In the given figure
Question: In the given figure, $\angle 1=\angle 2$ and $\frac{A C}{B D}=\frac{C B}{C E}$. Prove that ∆ACB ∆DCE. Solution: We have: $\frac{A C}{R D}=\frac{C B}{C F}$ $\Rightarrow \frac{A C}{C B}=\frac{B D}{C E}$ $\Rightarrow \frac{A C}{C B}=\frac{C D}{C E}($ Since, $B D=D C$ as $\angle 1=\angle 2)$ Also, $\angle 1=\angle 2$ i.e, $\angle D B C=\angle A C B$ Therefore, by SAS similarity theorem, we get: $\triangle A C B \sim \triangle D C E$...
Read More →In an isosceles ∆ ABC, the base AB is produced both ways to P and Q, such that AP × BQ = AC2.
Question: In an isosceles ∆ABC, the baseABis produced both ways toPandQ, such thatAPBQ=AC2.Prove that ∆ACP ∆BCQ. Solution: Disclaimer: It should be $\triangle A P C \sim \triangle B C Q$ instead of $\triangle A C P \sim \triangle B C Q$ It is given that $\triangle \mathrm{ABC}$ is an isosceles triangle. Therefore,CA=CB $\Rightarrow \angle C A B=\angle C B A$ $\Rightarrow 180^{\circ}-\angle C A B=180^{\circ}-\angle C B A$ $\Rightarrow \angle C A P=\angle C B Q$ Also, $A P \times B Q=A C^{2}$ $\Ri...
Read More →What do you mean by activity and selectivity of catalysts?
Question: What do you mean by activity and selectivity of catalysts? Solution: (a) Activity of a catalyst: The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active. (b) Selectivity of the catalyst: The ability of the catalyst to direct a re...
Read More →Give four examples of heterogeneous catalysis.
Question: Give four examples of heterogeneous catalysis. Solution: (i)Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as a catalyst. $2 \mathrm{SO}_{2(g)} \stackrel{\mathrm{Pr}_{(0)}}{\longrightarrow} 2 \mathrm{SO}_{3(g)}$ (ii)Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron. $\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \stackrel{\mathrm{Fe}_{(0)}}{\longrightarrow} 2 \mathrm{NH}_{3(g)}$ This process is called t...
Read More →The perimeter of the sector OAB shown in the following figure, is
Question: The perimeter of the sectorOABshown in the following figure, is (a) $\frac{64}{3} \mathrm{~cm}$ (b) $26 \mathrm{~cm}$ (c) $\frac{64}{5} \mathrm{~cm}$ (d) $19 \mathrm{~cm}$ Solution: We know that perimeter of a sector of radius $l=2 r+\frac{\theta}{360} \times 2 \pi r$.....(1) We have given sector angle and radius of the sector and we are asked to find perimeter of the sector OAB. Therefore, substituting the corresponding values of the sector angle and radius in equation (1) we get, Per...
Read More →Action of soap is due to emulsification and micelle formation.
Question: Action of soap is due to emulsification and micelle formation. Comment. Solution: The cleansing action of soap is due to emulsification and micelle formation. Soaps are basically sodium and potassium salts of long chain fatty acids, R-COO-Na+. The end of the molecule to which the sodium is attached is polar in nature, while the alkyl-end is non-polar. Thus, a soap molecule contains a hydrophilic (polar) and a hydrophobic (non-polar) part. When soap is added to water containing dirt, th...
Read More →A vertical stick of length 7.5 m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long.
Question: A vertical stick of length 7.5 m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower. Solution: Let AB be the vertical stick and BC be its shadow.Given:AB = 7.5 m, BC = 5 m Let PQ be the tower and QR be its shadow.Given:QR = 24mLet the length ofPQbexm. In $\triangle A B C$ and $\triangle P Q R$ we have: $\angle A B C=\angle P Q R=90^{\circ}$ $\angle A C B=\angle P R Q$ (Angular elevation of the Sun at the same time) ...
Read More →What is demulsification?
Question: What is demulsification? Name two demulsifiers. Solution: The process of decomposition of an emulsion into its constituent liquids is called demulsification. Examples of demulsifiers are surfactants, ethylene oxide, etc....
Read More →What are emulsions? What are their different types?
Question: What are emulsions? What are their different types? Give example of each type. Solution: The colloidal solution in which both the dispersed phase and dispersion medium are liquids is called an emulsion. There are two types of emulsions: (a) Oil in water type: Here, oil is the dispersed phase while water is the dispersion medium. For example: milk, vanishing cream, etc. (b) Water in oil type: Here, water is the dispersed phase while oil is the dispersion medium. For example: cold cream,...
Read More →Explain what is observed
Question: Explain what is observed (i)When a beam of light is passed through a colloidal sol. (ii)An electrolyte, NaCl is added to hydrated ferric oxide sol. (iii)Electric current is passed through a colloidal sol? Solution: (i)When a beam of light is passed through a colloidal solution, then scattering of light is observed. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidal solution. (ii)When NaCl is added to ferric oxide sol, it diss...
Read More →If the perimeter of a semi-circular
Question: If the perimeter of a semi-circular protractor is 36 cm, then its diameter is(a) 10 cm(b) 12 cm(c) 14 cm(d) 16 cm Solution: We know that perimeter of a semi-circle of radius $r=\frac{1}{2}(2 \pi r)+2 r$......(1) We have given the perimeter of the semi-circle and we are asked to find the diameter of the semi-circle. Therefore, substituting the perimeter of the semi-circle in equation (1) we get, $36=\frac{1}{2}(2 \pi r)+2 r$ Multiplying both sides of the equation by 2 we get, $72=2 \pi ...
Read More →How are colloids classified on the basis of
Question: How are colloids classified on the basis of (i)Physical states of components (ii)Nature of dispersion medium and (iii)Interaction between dispersed phase and dispersion medium? Solution: Colloids can be classified on various bases: (i)On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids, or gases, we can have eight types of colloids. (ii)On the basis of the dispers...
Read More →How are colloids classified on the basis of
Question: How are colloids classified on the basis of (i)Physical states of components (ii)Nature of dispersion medium and (iii)Interaction between dispersed phase and dispersion medium? Solution: Colloids can be classified on various bases: (i)On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids, or gases, we can have eight types of colloids. (ii)On the basis of the dispers...
Read More →In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC.
Question: In the given figure,DBBC,DEABandACBC. Prove that $\frac{B E}{D E}=\frac{A C}{B C}$. Solution: In $\triangle B E D$ and $\triangle A C B$, we have: $\angle B E D=\angle A C B=90^{\circ}$ $\because \angle B+\angle C=180^{\circ}$ $\therefore B D \| A C$ $\angle E B D=\angle C A B$ (Alternate angles) Therefore, by AA similarity theorem, we get: $\triangle B E D \sim \triangle A C B$ $\Rightarrow \frac{B E}{A C}=\frac{D E}{B C}$ $\Rightarrow \frac{B E}{D E}=\frac{A C}{B C}$ This completes t...
Read More →What are enzymes?
Question: What are enzymes? Write in brief the mechanism of enzyme catalysis. Solution: Enzymes are basically protein molecules of high molecular masses. These form colloidal solutions when dissolved in water. These are complex, nitrogenous organic compounds produced by living plants and animals. Enzymes are also called biochemical catalysts. Mechanism of enzyme catalysis: On the surface of the enzymes, various cavities are present with characteristic shapes. These cavities possess active groups...
Read More →What is the difference between multimolecular and macromolecular colloids?
Question: What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids? Solution: (i)In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than 1 nm. The molecules in the aggregate are held together by van der Waals forces of attraction. Examples of such colloids include gold sol and sulphur sol. (ii)In macro-molecular ...
Read More →ABCD is a parallelogram and E is a point on BC.
Question: ABCDis a parallelogram andEis a point onBC. If the diagonalBDintersectsAEatF, prove thatAFFB=EFFD. Solution: We have: $\angle A F D=\angle E F B \quad$ (Vertically Opposite angles) $\because \mathrm{DA} \| \mathrm{BC}$ $\therefore \angle D A F=\angle B E F \quad$ (Alternate angles) $\triangle D A F \sim \triangle B E F$ (AA similarity theorem) $\Rightarrow \frac{A F}{E F}=\frac{F D}{F B}$ or, $A F \times F B=F D \times E F$ This completes the proof....
Read More →If the sum of the areas of two circles with radii
Question: If the sum of the areas of two circles with radii $r_{1}$ and $r_{2}$ is equal to the area of a circle of radius $r$, then $r_{1}^{2}+r_{2}^{2}$ (a) $r^{2}$ (b) $=r^{2}$ (c) $\mathrm{r}^{2}$ (d) None of these Solution: We have given area of the circle of radiusr1plus area of the circle of radiusr2is equal to the area of the circle of radiusr. Therefore, we have, $\pi r_{1}^{2}+\pi r_{2}^{2}=\pi r^{2}$ Cancelling, we get $r_{1}^{2}+r_{2}^{2}=r^{2}$ Therefore, $r_{1}^{2}+r_{2}^{2}=r^{2}$...
Read More →What are lyophilic and lyophobic sols?
Question: What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated? Solution: (i) Lyophilic sols: Colloidal sols that are formed by mixing substances such as gum, gelatin, starch, etc. with a suitable liquid (dispersion medium) are called lyophilic sols. These sols are reversible in nature i.e., if two constituents of the sol are separated by any means (such as evaporation), then the sol can be prepared again by simply mixing the dispersion...
Read More →Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Question: Discuss the effect of pressure and temperature on the adsorption of gases on solids. Solution: Effect of pressure Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure. Effect of temperature Adsorption is an exothermic process. Thus, in accordance with Le-Chateliers principle, the magnitude of adsorption decreases with an increase in temperature....
Read More →The ratio of the areas of a circle and an equilateral
Question: The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal, is (a) $\pi: \sqrt{2}$ (b) $\pi: \sqrt{3}$ (c) $\sqrt{3}: \pi$ (d) $\sqrt{2}: \pi$ Solution: We are given that diameter and side of an equilateral triangle are equal. Let d and a are the diameter and side of circle and equilateral triangle respectively. $\therefore d=a$ We know that area of the circle $=\pi r^{2}$ Area of the equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$ ...
Read More →How are the colloidal solutions classified on the basis of physical states of
Question: How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium? Solution: One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems....
Read More →P and Q are points on the sides AB and AC, respectively, of ∆ ABC. If AP = 2 cm, PB = 4 cm,
Question: PandQare points on the sidesABandAC, respectively, of ∆ABC. IfAP= 2 cm,PB= 4 cm,AQ= 3 cm andQC= 6 cm, show thatBC= 3PQ. Solution: We have: $\frac{A P}{A B}=\frac{2}{6}=\frac{1}{3}$ and $\frac{\mathrm{AQ}}{\mathrm{AC}}=\frac{3}{9}=\frac{1}{3}$ $\Rightarrow \frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$ In $\triangle A P Q$ and $\triangle A B C$, we have : $\frac{A P}{A B}=\frac{A Q}{A C}$ $\angle A=\angle A$ Therefore, by AA similarity theorem, we get: $\triangle A P Q...
Read More →Why is adsorption always exothermic?
Question: Why is adsorption always exothermic? Solution: Adsorption is always exothermic. This statement can be explained in two ways. (i)Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic. (ii)ΔHof adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ΔSis...
Read More →Why is adsorption always exothermic?
Question: Why is adsorption always exothermic? Solution: Adsorption is always exothermic. This statement can be explained in two ways. (i)Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic. (ii)ΔHof adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ΔSis...
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