Test Papers – ICSE – Class – X
Std 11, Physics,Vectors,Ch-08 String Tension problems
Consider the given scenario in which block, block is suspended, scenario in which a block is suspended with the help of a string. The mass of the block is 10 kg. Listen carefully, this 10 kg block will experience a downward force due to the gravity, I think we all know that. So there will be a force in downward direction, which is known as the weight and that weight is equal to mass of the object multiplied by the famous acceleration due to gravity. So, this 10 kg block will experience, so here we are taking g as 10 not 9.8, okay. So this block is having weight 100 Newton, weight is the force so its unit must be Newton, so 100 Newton force will be applied on the block. It means the ground is pulling this object downward with 100 Newton force. But the block is at rest, can we see that, the block is at rest. And if the block is at rest that means what is the total force on the block. Must be zero, total force zero means what, that if somebody is pulling downward that means something is pulling upward. And which object can pull this block upward, the only option is string. So first of all, we will see the force of string, the force of the string, we call it as Tension and we represent it as T. And what is the technique to find the direction of Tension in a string. Where will the string put the Tension, there is one technique to find out. Its name is ‘Cut and Pull’. Think how, virtually we will cut this string and then you pull the end. And now tell where will you pull it? Don’t use physics in this, use logic used in daily life. Where will you pull the end, where will the Tension force come? Downward. Wherever you will pull the string there the Tension will come, done, where will you pull this end. There is only one option. Does anybody have any other options? Can anybody pull it down, it has to be upward. So T and T, T or 100, they must be equal because the block is in equilibrium or the block is at rest. Equilibrium means net force is zero, rest means net force zero. So the T must be equal to 100 Newton. With the help of this particular example, I want to just give you this idea, how to find the Tension in the string, done, technique Cut and Pull.
Let us consider one more scenario. Let us consider one more case in which a block is at rest. The block is at the rest with the help of three different strings. Then find Tension in each string, okay.
Std 11, Physics, Vectors, Ch-07 Cross Product continued
Std 11, Vectors, 07 cross product, continued
Let us take one more example on cross product. Write down the problem, find the value of A cross B if A and B vectors are together, if we already know A and B vector then lets us find the value of A cross B. okay. Let’s try. We know that A cross B, we can find the value by using the technique of determinant. In determinant we write I cap minus J cap plus K cap. A is the first vector, so let us write the coefficient of the first vector one two and three , B is the second vector six minus two and minus three, lets us take a break , is there a doubt in writing this determinant? Anyone? Sure, if you don’t get it in your exams then it won’t be right.
So now let us find the coefficient of I cap, when we are calculating the value of I cap, this row and this column will be removed, and probably I must have definitely told you; you have to take care of this cross; two into minus three is minus six; minus two into minus three is minus six, isn’t that right? This plus minus errors you have to see sincerely, second minus J cap this row and this column will disappear one minus three product will be minus three, minus three into six is equal to eighteen plus K cap, this row and this column will not be used only these four values are used one into minus two, minus six into two twelve when you solve it, it will become zero I cap plus twenty one J cap minus fourteen K cap so this will be the result of our A cross B.
Let us take one more problem on the cross product.
Let us consider two vectors A and B, if A dot B is equal to zero then what will be the value of magnitude of A cross B, if magnitude of A vector is two and magnitude of B vector is three? So what do we have here is A dot B is equal to zero, so we need to find the value of magnitude of A cross B where in you know the value of A and B. Okay. Done.
Let us try this, if A cross B is equal to zero, what we can find from this particular equation, the angle between the vectors is ninety degrees or pi by two, read out, the magnitude of A cross B is AB sine theta, you know that formula, magnitude of A cross B or the magnitude of their result, and you are creating the cross product. So this will become magnitude of first vector, magnitude of second vector and the angle between the two which is pi by two, and that is equal to six. Any doubts?
Try this particular problem if A dot B is equal to the magnitude of A cross B then find the angle between the A vector and B vector.
A dot B is equal to the magnitude of A cross B let’s try.
Std 11, Physics, Vectors, Ch-06 Cross Product
In last lecture we were dealing, we were discussing about the position vectors. Let us consider a point at certain coordinates x, y and z. And then the position of A point with respect to origin is defined as x coordinate i cap, y coordinate j cap and z coordinate k cap. This is position vector with respect to origin.
Similarly using the same idea we can define displacement vector which we are going to use in kinematics. Displacement is, let’s say the particle is initially at point A, the initial point, having the coordinate x1, y1, z1. So the initial position of the particle can be written as x1 icap, y1 j cap and z1 k cap. Particle will move from point A to point B in any path. So B must be the final part of the particle. So this will represent the position of the final point with respect to origin and that will be x2 i cap, y2 j cap and z2 k cap.
But a vector from A to B, a vector from the initial point to the final point will be the displacement vector. Position vector of final point with respect to initial point or from initial point will be defined as the displacement vector.
Now we have three vectors, 1, 2 and 3. By using the triangle of vector addition we can say that out of these vectors, r final is the resultant one, or r final is the vector summation of these two vectors. So we will put it like that r initial vector plus displacement vector is equal to r final vector. So D can be written as final position vector minus initial position vector will be defined as the displacement vector. So finally it will come out to be x2 minus x1 icap, y2 minus y1 j cap, Z2 minus z1 k cap. So this is the displacement vector of the particle. And then we use the same idea to calculate work done by force in this particular problem.
Particle is moving under the effect of this given force. It starts its motion from this initial position vector to this final position vector. Work done is defined as the dot product of force under displacement. So first let us calculate the displacement vector from these two. That displacement vector will be written as the final position vector minus initial position vector, okay. So x2 minus x1 i cap, y2 minus y1 j cap, z2 minus z1 k cap. So, the displacement is i cap to j cap plus k cap. So this is our displacement vector, okay, this is how we define the displacement vector. And now we want to calculate the work done by force, so work done by force will be the dot product of force with the displacement. We know that if the vectors are written in terms of i, j, k, the dot product is Ax, Bx, Ay, By, Az, Bz. Ax, Bx 1 into 6 plus Ay, By 4 into 2, plus Az, Bz 2 into 1. So finally it will be 6 plus 8 plus 2 so it will be 16. So what is the unit of work, it is Joules. So total work done by this particular force is equal to 16 Joules. Any doubt? So this how we define the work done.
See it is an application of dot product only, what was the learning point in this problem, you should know by this particular example you should able to see how to write position vector, how to write displacement vector and just the dot products of force and displacement. You are not supposed to know this particular formula, right now even if you do not know that work done is F dot D that is perfectly alright. Because whenever that kind of problem will come into the picture we will give you the formula that work done is force dot product with the displacement.
Let’s take one more good example based on the application of dot product that we have already seen. A particle is moving under the effect of the given force vector and having the given velocity vector at any time T. Here is the force vector and here is the velocity vector. A particle is moving because of this particular force. And at certain time of its motion its velocity vector was this one. Then find the power given by the force then find the power delivered or given by the force if the power is defined as F dot V. So power at certain instant, the instantaneous power is defined as F dot V. So you are not supposed to know this particular formula right now but you should know how to use the dot product. So the purpose of this particular problem is to allow yourself to do some practice on the dot product. So if with force velocity you do dot product then you will get power, so power will be, we know that. Ax Bx plus Ay By. It will be 14. So what is the unit, either Joule per second or watt. Power delivered by this particular force is, okay, any doubt.
Let’s see one more good problem on the application of dot product.
Let’s say I am defining a vector A, something like that, and I am asking you find the angle made by this vector with x axis, y axis and z axis, straightforward you should get this result in your mind. What will be the angle made with…….
Std 11, Physics, Vectors, Ch-05 Dot Product
So far, okay, so far we have discussed the addition and subtraction of two vectors. Now let’s see one more mathematical operation which is very much important for the understanding of various aspects of physics and that is the product of two vectors. So let us discuss the product of two vectors.
Depending upon the end result the product of vectors is divided into two categories, okay. One is scalar product and the other one is the vector product. Why are they given names scalar and vector, that is very much easy to analyze? That means these kinds of products will give you a scalar result when a product of two vectors resulting into a scalar quantity, the product is known as the scalar product and when the result is a vector quantity it is a vector product. To represent them separately for scalar we use dot between two vectors, A dot B so it is also known as dot product. And to represent the vector product we use cross, A cross B so it is known as cross product, okay. So they are also known as scalar dot product and the cross product. So we will consider first the scalar product and then we will move to the vector product.
So what is the topic of today’s lecture, that is the dot product of two vectors, done.
Dot product. Dot product of vectors is defined as A dot B, A dot B, now I have put this dot, right, so that is why it is also known as the dot product. And what will be the result? Result will be a scalar quantity, it will be C and C will not be vector. So C will be a scalar quantity.
And if C is a scalar quantity it will carry only magnitude. It won’t have any direction. So what is the magnitude of C, magnitude of C is defined as A B cos theta. What is A? The magnitude of A vector. And what is B? Magnitude of B vector. And what is the theta? It is angle between A vector and B vector. And as we use the word ‘angle’ then what thought should come to mind, tail to tail. How to define the angle between two vectors? Tail to tail. So the dot product of two vectors quantities, A dot B, how do we write this, A dot B, and what will be the result? A dot B = to AB cos theta. So I have removed C from this, so what is A dot B, so it is AB cos theta. And it depends on three things, magnitude of first vector, magnitude of second vector and the angle between these two vectors. This is how we define the dot product, that’s all.
Now we will see its application, several possible cases and several possible applications and to use it in different, different problems. Done? Any doubt? Okay. So let’s write a few cases, some important points, note down. First one….
Std 11, Physics,Vectors,Ch-04, Addition and Subtraction of two Vectors
Let us consider a vector a. let’s say this is a vector a. Then what is the meaning of vector minus a. minus a that means a vector is multiplied by minus one. That means the vector is rotated by an angle of 180 degrees. So if a vector is multiplied by minus one that means it is rotated by an angle of 180 degrees. It’s just became opposite in direction. What happened? Just opposite. That means vector is rotated by 180 degrees. Done? Second example, say I am taking one vector b and if I am writing 2b. What does it mean? 2b vector that means I am increasing its magnitude, we have doubled its magnitude. Fine. So what does minus 2b mean? We have doubled the magnitude as well as directed it in an opposite manner. Done? Any doubts? Then note down. If a vector is multiplied, If a vector is multiplied, with a positive number, If a vector is multiplied with a positive number its magnitude changes. But its direction does not change. On the other hand, On the other hand if a vector is multiplied by, if a vector is multiplied by a negative number both its magnitude and direction changes. And draw these diagrams. a vector, minus a vector. Multiplied by minus one, direction has been changed, for the moment there is no change in magnitude. But if I multiply it by -2, then both direction and magnitude will change. Please draw all these diagrams.
Let’s take one better thing. Let’s us consider a vector and b vector. And angle between them is theta. There are two vectors and angle between them is theta. Okay? Now let’s see, if the angle between the vectors a and b is theta, then what should be the angle between a and –b vector? See here is the, a and b vectors, as I have kept here the blue marker. And if I multiplying it by minus b, what does it mean? That means I am rotating it by 180 degrees. So when the vector was over here, so what was the angle? Sir that was the angle theta, wasn’t it? And when I am changing it to minus b, then what would be its angle? 180 degree. So the vector came here and now shouldn’t we count this as an angle or not? If this angle is theta, so what about this angle my dear? 180 minus theta. Done? So that is a very important T&T, if the angle between a and b is theta, so then angle between a and –b is, π minus theta. 180 minus theta. So that means if this angle is 30 degrees, so how much would be this angle? 150 degrees. Please note down. If the angle between a and b is theta, so angle between a and –b will be, 180 minus theta. So now please understand how wonderfully we will use this concept. In our last class what we learnt is that if a vector is added to b vector it will give you the resultant vector and magnitude of resultant vector was what? Root of a square plus b square plus 2ab cos theta. What was the theta? Angle between the two vectors. Okay. So this was the resultant of addition. This was the resultant of addition. What if I am writing a vector minus b vector? So there would be a resultant of this as well. Now if the resultant of addition is r, so then the resultant of subtraction would be different. Right? So let us find, put the heading as subtraction of two vectors. So what we are planning, here we are planning to work out the value of a vector minus b vector.
Std 11, Physics, Vectors, Ch-03 Addition of two vectors
Addition of two vectors. Case one. Case one. If we have magnitudes of two vectors. Magnitude of two vectors and angle between them. So we are planning to work out the addition or we are planning to work out about the resultant force, resultant vector in fact, if you know the magnitude of two vectors and the angle between them. Say I know one vector a having magnitude a and the another vector is b having magnitude b and the angle between them theta. If I want to find the resultant of a and b, that mean I want to find a vector plus b vector and their resultant. Done? We have seen the triangle law of vector addition. What we do is, we either shift one or both the vectors parallel, so let’s shift b parallel to itself and rearrange it like that. This is a vector and this is b vector. Now class, see carefully. If I draw an extended line here, this angle will also be the theta. See carefully because I am shifting b parallel, so the new b vector is also is parallel to the previous one, so that this angle is theta, this angle will also be the theta. Any doubts? Okay. Now we know that this is the resultant vector. This is the resultant vector. Say the resultant is having the magnitude r. so what we are planning is actually, we are planning to find the magnitude of resultant vector. We want to find a + b; r vector. So how we get the complete information about the r vector? With the magnitude of r vector and with the direction of r vector. Done? So first we will try to work out the magnitude of r vector. Okay. See we will use the properties of triangle. Now class see it very carefully. An extended line, this point here, from here draw a perpendicular. Perpendicular is 90 degrees. Okay? You can notice here carefully. Now see lets concentrate on two different angles. Triangles. X, Y, Z and T. See carefully triangle XTY is right angled triangle. Similarly triangle ZTY is also the right angled triangle. And in these two triangles we can directly use the ordinary ideas that we know about a right angled triangle. Done? Okay. Let’s take triangle ZTY. Can I write ZT as b cos theta. This is b vector having magnitude b. this can be considered as the horizontal component of b vector, and the horizontal component of b vector will become b cos theta. See it carefully. Done? Only concentrate on b vector and think about the horizontal component of b vector. If this is b vector this is angle theta, so its horizontal component will be b cos theta. Done. Similarly YT will be the vertical component and it will be sine theta. Okay. Class now see carefully.
