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Yes, it is possible to get admission in Ranchi University BA without CET UG scores. Students who have not taken the CUET exam can apply via Class 12 results. However, students must note that first preference is given to students with CUET UG scores. Those who wish to apply for the course can visit the official website of the Chancellor's Portal.

Candidates looking for course admission at Cheran College of Nursing can join BSc and MSc course based on merit. Candidates must. Aspirants also need to meet the eligibility criteria to apply for a course. The basic eligibility for degree courses is to complete Class 12, whereas for MSc courses, students need to complete graduation. 

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: D

Two or more compounds with the same formula but distinct structures and characteristics are known as isomers.

So, in this case compounds 

[Co (SO₄) (NH₃)₅]Br and  [Co (SO₄) (NH₃)₅]Cl show no isomerism

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: (A)

Linkage isomerism arises in a coordination compound containing ambidentate ligand.

A simple example is by complexes containing the thiocyanate ligand, NCS-which may bind through the nitrogen to give M-NCS or through sulphur to give M-SCN.

This is a multiple choice answer as classified in NCERT Exemplar

(d) For motion to be in SHM acceleration of the particle must be proportional to negative of displacement.

a $\propto -y$ , so y has to linear.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: A

CFSE for octahedral and tetrahedral complex is related as

Δt = $\frac{4}{9}$ Δ₀

Where Δ₀= CFSE for octahedral complex

Δ₀= CFSE for tetrahedral complex

Δ₀=1800 cm⁻¹

Δt =49×18000=8000 cm⁻¹

Knowing the TN 12th syllabus helps you understand what topics will come in the exam. It helps in planning your studies better and avoiding unnecessary topics. You can also focus more on important chapters that carry more weightage. This saves time and improves your exam preparation. It also helps while practicing model papers and sample tests.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: A

This is due to the complex's octahedral structure and (cis-trans) isomerism. Two Cl ligands are next to each other in trans isomerism. Although these isomers have the same structure, they are not identical.

Possible isomers are:

This is a multiple choice answer as classified in NCERT Exemplar

(b) y = sin³wt

= [3sinwt-4sin3wt]/4

dy/dt= [ $\frac{d}{dt}\left(3sinwt\right)-\frac{d}{dt}\left(4sin3wt\right)$ ]/4

4dy/dt=3wcoswt-4 [3wcoswt]

4 $\times \frac{d^{2}y}{d{t}^2}=-3w^{2}sinwt+12wsin3wt$

$\frac{d^{2}y}{d{t}^2}$ = - $\frac{3w^{2}sinwt+12w^{2}sin3wt}{4}$

$\frac{d^{2}y}{d{t}^2}$  is not proportional to y. hence it is not SHM.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: C

The chelate effect is when a chelating ligand coordinates the stabilisation of a molecule. Any ligand that binds to metal forms a ring with 5 or 6 members, which is the most stable.

Here, CO, CN, H₂O is monodentate ligand [Fe (C₂O₄)₃]^3-  is an oxalate ligand. 

The admission process for Ranchi University BA courses begins with the application process. Students interested in enrolling must apply for their preferred course and institute by visiting the official website of Chancellor's Portal. Once the application window is closed, the admission committee shortlists students based on marks obtained in CUET UG/Class 12.

The shortlisted applicants must report to the allocated university/institute and get their documents verified. After successful verification, candidates are required to pay the admission fee to secure their admission.

Yes, Ahmedabad University application 2025 is still open to UG, PG, and doctoral courses. Online or onsite submission of application is possible by the students via the official university site. The admissions are also based on merit and on entrance basis depending on the course. 

Subjects like Mathematics, Physics, Chemistry, Biology can be high scoring if you understand the concepts well. Regular practice, revision, and solving previous year question papers can help you score better. Subjects like English, Tamil also carry good marks if your grammar and writing skills are strong. Focus on your strengths but don't ignore weak areas.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option:  A

Diamminedichloridoplatinum (II) is also known as azane 

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: D

When 1 mol [CrCl₃ (H₂O)₃]6H₂O is treated with excess of AgNO₃, 3 mol of AgCl is produced, i.e., [CrCl₃ (H₂O)₃]6H₂Ois dissociated in aqueous solution and all three chloride comes in solution.

[Cr (H₂O)₆]Cl₃→ [Cr (H₂O)₆]³⁺  + 3Cl^- 

To manage time well, make a proper study timetable and follow it daily. Start with difficult subjects when your mind is fresh. Divide your TN 12th syllabus into small parts and try to complete each part before the deadline. Keep enough time for revision and practice sample papers. Avoid distractions like mobile phones during study time.

The RIE CEE exam will be conducted on June 29, 2025. The test will be conducted by NCERT in offline mode. 

There are about 40+ top Arts colleges in Tamil Nadu. Of these,  30 colleges are privately owned,  5 colleges are owned by public/government organisations, and public-private entities own 2 colleges. Students can secure admission as per their merit. Some of the best BA colleges in Tamil Nadu include MCC, Loyola College, Stella Maris College, Women's Christian College, PSGCAS, Ethiraj College for Women, Dr. N.G.P. Arts and Science College, St.Joseph's College, Tiruchirappalli,  and many others.

This is a multiple choice answer as classified in NCERT Exemplar

Velocity =dy/dt= $\frac{d}{dt}\left. [3cos\left(\frac{\pi }{4}-2wt\right)\right]$

= 3 (-2w) [-sin ( $\frac{\pi }{4}-2wt$ )]

= 6wsin $\left. [\frac{\pi }{4}-2wt\right]$

Acceleration a = dv/dt= $\frac{d}{dt}\left. [6wsin\left(\frac{\pi }{4}-2wt\right)\right]$

= -4w² [3cos ( $\frac{\pi }{4}-2wt$ )]

A = -4w²y hence acceleration is directly proportional to displacement so it follows SHM

w’= 2w

2 $\frac{\pi }{T}=2w$

T’= 2 $\pi /2w$ = $\frac{\pi }{w}$