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Map scales can be represented as

(a) Verbal or statement scale—i.e. the scale is stated in words as 1 cm = 5 km or 1 cm to 5 km. It means 1 cm on the map is equal to 5 km on ground.

(b) Representative fractions — In this system, the numerator expresses the distance on map and denominator represents the actual distance on ground. Both should have same units i.e. 1 cm on map represents 50,000 cm on ground. R.F = distance on map cm / distance or ground cm

(A) December 22

The same two places on the ground scale.

Snow forms in the clouds that are below freezing. Andhra Pradesh is in tropical belt. To get snow the temperatures in Andhra Pradesh are not enough cold. So Andhra Pradesh does not receive any snowfall during winter season.

B) the northern region to the equator

A legend or key is an essential feature of any map. It explainsthe colours, signs, and symbols used in the map. It uses different colours to show the height or depth of an area above or below sea level respectively. It is provided near the top or the bottom of the map, either on the left-hand or right-hand side.

Delhi is located between 28°22″ N. latitude and 28°54″ N. latitude. It is in Temperate Belt. It records low temperatures but there is no snowfall.

The polar region is called the ‘Land of the midnight sun”.

Snow Fall: In the extreme north and on high altitudes there is snowfall instead of rainfall.

Seasons: Seasons occur on the earth due to the differences in temperatures.

There are 3 temperature belts on the earthTorrid, Temperate and Frigid Zones.

The curvature of the Earth: Earth The earth’s surface being curved.

Temperature belts: 

There are three temperature belts on the earth. 

1. Tropical belt 
2. Temperate belt 
3. Polar belt

1 km = 1000 m, 1 m = 100 cm

100 km = 100 × 1000 × 100 = 1,00,00,000 cm

Since RF = Distance on the map in cm

RFscale = 1 / 1,00,00,000 = 1:100,00,000

The factors that influence the order of seasons are:

1. The spherical shape of the Earth and the curvature of its surface. 

2. Daily rotation of the Earth on its own Axis. 

3. The tilt of the Axis of rotation compared to the plane on which the Earth moves 

4. The Earth’s movement around the Sun once a year (revolution).

The earth is rotating in such a speed with all its – atmosphere, human, animal and plant kingdoms. So we don’t feel this.

The proportion which exists between the map and actual surface of the Earth is called the scale. A scale can be expressed in three ways

1. by a statement

2. by representative fraction

3. linear scale or graph

We use twine method to measure the length of a river. In the twine method, a twine is placed along the feature to be measured from one end to the other, carefully following all the curves and bends. The length of the twine is then measured in centimetres or inches using a ruler or linear scale. Thereafter, this length is converted into kilometres or miles using the scale of the given map.

A variety of colourful signs and symbols are used on maps to show natural and manmade features on maps. These signs and symbols give plenty of information and are simple to draw and understand. Some of these symbols are internationally recognized as they have been determined by convention, i.e., these symbols have been agreed upon and accepted internationally. Therefore, they are also called conventional signs and symbols.

The numerator of a representative fraction is always 1.

Scale on a map is the distance shown in the map. The scale is given just below the map. Scale helps us to find out the correct distance between various points on a map. In a scale there is always a proportion between the dimension of the map and the actual area they represent.

The factors are: 

a) The spherical shape of the Earth and the curvature of its surface. 

b) Daily rotation of the Earth on its own Axis.

The meridian of longitude 82^∘30' East, that passes through Allahabad is taken as Indian standard meridian.

The tropic of Cancer passes through the central part of India. 

The meridian of longitude 82^∘ 30' East, which passes through Allahabad is considered as the standard meridian of India for time.

Area wise India is the seventh largest country in the world.

The difference between the longest and shortest day near Kanyakumari is about 45 minutes.

Correct answer is (b) animals

Correct answer is (c) water

Correct answer is (c) whip

Correct answer is (c) middle

This lesson is about the special knowledge given to animals by God. The instinct in animals had saved the lives of men. One morning late in the autumn, John harnessed Black Beauty, the horse to the new cart. The Master, John and Black Beauty proceeded on a long journey. It was very windy and there was a great deal of rain. When they came to the entrance of the toll gate, the man at the toll gate warned them that the river was rising fast. Suddenly an oak tree came crashing down in front of them. The river was flooding and it grew darker. Black Beauty sensed danger, the moment its feet touched the bridge. It made a dead stop. The man at the other end of the toll gate flashed a torch and said that the bridge was broken in the middle. John praised Black Beauty for saving their lives. If the horse had not been wiser, they would have drowned. John gave a good supper and a thick bed of straw to the horse. Black Beauty was grateful for everything, as it was tired

1. John took Black Beauty to the stable.

2. He gave him a good supper and a thick bed of straw that night.

3.Black Beauty was grateful for everything because it was tired.

1. One morning in the autumn they went on a long journey.

2. John harnessed Black Beauty to the new Cart.

3. Yes, the cart was light to pull.

According to the law of conservation of angular momentum, the total angular momentum of a system of particles is constant when the net external torque on the system is zero.

The S.l. unit of angular momentum is Kg m² S^-1.

When b→0, the density becomes uniform and hence the centre of mass is at x = 0.5. Only option (a) tends to 0.5 as b→0.

Vector r = (vector m₁r₁ + m₂r₂/m₁ +m₂).

Given,

Total momentum = (18 kgm/s)\(\hat{I},\)

Velocity of centre of mass = (3 m/s)\(\hat{I},\)

Mass of one object = 4 kg,

Velocity of the objet = (1.5 m/s)\(\hat{I},\)

Let m be the mass of other object and be the velocity.

Now we know total momentum = Total mass × velocity of centre of mass

(18 kgm/s)\(\hat{I},\)= ( + 4)(3 m/s)\(\hat{I},\)

or m = 2 kg

Now, v_cm = \(\frac{m_1v_1+m_2v_2}{m_1+m_2}\)

Or, 3i = (4×1.5i + 2v)/6

So, 18i = 6i + 2v

Or, v = 6i m/s.

Given (1.999)⁵

But the integer nearest to 1.999 is 2,

So, 1.999 = 2-0.001

∴, a = 2 and h = -0.001

Hence, (1.999)⁵ = (2+(-0.001))⁵

Let the function becomes,

f(x) = x⁵………(i)

Now applying first derivative, we get

f’(x) = 5x⁴……….(ii)

Now let f(a+h) = (1.999)⁵

Now we know,

f(a+h) = f(a)+hf’(a)

Now substituting the function from (i) and (ii), we get

f(a+h) = a⁵+h(5a⁴)

Substituting the values of a and h, we get

f(2+(-0.001)) = 2⁵+( -0.001) (5(2⁴))

⇒ f(1.999) = 32+(-0.001)(5(16))

⇒ (1.999)⁵ = 32+(-0.001)(80)

⇒ (1.999)⁵ = 32-0.08

⇒ (1.999)⁵ = 31.92

Hence the approximate value of (1.999)⁵ = 31.92.

y = 4x³ – 2x⁵ dy 

\(\frac{dy}{dx}\)= 12x² – 10x⁴dx 

Let (a, b) be the point on the curve at which the tangent passes through the origin. 

∴ Equation of tangent is 

y – b = (12a² – 10a²) (x – a) 

but this passes through the origin 

∴ 0 – b = (12a² – 10a⁴) (-a) 

b = 12a³ – 10a⁵ ….(1) 

Also from the equation b = 4a³ – 2a⁵ …. (2) 

from (1) and (2) 12a³ – 10a⁵ = 4a3 – 2a⁵ 8a³ = 8a⁵ 

a^{3 (}1 – a²) = 0 ⇒ a = 0, a = + 1 

when a = 0, b = 0, (0, 0) 

a = 1,b= 12(1)- 10(1) = 2, (1,2) 

a = -1, b = 12 (-1) -10 (-1) = -2, (-1, -2) 

Hence the required points are (0,0), (1,2), (-1,-2).

(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5) dy 

\(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10 dx 

slope at (0,5) = -10 

∴ Equation of tangent at (0, 5) is 

y – 5 = -10 (x – 0) 

y – 5 = -10 x 10 x + y – 5 = 0 

slope of the normal at (0,5)

(0, 5) = \(\frac{-1}{-10} = \frac{1}{10}\) 

∴ equation of normal is 

y – 5 = \(\frac{1}{10}\)(x – 0) = 10y – 50 = x 

x – 10y + 50 = 0

(ii) \(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10 dx 

slope of the tangent at x = 1

\(\frac{dy}{dx}\) = 3x² – 6x – 9, slope of the tangent 

since the tangent is parallel to x-axis \(\frac{dy}{dx}\)= 0 

3x² – 6x – 9 = 0 

3(x + 1) (x – 3) = 0, x = 3, x = -1 

when x = 3, y = 27 – 27 – 27 - 7 = -20 

when x = -1, y = -1 - 3 + 9 + 7 = 12 

The points at which the tangent parallel to x – axis are (3, -20) and (-1, 12).

(i) f(x) = |x + 2| – 1 

f(x) = |x + 2| – 1 ≥ -1 

minimum value is – 1 when x + 2 = 0, x = -2 

however it has no maximum value. 

(ii) g(x) = – | x + 1| + 3 = 3,-1 x + 1| 

g(x) < 3 v x + 1 = 0 

∴ max. value is 3 

when x = – 1 

how ever no minimum value. 

(iii)  h(x) = sin (2x) + 5

maximum value of sin 2x = 1 and minimum value is -1 

f(x) = 1 + 5 is 1 + 5 

∴ max. h (x) = 6 and 

min. h (x) = 4. 

(iv) f(x) = |sin 4x + 3| 

-1 < sin 4x < 1 ⇒ 3 -1 < sin 4x + 3 < + 1 + 3 + 2′< sin 4 x + 3 < 4 

2 < | sin 4x + 3 | < 4 

f(x) > 2 and f (x) < 4 

min. f(x) = 2 when sin 4x + 3 = 0 

max f (x) = 4 when sin 4x + 3 = 0 

∴ minimum value is 2 at sin 4x = -1 

maximum value is 4 at sin 4x = 1 

(v) h(x) = x + 1, x ∈ (-1,1) 

given that x ∈ (-1, 1) 

i.e. -1 < x < 1 

-1 + 1< x + 1 < 1 + 10 < x + 1< 2 

∴ x + 1 > 0 or x + 1 < 2 

x + 1 > 0 so no minimum value 

x + 1 < 2, so no maximum value.

(i) f(x) = e^x 

f’(x) = e^x 

f'(x) > 0 ∀ x ∈ R 

hence function has no critical point There is no point at which the function is maximum or minimum. 

(ii) g(x) = log x, x > 0 

g'(x) = \(\frac{1}{x}\), where x > 0 

hence the function has no critical point 

∴ There is no point at which the function is maximum or minimum. 

(iii) h(x) = x³+ x²+ x +1 

h’(x) = 3x² + 2x + 1 

h’(x) = 0 ⇒ 3x² + 2x + 1 = 0, 

x has no real value, hence there is no critical point. 

∴ For no point the function has max. or min. value.

(i) 3-methylphenol

(iii) 2-methoxypropane

(iii) b < c < a

Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on an object. The mass of a substance is constant whereas its weight may vary from one place to another due to change in gravity. The mass of a substance can be determined very accurately by using an analytical balance.