JEE Main Test 147

Calculation:

Work done under a conservative force such as gravitation is only dependent on displacement between final & initial points.

From figure, it is clear that the block of mass m, rests on inclined plane due to frictional force. Hence,

The correct answer is Statement I is correct and Statement II is incorrect.
Concept :

• In complex, the magnetism depends on the unpaired electron present in the metal.
• If the unpaired electron is present then the complex is paramagnetic. 
• If no unpaired electron is there then the complex is diamagnetic.

Again crystal field splitting ( Δ_o ) depends on -

• The oxidation state of metal: the higher the oxidation state high is the magnitude of Δ_o.
• Nature of ligand: for strong field ligand magnitude of  Δ_o is higher.
• Principal quantum no. of metal cation: going down the group the magnitude of  Δ_o increases.
• No. of ligand: with increasing the no. of ligands the magnitude of  Δ_o increases.

Again in potassium ferrocyanide, Fe is in a +2 oxidation state, and in potassium ferricyanide Fe is in a +3 oxidation state

Now increasing the oxidation state the crystal field splitting increases

So, crystal field splitting in potassium ferricyanide is greater than in potassium ferrocyanide

Conclusion :

So, Both Statement I is correct and Statement II is incorrect.

Concept:

• Electrophilic aromatic substitution (EAS) is where benzene acts as a nucleophile to replace a substituent with a new electrophile. That is, benzene needs to donate electrons from inside the ring. An electrophile attacks the region of high electron density.
• Activating group is that which increases the rate of an electrophilic aromatic substitution reaction, relative to hydrogen.CH₃ is a perfect example of an activating group; when we substitute a hydrogen on benzene for CH₃, the rate of nitration is increased.
• A deactivating group, on the other hand, decreases the rate of an electrophilic aromatic substitution reaction, relative to hydrogen. The trifluoromethyl group, F₃, drastically decreases the rate of nitration when substituted for a hydrogen on benzene.
• This definition is ultimately based on experimental reaction rate data.  It doesn’t tell us why each group accelerates or decreases the rate. “Activating” and “deactivating” just refers to the effect of each substituent on the rate, relative to H.
• CH₃ is an activating group because of its +I effect, −Cl and −NO₂ are deactivating due to their −I and −M effect. −CH₃ group, having +I effect, increases the electron density in the benzene ring whereas −Cl group having −I effect decreases the electron density the benzene ring.

→ Thus, the increasing order of reactivity of the given compounds towards aromatic electrophilic substitution reaction is

D (-M) < A (-I) < C (+I) < B (+M)

Correct answer: 2)

Concept:

• Any sugar is said to be a reducing sugar if it is capable of acting as a reducing agent because it has a free aldehyde group or a free ketone group.
• The aldehyde group of aldoses is very susceptible to oxidation, whereas ketoses are less so, but can easily be oxidized if, like fructose, they contain an α-hydroxyl and can tautomerize to an aldose.
• Most monosaccharides are reducing sugars.
• This includes all of the common ones galactose, glucose, fructose, ribose, xylose, and mannose. 
• Some disaccharides, such as lactose and maltose are reducing sugars since they have at least one anomeric carbon-free, allowing that part of the sugar to linearize and yield an aldose.

Explanation:

• Sugar is a substance that is crystalline, sweet in taste, and soluble in water.
• A compound that acts as a reducing agent because of the presence of either ketoses or aldoses group forms a reducing sugar.
• All monosaccharides (cannot be hydrolyzed further) come under reducing sugar. 
• Disaccharides which have aldehydic and ketonic groups free, act as reducing sugars.
• Reducing sugars are those species with free - CHO group
• (I) Sucrose - Non-reducing sugar
• (II) Ribose - Reducing sugar
• (III) Lactose - Reducing sugar
• (IV) Fructose - Reducing sugar

Conclusion:

• Thus, Ribose, Lactose, and Fructose are reducing sugar.

The correct answer is Both A and R are true and R is the correct explanation of A

Concept:

Mixtures are combinations of two or more substances that are physically combined but not chemically bonded. In a mixture, the substances retain their individual properties and can be separated by physical means. Mixtures can vary widely in composition and properties, depending on the types and proportions of substances involved. There are two main types of mixtures:

• Homogeneous Mixtures: Also known as solutions, homogeneous mixtures have a uniform composition throughout. The individual components are evenly distributed and not easily distinguishable by the naked eye.
  • Examples include salt dissolved in water (creating a saline solution
• Heterogeneous Mixtures: Heterogeneous mixtures have a non-uniform composition, meaning that the individual components are not evenly distributed. The different substances may exist as distinct phases or layers within the mixture.
  • Examples include a mixture of sand and water.

Explanation:

• Chromatography is a separation technique based on differential affinities of components in a mixture for a stationary phase and a mobile phase. In chromatography, the mixture to be separated is dissolved in a solvent (the mobile phase) and passed through a stationary phase. The stationary phase can be a solid or a liquid, depending on the type of chromatography being used.
• The unique interactions between the components and the stationary and mobile phases enable their separation. Each component interacts differently with the stationary phase (which may be a solid support or a liquid) and the mobile phase (usually a solvent), leading to varying rates of movement and ultimately separation of the components.

Therefore, both the assertion and reason are true, and the reason supports the assertion.

Concept: