Step 1: Start with the formula for potential energy in S.H.M.

The potential energy PE is given by:

PE = (1/2) mω²x²,

where:

• m is the mass of the particle,
• ω is the angular frequency,
• x is the displacement.

From the question, the potential energy is given as:

PE = 0.1π²x².

Equating:

(1/2) mω²x² = 0.1π²x².

Cancel x² from both sides:

(1/2) mω² = 0.1π².

Substitute m = 20 g = 0.02 kg:

(1/2) (0.02) ω² = 0.1π².

Simplify:

0.01 ω² = 0.1π².

ω² = 10π².

Step 2: Solve for angular frequency and frequency.

Take the square root:

ω = √(10π²) = π√10.

The frequency f is given by:

f = ω / (2π).

Substitute ω = π√10:

f = (π√10) / (2π) = √10 / 2.

Using √10 ≈ 3.162:

f = 3.162 / 2 = 1.581 Hz.

Thus, the frequency of S.H.M is 1.581 Hz.

The radiant energy E emitted by a star is given by:

E = σ ε A T⁴.

For spherical stars:

A ∝ R²,

where R is the radius. Hence:

E ∝ R²T⁴.

Step 1: Write the energy ratio.

For two stars, the ratio of radiant energies is:

E₁ / E₂ = (R₁²T₁⁴) / (R₂²T₂⁴).

Given:

E₁ = 3 × E₂.

Substitute:

(3 ⋅ E₂) / E₂ = (R₁² ⋅ T₁⁴) / (R₂² ⋅ T₂⁴).

Simplify:

3 = (R₁² (6000)⁴) / (R₂² (2000)⁴).

Step 2: Simplify the temperature ratio.

The ratio of temperatures is:

T₁ / T₂ = 6000 / 2000 = 3.

(6000/2000)⁴ = 3⁴ = 81.

Substitute back:

3 = (R₁² / R₂²) ⋅ 81.

Simplify:

(R₁² / R₂²) = 3 / 81 = 1 / 27.

Multiply through by 81:

(R₁² / R₂²) = 3 / 81 = 1/27

(R₁² / R₂²) = 1/ (1/3)⁴ = 81

(R₁² / R₂²) = 3 * 81

(R₁² / R₂²) = 3 * 81= 121

Step 3: Take the square root.

(R₁ / R₂) = √243 ≈15

10000 = (R₁²/R₂²)(1/81)

(R₁² / R₂²)= 10000*81 = 900²

(R₁ / R₂) = √900²

R₁ / R₂ = 900:1.

Thus, the ratio of their radii is:

R₁ : R₂ = 900:1.

The wavelength (λ) of a wave is the distance between two consecutive crests. When there are 11 crests, the distance between them corresponds to 10 wavelengths:

10 λ = 1 m.

Thus, the wavelength is:

λ = 1 / 10 = 0.1 m.

The frequency (f) of a wave is given by the formula:

f = v / λ,

where:

• v = 30 m/s is the speed of the wave,
• λ = 0.1 m is the wavelength.

Substitute the values:

f = 30 / 0.1 = 300 Hz.

Thus, the frequency of the wave is 300 Hz.

Step 1: Recall the frequency of a closed organ pipe.

The fundamental frequency of a closed organ pipe is given by:

f_closed = v / (4L_closed),

where:

• v is the speed of sound in air,
• L_closed is the length of the closed organ pipe.

Substitute L_closed = 20 cm = 0.2 m:

f_closed = v / (4 ⋅ 0.2) = v / 0.8.

Step 2: Recall the frequency of the second overtone of an open pipe.

For an organ pipe open at both ends, the frequency of the second overtone (third harmonic) is:

f_{open, overtone} = (3v) / (2L_open),

where L_open is the length of the open pipe.

Step 3: Equate the frequencies.

Given that the fundamental frequency of the closed pipe is equal to the second overtone of the open pipe:

v / 0.8 = (3v) / (2L_open).

Simplify by canceling v from both sides:

1 / 0.8 = 3 / (2L_open).

Rearrange to find L_open:

L_open = (3 ⋅ 0.8) / 2 = 1.2 m.

Thus, the length of the open organ pipe is 1.2 m.

The shunt resistance S is connected in parallel with the galvanometer resistance G to extend its range. The total current I is divided into two parts:

• I_g: The current passing through the galvanometer.
• I_s: The current passing through the shunt.

From the given data:

I_g = 8% × I = 0.08I.

Thus, the current through the shunt is:

I_s = I - I_g = I - 0.08I = 0.92I.

Step 1: Use the relationship between S and G.

The voltage across S is equal to the voltage across G:

I_s ⋅ S = I_g ⋅ G.

Substitute I_s = 0.92I and I_g = 0.08I:

0.92I ⋅ S = 0.08I ⋅ G.

Simplify by canceling I:

0.92S = 0.08G.

Solve for S:

S = (0.08G) / 0.92.

Step 2: Simplify the fraction.

S = (8G) / 92 = (2G) / 23.

Thus, the shunt resistance is:

S = (2G) / 23.

The electrostatic potential energy of a system of point charges is given by:

U = ∑_{i < j} (kq_iq_j) / r_ij,

where:

• k is the Coulomb constant,
• q_i and q_j are the charges,
• r_ij is the distance between the charges.

Step 1: Identify charge pairs.

In the given system, there are three charges +q at the vertices of an equilateral triangle of side a. The potential energy is due to the pairwise interactions between the charges. The charge pairs are:

1. Between charges at vertices 1 and 2,
2. Between charges at vertices 2 and 3,
3. Between charges at vertices 3 and 1.

Step 2: Calculate the potential energy for one pair.

The potential energy for one pair of charges is:

U_pair = (kq²) / a.

Step 3: Sum the potential energy for all pairs.

Since there are three pairs of charges in an equilateral triangle:

U_total = 3 ⋅ U_pair = 3 ⋅ (kq²) / a.

Thus:

U_total = 3kq² / a.

The time period of a simple harmonic motion (SHM) is given by:

T = 2π √(m / k),

where:

• T is the time period,
• m is the mass,
• k is the spring constant.

Step 1: Initial time period.

The time period with mass m is:

T₁ = 2 = 2π √(m / k).

Squaring both sides:

T₁² = (4π²m) / k.

Rearranging for m:

m = (kT₁²) / (4π²).

Step 2: New time period with added mass.

The new time period is T₂ = 5 s, and the total mass becomes m + 0.04:

T₂² = (4π²(m + 0.04)) / k.

Rearranging:

m + 0.04 = (kT₂²) / (4π²).

Step 3: Subtract initial from new equation.

From the two equations:

(kT₂² / 4π²) - (kT₁² / 4π²) = 0.04.

Factorize:

(k / 4π²)(T₂² - T₁²) = 0.04.

Substitute T₁ = 2 s and T₂ = 5 s:

(k / 4π²)(5² - 2²) = 0.04.

Simplify:

(k / 4π²) ⋅ (25 - 4) = 0.04.

(k / 4π²) ⋅ 21 = 0.04.

Solve for k / (4π²):

k / (4π²) = 0.04 / 21.

Step 4: Substitute back to find m.

Using the expression for m:

m = (kT₁²) / (4π²).

Substitute T₁ = 2 s:

m = (0.04 / 21) ⋅ 4.

Simplify:

m = (0.04 ⋅ 4) / 21 = 0.00761 kg.

Convert to grams:

m = 7.64 g.

Thus, the mass is 7.64 g.

The term 'root pressure theory' was coined by J. Priestley. This theory explains the process by which pressure is generated in the roots of plants, pushing water upward through the xylem vessels. Root pressure is a result of osmotic forces created by active transport of ions into the root xylem.

Key Points:

• Root pressure is most significant during the night or in conditions of low transpiration.
• It contributes to the upward movement of water in small plants, but it is insufficient to explain water transport in tall trees.

In Drosophila, the normal wing phenotype is determined by the presence of at least one dominant Vg⁺ allele. The recessive alleles Vgⁿⁱ, Vg^no, Vg^st, Vg, etc., cause abnormal wing development only when present in homozygous recessive combinations.

Step 1: Evaluate each genotype.

1. Vg⁺Vg⁺: Both alleles are wild type, so the wings are normal. Normal wings.
2. Vg⁺Vgⁿⁱ: One dominant Vg⁺ allele is sufficient for normal wings. Normal wings.
3. Vg⁺Vg^no: One dominant Vg⁺ allele is sufficient for normal wings. Normal wings.
4. Vg⁺Vg^st: One dominant Vg⁺ allele is sufficient for normal wings. Normal wings.
5. Vg⁺Vg: One dominant Vg⁺ allele is sufficient for normal wings. Normal wings.

Step 2: Count the genotypes.

All five genotypes contain at least one dominant Vg⁺ allele and hence represent normal wings.

Genotypes representing normal wings = (i), (ii), (iii), (iv), (v).

Thus, the correct answer is:

(D) (i), (ii), (iii), (iv), and (v).

Statement I: Cell wall is freely permeable.

The cell wall in plants is freely permeable to water and solutes. This allows the movement of substances through the apoplast pathway, ensuring easy transport across the cell wall. Hence, Statement I is correct.

Statement II: Plasma membrane is selectively permeable.

The plasma membrane regulates the entry and exit of substances into the cell. It is selectively permeable, allowing only specific molecules (e.g., ions, nutrients) to pass through, while restricting others. Hence, Statement II is correct.

Conclusion:

Since both statements are correct, the correct answer is:

(B) Both Statement I and Statement II are correct.

Friedrich Miescher was the first scientist to discover DNA in 1869. He isolated a substance from the nuclei of pus cells, which he callednuclein.
Later, it was identified as DNA. Although Miescher did not fully understand its function, his discovery laid the foundation for future studies on genetic material.

Key Points:

• Watson and Crick discovered the double-helix structure of DNA.
• Rosalind Franklin contributed significantly to the structural discovery through X-ray diffraction studies.
• Griffith's experiments demonstrated the principle of transformation, showing that DNA is the genetic material.

Thus, the discovery of DNA itself is credited to F. Miescher.

RNA (Ribonucleic Acid) differs from DNA in its nucleotide composition. The nitrogenous bases in RNA include:

• Adenine (A),
• Guanine (G),
• Cytosine (C),
• Uracil (U).

Thymine (T) is not present in RNA. Instead, Uracil (U) takes its place and pairs with Adenine during transcription.

Key Points:

• Thymine is exclusively found in DNA.
• Uracil is specific to RNA and is chemically similar to Thymine but lacks a methyl group.

Thus, the correct answer is Thymine.

The dipole moment of a molecule depends on the molecular geometry and the net vector sum of individual bond dipole moments.

Analysis of each option:

1. CCl₄: The molecule is tetrahedral and symmetric. The individual bond dipoles cancel out, resulting in a net dipole moment of zero.
2. CO₂: The molecule is linear and symmetric. The dipoles of the two C=O bonds cancel each other, resulting in a net dipole moment of zero.
3. BF₃: The molecule is trigonal planar and symmetric. The dipoles of the three B-F bonds cancel each other, resulting in a net dipole moment of zero.

Since all the given molecules have zero dipole moments, the correct answer is (D) None of these.

Reduction involves the gain of electrons. The given reaction can be written as:

Cr³⁺ + 3e^- → Cr⁰.

From the reaction:

• Each Cr³⁺ ion requires 3 electrons (3e^-) to be reduced to Cr⁰.
• Therefore, for 1 mole of Cr³⁺, 3 moles of electrons are required.

Key Points:

• The oxidation state of chromium changes from +3 in Cr³⁺ to 0 in Cr⁰.
• The number of moles of electrons required equals the change in oxidation state.

Thus, the answer is 3 moles of e^-.

Bakelite is a thermosetting polymer and one of the first synthetic plastics. It is synthesized through a condensation reaction between the following monomers:

• Phenol (C₆H₅OH), and
• Formaldehyde (HCHO).

The reaction involves the formation of a phenol-formaldehyde network, resulting in a hard, durable polymer known as Bakelite.

Key Points:

• Bakelite is widely used in electrical insulators and household items due to its high mechanical strength and heat resistance.
• Phenol and formaldehyde undergo a condensation reaction to form a polymer with cross-linked structures.

Thus, the monomers of Bakelite are Phenol and Formaldehyde.