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How Do You Measure Density

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How Do You Measure Density

  • \(\text{Density}=\frac{\text{Mass}}{\text{Volume}}\text{    or   d}=\frac{\text{M}}{\text{V}}\)
  • SI unit of density is kg/m3 and CGS unit of density is g /cm3
  • 1 gm/cm3 = 1000 kg/m3
  • Different substances have different densities which gets affected by temperature.
  • Respective density of a substance decreases on heating due to the expansion of the substance.
  • Water has anomalous expansion. When water is cooled at 4ºC, its volume decreases but on further cooling its volume starts increasing.
  • The density of water is maximum at 4ºC.

Relative Density

Relative density of a substance
\( =\frac{\text{density}\,\text{of}\,\text{subs}\,\text{tance}}{\text{density}\,\text{of}\,\text{water}\,\text{at}\,\text{4 }\!\!{}^\circ\!\!\text{ C}} \)
\( =\frac{\text{mass}\,\text{of}\,\text{unit}\,\text{volume}\,\text{of}\,\text{subs}\,\text{tance}}{\text{mass}\,\text{of}\,\text{unit}\,\text{volume}\,\text{of}\,\text{water}\,\text{at}\,\text{4 }\!\!{}^\circ\!\!\text{ C}} \)
\( =\frac{\text{mass}\,\text{of}\,\text{certain}\,\text{volume}\,\text{of}\,\text{subs}\,\text{tance}}{\text{mass}\,\text{of}\,\text{same}\,\text{volume}\,\text{of}\,\text{water}\,\text{at}\,\text{4 }\!\!{}^\circ\!\!\text{ C}} \)
\( =\frac{\text{weight}\,\text{of}\,\text{certain}\,\text{volume}\,\text{of}\,\text{subs}\,\text{tance}}{\text{weight}\,\text{of}\,\text{same}\,\text{volume}\,\text{of}\,\text{water}\,\text{at}\,\text{4 }\!\!{}^\circ\!\!\text{ C}} \)

Unit of Relative Density:
It has no units.
Note: density of water in CGS system is 1 gm/cm3.

Relative Density for solids:
\(\text{Relative Density}=\frac{{{W}_{1}}}{{{W}_{1}}-{{W}_{2}}}\)
Where W1 = weight of solid body in air
W2 = weight of solid body in water

Relative Density for liquids:
\(\text{Relative Density}=\frac{W-W”}{W-W’}\)
Where,
W’ = weight of the body fully immersed in water
W” = weight of the body fully immersed in liquid.
W = weight of the body in air

Relative Density Example Problems With Solutions

Example 1:    A body weighs 30 N in air and 26 N when fully immersed in water. Calculate its relative density.
Solution:    Given: Weight of body in air, W1 = 30 N
Weight of body in water, W2 = 26 N
\(\text{(iii) Relative Density}=\frac{{{W}_{1}}}{{{W}_{1}}-{{W}_{2}}}\)
\(=\frac{30}{30-26}=7.5\)

Example 2:    Relative density of copper is 8.8. What is its density in SI system?
Solution:     
\(\text{Relative density}=\frac{\text{density}\,\text{of}\,\text{copper}}{\text{density}\,\text{of}\,\text{water}}\)
∴ Density of copper = R.D. × density of water
= 8.8 × 103 kg/m3

Example 3:    A solid weighs 60 gmf in air and 52 gmf when completely immersed in water. Calculate the following
(i) upthrust,
(ii) volume of the solid,
(iii) relative density of the solid
Solution:     Given:
Weight of solid in air = 60 gmf
Weight of solid in water = 52 gmf
∴ Loss of weight in water = 60 – 52 = 8 gmf
(i) ∴ Upthrust = loss of weight in water = 8 gmf
(ii) Since density of water is 1 gm/cm3, and weight of equal volume of water = 8 gmf
∴ Volume of solid = 8 cm3
\(\text{(iii) Relative Density of solids}=\frac{{{W}_{1}}}{{{W}_{1}}-{{W}_{2}}}\)
\(=\frac{60}{60-52}=7.5\)

What Is The Normal Force

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What Is The Normal Force

The external agent which tends to set a body in motion or which changes the speed and direction of motion of a body or which can change the shape of a body is called force.
SI unit of force is Newton.

Types of Forces:

(a) Non Contact:
These are the forces in which contact between two objects is not necessary. Gravitational force between two bodies and electrostatic force between two charges are two examples of field forces.
(b) Contact Forces:
Two bodies in contact exert equal and opposite forces on each other. If the contact is frictionless the contact force is perpendicular to the common surface and known as normal reaction.
Contact-forces
If, however, the objects are in rough contact and move (or have a tendency to move) relative to each other without losing contact then frictional force arise which oppose such motion. Again each object exerts a frictional force on the other and the two forces are equal and opposite. This force is perpendicular to normal reaction. Thus, the contact force (F) between two objects is made up of two forces.
(i) Normal reaction (N) (ii) Force of friction (f)
and since these two forces are mutually perpendicular.
\(\text{F}=\sqrt{{{\text{N}}^{\text{2}}}\text{+ }{{\text{f}}^{\text{2}}}}\)

 

What Is Circular Motion

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Circular Motion

When a body moves in such a way that its distance from a fixed point always remains constant, then its motion is said to be the circular motion.

Uniform circular motion:
If the radius vector sweeps out equal angles in equal times, then its motion is said to be uniform circular motion.
Uniform-circular-motion
In uniform circular motion speed remains constant.
Linear velocity, being a vector quantity, its direction changes continuously.
The direction of velocity is along the tangent at every point.

Angular velocity:
\(\omega =\frac{\Delta \theta }{\Delta t}\)
A vector quantity
Direction is perpendicular to plane of rotation
Note: If the particle is revolving in the clockwise direction then the direction of angular velocity is perpendicular to the plane downwards. Whereas in case of anticlockwise direction, the direction will be upwards.
Unit is Radian/sec.
In uniform circular motion the direction of angular velocity is along the axis of rotation which is constant throughout.
Angular velocity remains constant in magnitude as well as in direction.
v = rω where r = radius of the circle.

Centripetal acceleration:
In uniform circular motion the particle experiences an acceleration called the centripetal acceleration.
\({{a}_{c}}=\frac{{{v}^{2}}}{r}\)
The direction of centripetal acceleration is along the radius towards the centre.

Centripetal force:
Always acts towards centre.
Centripetal force is required to move a particle in a circle.
Because Fc is always perpendicular to velocity or displacement, hence the work done by this force will always be zero.
Note:
Circular motion in horizontal plane is usually uniform circular motion.
Remember that equations of motion are not applicable for circular motion.

Time period:
It is the time taken to complete one complete revolution.
In one revolution, angle subtended is 2π and if T is time period, then the angular velocity is given by
\(\omega =\frac{2\pi }{T}\text{         or         }T=\frac{2\pi }{\omega }\)

Frequency:
Frequency is defined as the number of revolutions per second.
\(i.e.~\text{    n}=\frac{1}{T}=\frac{\omega }{2\pi }\)

Uniform Circular Motion Example Problems With Solutions

Example 1:    A particle moves in a circle of radius 2 m and completes 5 revolutions in 10 seconds. Calculate the following
(i) Angular velocity and
(ii) Linear velocity.
Solution:    Since, it completes 5 revolutions in 10 seconds.
∴ Time period = 10/5 = 2s
(i) Now angular velocity,
\(\omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi \text{ rad/s}\)
(ii) Linear velocity is given by
v = rω = 2π
∴ v = 2π m/s

Example 2:    The length of second’s needle in a watch is 1.2 cm. Calculate the following :
(i) Angular velocity and
(ii) Linear velocity of the tip of the needle.
Solution:    (i) We know that the second’s needle in a watch completes one revolution in 60 seconds.
∴ Time period, T = 60 s
Angular velocity,
\( \omega =\frac{2\pi }{T}=\frac{2\pi }{T}=\frac{\pi }{30}\text{ rad/s}~ \)
(ii) Length of the needle = 1.2 cm = Radius of the circle
Linear velocity of the tip of the needle is given by
v = rω
\( v=1.2\times \frac{\pi }{30}=~\frac{\pi }{25} \)
\( v=\frac{\pi }{2s}=1.266\times {{10}^{-1~}}\text{cm/sec}\text{.} \)

Example 3:    Earth revolves around the sun in 365 days. Calculate its angular velocity.
Solution:    Time period,  T = 365 days
= 365 × 24 × 60 × 60 seconds
∴ Angular velocity
\(\omega =\frac{2\pi }{T}=\frac{2\pi }{365\times 24\times 60\times 60}\)
= 1.99 × 10-7 rad/s.

Image Formation by Concave and Convex Lenses

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Image Formation By Convex Lens in Different Cases

Case 1: Object at Infinity
A point object lying on the principal axis:
Rays come parallel to the principal axis and after refraction from the lens, actually meet at the second principal focus F2.
Image-Formation-By-Concave-And-Convex-Lenses-1
Fig. Convex lens point object at infinity, image at focus.
The image is formed at focus F2. It is real and point sized.

A big size object with its foot on the principal axis:
Parallel rays come inclined to the principal axis. Image of foot is formed at the focus.
Image is formed at the second principal focus F2. It is real inverted and diminished (smaller in size than the object). (Fig.)
Parallel rays from infinity
Image-Formation-By-Concave-And-Convex-Lenses-2
Fig. Convex lens : big size object at infinity, image at focus

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Case 2: Object at distance more than twice the Focal Length
Real object AB has its image A’B’ formed between distance f and 2f.
The image is real inverted and diminished (smaller in size than the object)
Image-Formation-By-Concave-And-Convex-Lenses-3
Fig. Convex lens : object beyond 2f, image between f and 2f.

Case 3: Object at distance twice the Focal Lengths
Real object AB has its image A’B’ formed at distance 2f.
Image-Formation-By-Concave-And-Convex-Lenses-4
Fig. Convex lens : object at distance 2f, image at distance 2f.
The image is real, inverted and has same size as the object.

Case 4: Object at distance more than Focal Length and less than twice is Focal Length
Real object AB has its image A’B’ formed beyond distance 2f.
Image-Formation-By-Concave-And-Convex-Lenses-5
Fig. Convex lens : object at distance between f and 2f, image beyond 2f.
The image is real inverted and enlarged (bigger in size than the object).

Case 5: Object at Focus
Real object AB has its image formed at infinity.
Image-Formation-By-Concave-And-Convex-Lenses-6
Fig. Convex lens : object at focus, image at infinity.
The image is imaginary inverted (refracted rays to downward) and must have very large size.

Case 6: Object between Focus and Optical Centre
Real object AB has its image A’B’ formed in front of the lens.
Image-Formation-By-Concave-And-Convex-Lenses-7

Image Formation By Concave Lens in Different Cases

Case 1: Object at infinity
A point object lying on the principal axis:
Rays come parallel to the principal axis and after refraction from the lens, appears to come from the second principal focus F2.
Image-Formation-By-Concave-And-Convex-Lenses-9
Fig. Concave lens point object at infinity, image at focus.
The image is formed at focus F2. It is virtual and point sized (fig.)

A big size object with its foot on the principal axis:
Parallel rays come inclined to the principal axis. Image of foot is formed at focus.
The image is formed at the second principal focus F2.
It is virtual–erect and diminished (fig.)
Image-Formation-By-Concave-And-Convex-Lenses-10
Fig. Concave lens : big size object at infinity image at focus.

Case 2: Object at a Finite Distance
Real object AB has its image A’B’ formed between second principal focus F2 optical centre C.
The image is virtual–erect and diminished.

 

What Is Arithmetic Mean

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What Is Arithmetic Mean

If three or more than three terms are in A.P., then the numbers lying between first and last term are known as Arithmetic Means between them.i.e.
The A.M. between the two given quantities a and b is
A so that a, A, b are in A.P.
i.e. A – a = b – A
\( \Rightarrow A=\frac{a+b}{2} \)
Note: A.M. of any n positive numbers a1, a2 ……an is
\( A=\frac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+…..{{a}_{n}}}{n} \)
n AM’s between two given numbers
If in between two numbers ‘a’ and ‘b’ we have to insert n AM A1, A2, …..An then a, A1, A2, A3….An, b will be in A.P. The series consist of (n + 2) terms and the last term is b and first term is a.
a + (n + 2 – 1) d = b
\( d=\frac{b-a}{n+1} \)
A1 = a + d, A2 = a + 2d, ……  An = a + nd       or
An = b – d
Note:
(i) Sum of n AM’s inserted between a and b is equal to n times the single AM between a and b i.e.
\( \sum\limits_{r\,=\,1}^{n}{{{A}_{r}}}=nA\text{ Where }A=\frac{a+b}{2} \)
(ii) between two numbers
\( =\frac{sum\,of\,m\,AM’s}{sum\,of\,n\,AM’s}=\frac{m}{n} \)

Arithmetic Mean Examples

Example 1:    If 4 AM’s are inserted between 1/2 and 3 then find 3rd AM.
Solution.   Here
\( d=\frac{3-\frac{1}{2}}{4+1}=\frac{1}{2} \)
∴  A3 = a + 3d
\( \Rightarrow \frac{1}{2}+3\times \frac{1}{2}=2 \)

Example 2:    n AM’s are inserted between 2 and 38. If third AM is 14 then find n.
Solution.    Here 2 + 3d = 14 ⇒ d = 4
\( \therefore 4=\frac{38-2}{n+1} \)
⇒ 4n + 4 = 36
⇒ n = 8