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A piece of solid weighs \(120 \mathrm{~g}\) in air, \(80 \mathrm{~g}\) in water and \(60 \mathrm{~g}\) in liquid the relative density of the solid and that of the solid and that of the liquid are respectively. (A) 3,2 (B) \(2,(3 / 4)\) (C) \((3 / 4), 2\) (D) \(3,(3 / 2)\)

Short Answer

Expert verified
The relative density of the solid is 3, and the relative density of the liquid is 2. The answer is (A) 3, 2.

Step by step solution

01

Understanding Archimedes' principle

Based on Archimedes' principle, when an object is submerged in a fluid, it experiences a buoyant force equal to the weight of the fluid it displaces. This force acts opposite to the force of gravity and reduces the apparent weight of the object when submerged in the fluid. The buoyant force can be calculated using the formula: \(Buoyant \ force = weight_{object \ in \ fluid} - weight_{object \ in \ air}\).
02

Calculate the buoyant force in water

We will calculate the buoyant force experienced by the solid in water using the given weights: Buoyant force in water = apparent weight of the solid in water - weight of the solid in air Buoyant force in water = \(80 \mathrm{~g} - 120 \mathrm{~g} = -40 \mathrm{~g}\)
03

Calculate relative density of solid

Relative density or specific gravity is the ratio of the density of a substance to that of a reference substance (water in this case). Using the calculated buoyant force in water, we can find the relative density of the solid: Relative density of solid = \(\frac{Weight \ of \ solid \ in \ air}{Weight \ of \ solid \ in \ air - Weight \ of \ solid \ in \ water}\) Relative density of solid = \(\frac{120 \mathrm{~g}}{40 \mathrm{~g}} = 3\)
04

Calculate the buoyant force in liquid

Now we will calculate the buoyant force experienced by the solid in the liquid using the given weights: Buoyant force in liquid = apparent weight of the solid in liquid - weight of the solid in air Buoyant force in liquid = \(60 \mathrm{~g} - 120 \mathrm{~g} = -60 \mathrm{~g}\)
05

Calculate relative density of liquid

We can find the relative density of the liquid in the same way as we found for the solid: Relative density of liquid = \(\frac{Weight \ of \ solid \ in \ air}{Weight \ of \ solid \ in \ air - Weight \ of \ solid \ in \ liquid}\) Relative density of liquid = \(\frac{120 \mathrm{~g}}{60 \mathrm{~g}} = 2\) So the answer is (A) relative density of the solid = 3, and relative density of the liquid = 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyant Force
Understanding the concept of buoyant force is key to solving many physics problems related to objects immersed in fluids. As per Archimedes' Principle, when an object is submerged in a fluid, it experiences an upward force called buoyant force. This force is equal to the weight of the fluid displaced by the object. Because the force acts in the opposite direction to gravity, it makes the object feel lighter in the fluid. When calculating the buoyant force on an object, use the formula:
  • Buoyant Force = Weight of Object in Air - Apparent Weight in Fluid
For instance, in a given problem, if a solid weighs 120 g in air but only 80 g in water, the buoyant force exerted by the water on the solid can be calculated as 120 g - 80 g = 40 g. This means that the fluid is pushing up on the solid with 40 g, making it appear 40 g lighter.
Relative Density
Relative density, also known as specific gravity, is a dimensionless quantity that describes the ratio of the density of a substance to that of a reference material. Water is commonly used as the reference material for liquids and solids at room temperature. Relative density helps us understand how dense a material is compared to water.
  • Formula: Relative Density = \(\frac{Weight \ in \ Air}{Weight \ in \ Air - Apparent \ Weight \ in \ Fluid}\)
When you determine the relative density of a solid based on its different weights in air and in water, it shows how much denser the solid is compared to water. In this exercise, if a solid's weight in air is 120 g and its apparent weight in water is 80 g, the relative density is calculated as \(\frac{120}{120-80} = 3\). This implies the solid is three times denser than water.
Specific Gravity
Specific gravity is another term for relative density. It is a measure of the ratio of a material's density to a reference density. For solids and liquids, this reference is often the density of water at a specific temperature. Specific gravity provides insights into whether an object will sink or float when placed in water.
  • Specific Gravity Formula: \(\frac{Density \ of \ Substance}{Density \ of \ Water}\)
Specific gravity can be particularly useful in identifying substances, as it reflects how the object's density compares to water, independent of units. If a liquid has a specific gravity of 2, as seen in the problem, it means that the liquid is twice as dense as water.
Density Calculation
Density is a key concept in physics that defines how much mass is contained in a given volume. It is typically expressed in units like grams per cubic centimeter (g/cm³). Calculating the density of materials is essential for understanding their properties and behaviors in different environments.For practical purposes, density can be calculated using the formula:
  • Density = \(\frac{Mass}{Volume}\)
In practical exercises, like the one discussed, specific gravity (or relative density) is used to understand density without directly measuring mass or volume. The calculated relative density tells us how the density of a material compares to water, which provides an indirect yet effective way to understand its characteristics. In this context, the density of the solid and the liquid was inferred through changes in apparent weight and buoyant force.

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Most popular questions from this chapter

A \(2 \mathrm{~m}\) long rod of radius \(1 \mathrm{~cm}\) which is fixed from one end is given a twist of \(0.8\) radians. What will be the shear strain developed ? (A) \(0.002\) (B) \(0.004\) (C) \(0.008\) (D) \(0.016\)

A vessel contains \(110 \mathrm{~g}\) of water the heat capacity of the vessel is equal to \(10 \mathrm{~g}\) of water. The initial temperature of water in vessel is \(10^{\circ} \mathrm{C}\) If \(220 \mathrm{~g}\) of hot water at \(70^{\circ} \mathrm{C}\) is poured in the vessel the Final temperature neglecting radiation loss will be (A) \(70^{\circ} \mathrm{C}\) (B) \(80^{\circ} \mathrm{C}\) (C) \(60^{\circ} \mathrm{C}\) (D) \(50^{\circ} \mathrm{C}\)

Eight drops of a liquid of density 3 and each of radius a are falling through air with a constant velocity \(3.75 \mathrm{~cm} \mathrm{~S}^{1}\) when the eight drops coalesce to form a single drop the terminal velocity of the new drop will be (A) \(15 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(2.4 \times 10^{-2} \mathrm{~m} / \mathrm{s}\) (C) \(0.75 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(25 \times 10^{-2} \mathrm{~m} / \mathrm{s}\)

Radius of a soap bubble is \(\mathrm{r}^{\prime}\), surface tension of soap solution is \(\mathrm{T}\). Then without increasing the temperature how much energy will be needed to double its radius. (A) \(4 \pi r^{2} T\) (B) \(2 \pi r^{2} T\) (C) \(12 \pi r^{2} T\) (D) \(24 \pi r^{2} T\)

8000 identical water drops are combined to form a bigdrop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is (A) \(1: 10\) (B) \(1: 15\) (C) \(1: 20\) (D) \(1: 25\)

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