NEET Part Test - 1 - NEET MCQ

To find the height above the Earth's surface where the gravitational acceleration is one-fourth of its value at the surface, we can use the formula for gravitational acceleration:

Thus, the height at which the gravitational acceleration is one-fourth of that at the Earth's surface is R_e. 

Heat extracted from source Q1= 700 cal
Heat rejected to the sink Q2= 500 cal
Temperature of source T1= 150 +273= 423K
Temperature of sink T2 to be found:
 Q2/Q1 = T2/T1
So,
T2 = (Q2/Q1) x T1
= (500 x 423)/700
= 302.02 K
= 29.02 degree C.
 

According to Carnot theorem no heat engine working between two given temperatures of source and sink can be more efficient than a perfectly reversible engine working between the same two temperatures, by the statement of Carnot theorem.

We know that dq = dU + dW
And as dU =  60 and dq = 140J
We get dW = 140 - 60 = 80J

The work done against the force of friction 
= μR × displacement = 0.2 × 2 × 9.8 × 2 (in one second)
= (0.2 × 2 × 9.8 × 2) × 5 (in 5 second)
= 39. 2J 
Heat generated = 39.2/4.2 = 9.33 cal

Mass of compound taken = 0.48 g
Mass of boron in sample = 0.192 g
Mass of oxygen in sample = 0.288 g
Percentage of B = 0.192/0.48 x 100 = 40%
Percentage of O = 0.288/0.48 x 100 = 60%

For element Fe, mole of atoms = 69.9/56 = 1.25
For element O, Mole of atoms = 30.1/16 = 1.88
Mole ratio of Fe = 1.25/1.25 = 1
Mole ratio of O = 1.88/1.25 = 1.5
Simplest whole number ratio of Fe and O = 2, 3
Empirical formula of compoimd = Fe₂O₃
Molecular mass of Fe₂O₃ = 160

Molecular formula = Fe₂O₃

(K.E.)₁ = hv₁ - hv₀
(K.E)₂ = hv₂ - hv₀ 
As (K.E)₁ = 2 x (K.E.)₂
∴ (hv₁ - hv₀)0) = 2(hv₂ - hv₀)
or v₀ = 2v₂ - v₁
= 2 x (2 x 10¹⁶) - (3.2 x 10¹⁶) 
= 0.8 x 10¹⁶ Hz or 8 x 10¹⁵ Hz

Mass number of the element = 81
i.e., p + n = 81
Let the number of protons be x.
Number of neutrons 


Symbol of the element

C₆H₁₀(l) + H₂(g) → C₆H₁₂(l) , ∆hydrogenationH° = -119 kJ mol-1
6C(s) + 6H₂(g) → C₆H12(l) , ∆_fH° = -156 kJ mol-1
6C(s) + 3H₂(g)  → C₆H₆(l) , ∆fH° = +49 kJ mol-1
C6H6(l) → C₆H₁₂(l) , ∆hydrogenationH°
(∆_{hydrogenation}H°)_Actual = -156-49 = -2055 kJ mol-1
Hypothetical heat of        = (heat of hydrogenation = 3 × (Heat of hydrogenation 
hydrogenation of benzene   of cyclohexatriene) of cyclohexene)
= 3(-119) = -357 kJmol-1
Here, we have assumed that there are only double bonds in benzene(Kekule structure) and in hydrogenating it, energy released will be roughly 3 times of hydrogenating cyclohexene
Clearly, heat of hydrogenation, for theoretical benzene is more negative and thus it is less stable. So, resonance energy should be negative(since actual benzene is stable)
Resonance energy = -357 -(-205) = -152 kJ mol^-1

The reaction is 
C(diamond)     →     C(graphite)      ∆H = (94-91) = 3 kcal mol-1
∆S = ∆H/T
∆H = (94-91)×2.4×103/12   
= 600 kcal
∆S = 600/298 = 2.013 kcal K^-1

For precipitation, Qrkn > Ksp, then tto establish equilibrium, ions will combine to give molecule as ppt.
So applying the above concept,
 1.8×10^-10 < [Ag⁺][Cl^-]
 1.8×10^-10 < 10^-5 × [Cl^-]
Or [Cl-] > 1.8×10^-5
By seeing the option, only option d satisfies this condtion.

We know that K_p depends only on Temperature. As the temp. remains constant , K_p also remains constant.
α depends on the concentration of the reactant
So as the volume of the reaction container is halved, α also changes
So option d is correct.

The discovery of zero group of elements (inert gases) gave a death blow to the Newland's Law. 

According to Newlands Law of octaves, when the elements are arranged in the increasing order of atomic weights, the properties of every eight element are similar to the first one.

Non-keratinised stratified squamous epithelium covers moist surfaces like buccal cavity, pharynx, vagina, cervix, etc. Adipocytes store fat in adipose tissue. Urinary bladder is lined by transitional epithelium.

A cartilage cell is present in a fluid filled space, the cartilage lacuna. Matrix consists essentially of water. proteoglycans (proteins and carbohydrates), some lipids. collagen, non collagenous protein and collagen fibres. Mature cells are larger in size, more rounded, but still have a few surface projections and are known as chondrocytes. The free surfaces of hyaline cartilage are covered by a fibrous membrane called the perichondrium but it is not present in fibrous cartilage.

The given figure represents areolar tissue.

This type of connective tissue contains various components:

• Macrophage: A type of immune cell that engulfs and digests cellular debris and pathogens.
• Collagen fibres: Strong and flexible proteins that provide structural support.
• Fibroblast: Cells that produce and maintain the extracellular matrix and collagen.
• Mast cell: Cells involved in the inflammatory response and allergic reactions.

Understanding these components helps in recognising the role of areolar tissue in the body.