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(d) all will take same time.

Explanation: 

Since pure rolling does not occur, the objects slide. 

Therefore on each of them, a frictional force acts against the motion. 

Now, the normal force on each of the objects =mg.cosθ. 

Frictional force = µ*mg.cosθ 

Net force along the incline = mg.sinθ -µ*mg.cosθ 

= m(g.sinθ -µg.cosθ) 

Acceleration of each of the block 

= m(g.sinθ -µg.cosθ)/m = (g.sinθ -µg.cosθ) 

Since the accelerations and the initial velocities are same the time taken to reach the bottom will be the same for all objects. 

Time can be calculated as in the previous explanation.

2³ or 3² 

It can be written as 

2³ =2×2×2=8 

3² =3×3 Hence, 9 is greater than 8 

i.e 3² >2³

Correct match  (A) → (q); (B) → (t); (C) → (p), (r); (D) → (s)

(A) - (q) 3x + 2y + 6z = 70

(B) - (t) 2x - 3y + 6z = 35

(C) - (p),(r) 

(p) 2x + 3y + 2√3z + 11 = 0

(r) 2x + 3y + 2√3z + 1 = 0

(D) - (s) 2x - y + 3z + 8 = 0

(A) → (s), (B) → (p), (C) → (t), (D) → (q)

Explanation :

(A) → (s)

(A) The distance of a point from z-axis is √(x² + y²). The distance of the point from xy-plane is |z|. Therefore x² + y² = z² or x² + y² − z² = 0

(B) Since the sphere touches yz-plane, its radius is |x|. Hence, the equation of the sphere is 

(x - 3)² + (y + 1)² + (z - 2)² = 3²

x² + y² + z² - 6x + 2y - 4z + 5 = 0

(C) The locus is

(x - 2)² + (y + 1)² + (z - 3)² = 16

x² + y² + z² -4x + 2y - 6z - 2 = 0

(D) We have

√(x² + z²) = √((x - 2)² + (y - 1)² + (z + 1)²)

x² + z² = (x - 2)² + (y - 1)² + (z + 1)²

y² - 4x - 2y + 2z + 6 = 0

Mesh analysis provides another general procedure for analyzing circuits, using mesh current as the circuit variables. 

Definition: 

Mesh is a loop which does not contain any other loop within it.

Steps to Determine Mesh Currents:

Step 1: Determine the number of meshes n.

Step 2: Assign mesh current i₁, i₂, , i_n, to the n meshes.

• The direction of the mesh current is arbitrary-(clockwise or counterclockwise)- and does not affect the validity of the solution
• For convenience, we de ne currents flow in the clockwise (CW) direction.

Step 3: From the current direction in each mesh, denote the voltage drop polarities.

Step 4: Apply KVL to each of the n meshes. Use Ohm's law to express the voltages in terms of the mesh current.

Step 5: Solve the resulting n simultaneous equations to get the mesh current.

\(i_{1}=0.222A\)  and  \(i_{2}=0.888A.\)

n(A ∩ B) = 0 

∵ A and B are disjoint net 

∴ n(A ∪ B) = n(A) + n(B) = 15 + 10 = 25.

voltage across 4Ω

V_x = 0.01V.

According to KCL, incoming currents= outgoing currents

I₁ + I₃ = I₂ + I₄

Or, I₃ = -I₁ + I₂ + I₄ = -3 + 4 + 2 = 3 A.

i_{1 =} 0.0196A;

i₂ = 0.01225A;

i₃ = 0.00735A;

i₄ = 0.0196A   and  V₁= 0.098V

Correct answer is (b) 1/4

When two coins are tossed the number of outcomes = 4 {HH, TT, HT, TH}

The favorable outcomes = 1 {HH}

∴ Probability of getting two head

\(=\frac{\text{Favourable outcomes}}{\text{Total no. of outcomes}}\)

\(=\frac{1}{4}\)

Selection of 4 cards from 52 = 52C₄

1. There are 4 suits in a pack of 52 playing cards. They are Club, Spade, Diamond, and Heart. Selecting 4 from each can be done in,

= 13C₄ + 13C₄ + 13C₄ + 13C₄

= 4 × 13C₄ = 4 × \(\frac{13 \times 12 \times 11 \times 10}{1 \times 2 \times 3 \times 4}\) = 2860.

2. There are 12 face cards in a pack of 52 playing cards. Selection 4 cards can be done in 12C₄ = 495.

3. There are 26 red cards and 26 black in a pack of 52 playing cards. Selection of 2 cards should be done from each colour, this can be done in 26C₂ × 26C₂ = (325)² = 105625.

4. Selection of 4 cards from same colour = 26C₄ + 26C₄ = 29900

1. According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions, H⁺(aq) and bases are substances that produce hydroxyl ions, OH^- (aq). For example, HCl is an Arrhenius acid and NaOH is an Arrhenius base. 

2. In aqueous solution the proton bonds to the oxygen atom of a solvent water molecule to give trigonal pyramidal hydronium ion, H₃O⁺ (aq). This is because a bare proton, H+ is very reactive and cannot exist freely in aqueous solutions.

1. An acidic buffer is a buffer solution having pH less than 7. It is prepared by mixing a weak acid and its salt formed with a strong base.

2. Mixture of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) is an example for an acidic buffer. Its pH is around 4.75.

1. A basic buffer is a buffer solution having pH greater than 7. It is prepared by mixing a weak base and its salt formed with a strong acid. 

2. Mixture of ammonium hydroxide (NH₄OH) and ammonium chloride (NH4CI) is an example for a basic buffer. Its pH is around 9.25.

1. pH = – log[H⁺] = – log[3.8 × 10^-3] = 2.42 Since pH is less than 7, it is an acidic solution,

2. HCOOH is more acidic.

K_a value is directly proportional to the acid strength, i.e., greater the K_a value, stronger is the acid.

1. An aqueous solution of (NH₄) SC₄ is acidic in nature. 

2. (NH₄)₂SO₄ is formed from weak base, NH₄OH, and strong acid, H₂SO₄ .

In water, it dissociates completely

(NH₄)₂ SO₄ (aq) → 2NH⁺₄ (aq) + SO^2-₄ (aq)

NH⁺₄ ions undergoes hydrolysis to form NH₄OH and H⁺ ions.

NH₄ +(aq) + H₂O(l) ⇌ NH₄OH(aq) + H⁺(aq)

NH₄OH is a weak base and therefore remains almost unionised in solution. This results in increased H⁺ ion concentration in solution making the solution acidic.

a. GMM of (NH₄)₂ SO₄ 

= (14+4) × 2 + 32 + 4 × 16 = 36 + 32 + 64 

= 132 g

b. Mole = \(\frac{Mass}{GMM}\) = \(\frac{1.32}{132}\) = \(\frac{1}{100}\) = 0.01

number of molecules

= mol x 6.022 x 10²³ = 0.01 x 6.022 x 10²³

number of atoms

= 0.01 x 6.022 x 10²³ x 15 = 0.15 x 6.022 x 10²³(because, the number of atoms in 1 molecules of (NH₄)₂SO₄ is 15)

Triethyl aluminium and titanium tetrachloride inert solvent.

The colour of a particular coordination compound depends on the magnitude of the crystal field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand. In case of [Fe(CN)6]⁴⁻ and [Fe(H₂O)₆]²⁺, the colour differs because there is a difference in the CFSE. Now, CN− is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that the absorption of energy for the intra d-d transition also differs. Hence, the transmitted colour also differs.

(4) CH₃(CH₂)₁₆CH₃

Sulphonates and quaternary ammonium salt of long chain fatty acids are used as synthetic detergent option (4) is just a hydrocarbon chain which is insoluble in water

Two oxygen atoms are involved in peroxide bonding in the H₂S₂O₈ structure, so the oxidation number would be -1each. The remaining atoms of oxygen would have a charge of -2. We can therefore determine the state of oxidation of sulphur (x) as shown in:

H₂S₂O₈

⇒ 2+2x+[-2] × 6+[-1] × 2

⇒ 2x – 12

⇒ 2x – 12 = 0

⇒ 2x = 12

⇒ x = 12/2

x = 6

(i) Vit. B-12, it is a antipernicious anemia factor.
(ii) Determining and estimation of metal.
(iii)EDTA is used in lead poisoning.

(i) CO has high value of crystal field splitting energy than Cl.
(ii) Ni has d⁸ conguration which does not affect by field strength of ligand.
(iii) In this case, Ti⁺³ has t_2g 1 e_g ⁰ configuration. It can perform d-d transition.