Given, vec(A) =-2i+2j-k

vec(B )= 3i+6j+2k.

Now A cross B =

| i                j              k  |

| -2             2              -1|

| 3              6               2 |

=> A cross B = i(2*2-[-1*6])-j(-2*2-[-1*3])+k(-2*6-[2*3])

=> A cross B = 10i +j - 18k

I hope this answer helps. All the very best for your future endeavors!

Hello candidate,

Using the third equation of motion, that the square of final velocity is equal to the arithmetic sum of the square of initial velocity and 2 multiplied by the acceleration into the distance.

So, taking the initial velocity is equal to zero the expression becomes v^2= u^2+ 2*alpha*distance. So, the expression is perfectly correct.

Hope That this information helps you!!

Have a good day.

Dear aspirant,

The basic formula of work done by a force is equal to force into displacement but in vector it F.S cos theta. So if a force and displacement graph is given to you and so to find the work done by the force you need to find the area of the force displacement graph.

Hello Candidate,

The galvanometer is a simple electrochemical device which is used to measure the supply of current in the circuit or in a loop.

Galvanometer consists of several divisions including main scale divisions and small scale division for accurate measurement of the current supply, when the fluctuations age produced in the device upon changing the voltage and other electrical parameters.

Hope this information helps you!!

Hello candidate,

Projectile in general words can be referred to the path taken by any object while it is in motion in air before hitting the ground surface.

The formula for horizontal range of a projectile is given as, v^2 × sin(2 theta)/ gravity, for the maximum range the value of horizontal range is equal to v^2/g. The formula for maximum height is equal to v^2/2g.

Hope that this information was helpful for you!!

Hello Dear,

As you have not mentioned the board of the examination. Every board has different pattern of examination. If your board is cbse then you can download the papers form the website: http://www.mycbse.in  Please elaborate your question and feel free to ask your doubts here for quick resolution of your query

Good Luck