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Correct Answer is: (b) QE₀/mω²

For a particle undergoing SHM with an amplitude A and angular frequency ω, the maximum acceleration = ω²A.

Here, the maximum force on the particle = QE₀.

∴ its maximum acceleration = QE₀/m = ω²A

or A = QE₀/mω².

1. The organisms: 

A - Tape worm 

B – Ascaris 

C – Earthworm 

2. Features: 

• Tapeworm – Body is dorsoventrally flattened and segmented. 
• Ascaris – Body is cylindrical and vermiform 
• Earthworm – Body is divided into similar segments and clitellum is present.

1. Figure C is the body cavity of earthworm. Because it is a true coelom 

2. The names of germ layers 

• a-ectoderm 
• b-endoderm

1. Radula: File like rasping organ for feeding.

 Phylum: Mollusca 

2. Flame cells: Osmoregulation and excretion 

Phylum: Platyhelminthes 

3. Parapodia: help in swimming 

Phylum: Annelida

1. Characters: 

• a – Gillslits are separate and without operculum 
• b – Placoidscale 
• c – Mouth is located ventrally 
• d – heterocercal caudal fin

2. Chondrichthyes

1. Cnidaria 

2. Alternation of sexual and asexual forms of organism: ie., Polyp asexually produce medusa, medusa sexually product polyp.

1. Echinodermata 

• eg: Starsh 

2. Class – Cyclostomata 

• eg: Petromyzon 

3. Annelida 

• eg: Earthworm

(i) B – Nereis 

(ii) B – Ctenoplanna

C – Ctenophora 

(iii) A – radula

1. Annelida 

2. Mollusca 

3. Platyhelminthes 

4. Arthropoda

1. Ctenophora  

2. Echinodermata

1. Organism in (A) and (B) 

A – Bony fish – Osteichthyes 

B – Cartilagenous fish – Chondrichthyes

2.

Porifera → Cnidaria → Ctenophora→ Platyhelminthes → Aschelminthes → Annelida → Arthropoda → Mollusca → Echinodermata

1. Exocoetus – Oesteichthyes 

2. Trygon – Chondrichthyes

(b) Phylum Arthropoda

1. Porifera – Choanocytes 

2. Coelenterata – Cnidoblast 

3. Platyhelminthes – Flame cells 

4. Annelida – Hirudin 

5. Arthropoda – Jointed legs 

6. Mollusca – Radula 

7. Echinodermata -Dermal ossicles 

8. Chordata – Notochord

1. Cnidoblastornematocyst 

2. Cnidaria 

3. Capture of Prey 

• Defense organ.

1. Chordata 

2. Notochord, Dorsal nerve cord, Pharyngeal gill slits, Post anal tail 

3. In Urochordatarotochond is present only in larval tail. In Cephalochondata, notochord extends from head to tail region is persistent throughout their life. In vertebrata, the notochord is replaced by a cartilagenous or bony vertebral column in the adults.

Tapeworm show false segmentation. In true segmentation number of segments is fixed and age of the segments are all same.

(a) Hydra/Obelia 

(b) Shark 

(c) Spongilla 

(d) Hydra/Obelia

1. Cnidaria 

2. Mollusca 

3. Annelida 

4. Arthropoda

Answer: 100 ppm

Hardness due to CaCO₃ = (10⁻³ × 10³) × 100 

= 100 ppm

Correct answer is 03

x = \(\frac{3R}{8}\) = 3cm

(3) Inhibits the action of histamine receptor 

Seldane is an antihistamine drugs it inhibits the action of histamine receptor.

Solution: A/√2 from the mean position. where A is amplitude.
We know
Potential energy = (1/2) kx² = (1/2) mω ²x²
Kinetic energy = (1/2) mv² = (1/2) m[ω √ (A²-x²)]² = (1/2) mω ²(A²-x²)
Where x is the required position about the mean position,
A is amplitude of SHM,
m is mass of particle executing SHM,
ω is the angular frequency of the particle.

Given, P.E. = K.E. & A = 10√2
so, (1/2) mω ²x² = (1/2) mω ²(A²-x²)
so, x² = A²/2
so, x = A/√2

so, x = 10√2​/√2​ = 10cm

Hence, at distance x=10cm from the mean position, the K.E is equal to P.E

Solution:

Given that,Tension in string, T = 40N , mass of string, m = 5gm = 0.005 kg ,length of the string, l = 10 m ,  v = ?

We know that velocity of a transverse wave on a string is given by,  v = √(T/μ)

where, t is tension in string & μ is the mass per unit length

μ = mass/length = 0.005/10 = 0.0005 kg/m

v = √(40/0.0005) = √80000 = 2√2 x 10² m/sec

Hence, Velocity v = 2√2 x 10² m/sec

Derry’s mother did not have a good opinion about Mr. Lamb as she had heard many things about him. She did not want Derry to go and meet him. She believed Mr. Lamb to be a dangerous person who lived alone in a big home.

Solution:

Let the unknown frequency be f.

So, before filling f – 310 = 4 Hz => f = 314 Hz

And, after filling, 310 – f = 4 Hz => f = 306 Hz

Thus, the required frequency is 314 Hz.

Correct option is C) brings about

Correct option is B) brought on

1. Great people show their best worth when their soul is touched to their innermost depth.

2. When a lion is severely wounded, he gives out the most frightening roars.

3. Surrounding darkness can stop our marching feet.

4. A friend or a lover can turn faithless.

5. Friendship fail and love betray : Personification; And clotted darkness : Metaphor.

1. A citizen of India should be armed with technology and knowledge.

2. By working hard we can transform our nation into a developed nation.

3. Every Indian citizen should have a vision to see this great nation, as a developed country with vigorous economic growth based on moral values.

4. The ignited soul is the most powerful resource to make India a developed nation.

5. We should keep the lamp of knowledge burning to achieve this vision.

1. A citizen of India should be armed with technology and knowledge.

2. By working hard we can transform our nation into a developed nation.

3. Every Indian citizen should have a vision to see this great nation, as a developed country with vigorous economic growth based on moral values.

4. The ignited soul is the most powerful resource to make India a developed nation.

5. We should keep the lamp of knowledge burning to achieve this vision.

Correct option is B) brought back

-According to the poet, we should stop all our activities for a moment. Let us give up using all our machines and engines for a short while because most of our troubles are due to our hurry and becoming rash. We should do introspection and keeping still will pave our path for peace.

a) Both assertion and reason are correct and reason is the correct explanation of assertion.