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Guys, it's urgent, plz answer them immediately...

Q1. A block weighing 1.0 kg is in the shape of a cube of length 10 cm. It is kept on a horizontal table. Find the pressure on the portion of the table where the block is kept. (ans. 1000Pa)


Q2. Find the thrust acting on the human body due to atmospheric pressure. Take the surface area of a man of middle size to be 1.5m2and atmospheric pressure (1atm) =1.013×105Pa.(ans.15.2 ton wt)

Q3. Calculate the mass of a body whose volume is 2 m3and density 0.52 g/cm3. (ans. 1040 kg)

Q4. A dining hall has dimensions 50m× 15m × 3.5m. Calculate the mass of air in the hall. Given, density of air =1.30kg/m3. (ans. 3412.5 kg)

Q5. A thread of mercury of 10.2 g is in a tube of uniform cross section 0.1cm3. Calculate the length of thread. The density of mercury is 13.6g/cm3. (ans. 7.5cm)

Q6. A cubical block of water is dipped completely in water. Each edge of the block is 1cm in length. Find the buoyant force acting on the block. (ans. 10-2N)

Q7. A body of mass 2.0 kg and density 8000 kg/m3is completely dipped in a liquid of density 800 kg/m3. Find the force of buoyancy on it. (ans. 2N)

Q8. A piece of iron of density 7.8 × 103kg/m3and volume 100 cm3is totally immersed in water. Calculate (a) the weight of the iron piece in air (b) the upthrust and (c) apparent weight in water. (ans. (a) 7.8N (b) 1N (c) 6.8 N)

Q9. A solid body of mass 150g and volume 250cm3is put in water. Will the body float or sink.

Q10. A solid of density 5000kg/m3weights 0.5 kg in air. It is completely immersed in water of density 1000kg/m3.
(a)Calculate the apparent weight of solid in water.(ans. 0.4 kg)
(b)What will be its apparent weight if water is replaced by a liquid of density 8000kg/m3? (ans. 0)

Q11. The mass of a block made of certain material is 13.5 kg and its volume is 15 × 10-3m3. Will the block float or sink in water. Give reason for your answer.

Q12. (a) What is the density of air in NTP? (b)What is the unit of relative density?

Q13. (a) When does a body sinks in a fluid?
(b)Why does a balloon filled with hydrogen gas rise up against gravity?
Q14. (a) Which has greater density: 1 kg of iron or 2 kg of iron?
(b)If a hollow sphere and a solid sphere are both made of the same amount of iron, which sphere has greater average density?

Q15. (a). A body weighs 10 N in air and 8 N when fully immersed in water. How much is the buoyant force acting on the body?
(a) Why are the buoys making the channel in a river are hollow spheres?
Anirudh Bhat & 2 others asked a question
Subject: Science, asked on on 27/10/13
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Subject: Science, asked on on 14/5/14
Gravitation - An Overview

The picture that opens this lesson shows our location in the Milky Way galaxy. We are about 26,000 light years away from the centre of the galaxy. We are residing in this location from the very beginning of the Universe. This shows that there exists a powerful force that holds us in our place since the inception of the Universe. This force binds everything from stars togalaxiestosuperclustersand is known as theGravitational force.

Universal Law of Gravitation

It is common to see things falling to the ground. The falling of a body to the ground is attributed to Earths attraction for it. In fact, the weight of a body is expressed in terms of this force of attraction.

Newtons experiments showed that Earths attraction, when at a constant distance from another body, varied directly with the mass of the other body. However, this was only a partial expression of the general law of gravitation. This law states thatevery particle of matter attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

All planets in the solar system are bound to revolve in their fixed orbits by the gravitational attraction of the sun. The same force of gravity acts between Earth and the moon, making the moon revolve around Earth in its fixed orbit.

Falling of a ball from a heightFalling of raindrops from cloudsRevolution of planets around the sunGravitational Force - Mathematical Form

Let two objectsIandII,of massesM1andM2respectively, be placed at a distancedfrom each other. As per the law of gravitation, the following two assertions can be made about the force of gravity (F) between the two objects.

(a)The force of gravity between the two objects is directly proportional to the product of their masses. This is expressed as:

FM1×M2(By Product rule)

(b)The force of gravity between the two objects is inversely proportional to the square of the distance between them. This is expressed as:

F∝ (1/d2)(By Inverse-Square rule)

On combining both the equations, we obtain:

F⁢ ∝M1M2d2⁢ or⁢ F⁢ =GM1M2d2 " width="195" height="31" src="https://img-nm.mnimgs.com/img/study_content/editlive_lp/62/2013_02_09_10_27_23/mathmlequation3568840687176322813_1624553838655146822.png" /

Where, G is a constant calledUniversal Gravitational ConstantorNewtons constant.


Universal Gravitational Constant

Universal Gravitational Constant (G) is a constant of proportionality. Its value is constant at all places in the universe. Its value does not depend on the medium between two bodies.

SI unit of G

The force of gravity (F)between two objects of massesM1andM2,which are at a distancedfrom each other, is given as:

F⁢ = GM1M2d2⁢  or⁢  G⁢ =Fd2M1M2 " src="https://img-nm.mnimgs.com/img/study_content/editlive_lp/62/2013_02_09_10_27_23/mathmlequation6404826584263522868.png" /

On substituting the SI units of the various quantities in this equation, we obtain: G = Nm2/ kg2

Therefore, the SI unit of G is Nm2/kg2.

Value of G

Henry Cavendish found the value ofUniversal Gravitational Constant,G with the help of a very sensitive balance. Its value is6.673 × 1011Nm2/kg2

Consider two bodies, each having the mass 1 kg. They are placed at a distance 1m from each other. Using the value of G, the force of attraction is given as:


Universal Gravitational Constant

CAN U SIMPLIFY THIS EXPRESSION



Determination of 'G' byHenryCavendish

The value of universal gravitational constant (G) was first determined by Henry Cavendish through the torsion bar experiment. The apparatus of this experiment comprises two pairs of spheres. Each pair of spheres forms a dumbbell having a common axis, as shown in the figure. One of the dumbbells is suspended from a quartz fibre. It rotates freely when the fibre is twisted. The position of a reflected light spot from a mirror attached to the fibre gives the measure of the amount of twists. The second dumbbell can be swivelled in such a way that each of its spheres is close to one of the spheres of the other dumbbell. The gravitational attraction between the two pairs of spheres twists the fibre and the magnitude of the force of gravity is calculated by measuring the amount of twists in the fibre.

The value of G, as determined by Cavendish, came out to be6.67 × 10-11Nm2/kg2.


Universal Law of Gravitation

Gravity in the Earth-Moon System

Universal Law of Gravitation
Importance of the Universal Law of Gravitation

The universal law of gravitation helps us understand several natural phenomena. Some of these are given below.

Dropped objects fall toward the ground.

Earth pulls all objects toward itself through the gravitational force. Hence, when any object is dropped, it falls toward the ground.

The moon revolves around Earth.

The moon is attracted by Earths gravitational force. This keeps the moon revolving around Earth in its orbit of movement.

The planets revolve around the sun.

The gravitational attraction between the sun and the planets binds the planets in their respective orbits around the sun.

High and low tides occur on Earths surface.

The water present on Earths surface (in oceans, seas, etc.) is attracted by the gravitational forces of the sun and moon. Hence, the level of water in the seas and oceans rises and falls depending on the relative positions of the sun and moon. This causes high and low tides on Earth.


Formation of Tides

The water present on Earths surface (in oceans, seas, etc.) is attracted by the gravitational forces of the sun and moon. Hence, the level of water in the seas and oceans rises and falls depending on the relative positions of the sun and moon. This causes high and low tides on Earth.


Keplers Laws of Planetary Motion

A German astronomer Kepler concluded the orbits of the planets to be circular. the laws which he gave are:

First Law: The orbits of the planets are in the shape ofellipse, having the sun at one focus.

In the figure, the sun is not at the centre of the ellipse. It is at one of the foci marked X. The planet follows the ellipse in its orbit. This means that the distance between a planet and the sun constantly changes as the planet revolves in its orbit.

Second Law: The area swept over per hour by the radius joining the sun and the planet is the same in all parts of the planets orbit.

In the figure, the imaginary line joining the sun and the planet sweeps out equal areas in equal times. The planet moves faster when it is nearer to the sun. Thus, a planet executes elliptical motion with constantly changing speed as it moves around the sun in its orbit. The point of nearest approach of the planet to the sun is termed perihelion and the point of greatest separation is termed aphelion.

Third Law: The squares of the periodic times of the planets are proportional to the cubes of their mean distances from the sun.

It implies that the time taken by a planet to revolve around the sun increases rapidly with the increase in the radius of its orbit.


Keplers Laws of Planetary Motion
Newton and the Inverse-Square Rule

Newtons universal law of gravitation states that the force between two bodies is inversely proportional to the square of the distance between them. Hence, this law is also known as the inverse-square rule. Mathematically, this can be expressed as:

F∝ 1/r2...(1)

Newton used Keplers third law of planetary motion to arrive at the inverse-square rule. He assumed that the orbits of the planets around the sun are circular, and not elliptical, and so derived the inverse-square rule for gravitational force using the formula forcentripetal force. This is given as:

F=mv2/r2...(2) where,mis the mass of the particle,ris the radius of the circular path of the particle andvis the velocity of the particle.

Newton used this formula to determine the force acting on a planet revolving around the sun. Since the massmof a planet is constant, equation (2) can be written as:

Fv2/r...(3)

Now, if the planet takes timeTto complete one revolution around the sun, then its velocityvis given as:

v= 2r/T...(4) where,ris the radius of the circular orbit of the planet

or,vr/T...(5) [as the factor 2 is a constant]

On squaring both sides of this equation, we get:

v2r2/T2...(6)

On multiplying and dividing the right-hand side of this relation byr, we get:

v2⁢ ∝r2T2⁢ ×rr⁢ or⁢   v2⁢ ∝r3T2⁢×1r⁢  ...(7)  " src="https://img-nm.mnimgs.com/img/study_content/editlive_lp/62/2013_02_09_10_27_23/mathmlequation8727827952146380765.png" /

According to Keplers third law of planetary motion, the factor r3/ T2is a constant. Hence, equation (7) becomes:

v2∝ 1/r...(8)

On using equation (8) in equation (7), we get:

F⁢ ∝ 1r⁢ ×1r⁢ or  F⁢ ∝1r2 " /

Hence, the gravitational force between the sun and a planet is inversely proportional to the square of the distance between them.



Newton and the Inverse-Square Rule

Example 1:

An imaginary planetP, with an orbit of radiusR, completes one revolution around a star in 64 days. Another planetQhas an orbit of radius 4R. How much time will it take to complete one revolution around the same star?

Solution:

According to Keplers third law of planetary motion:

(1)

Where,is the time period of revolution ofis the radius of the orbit ofP

(2)

Where,is the time period of revolution ofQandis the radius of the orbit ofQ

On dividing (1) by (2), we get:



Keplers Laws and the Inverse-Square Rule
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