nutrition in animals for class 7 Important questions

Most Important Questions

  Q 1 What do you understand by animal nutrition? 
  Q 2 Define digestion.  
  Q 3 Name the type of food and mode of feeding of the following animals: (a) Ant (b)Mosquito
  Q 4 Name the process of taking in food? 
  Q 5 Name the glands present inside the oral cavity. 
  Q 6 What helps our tongue to taste the food? 
  Q 7 What do you understand by alimentary canal or digestive tract? 0
  Q 8 What constitutes the Digestive System? 
  Q 9 What are the two components of the digestive system? 
  Q 10 What is the difference between milk teeth and permanent teeth? 
  Q 11 How many types of teeth are present in humans? Name them. 
  Q 12 Which teeth do you use for piercing and tearing? Write the number of these teeth. 
  Q 13 What is the function of premolars and molars? Write the number of both teeth in each jaw. 
  Q 14 List the functions of the tongue. 
  Q 15 How many pairs of salivary glands are present in humans and what is the function of saliva? 
  Q 16 What are the main steps of digestion in humans? 
  Q 17 Write short notes on ---   (a) The stomach (b) Liver 
  Q 18 What are villi? What is their function? 
  Q 19 What do you understand by rumen and rumination? 
  Q 20 Why do we get instant energy from glucose? 
  Q 21 Name the type of carbohydrate that can be digested by ruminants but not by humans. 
  Q 22 What is the mode of feeding in Python? 
  Q 23 Name the largest gland in the human body. 
  Q 24 Write a short note on structure and feeding of the amoeba. 
   Q 25 Write one similarity and one difference between the nutrition in amoeba and human beings. 
  Q 26 Draw labelled diagrams of ---    
           (a) Human digestive system                                                                           
           (b) Amoeba



Dimensional analysis practice worksheet

This post is about Dimensional analysis practice worksheet practice these problems and try to solve them in one hour.

Question 1 The parallax of a heavenly body measured from two points diametrically opposite on the equator of the earth is 1.0 minute. If the radius of the earth is 6400 Km, find the distance of heavenly body from the center of the earth in AU. Take 1AU=1.5×1011m.
Question 2 A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the gauge arbitrarily by increasing the number of divisions on the circular scale?
Question 3 Assuming that the frequency ($\nu$) of the vibrating string depends upon the load (F) applied , length of the string (l) and mass per unit length (m), prove that  
$\nu = \frac{1}{2l} \sqrt{\frac{F}{m}}$
Question 4 What are the dimensions of
(a) gravitational constant
(b) work
(c) angular velocity
(d) pressure
Question 5 Write the dimensions of a/b in the relation
$F=a\sqrt{x}+bt^{2}$
Where F is force, x is distance and t is time.
Question 6 Check the correctness of the relation
$\rho =\frac{3g}{4\pi rG}$
Where symbols have usual meaning.
Question 7 The coefficient of viscosity ($\eta$) of a gas depends upon mass m, the effective diameter d, and mean speed v of the gas molecules. Use dimensional analysis to find the relation between them.
Question 8 The potential difference of V=20±1 Volt , when applied across a resistance gives a current of 2.5±0.5 ampere. Find the resistance with limits of error.
Question 9 Convert 4.29 light years into parsecs. Calculate the parallax of a star at the distance when viewed from two locations of earth six months apart in its orbit around the sun.
Question 10 In an experiment, on determining the density of a rectangular block, the dimensions of the block are measured with a Vernier calipers (with a least count of 0.01 cm) and its mass is measured with a beam balance of least count of 0.1 g. How do we report our result for the density of the block.
Question 11 It is required to find the volume of a rectangular block. A Vernier calipers is used to measure the length, width and height of the block. The measured values are  1.37 cm, 4.11 cm, and 2.56 cm respectively.

Answer the following questions

Question 1 What is relative error or percentage error?
Question 2 What is Parallax error?

Question 3 How errors are propagated or combined?

Download Physics formulas and concepts pdf

Download free concise physics formula book containing class 11 and class 12 physics formulas


For more visit this link
Download Physics formulas and concepts pdf

How to do well in physics Board exams


So when there are few days left for your board exams , you might be thinking how to do well in physics Board exams or how do you memorise your physics formulas so that you can write all the answers that are asked in your exams easily. It is important to memorise physics formulas for exams because memorising them will increase your speed and by the end of the time you will be able to complete all your questions and still have time to revise and check for the mistakes in all of your paper.



For more visit this link
How to do well in physics Board exams

Collection of CBSE books in electronic format


CBSE  has posted all the CBSE books in electronic format on it site  http://cbse.nic.in
So  all the books from CBSE board can be viewed at below link
http://cbse.nic.in/ecbse/welcome.html
You don’t need to look at other site to ge...
For more visit this link
Collection of CBSE books in electronic format

Class 10 maths quadratic equations

This page is about class 10 maths chapter 4 : Quadratic equations along with NCERT solutions for this chapter

Quadratic Polynomial

P(x) = ax2 +bx+c   where a≠0
Quadratic equation
ax2 +bx+c   =0     where a≠0
Solution or root of the Quadratic equation
A real number α is called the root or solution of the quadratic equation if
2 +bα+c=0
Some other points to remember
  • The root of the quadratic equation is the zeroes of the polynomial p(x).
  • We know from chapter two that a polynomial of degree can have max two zeroes. So a quadratic equation can have maximum two roots
  • A quadratic  equation has no real roots if b2- 4ac < 0
To know more about how to solve quadratic equation you can visit this link Quadratic equation concepts

 Wait there is more
We have few more links that you can follow to get more study material on this chapre

  1. Important questions
  2. NCERT book solutions
  3. Revision sheet

Recommended Maths books for class 10

These are the links of few books for class 10 maths from amazon.in
  1. IIT Foundation & Olympiad Explorer Class-10 Maths (PB).
  2. NCERT Solutions - Mathematics for Class X
  3. Mathematics Class - 10 (R.D. Sharma)
  4. New Pattern Class 10 Boards + PMT/IIT Foundation (Science + Maths) - Set of 4 books
  5. Full Marks Mathematics Class 10 :Set of Term 1 & 2
  6. Mathematics foundation course for Boards /JEE/PETs/ NTSE

CBSE Class 9 maths (notes, extra questions and downloads)

CBSE Class 9 maths notes with extra questions and downloads

Table given below contains all the CBSE Class 9 maths material I have prepared. These links I am sure would prove helpful to you and you can practice extra questions to improve yourself. Hope you like them.
S.No. Chapter Study Material Questions Revision Notes
1 Number System notes questions download
2 Polynomials notes questions download
3 Coordinate Geometry notes questions download
4 Euclid Geometry notes questions download
5 Geometry notes questions download
6 Triangles notes questions download
7 Heron Formula notes questions download
8 Quadilaterals notes questions download
9 Area of Parallelograms notes questions download
10 Probability notes questions download
11 Statistics notes questions download
12 Surface Area and Volume notes questions download

Parabolic motion (motion in two dimension)

Question
A stone is thrown from ground level over horizontal ground. It just clears three walls, the successive distances between them being r and 2r. The inner wall is 15/7 times as high as the outer walls which are equal in height. The total horizontal range is nr, where n is an integer. Find n.
Solution
Let us just assume that both the outer walls are equal in height say \(h\) and they are at equal distance \(x\) from the end points of the parabolic trajectory as can be shown below in the figure.

Now equation of the parabola is
 \(y = bx - c{x^2}\)                                       (1)
 \(y = 0\) at \(x = nr = R\)
where \(R\) is the range of the parabola.
Putting these values in equation (1) we get
\(b = cnr\)                                                       (2)
Now the range \(R\) of the parabola is
\(R = a + r + 2r + a = nr\)
This gives
\(a = \left( {n - 3} \right)\frac{r}{2}\)               (3)
The trajectory of the stone passes through the top of the three walls whose coordinates are
\(\left( {a,h} \right),\left( {a + r,\frac{{15}}{7}h} \right),\left( {a + 3r,h} \right)\)
Using these co-ordinates in equation 1 we get
\(h = ab - c{a^2}\)                                           (4)
\(\frac{{15}}{7}h = b(a + r) - c{(a + r)^2}\)                              (5)
\(h = b(a + 3r) - c{(a + 3r)^2}\)                                      (6)
After combining (2), (3), (4), (5) and (6) and solving them we get n = 4.

How to add vectors using components

Using this concept map you can solve problems of vector addition involving two or more vectors step by step.



Advantages of potential formulation in electrostatics

We already know about electric field and electric potential. We also know that electrostatic field is completely characterized by vector function E(r). The electric field depicts the force exerted on other electrically charged objects by the electrically charged particle the field is surrounding. Now a question arises why do we need introduction of electric potential when we already have electric field for the description of electric force between charges.
Firstly, the concept of electric potential is very useful not only in physics but as well as in engineering .This is because if we know the potential we can easily calculate the work  done by field forces when a charge is displaced from point 1 to point 2 that is
\[{W_{12}} = q({\varphi _1} - {\varphi _2})\]
where \({\varphi _1}\)  and \({\varphi _2}\) are the potentials at points 1 and 2. This means that required work is equal to the decrease in the potential energy of charge q when it is displaced from point 1 to 2.
Calculation of the work of field forces with the help of above mentioned formula is not just simple but the only possible method in some cases.
Secondly in some cases of electrostatic field calculation it is often easier to first calculate the potential and then find the gradient of potential  \({\varphi}\) to calculate the value of electric field intensity E. Also for calculating \({\varphi}\) we only need to evaluate one integral but for calculation of E we must take three integrals all for x, y, and z directions since E is a vector quantity.
But we must note that for problems with high symmetry we must directly calculate E using Gauss's Theorem which is much simpler way to find electric field intensity when charge distribution is symmetrical.

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