Dr KN Modi University, Tonk All Questions

Hii Meghana,

It depends on the exam pattern . If the given question has partial marking then you will get partial marking in the above case . If there is no partial marking then you will loose marks, in some cases even there is a chance of negative marking also.

Hello aspirant,

Hope you are doing well.

As per your question regarding ICAR exam if you are eligible for exam then definitely you can give the ICAR exam. Here Eligibility mainly concerns with that you should have passed 10+2 with a minimum percentage of 50 percentage as per general category

Dear Julia ,

If you are join XYZ college and complete your 1st year course but if you will take admission in another college without informing XYZ college then it will trouble in future because you have already student of this college or university with particular registration number and it

Hey Jaywadmare, the anwer of your question is 34/3.

Also go through following explanation.

Explanation : a^2 = d ^12

b^3 = d ^12

c^9 = d ^12

after that logd ( d^12) 1/2 + 1/3 + 1/9 = logd (d^12)^ 51/54

= logd (d ^12 * 17/18)

= 1

In a triangle ABC cos(A+B)= cosAcosB-sinAsinB. cos(A-B)= cosAcosB+sinAsinB.

there by, by adding these two and multiplying the result by 2 times you get the desired answer. 4cosAcosB = 2[cos(A+B) + cos (A-B)]. This can also be done by interchanging the signs and the answer will remain the same.

Hey!

NO, I really dont think modern ABC is the appropriate book for NEET prep… Though it can help you with the 12 grade exams.

For NEET-UG,

1st, For the KNOWLEDGE part -primarily you should go for NCERT first surely and after that for extra knowledge Allen packages or Trueman’s

Using distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2) to find distance between two points A(x1, x2) and B(y1, y2), we can find the distance between A and B, B and C and A and C respectively.

We have, AB = sqrt((3-1)^2 + (5-k)^2)

= sqrt(2^2 +

Greetings Aspirant,

If the magnetic field along AB is  B , the force acting on BC=F=BiL

Now the length of the side AC is  Root"2L"

Also the current reverses the direction in this section as compared to that in BC.

Hence force on AC= Root"2"L( B× I )

= −

Since A, B , C are the angles of a triangle and we know that the sum of all angles in a triangle is 180 degrees .

Consider A =5x

B=1x

C=6x

5x+1x+6x=180

12x=180

x=15 degrees

Angles A=5 X 15=75 degrees

B=15 degrees & C= 6X15= 90 degrees

By sine

Hi

As per your question the solutions is:

It is for the reason that ΔABC ~ ΔPQR

We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR

...(i)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)

Since AD and PM