2\_\_\_\_\_\_\left( 2 \right)$
We know that the density of water is 1000 and also g value is nearly equal to 10.
Substituting these values in equation 2, we get,
$V \times 1000 \times 10 = 196.2$
$ \Rightarrow V = \dfrac{{196.2}}{{{{10}^4}}}{m^3} = 1.962 \times {10^{ - 2}}{m^3}$
Substituting the values of V and g values in equation 1, we get
$1.962 \times {10^{ - 2}} \times d \times 10 = 392.4$
$ \Rightarrow d = 2000$
Specific gravity is given by the formula,
$\text{Specific Gravity} = \dfrac{{\text{density of the object}}}{{\text{density of the water}}}$
$ \Rightarrow \text{Specific Gravity} = \dfrac{{2000}}{{1000}} = 2$
Thus the volume of the stone is \[1.962 \times {10^{ - 2}}{m^3}\] and the specific gravity is 2.
Note: 1. Weight is defined as the force exerted on the object by gravity. The amount of matter contained in the object is called mass. The weight is concerned with gravity and mass while the mass is independent of gravity.
2. The density is defined as the amount of matter that is contained in a unit volume. The density is concerned with mass and volume.
3. If the value of the specific gravity is one, then the object will not sink or float.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

