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DU admissions offer Sports and ECA quota admissions to students who excel at their chosen sport or extra-curricular skill. According to the official guidelines, the weightage allotted for the CUET test is 25%, while the certificates and trials as part of the Sports/ECA quota have a weightage of 75%. Hence, students must be properly prepared for both the stages of admission.

I think you are not able to get admission in BAMS in this mark 

You get admission in private BAMS college 

The following are the possible ways:

I.   Kolkata – Mumbai – Nagpur

II.   Kolkata – Lucknow – Jammu – Nagpur – Mumbai

There are two possible ways.

No, Madan Mohan Malviya University of Technology does not conduct any entrance exam for admission to its BBA course. The university only accepts valid CUET UG marks. Students who have appeared for the exam and also fulfil the eligibility criteria can apply for the course. The applications can be submitted via the university's official website.

$\theta =\frac{60°}{2}=30°$

$R_{eq}=3\Omega$

Once the MMMUT BBA counselling process is over, students who receive the admission offers must pay the first-year fees. In general, at this stage, students are required to pay an admission fees. Along with this, students must also pay some part of the tuition fees. Those who wish to learn more can visit the official website of the university.

Kindly go through the solution 

 $\frac{T_1}{400}=\frac{300}{240}\Rightarrow T_1=500K$

The MMMUTBBA admission process begins with the submission of the application form. To be able to apply successfully, students must follow the below-mentioned steps:

1. Go to the university's official website.

2. Click on the 'Apply Now' option under the 'Admissions' tab.

3. Fill in the details to register and generate login credentials.

4. Log in to fill out the application form.

5. Upload the documents, pay the application fees, and submit the form.

  $R_{eq}=6?$

$\therefore i=\frac{12v}{6\Omega }\Rightarrow 2A$ (through battery)

$\therefore v_{15\Omega }=6V$

When switch is open,

$R_{eq}=\frac{3R}{2}$              

When switch is closed,

$R_{eq}^{\text{'}}=2\times \frac{R\times 2R}{R+2R}=\frac{4R}{3}$           

$\frac{R_{eq}}{R_{eq}^{\text{'}}}=\frac{3R/2}{4R/3}=\frac{9}{8}=\frac{x}{8}$

$\Rightarrow x=9$

MMMUT facilitates admission to its BBA course via a counselling process. The application portal for counselling is open. Candidates seeking admission must apply and participate in the counselling process. Moreover, as per the admission schedule, the last date to apply was 23 Jul, 2025. For more information, students can visit the university's official website.

Since Task T1 does not have either the first or last Task on Tuesday Task T1 must have the first or last Task on Monday. So, option (a) or option (b) is true but neither one must be true.

On Tuesday since Task T3 is first Task T7 must be second but Task T2 cannot follow Task T7. So Task T2 must have been done on Monday, hence (c) is eliminated. Since Task T4 has to be done on Tuesday Task T6 must have been done on Monday.

The overall SIMS Pune average package offered in 2025 placements is yet to be released on the official website. Meanwhile, check out the top 10%, 25% and 50% average packages offered during SIMS Pune placements 2025 in the table below:

Since Task T1, T4 and T5 has to be done on Monday. Therefore Task T2, T3, T6, T7 are to be done on Tuesday. Task T3 will be the first Task.

Task T7 must be the second Task according to the condition.

Task T6 must be the third Task and Task T2 will be the last Task according to the condition.

No, admission to the MMMUT BBA course is not possible without a valid CUET UG score. Thus, candidates seeking admission to the BBA course at the university must take the CUET UG exam. The candidate has to opt for the following subject papers in the exam:

Time taken by first drop to reach the ground,

$t_1=\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{2\times 9.8}{9.8}}=\sqrt[]{2}s$              

Time for which second drop has fallen,

$t_2=t_1-\frac{t_1}{2}=\sqrt[]{2}-\frac{1}{\sqrt[]{2}}s$

Height of the second drop

$=9.8-\frac{1}{2}\times 9.8\times {\left(\sqrt[]{2}-\frac{1}{\sqrt[]{2}}\right)}^2=7.35m$