Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Question: Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br. Solution: Mg: There are two valence electrons in Mg atom. Hence, the Lewis dot symbol for Mg is: $\ddot{M}_{g}$ Na: There is only one valence electron in an atom of sodium. Hence, the Lewis dot structure is: Na* B: There are 3 valence electrons in Boron atom. Hence, the Lewis dot structure is: ${ }^{\circ} \cdot{B} \cdot$ Q: There are six valence electrons in an atom of oxygen. Hence, the Lewis dot struct...
Read More →The escape speed of a projectile on the earth’s surface is 11.2 km s–1.
Question: The escape speed of a projectile on the earth's surface is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. Solution: Escape velocity of a projectile from the Earth, $v_{\mathrm{esc}}=11.2 \mathrm{~km} / \mathrm{s}$ Projection velocity of the projectile, $v_{p}=3 v_{e s c}$ Mass of the projectile $=m$ Velocity of the projectile far away from the ...
Read More →Prove the following identities,
[question] Question. Prove the following identities, where the angles involved are acute angles for which the following expressions are defined. (i) $(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$. (ii) $\frac{\cos \mathbf{A}}{\mathbf{1}+\sin \mathbf{A}}+\frac{\mathbf{1}+\sin \mathbf{A}}{\cos \mathbf{A}}=2 \sec \mathrm{A}$. (iii) $\frac{\tan \theta}{\mathbf{1}-\cot \theta}+\frac{\cot \theta}{\mathbf{1}-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$ (i...
Read More →Given the dental formula of human beings
Question: Given the dental formula of human beings Solution: Thedental formula expresses the arrangement of teeth in each half of the upper jaw and the lower jaw. The entire formula is multiplied by two to express the total number of teeth. The dental formula for milk teeth in humans is: $\frac{2102}{2102} \times 2=20$ Each half of the upper jaw and the lower jaw has 2 incisors, 1 canine, and 2 molars. Premolars are absent in milk teeth. The dental formula for permanent teeth in humans is: $\fra...
Read More →Explain the formation of a chemical bond.
Question: Explain the formation of a chemical bond. Solution: A chemical bond is defined as an attractive force that holds the constituents (atoms, ions etc.) together in a chemical species. Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory, and molecular orbital theory. A chemical bond formation is attributed to the tendency of a system to attain stability. It was observed tha...
Read More →Describe the process of digestion of protein in stomach.
Question: Describe the process of digestion of protein in stomach. Solution: The digestion of proteins begins in the stomach and is completed in the small intestine. The digestive juice secreted in the gastric glands present on the stomach walls is called gastric juice. The food that enters the stomach becomes acidic on mixing with this gastric juice. The main components of gastric juice are hydrochloric acid, pepsinogen, mucus, and rennin. Hydrochloric acid dissolves the bits of food and create...
Read More →A rocket is fired vertically with a speed of 5 km s–1 from the earth’s surface.
Question: A rocket is fired vertically with a speed of $5 \mathrm{~km} \mathrm{~s}^{-1}$ from the earth's surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth $=6.0 \times 10^{24} \mathrm{~kg}$; mean radius of the earth $=6.4 \times 10^{6} \mathrm{~m}$; $\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$. Solution: $8 \times 10^{6} \mathrm{~m}$ from the centre of the Earth Velocity of the rocket, $v=5 \mathrm{~km} / \math...
Read More →Analyse the events during every stage of cell cycle and notice how the following two parameters change
[question] Question. Analyse the events during every stage of cell cycle and notice how the following two parameters change (i) Number of chromosomes (N) per cell (ii) Amount of DNA content (C) per cell [/question] [solution] Solution: During meiosis, the number of chromosomes and the amount of DNA in a cell change. (i) Number of chromosomes (N) per cell During anaphase I of the meiotic cycle, the homologous chromosomes separate and start moving toward their respective poles. As a result, the bi...
Read More →State the role of pancreatic juice in digestion of proteins.
Question: State the role of pancreatic juice in digestion of proteins. Solution: Pancreatic juice contains a variety of inactive enzymes such as trypsinogen, chymotrypsinogen, and carboxypeptidases. These enzymes play an important role in the digestion of proteins. Physiology of protein-digestion Theenzyme enterokinase is secreted by the intestinal mucosa. It activates trypsinogen into trypsin. TrypsinogenTrypsin + Inactive peptide Trypsin then activates the other enzymes of pancreatic juice suc...
Read More →Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is:
Question: Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is: (a) F Cl O N (b) F O Cl N (c) Cl F O N (d) O F N Cl Solution: The oxidizing character of elements increases from left to right across a period. Thus, we get the decreasing order of oxidizing property as F O N. Again, the oxidizing character of elements decreases down a group. Thus, we get F Cl. However, the oxidizing character of O is more than that of Cl i.e., O C...
Read More →Answer briefly:
Question: Answer briefly: (a) Why are villi present in the intestine and not in the stomach? (b) How does pepsinogen change into its active form? (c) What are the basic layers of the wall of alimentary canal? (d) How does bile help in the digestion of fats? Solution: (a)The mucosal wall of the small intestine forms millions of tiny finger-like projections known as villi. These villi increase the surface area for more efficient food absorption. Within these villi, there are numerous blood vessels...
Read More →Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is:
Question: Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is: (a) B C Si N F (b) Si C B N F (c) F N C B Si ( d) F N C Si B Solution: The non-metallic character of elements increases from left to right across a period. Thus, the decreasing order of non-metallic character is F N C B. Again, the non-metallic character of elements decreases down a group. Thus, the decreasing order of non-metallic characters of C and Si are C Si. However, Si is less non-...
Read More →Can there be DNA replication without cell division?
[question] Question. Can there be DNA replication without cell division? [/question] [solution] Solution: There can be DNA replication without cell division. During cell division, the parent cell gets divided into two daughter cells. However, if there is a repeated replication of DNA without any cell division, then this DNA will keep accumulating inside the cell. This would increase the volume of the cell nucleus, thereby causing cell expansion. An example of DNA duplication without cell divisio...
Read More →Assuming the earth to be a sphere of uniform mass density,
Question: Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface? Solution: Weight of a body of mass $m$ at the Earth's surface, $W=m g=250 \mathrm{~N}$ Body of mass $m$ is located at depth, $d=\frac{1}{2} R_{e}$ Wher $R_{c}=$ Radius of the Earth Acceleration due to gravity at depth $g(d)$ is given by the relation: $g^{\prime}=\left(1-\frac{d}{R_{e}}\right) g$ $=\left(1-\frac{R_{e}}{2 \tim...
Read More →Can there be mitosis without DNA replication in S phase?
[question] Question. Can there be mitosis without DNA replication in S phase? [/question] [solution] Solution: Mitotic cell division cannot take place without DNA replication in S phase. Two important events take place during S phase – one is the synthesis or duplication of DNA and the other is the duplication of the centriole. DNA duplication is important as it maintains the chromosome number in the daughter cells. Mitosis is an equational division. Therefore, the duplication of DNA is an impor...
Read More →A body weighs 63 N on the surface of the earth.
Question: A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth? Solution: Weight of the body, $W=63 \mathrm{~N}$ Acceleration due to gravity at heighthfrom the Earths surface is given by the relation: $g^{\prime}=\frac{g}{\left(\frac{1+h}{R_{e}}\right)^{2}}$ Where, g = Acceleration due to gravity on the Earths surface Re = Radius of the Earth For $h=\frac{R_{e}}{2}$ $g^{\prime}=\frac{g}{\left(1+\f...
Read More →A car weighs 1800 kg.
[question] Question. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. [/question] [solution] solution: Mass of the car, m = 1800 kg Distance between the front and back axles, d = 1.8 m Distance between the C.G. (centre of gravity) and the back axle = 1.05 m The various forces acting on the car are shown in the following figur...
Read More →Choose the correct answer among the following:
Question: Choose the correct answer among the following: (a) Gastric juice contains (i) pepsin, lipase and rennin (ii) trypsin lipase and rennin (iii) trypsin, pepsin and lipase (iv) trypsin, pepsin and renin (b)Succus entericus is the name givento (i) a junction between ileum and large intestine (ii) intestinal juice (iii) swelling in the gut (iv) appendix Solution: Answer(a): (i)Pepsin, lipase, and rennin Gastric juice contains pepsin, lipase, and rennin. Pepsin is secreted in an inactive form...
Read More →Considering the elements B, Al, Mg, and K, the correct order of their metallic character is:
Question: Considering the elements B, Al, Mg, and K, the correct order of their metallic character is: (a) $\mathrm{B}\mathrm{Al}\mathrm{Mg}\mathrm{K}$ (b) $A IM gBK$ (c) $M gA IKB$ (d) $KM gA IB$ Solution: The metallic character of elements decreases from left to right across a period. Thus, the metallic character of Mg is more than that of Al. The metallic character of elements increases down a group. Thus, the metallic character of Al is more than that of B. Considering the above statements, ...
Read More →Discuss with your teacher about
[question] Question. Discuss with your teacher about (i) haploid insects and lower plants where cell-division occurs, and (ii) some haploid cells in higher plants where cell-division does not occur. [/question] [solution] Solution: (i) In some insects and lower plants, fertilization is immediately followed by zygotic meiosis, which leads to the production of haploid organisms. This type of life cycle is known as haplontic life cycle. (ii) The phenomenon of polyploidy can be observed in some hapl...
Read More →A Saturn year is 29.5 times the earth year.
Question: A Saturn year is $29.5$ times the earth year. How far is the Saturn from the sun if the earth is $1.50 \times 10^{8} \mathrm{~km}$ away from the sun? Solution: Distance of the Earth from the Sun, $r_{\mathrm{e}}=1.5 \times 10^{8} \mathrm{~km}=1.5 \times 10^{11} \mathrm{~m}$' Time period of the Earth $=T_{e}$ Time period of Saturn, $T_{S}=29.5 T_{e}$ Distance of Saturn from the Sun $=r_{\mathrm{s}}$ From Kepler's third law of planetary motion, we have $T=\left(\frac{4 \pi^{2} r^{3}}{\ma...
Read More →How will you ‘weigh the sun’, that is estimate its mass?
Question: How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun is $1.5 \times 10^{8} \mathrm{~km}$. Solution: Orbital radius of the Earth around the Sun, $r=1.5 \times 10^{11} \mathrm{~m}$ Time taken by the Earth to complete one revolution around the Sun, $T=1$ year $=365.25$ days $=365.25 \times 24 \times 60 \times 60 \mathrm{~s}$ Universal gravitational constant, $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ Thus, ma...
Read More →Which one of the following statements is incorrect in relation to ionization enthalpy?
Question: Which one of the following statements is incorrect in relation to ionization enthalpy? (a) Ionization enthalpy increases for each successive electron. (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration. (c) End of valence electrons is marked by a big jump in ionization enthalpy. (d) Removal of electron from orbitals bearing lowernvalue is easier than from orbital having highernvalue. Solution: Electrons in orbitals b...
Read More →Choose the correct option. Justify your choice :
[question] Question. Choose the correct option. Justify your choice : (i) $9 \sec ^{2} A-9 \tan ^{2} A=$ (A) 1 (B) 9 (C) 8 (D) 0 (ii) $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$ (A) 0 (B) 1 (C) 2 (D) – 1 (iii) $(\sec A+\tan A)(1-\sin A)=$ (A) sec A (B) sin A (C) cosec A (D) cos A (iv) $\frac{\mathbf{1}+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}=$ (A) $\sec ^{2} \mathrm{~A}$ (B) $-1$ (C) $\cot ^{2} A$ (D) $\tan ^{2} \mathrm{~A}$ [/question] [solution] Sol...
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