1. Dimensional analysis of the answers

Dimensional analysis is very handy in almost all the entrance examinations. You will even get questions from IIT JEE which can be solved by using this trick. This is not a short cut; it’s what you learnt in the first Chapter of Physics in Plus one. The method is very easy and can be applied in Physics and Chemistry questions; Let us understand it with an example.

*Q. A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the same dimensions and of low modulus of rigidity ɳ such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn, block A executes a small oscillation. The time period of oscillation is: (IIT JEE 1992)*

(a)2π MɳL (b) 2π Mɳ/L

(c) 2π ML/ɳ (d) 2π M/ɳL

Sol:

How’s the question? A little bit lengthy and tougher? How should it be when it’s an IIT JEE question? But the question is not that much tough if you are going in the indirect method. And often the bright students will use these indirect steps first before doing the direct method.

What’s the indirect method? It’s none other than dimensional analysis. Please read the last sentence of question. We are asked to find out the time period of oscillation. The dimensional formula of Time period or time is [T]. So the dimension of our answer should be [T].

You need to know the dimension of other given information in the problem. The dimension of Mass M is [M] and length L is [L] . Some of you may find it difficult to find the dimension of modulus of rigidity ɳ . But it’s very easy to find if you know what modulus of rigidity is. I will tell the direct formula here;

Modulus of rigidity, ɳ =-F/lx

Therefore dimensional formula of ɳ =[MLT^-2]/[L][L] =[ML^-1 T^-2]

Now check for the dimension of Time period in each option. Let us take the first option. The final dimensional formula of that answer comes to be [MT^-1].

Option (b) comes out to be [ML^-1 T^-1]

Option (c ) comes out to be [LT]

Now check option (d). It’s [M]/[ ɳ ][L]

= [M]/[ML^-1 T^-2][L]

= [T]

How’s it? You can solve lots of questions like this in many entrance examinations. To solve problems like this, you should need some practice with these types of questions. Take some previous years’ question papers and solve. You will definitely find questions like this.

(a)2π MɳL (b) 2π Mɳ/L

(c) 2π ML/ɳ (d) 2π M/ɳL

Sol:

How’s the question? A little bit lengthy and tougher? How should it be when it’s an IIT JEE question? But the question is not that much tough if you are going in the indirect method. And often the bright students will use these indirect steps first before doing the direct method.

What’s the indirect method? It’s none other than dimensional analysis. Please read the last sentence of question. We are asked to find out the time period of oscillation. The dimensional formula of Time period or time is [T]. So the dimension of our answer should be [T].

You need to know the dimension of other given information in the problem. The dimension of Mass M is [M] and length L is [L] . Some of you may find it difficult to find the dimension of modulus of rigidity ɳ . But it’s very easy to find if you know what modulus of rigidity is. I will tell the direct formula here;

Modulus of rigidity, ɳ =-F/lx

Therefore dimensional formula of ɳ =[MLT^-2]/[L][L] =[ML^-1 T^-2]

Now check for the dimension of Time period in each option. Let us take the first option. The final dimensional formula of that answer comes to be [MT^-1].

Option (b) comes out to be [ML^-1 T^-1]

Option (c ) comes out to be [LT]

Now check option (d). It’s [M]/[ ɳ ][L]

= [M]/[ML^-1 T^-2][L]

= [T]

How’s it? You can solve lots of questions like this in many entrance examinations. To solve problems like this, you should need some practice with these types of questions. Take some previous years’ question papers and solve. You will definitely find questions like this.

Some other examples are;

*1. A body of mass M is situated in a potential field U(x)= U(1-Xcosαx) where U and α are constants. The time period of small oscillation will be.*

(a) 2π M/Uα^2

(b) 2π U/M α^2

(c) 2π Uα^2/M

(d) 2π M Uα^2

(a) 2π M/Uα^2

(b) 2π U/M α^2

(c) 2π Uα^2/M

(d) 2π M Uα^2

2. Substitution Method

Substitution method is another important trick which you can apply widely in Mathematics. There are many such topics in Mathematics where you can apply this method. Some of the main portions are; Coordinate geometry, Straight lines, Complex numbers; Sequence and series et.

Let us take one example to know the method.

*Q. The straight line (2+α)X + (1+ α)Y=5+ α , for different values of α passes through the fixed point*

(a) (2,9)

(b) (2, -9)

(c ) (-2,-9)

(d ) (-2, -9)

Sol:

This is very easy to solve, just put the value of α=0. Now the equation is simplified to

2X+Y=5. Look at the option and just put the value of x in each case in the equation and find the value of y. Check the value of y is the one given in option. For eg, take the first option. Here the value of x is 2. Put it in equation and you will get the value of y=1. But in the option it’s given as 9. So the option is wrong. From here itself, we find that second option too is wrong.

Now put x=-2 . This gives the value of y=-9. Hence, the correct answer is (d) (-2, -9).

(a) (2,9)

(b) (2, -9)

(c ) (-2,-9)

(d ) (-2, -9)

Sol:

This is very easy to solve, just put the value of α=0. Now the equation is simplified to

2X+Y=5. Look at the option and just put the value of x in each case in the equation and find the value of y. Check the value of y is the one given in option. For eg, take the first option. Here the value of x is 2. Put it in equation and you will get the value of y=1. But in the option it’s given as 9. So the option is wrong. From here itself, we find that second option too is wrong.

Now put x=-2 . This gives the value of y=-9. Hence, the correct answer is (d) (-2, -9).

This is one of the most effective methods to be adopted to solve entrance questions very easily. And even if you don’t know the theory, you can solve lots of questions just substitution as done above. Practice alone will make you a good problem solver of this kind.

Some other examples are;

*Q. If non-zero numbers a,b and c are in H.P, then the straight line x/a+ y/b +1/c=0 passes through the fixed point*

(a)(1,-.5)

(b) (1,-2)

(c )(-1,-2

(d ) (-1,2)

Sol:

Here you need to know only one thing; If a,b and c are in Harmonic Progression (HP), then their reciprocals are in AP, ie 1/a, 1/b and 1/c are in AP. Now just take an AP sequence. Let us take the least complicated one; 1,2,3 and substitute it in the above equation.

The new equation is; x+2y+3 =0.

Then substitute the value of x as given in each option and check the y value given there also satisfies the option. Take x=1, then the y value should be -2 and hence the correct answer is (b). You can check with x=-1 also. For x=-1, y=-2.5. There’s no such option given. So the correct answer is (b) itself.

(a)(1,-.5)

(b) (1,-2)

(c )(-1,-2

(d ) (-1,2)

Sol:

Here you need to know only one thing; If a,b and c are in Harmonic Progression (HP), then their reciprocals are in AP, ie 1/a, 1/b and 1/c are in AP. Now just take an AP sequence. Let us take the least complicated one; 1,2,3 and substitute it in the above equation.

The new equation is; x+2y+3 =0.

Then substitute the value of x as given in each option and check the y value given there also satisfies the option. Take x=1, then the y value should be -2 and hence the correct answer is (b). You can check with x=-1 also. For x=-1, y=-2.5. There’s no such option given. So the correct answer is (b) itself.

3. METHOD OF ASSUMPTION.

This is another important method. In some of the questions, some unknown quantities will be given. What we will do there is to assign some values to these quantities without changing the conditions given. This is almost similar to the method of substitution.

*Q. If a(n) denotes the nth term in an Arithmetic Progression. Then*

1/a(1)a(2) +1/a(2)a(3) +………………………………+1/a(n-1)a(n)=

(a) n-1/a(1)a(n)

(b)n/a(1)a(n+1)

(c )n-2/a(1)a(n-1)

(d) n/a(1)a(n)

Sol:

Take a simple AP=1, 2, 3, 4…..let us take n=3.

Therefore our questions comes to be:1/1*2 +1/2*3 =4/6 =2/3= n-1/a(1)a(n).

Hence the correct option is (a).

1/a(1)a(2) +1/a(2)a(3) +………………………………+1/a(n-1)a(n)=

(a) n-1/a(1)a(n)

(b)n/a(1)a(n+1)

(c )n-2/a(1)a(n-1)

(d) n/a(1)a(n)

Sol:

Take a simple AP=1, 2, 3, 4…..let us take n=3.

Therefore our questions comes to be:1/1*2 +1/2*3 =4/6 =2/3= n-1/a(1)a(n).

Hence the correct option is (a).

4. METHOD OF PROBLEM SOLVING BY DRAWING GRAPHS AND DIAGRAMS

Solving problems by drawing graphs and diagrams is a widely appreciated method. It’s also very easy to do comparing to the conventional method. If you are strong with this tool, you will come across lots of questions from many engineering entrance exams including IIT JEE.

5. METHOD OF APPROXIMATION

Method of approximation comes handy when the options differ largely in magnitude. Many of the students don’t know about this method and hence devotes more time to some questions. You should practice this method to use the time efficiently.

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