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Fri October 31, 2014  

Ans: According to Newton's Second law of motion we know that: F = m × a Given mass m =2.5 kg So the acceleration of the body when a force of 16 N acts on it is: F = m × a  a = F / m    =16N / 2.5 kg    =6.4 m/s 2 The acceleration of the body when a force of 4 N acts on it is: a = F / m    = 4N / 2.5 kg    =1.6 m/s 2  

Thu October 30, 2014  

Ans: Annelida and Arthropoda are two separate phyla that belong to the Kingdom Animalia. Characteristics of phylum annelida : The body is cylindrical and divided into ring-like segments. In each segment, there is representation of different organ systems. Have a well-developed digestive system with an alimentary canal open at both the ends. It consists of mouth and anus. Have a true body cavity called coelom, present between the body wall and the digestive tube. It is filled with coelomic fluid. The body cavity is lined with coelomic epithelium. Presence of thin and moist layer called cuticle on the body. Examples – Earthworm, leech, nereis Characteristics of phylum arthropoda : They have jointed limbs, one pair each on some or on all body segments. They have an exoskeleton made of chitin. They cast off their exoskeleton during growth in early life, which is regrown. The casting off and the exoskeleton is collectively called moulting. They have a complex muscular system, with exoskeleton for attachment, striated muscles for rapid actions and smooth muscles for visceral organs. Lack cilia. Sense organs consist of well-developed eyes, gustatory organs, antennae, balancing organs and auditory organs. Excretory organs are Malpighian tubules or green glands. Examples – Crab, spider, mosquito, butterfly

Thu October 30, 2014  

Ans:  To prove R=2f:               Consider a ray of light AB, parallel to the principal axis, incident on a spherical mirror at point B.  The normal to the surface at point B is CB and CP = CB = R, is the radius of curvature.  The ray AB, after reflection from mirror will pass through F (concave mirror) or will appear to diverge from F (convex mirror) and obeys law of reflection, i.e., i = r.    From the geometry of the figure, if the aperture of the mirror is small, B lies close to P.  Therefore, BF = PF    or FC = FP = PF    or PC = PF + FC          = PF + PF    or R = 2PF  or R = 2f   Mirror formalae in spherical mirror:

  In the figure shown above, an object AB is placed at a distance u from the pole of the concave mirror of small aperture, just beyond the centre of curvature. Hence, its real, inverted and diminished image A’B’ is formed at a distance v in front of the mirror. A/C to Cartesian sign convention, Object distance (PB) = -u Image distance (PB’) = -v Focal length (PF) = -f Radius of curvature (PC) = -R It is clear from the geometry of figure, right angle ABP and A’B’P’ are similar.

Thu October 30, 2014  

Ans: Covalent bond exist in liquid or gaseous state as they have weak forces of attraction between the binding molecules.

Thu October 30, 2014  

Ans: River Zaskar is a tributary of river Indus. Though, river Nubra joins the river Shylok (tributary of river Indus), it is broadly considered as a tributary of river Indus.

Thu October 30, 2014  

Ans: 1. SSS-Side-Side-Side criterion of congruence   Two Triangles are congruent if all three sides in one triangle are congruent to the corresponding sides in the other.   2. ASA-Angle-Side-Angle criterion of congruence   Two Triangles are congruent if two angles and an included side in one triangle are congruent to the corresponding angles and side in the other.   3. RHS-Right Angle-Hypotenuse-Side criterion of congruence   Two right-angled triangles are congruent if the hypotenuses and one pair of corresponding sides are equal.

Thu October 30, 2014  

Ans: C o n s i d e r space t h e space f o l l o w i n g space f i g u r e. G i v e n space t h a t space angle A P B equals alpha T h e space l e n g t h s space o f space t h e space tan g e n t s space f r o m space a n space e x t e r n a l space p o i n t space a r e space e q u a l. triangle A P O space a n d space triangle B P O space a r e space c o n g r u e n t. H e n c e comma space angle A P O equals angle B P O equals alpha over 2 C o n s i d e r space t h e space t r i a n g l e space triangle A P O : sin alpha over 2 equals fraction numerator O A over denominator O P end fraction rightwards double arrow O P equals fraction numerator O A over denominator sin alpha over 2 end fraction rightwards double arrow O P equals r cos e c alpha over 2.... left parenthesis 1 right parenthesis N o w space c o n s i d e r space t h e space t r i a n g l e comma space triangle O P G : sin beta equals fraction numerator O G over denominator O P end fraction rightwards double arrow O G equals O P sin beta rightwards double arrow O G equals r cos e c alpha over 2 sin beta space space space space space space space open square brackets f r o m space e q u a t i o n space left parenthesis 1 right parenthesis close square brackets T h u s comma space h e i g h t space o f space t h e space c e n t r e space o f space t h e space b a l l o o n space i s space r sin beta cos e c alpha over 2  

Thu October 30, 2014  

Ans: Atomicity is defined as the number of atoms in a molecule of that element. For example: Oxygen molecule consists of two atoms of oxygen chemically bonded to each other. So the atomicity of oxygen is 2 and therefore it is a diatomic gas. Some solid elements like phosphorus and sulphur have more than two atoms in their molecules. For example: Phosphorus exists as P 4 molecules. Each phosphorus molecule consists of four atoms of phosphorus. So its atomicity is 4 and is tetra-atomic. Sulphur molecules consist of eight atoms per molecule. Its atomicity is 8 and so it is  polyatomic.

Thu October 30, 2014  

Ans: Balanced reaction is: 3BaCl 2 + 2Na 3 PO 4   →   Ba 3 (PO 4 ) 2 + 6NaCl 3 moles of BaCl 2 react with 2 moles of Na 3 PO 4 to give 1 mole of Ba 3 (PO 4 ) 2
The mole ratio between BaCl 2 and Na 3 PO 4 is
n(BaCl 2 ):n(Na 3 PO 4 )=3:2 The ratio of these compounds in the given mixture is 0.5/0.2 = 5:2 > 3:2. Hence, the BaCl 2 is in excess and the Na 3 PO 4 is the limiting reactant.
Now, 2 moles of Na 3 PO 4 give 1 mole of Ba 3 (PO 4 ) 2
So, 0.2 moles of Na 3 PO 4 will give (1/2) x 0.2 = 0.1 mole of Ba 3 (PO 4 ) 2
So, number of moles of Ba 3 (PO 4 ) 2 formed = 0.1

Thu October 30, 2014  

Ans: CaC0 3 + heat = CaO + CO 2   x                                1.62 100                             56

x=1.62 x 100/56 = 2.89 g

Cross multiply through the equation and solve for X grams of CaCO 3 we get 2.89 grams of CaCO 3  

Original reaction CaCl 2 + Na 2 CO 3 = CaCO 3 + 2 NaCl                                       x                                    2.89
                                   111                                  100     x = 2.89 x 111/100=3.2 g

Cross multiply again and solve for X grams of CaCl 2 , we get 3.2 grams CaCl 2  

% CaCl 2 in original mixture 3.2 grams CaCl 2 over l0 gms mixture = .32 or 32%

Thu October 30, 2014  

Ans: The molar mass of the salt is m(Na 2 SO 4 * xH 2 O) = 2*23+32+16*4 + x (2 + 16) = 142 + x*18 [g/mol]
Then let's construct the proportion:
13.4 / 6.3 = (142 + 18x)/18x
13.4 * 18x = (142 + 18x) * 6.3
241.2x = 894.6 + 113.4x
127.8x = 894.6
x = 7

Na 2 SO 4 * 7H 2 O

Thu October 30, 2014  

Ans: A substance which prevents the coagulation of blood or the formation of clots in the blood is called an anticoagulant. Heparin is a natural anticoagulant that prevents coagulation of blood in our body. 

Wed October 29, 2014  

Ans: C o n s i d e r space t h e space f o l l o w i n g space f i g u r e.

Wed October 29, 2014  

Ans: Thermal wind is the difference in wind over different layers of atmosphere due to horizontal temperature difference. It is not exactly wind but a wind shear which means difference in wind speed over a certain short distance in the atmosphere.   The phenomenon is governed by an equation which relates the geopotential thickness of the atmospheric level and different pressures at those levels. This is beyond the scope of grade X and hence cannot be elaborately explained. If needed this concept can be referred over different sources on the internet.

Wed October 29, 2014  

Ans: Go through the properties of p block elements and do comparative study. Properties are mostly asked. Important topics in p-block elemets are: Structures of  Oxoacids: phosphorus  (2007) , sulphur  (2007) , Structures of  Fluoride: sulphur  (2008) , xenon  (2008, 2009) , bromine  (2009) Interhalogen compounds  (2007) Basicity of group 15 elements  (2008, 2009) Most important topics/questions tructures of PCl 5 , H2SO 3 , H 2 SO 4 , H 2 S 2 O 8 , H 2 S 2 O 7 , HOCl, HClO 2 , HClO 3 , HClO 4 Complete the reactions: Conceptual questions: Draw the resonating structures of: a) NO
b) NO 2
c) N 2 O NF 3  is an exothermic compound whereas NCl 3  is not. NH 3  is a stronger base than PH 3 . Why? Arrange the following in order of the indicated property: F 2 , Cl 2 , Br 2 , I 2  in increasing order of bond dissociation enthalpy HF, HCl, HBr, HI in increasing order of acidic strength NH 3 , PH 3 , AsH 3 , SbH 3 , BiH 3  in increasing order of basic strength What are interhalogen compounds? Sulphur has very high boiling and melting point when compared to oxygen. In group 16 tendencies to show -2 oxidation state decreases on going down the group. In group 16 +4 oxidation state become more stable than +6 oxidation state on going down the group. Oxygen can show a maximum covalency of 4 and it can not form hexa valent compound. Also, refer to the revision notes provided by www.topperlearning.com

Wed October 29, 2014  

Ans: Imagine a life without numbers ? The subject is applied to everything in this world. The basic building block of mathematics is numbers, which enables us to compare things with one another and without which calculations are not possible.   let us imagine a world without numbers. Imagine a day without calendar and time. How would you know the time, since watches and calendars use numbers? You would miss even your own birthday, your friends birthdays since all the days will be same in your life.   Consider that you need a toy. How will you buy a toy since you donot know how to count ? Without mathematics, you cannot do trade and do any business. How will a chemist prepare medicines if he donot know how to weigh things. He would not have an accurate measurement of ingredients to be used. All the technological innovations (eg. electricity etc. ) that have occured and the luxuries (eg. car, ac etc.) which you enjoy would not be there.   Even persons who donot like / understand maths have to use mathematics. Mathematics is a core part of our life. Mathematics is not very difficult if you understand the basics. In our basic classes we had understood addition, subtraction, multiplication and division. The basic is applied over and over again to all the chapter such as integers, fractions, simple interest, linear equations, rational numbers, mensuration, geometry etc. It is just your attitude that makes the subject how it is. If your attitude is good, then you will enjoy the subject otherwise you will find it boring.

Wed October 29, 2014  

Ans: 4. space G i v e n space q u a d r a t i c space e q u a t i o n space i s space a x squared plus b x plus c equals 0 S u m space o f space t h e space r o o t s space equals minus b over a P r o d u c t space o f space t h e space r o o t s equals c over a S i n c e space s u m space o f space t h e space r o o t s space i s space e q u a l space t o space t h e space P r o d u c t space o f space t h e space r o o t s comma space w e space h a v e comma minus b over a equals c over a rightwards double arrow minus b equals c rightwards double arrow b plus c equals 0 H e n c e space t h e space c o r r e c t space o p t i o n space i s space left parenthesis 3 right parenthesis  5. space G i v e n space r o o t s space a r e space p space a n d space 1 over p ; S u m space o f space t h e space r o o t s equals p plus 1 over p equals fraction numerator p squared plus 1 over denominator p end fraction ; P r o d u c t space o f space t h e space r o o t s equals p cross times 1 over p equals 1 T h u s comma space t h e space q u a d r a t i c space e q u a t i o n space i s space x squared minus s u m space o f space t h e space r o o t s cross times x plus p r o d u c t space o f space t h e space r o o t s equals 0 T h a t space i s comma space t h e space e q u a t i o n space i s space x squared minus fraction numerator p squared plus 1 over denominator p end fraction cross times x plus 1 equals 0 rightwards double arrow space p x squared minus open parentheses p squared plus 1 close parentheses x plus p equals 0 H e n c e space t h e space c o r r e c t space o p t i o n space i s space left parenthesis 2 right parenthesis.  11. space G i v e n space q u a d r a t i c space e q u a t i o n space i s space x squared plus k open parentheses 4 x plus k minus 1 close parentheses plus 2 equals 0 G i v e n space t h a t space t h e space r o o t s space o f space t h e space a b o v e space e q u a t i o n space a r e space r e a l space a n d space e q u a l. H e n c e space t h e space v a l u e space o f space t h e space d i s c r i m i n a n t space i s space z e r o. T h a t space i s space D equals b squared minus 4 a c equals o R e w r i t i n g space t h e space g i v e n space q u a d r a t i c space a s space f o l l o w s : x squared plus 4 k x plus k squared minus k plus 2 equals 0 T h u s comma space b equals 4 k comma space a equals 1 space a n d space c equals k squared minus k plus 2 T h u s comma space D equals 16 k squared minus 4 open parentheses k squared minus k plus 2 close parentheses equals 0 rightwards double arrow 16 k squared minus 4 k squared plus 4 k minus 8 equals 0 rightwards double arrow 12 k squared plus 4 k minus 8 equals 0 rightwards double arrow 3 k squared plus k minus 2 equals 0 rightwards double arrow 3 k squared plus 3 k minus 2 k minus 2 equals 0 rightwards double arrow 3 k open parentheses k plus 1 close parentheses minus 2 open parentheses k plus 1 close parentheses equals 0 rightwards double arrow open parentheses 3 k minus 2 close parentheses open parentheses k plus 1 close parentheses equals 0 rightwards double arrow k equals 2 over 3 comma space k equals minus 1 H e n c e space t h e space c o r r e c t space o p t i o n space i s space left parenthesis 4 right parenthesis. Please one question per query.

Wed October 29, 2014  

Ans:   1.        
    1.       First find the longest continuous parent chain of carbons, and name the chain.        In the given compound the longest carbon chain is 7 carbon atoms.                2.       It is numbered from R – L because that way the branched carbon can have lowest number possible. 3.       So the IUPAC name of the compound is   3- methyl heptane

Wed October 29, 2014  

Ans: MgSO4 = 24 + 32 + 16*4 = 120g/mol  
=> MgSO4.xH2O = 120 + 18x g/mol  

The percentage of water is 51.2%  
=> 18x/(120 + 18x) = 51.2/100 => 1800x = 51.2*(120 + 18x)  

x is 6.99 ≈ 7  

Hence, the formula is MgSO4.7H2O

Wed October 29, 2014  

Ans: Insectivorous plants also called carnivorous plants grow in nitrogen deficient environment. They capture insects, spiders, crustaceans, mites and protozoans and obtain nitrogen. One of the most critical plant nutrients is nitrogen which is usually taken up by plants in the form of nitrates. Nitrogen is a nutrient that is easily leached out even from ordinary soils. Therefore, the plants that grow in these soils have evolved into carnivorous plants that capture and digest insects as a means of obtaining nitrates. 
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