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The beginning of a new plant from the seeds is called is called germination.

Oaks, pines and deodars.

Forests, grass land, mountains, ponds and ocean etc.

(1) Ostrich egg is the biggest cell in nature. 

(2) Gills are the respiratory organs of fish. 

(3) Frogs have Webbed feet which help them swim in water.

Birds and Mosquitoes.

(1) A Chicken hatched from an egg, grows into a hen or cock. 

(2) Animals which eat both plants and flesh are called Omnivores.

1. It shows growth 

2. Needs nutrition 

3. It reproduce 

4. It responds to stimuli 

5. It respires

Butter, Leather, Wool, Cooking oil, Salt, Apple.

Plough, mushroom, sweing machine, Radio, boat, water hyacinth, Earthworm. Plough, Sewing machine, Radio, Boat.

Car is an example of a non living things which shows two characteristics of living thing: 

1. Shows motion 

2. Creates sound

Living things have certain common characteristics — they need food, they respire and, excrete, respond to their environment, reproduce, grow and show movement.

1. The leaves in desert plants are either absent or very small. 

2. Leaves are converted into spines which help to reduce loss of water. 

3. The stems become thick, flat and green which help in photosynthesis. 

4. The stem is covered with waxy layer which helps to retain water. In some plants stem is spongy and stores water. 

5. The roots go very deep in the soil to absorb water.

(c) Let the required distance be x km. 

Difference in the times take at the two speeds = 12 min = \(\frac15\) hrs

Given,

\(\frac{x}{25}-\frac{x}{30} = \frac15\) ⇒ \(\frac{6x-5x}{150}\) = \(\frac15\) ⇒ x = 30 km.

Correct option is A) Non-uniform motion

Correct option is B) Curvilnear

Correct option is D) Both A and B

When a stone is tied to a string and whirled it around, it move in a circular path around the hand. That means the stone fallows a curved path with respect to the hand as a fixed centre or axis of rotation. So it is said to be in rotatory motion.

When a stone is tied to a string and whirled it around, we find rotatory motion in it.

Correct option is B) Periodic motion

Correct option is A) Translatory motion

Correct option is: (b) non uniform

Answer is (C)

2x/3 + 1 = 7x/15 + 3

Transposing 7x/15 to L.H.S and 1 to R.H.S, we obtain

2x/3 - 7x/15 = 3 -1

5 x 2x- 7x/15 = 2

3x/15 = 2

x/5 = 2

Multiplying both sides by 5, we obtain

x = 10

L.H.S = 2x/3 +1 = 2 x 10/3 + 1 

= 2 x 10+ 1 x 3/3 = 23/3

R.H.S = 7x/15 + 3 = 7 x 10/15 + 3 = 7 x2/3 + 3

= 14 + 3 x 3/3 = 23/3

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

(i) At first, 

In order to find volume, we will use the following formula:

Volume of a cylinder = \(\pi r^2h\)

Where, 

‘r’ = radius of the base 

‘h’ = height of the cylinder 

Hence,

Volume of the cylinder = \(\pi\)(7)²(50)

= \(\frac{22}{7}\times7\times7\times50\)

= 22 × 7 × 50

= 7700 cm³

Now,

In order to find curved surface area, we will use the following formula:

Curved surface area of cylinder = \(2\pi rh\)

Where, 

‘r’ = radius of the base 

‘h’ = height of the cylinder 

Hence, 

Curved surface area of cylinder

r = \(2\pi rh\)

= \(2\times\frac{22}{7}\times7\times50\)

= 22 × 2 × 50 

= 2200 cm²

Now, 

In order to find the total surface area we will use the following formula: 

Total surface area of cylinder = \(2\pi r(r+h)\)

= \(2\times\frac{22}{7}\times7(7+50)\)

= 22× 2× 57 

= 2508cm²

(ii) At first, 

In order to find volume we will use the following formula: 

Volume of a cylinder = \(\pi r^2h\)

Where, 

‘r’ = radius of the base 

‘h’ = height of the cylinder 

Hence, 

Volume of the cylinder = \(\pi(5.6)^2(1.25)\)

= \(\frac{22}{7}\times5.6\times5.6\times1.25\)

= 22 × 0.8 × 7 × 50 

= 123.2 cm³ 

Now, 

In order to find curved surface area we will use the following formula: 

Curved surface area of cylinder = \(2\pi rh\)

Where, 

‘r’ = radius of the base 

‘h’ = height of the cylinder 

Hence, 

Curved surface area of cylinder = \(2\pi rh\)

= \(2\times\frac{22}{7}\times5.6\times1.25\)

= 22 × 2 × 0.8 × 1.25

= 44 cm² 

Now, 

In order to find the total surface area we will use the following formula: 

Total surface area of cylinder = \(2\pi r(r+h)\)

Where, 

‘r’ = radius of the base 

‘h’ = height of the cylinder 

Hence, 

Total surface area of cylinder = \(2\pi r(r+h)\)

= \(2\times\frac{22}{7}\times5.6(5.6+1.25)\)

= 22 × 2 × 0.8 × 6.85 

= 241.12 cm²

(iii) At first, 

We will convert the radius into metre 

Radius = 14dm = 1.4m 

Now, 

In order to find volume we will use the following formula:

Volume of a cylinder = \(\pi r^2h\)

Where,

r’ = radius of the base 

‘h’ = height of the cylinder Hence, 

Volume of the cylinder = \(\pi (7)^2(50)\)

= \(\frac{22}{7}\times1.4\times1.4\times15\)

= 22 × 0.2 × 1.4× 1.5 

= 92.4cm³

Now, 

In order to find curved surface area we will use the following formula:

Curved surface area of cylinder = \(2\pi rh\)

Where, 

‘r’ = radius of the base 

‘h’ = height of the cylinder 

Hence, 

Curved surface area of cylinder = \(2\pi rh\)

= \(2\times\frac{22}{7}\times1.4\times1.5\)

= 22 × 2 × 0.2 × 1.5

= 132cm²

Now, 

In order to find the total surface area we will use the following formula: 

Total surface area of cylinder = \(2 \pi r(r+h)\)

Where, 

‘r’ = radius of the base 

‘h’ = height of the cylinder 

Hence, 

Total surface area of cylinder = \(2 \pi r(r+h)\)

= \(2\times\frac{22}{7}\times1.4(1.4+1.5)\)

= 22 × 2 × 0.2 × 2.9

= 144.32cm²

Correct option is (A) \(616\,cm^2\)

Total surface area of cylinder \(=2\pi rh+2\pi r^2\)

\(=2\pi r(r+h)\)

\(=2\times\frac{22}7\times7\times(7+7)\)   \((\because r=\frac D2=\frac{14}2\) = 7 cm & h = 7 cm)

\(=44\times14\) \(=616\,cm^2\)

Correct option is  A) 616 cm² 

Volume of the sphere = \(\frac{4}{3}πR^3\)

= \(\frac{4}{3}π{3}^3\)

Volume of the cone = \(\frac{1}{3}πr^2h\)

= \(\frac{1}{3}πr^2\times3\)

Volume of the cone = volume of the sphere

 \(\frac{1}{3}πr^2\times3\)= \(\frac{4}{3}π{3}^3\)

⇒ \(\frac{1}{3}\times{r}^2\times3\) = \(\frac{4}{3}\times{3}^3\)

⇒ r² = \(\frac{{4} \times{3} \times{3}\times{3}\times{3}}{{3}\times{3}}\)

⇒ r² = 36 cm 

⇒ r = 6 cm

Given,

Radius of the copper sphere = 3 cm

We know that,

Volume of the sphere = 4/3 π r³

= 4/3 π × 3³ ….. (i)

Also, given that the copper sphere is melted and recasted into a right circular cone

Height of the cone = 3 cm

We know that,

Volume of the right circular cone = 1/3 π r²h

= 1/3 π × r² × 3 ….. (ii)

On comparing equation (i) and (ii) we have,

4/3 π × 3³ = 1/3 π × r² × 3

r² = 36

r = 6 cm

Therefore, the radius of the base of the cone is 6 cm.

Correct option is (A) 180 cm³

Height of prism is 6 cm.

\(\because\) Base of prism is a triangle with sides 5 cm, 12 cm and 13 cm.

\(\therefore\) Base of prism is a right angled triangle

\((\because5^2+12^2=25+144=169=13^2)\)

\(\therefore\) Base area \(=\frac12\times5\times12=30\,cm^2\)

\(\therefore\) Volume of prism = Base area \(\times\) height of prism

\(=30\times6=180\,cm^3\)

Correct option is  A) 180 cm³

Correct option is (D) 1848 cm²

r = 14 cm

Total surface area of hemisphere \(=3\pi r^2\)

\(=3\times\frac{22}7\times14\times14=66\times28\)

= \(1848\,cm^2\)

Correct option is D) 1848 cm²

(i) Let f (x) = x⁶ - ax⁵ + x⁴ - ax³ + 3x-a + 2 be the given polynomial

From factor theorem,

If (x – a) is a factor of f (x) then f (a) = 0 [Therefore, x – a = 0, x = a]

f (a) = 0

(a)⁶ – a (a)⁵ + (a)⁴ – a (a)³ + 3 (a) – a + 2 = 0

a⁶ – a⁶ + a⁴ – a⁴ + 3a – a + 2 = 0 

2a + 2 = 0 

a = -1 

Hence, 

(x – a) is a factor f (x) when a = -1. 

(ii) Let, f (x) = x⁵ - a²x³ + 2x + a + 1 be the given polynomial 

From factor theorem, 

If (x – a) is a factor of f (x) then f (a) = 0 [Therefore, x – a = 0, x = a] 

f (a) = 0 

(a)⁵ – a² (a)³ + 2 (a) + a + 1 = 0 

a⁵ – a⁵ + 2a + a + 1 = 0 

3a + 1 = 0 

3a = -1 

a = \(\frac{-1}{3}\)

Hence, 

(x – a) is a factor f (x) when a = \(\frac{-1}{3}.\)

We know that, 

Length of the diagonal of the cuboid = \(\sqrt{1^2+b^2+h^2}\)

= \(\sqrt{12^2+9^2+8^2}\)

= \(\sqrt{144+81+64}\)

= \(\sqrt{289}\)

= 17 cm 

Therefore, option B is correct

(c) \(\bigg(\frac{1}{2}\bigg)^\frac13\)

Volume of the bigger cube = (10)³ cm³ = 1000 cm³ 

Volume of the smaller cube = 500 cm³ 

Required ratio = \(\frac{\text{Edge of smaller cube}}{\text{Edge of bigger cube}}\)

= \(\frac{(500)^{\frac13}}{(1000)^{\frac13}}\) = \(\bigg(\frac{(500)}{(1000)}\bigg)^\frac13\) = \(\bigg(\frac{1}{2}\bigg)^\frac13\)

(b) 234 cm²

When one cube is pasted on each of 5 faces of a cube, then 

Surface area of each cube pasted = 5x² 

= 5 × (3)² cm² = 45 cm² 

For 5 faces, total area = 5 × 45 cm² = 225 cm² 

But surface area of the bottom face where the cube is not pasted = x² = 32 = 9 cm² 

∴ Total area = 225 cm² + 9 cm² = 234 cm².

A bowler runs because it wants to give speed to the bowling ball and hence momentum , so that the bowler just leaves the ball when he reaches the line and the ball keeps on moving due to the momentum provided by running of bowler.

B) (A) is correct, (R) is incorrect.

For (a) (1/2)v_z² = gH         v_z = √2gH

Speed at ground = √v_z² + v_z² =  √v_z² + 2gH

For (b) also [(1/2)mv_s² + mgH] is the total energy of the ball when it hits the ground.

So the speed would be the same for both (a) and (b).

Using Resolution of forces IN and 2N and then applying laws of vector addition. Calculate for F₁ & F₂.

F₁ = \(\frac{1}{\sqrt2}\)N,

F₂ = \(\frac{3}{\sqrt2}\) N

in the case of the first body,

u = 0;

t = 5 s;

s = 125 m

Substituting these values in the equation,

s = ut + 1/2 at

125 = 0 × 5 + 1/2 × a × (5)²

1/2 × a × 25

∴ a = (2 x 125/25) = 10 ms^-2

F = m × a = 10 × 10 = 100 N

If the same force acts on another body of mass 15 kg, the amount of acceleration produced is,

a = F/m = 100/15

= 6.67 ms^-2.

Chemical fertilizers: The manure which are prepared in factories by using chemicals and is also called artificial manure.

Homolytic cleavage occurs under the conditions of High temperature.

(a) Statement-I and II are correct and statement-Il is correct explanation of statement-I.

(D) भाखड़ा-नाँगल, चंबल घाटी, नर्मदा घाटी, नागार्जुन सागर

D. Cellulose

PHBV is a bio-degradable

Correct Answer is: D. Cellulose

Biodegradable polymers are polymers which break down after its planned use. For example, PHBV, starch, cellulose, dextron, etc.

Correct Answer is: A. Benzene 

When phenol is heated with Zn dust reduction of phenol occurs and it gets converted into Benzene with ZnO as a by-product. And here's the detailed mechanism; Zn shows an oxidation state of +2.

(2) 21

f(x) = x² + 5

f(- 4) = (-4)² + 5 = 16 + 5 = 21