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Yes, many colleges may admit students for UG Forensic Medical Science based on Class 12 results in any human Science (science stream). You can see on their websites what there admission to Diploma or undergraduate courses on forensic medical Science will be without the need for an entrance exam. 

That said, there may be some reputable universities (or postgraduate programme) that may require you to take an internal exam or interview, so please carefully look at the eligibility when considering application to the institute.

The NEET PG, which is required for MD in Forensic Medicine is taken online as a computer-based test, without an entrance exam format through out India, as in any other subject as it is assess for medical graduates with questions across subjects including forensic medicine. 
For non-MBBS, and courses such as BSc for forensic medical Science (or MSc), the entrance exams (if entrance exam is given), will differ by institution and can be online or offline depending whether you can do it in whatever format from the university.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation-A = i+j

 B = i-j

 A.B=|A|B|cos $\theta$

(? +? ). (? -? ) = $\left(\sqrt[]{1+1}\right)\left(\sqrt[]{1+1}\right)$ cos $\theta$

Where $\theta$  is the angle between A and B

Cos $\theta$ = $\frac{1-0+0-1}{\sqrt[]{2\sqrt[]{2}}}$ = $\frac{0}{2}=0$

$\theta$  = 90^o

Yes, there is a negative marking scheme in BOPEE nursing exam. If the candidate choses correct option, he/she will be awarded +1 mark. However, if the candidate choses incorrect option -0.25 will be deducted per question. Hence, candidates must choose the options carefully to avoid the negative scoring.

The BOPEE BSc nursing exam carries 180 questions of 1 mark each. These 180 questions will be divided in three parts, i.e. 60 questions each in three sections. The exam will be OMR based, therefore candidates are advised to fill the choices carefully.

There is also a negative marking scheme which would lead to deduction in marks in case the candidate chose incorrect option. There is a negative marking of 0.25 mark per question for each incorrect response chosen at the time of answering.

The BOPEE BSc nursing exam will be conducted in offline mode. The question paper will carry 180 objective-type questions of 1 mark each. There will be three sections in the exam, namely Physics (60), Chemistry (60), Biology (60). Each section will be of 60 marks. Candidates will be provided 180 minutes (i.e. 3 hours) to attempt the question paper solutions. This means 1 minute per question will be allotted to the aspirants.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- given |A|=2 and |B|=4

a)|A $\times B$ |=AB cos $\theta$ = 0

so 2 $\times 4$ cos $\theta$ =0

so $\theta =90$  so it matches with ii

B) |A $\times$ B|= ABcos $\theta$ =8

2 $\times 4$ cos $\theta$ =8

So $\theta$ = 0 so it matches with option i

c) |A $\times$ B|= ABcos $\theta$ =4

so $\theta$ =60 so it matches with option iv

d) |A $\times$ B|= ABcos $\theta$ =-8

so $\theta$ =180 so it matches with option iii

The BOPEE BSc nursing application process for the current academic year has closed. However, candidates can expect the JKBOPEE BSc nursing registration to begin in March 2026 akin to this year. The JKBOPEE BSc nursing application forms are made available generally for a period of 30 days. Last year, the JKBOPEE BSc nursing application fee was INR 1000/-. However, a slight increase in the JKBOPEE BSc nursing application fee can be expected in the next year. The official confirmation of all the details of registration will be done at the official website, jkbopee.gov.in by the exam organizing body.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- given |A|=2 and |B|=4

a)|A $\times B$ |=AB sin $\theta$ = 0

so 2 $\times 4$ sin $\theta$ =0

so $\theta =0$ so it matches with iv

B) |A $\times$ B|= ABsin $\theta$ =8

2 $\times 4$ sin $\theta$ =8

So $\theta$ = 90 so it matches with option iii

c) |A $\times$ B|= ABsin $\theta$ =4

so $\theta$ =30 so it matches with option i

d) |A $\times$ B|= ABsin $\theta$ =4 $\sqrt[]{2}$

so $\theta$ =45 so it matches with option ii

Invertis University Bareilly is one of the leading private universities in Bareilly to offer quality education to students, mainly in the field of Engineering. The university has been accredited by NAAC and is a member of AIU. Further, Invertis University is approved by the State Government, Uttar Pradesh, UGC, NCTE, PCI,  BCI,  AICTE, and Institution Innovation Council of the Ministry of Education (MIC).

Explanation – here A and B vectors are joint by head and tail. So C= A+B

(a) from fig iv it is clear that c=a+b

(a) from fig iii it is clear that c+b=a so a-c=b

(b) from fig I it is clear that b=a+c so b-a =c

(c) from ii it is clear that -c= a+b so a+b+c=0

TNEA (Tamil Nadu Engineering Admissions) counselling is a centralized online process used for allotment to students seeking admission to BE/BTech and othe related courses in the state of Tamil Nadu. It is the way by which students get admitted to government, government-aided, and some self-financing engineering colleges in the state. 

Yes, BOPEE BSc nursing counselling will be held for admission to both government and private colleges. Seat matrix has been provided in the BOPEE official prospectus. Candidates can check the seats available in the location where they want to seek admission. There are total 2368 seats available in the BOPEE participating institutes for the BSc nursing programme.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – a) radius of earth =6400km= 6.4 $\times {10}^{6}m$

Time period = 1 day = 24 $\times 60\times 60$ = 86400s

Centripetal acceleration a= w²r= R(2 $\pi /T$ )²=4 $\pi$ ²R/T

= $\frac{4\times {\left(\frac{22}{7}\right)}^2\times 6.4\times {10}^6}{({24\times 60\times 60)}^2}$ = 0.034m/s²

b) time = 1yr=365 $\times 24\times 60\times 60$ days= 365=3.15 $\times {10}^{7}s$

centripetal acceleration = Rw²= $\frac{4{\pi }^{2}R}{T^2}$

^{$=\frac{4\times {\frac{22}{7}}^2\times 1.5\times {10}^{11}}{({3.15\times {10}^7)}^2}=5.97\times {10}^{-3}m/s$ 2}

$\frac{a_c}{g}=\frac{5.97\times {10}^{-3}}{9.8}=\frac{1}{1642}$

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- when it be at position P, drops a bomb to hit a target T

Let $\theta$

Speed of the plane =720km/h = 720 $\times 5/18$ = 200m/s

Altitude of the plane P’T = 1.5km= 1500m

If bomb hits the target after time t then horizontal distance travelled by the bomb PP’=u $\times t$ =200t

Vertical distance travelled by the bomb P’T=1/2gt²

1500 = ½ (9.8)t²

So t²= 1500/4.9, t = $\sqrt[]{\frac{1500}{4.9}}=17.49s$

PP’=200 (17.49)m=

tan $\theta$ =P’T/P’P=1500/200 (17.49)=0.49287= tan23⁰12’

A random number will be assigned to all the applicants for the breaker. In case all other tie-breaking criteria for determining rank in the TNEA merit list are unsuccessful, a 10-digit random number will be assigned to each applicant to break the ties. The higher the value of the random number, the higher the preference will be in the merit list. To access the TNEA random number, candidates must log in using their registration number and date of birth.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – due to air resistance particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper then rise. When we are neglecting air resistance path is parabola when we consider air resistance then path is asymmetric parabola.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – the boy throws the ball at an angle of 60.

Horizontal component of velocity 4cos $\theta$ = 10cos60

=10 (1/2)

=5m/s.

so horizontal speed of the car is same, hence relative velocity of car and ball in the horizontal direction will be zero.