A cone, a hemisphere and a cylinder stand
Question: A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2 : 3. Solution: Let r be the radius of the base. and h be the height. Here,h = r. Now, The ratio of their volumes will be Volume of cone : volume of hemisphere : volume of a cylinder $\frac{1}{3} \pi r^{2} h: \frac{2}{3} \pi r^{3}: \pi r^{2} h$ $V_{1}: V_{2}: V_{3}=\frac{1}{3} \pi r^{3}: \frac{2}{3} \pi r^{3}: \pi r^{3}$ Hence, $\quad V_{1}: V_{2}: V_{3}=1: 2...
Read More →Describe the following:
Question: Describe the following: (i)Acetylation (ii)Cannizzaro reaction (iii)Cross aldol condensation (iv)Decarboxylation Solution: (i)Acetylation The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dirnethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating agent...
Read More →How will you bring about the following conversions in not more than two steps?
Question: How will you bring about the following conversions in not more than two steps? (i)Propanone to Propene (ii)Benzoic acid to Benzaldehyde (iii)Ethanol to 3-Hydroxybutanal (iv)Benzene tom-Nitroacetophenone (v)Benzaldehyde to Benzophenone (vi)Bromobenzene to 1-Phenylethanol (vii)Benzaldehyde to 3-Phenylpropan-1-ol (viii)Benazaldehyde to -Hydroxyphenylacetic acid (ix)Benzoic acid tom- Nitrobenzyl alcohol Solution: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)...
Read More →1 mole of rigid diatomic gas performs a work
Question: 1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat $Q$ is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{x R}{8}$. The value of $x$ is Solution: (25) From thermodynamics law: $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W} \quad \ldots . .(1)$ $\mathrm{Q}=\mathrm{n} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}+\frac{Q}{5}$ $Q-\frac{Q}{5}=1 \times \frac{5}{2} R \times \Delta \mathrm{T}$ $\mathrm{Q}=\frac{25}{8} R \Delta \...
Read More →A metallic sphere of radius 10.5 cm
Question: A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained. Solution: The radius of spherer= 10.5 cm The volume of sphere $=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \pi \times(10.5)^{3}$ $=\frac{4}{3} \pi\left(\frac{21}{2}\right)^{3}$ $=\frac{4}{3} \pi \frac{441 \times 21}{8}$ $=\frac{441 \times 21}{6} \pi(\mathrm{cm})^{3}$ Letnbe the number of cones obtained when the sphere is recast in to small c...
Read More →In the given figure, LM CB and LN CD.
Question: In the given figure, LM CB and LN CD. Prove that $\frac{A M}{A B}=\frac{A N}{A D}$. Solution: $L M \| C B$ and $L N \| C D$ Therefore, applying Thales' theorem, we have: $\frac{A B}{A M}=\frac{A C}{A L}$ and $\frac{A D}{A N}=\frac{A C}{A L}$ $\Rightarrow \frac{A B}{A M}=\frac{A D}{A N}$ $\therefore \frac{A M}{A B}=\frac{A N}{A D}$ This completes the proof....
Read More →How will you prepare the following compounds from benzene?
Question: How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom (i)Methyl benzoate(ii)m-Nitrobenzoic acid (iii)p-Nitrobenzoic acid(iv)Phenylacetic acid (v)p-Nitrobenzaldehyde. Solution: (i) (ii) (iii) (iv) (v)...
Read More →Give simple chemical tests to distinguish between the following pairs of compounds.
Question: Give simple chemical tests to distinguish between the following pairs of compounds. (i)Propanal and Propanone (ii)Acetophenone and Benzophenone (iii)Phenol and Benzoic acid (iv)Benzoic acid and Ethyl benzoate (v)Pentan-2-one and Pentan-3-one (vi)Benzaldehyde and Acetophenone (vii)Ethanal and Propanal Solution: (i)Propanal and propanone can be distinguished by the following tests. (a)Tollens test Propanal is an aldehyde. Thus, it reduces Tollens reagent. But, propanone being a ketone do...
Read More →A hemisphere of lead of radius 7 cm
Question: A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base. Solution: Radius of hemispherer= 7 cm The volume of hemisphere $=\frac{2}{3} \pi r^{3}$ $=\frac{2}{3} \pi \times(7)^{3}$ $=\frac{2}{3} \pi \times 343$ $=\frac{686}{3} \pi \mathrm{cm}^{3}$ Since, The hemisphere cast into the right circular cone. The height of coneh= 49 cm Letxbe the radius of cone. Clearly, Volume of cone = volume of hemisphere $=14$ $x^{2}=14$ $x=\sqrt{1...
Read More →The corresponding sides of two similar triangles are in the ratio 2 : 3.
Question: The corresponding sides of two similar triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm2, find the area of the larger triangle. Solution: It is given that the triangles are similar.Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides. $\therefore \frac{48}{\text { Area of larger triangle }}=\frac{2^{2}}{3^{2}}$ $\Rightarrow \frac{48}{\text { Area of larger triangle }}=\frac{4}{9}$ $\Rightarro...
Read More →Arrange the following compounds in increasing order of their property as indicated:
Question: Arrange the following compounds in increasing order of their property as indicated: (i)Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyltert-butyl ketone (reactivity towards HCN) (ii)CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength) (iii)Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) Solution: (i)When HCN reacts with a compound, the attacking species is a nucleophile, CN. Therefore, as the negative charg...
Read More →Solve this
Question: The temperature $\theta$ at the junction of two insulating sheets, having thermal resistances $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ as well as top and bottom temperatures $\theta_{1}$ and $\theta_{2}$ (as shown in figure) is given by : $\frac{\theta_{1} R_{2}+\theta_{2} R_{1}}{R_{1}+R_{2}}$$\frac{\theta_{1} R_{2}-\theta_{2} R_{1}}{R_{2}-R_{1}}$$\frac{\theta_{2} R_{2}-\theta_{1} R_{1}}{R_{2}-R_{1}}$$\frac{\theta_{1} R_{1}+\theta_{2} R_{2}}{R_{1}+R_{2}}$Correct Option: 1 Solution: (1) Te...
Read More →In an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD.
Question: In an trapeziumABCD, it is given thatAB∥CDandAB= 2CD. Its diagonalsACandBDintersect at the pointOsuch that ar(∆AOB) = 84 cm2. Find ar(∆COD). Solution: In ∆AOB and ∆COD, we have: $\angle A O B=\angle C O D($ Vertically opposite angles $)$ $\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alternate angles as $\mathrm{AB} \| \mathrm{CD}$ ) Applying AA similiarity criterion, we get: $\triangle \mathrm{AOB} \sim \triangle \mathrm{COD}$ $\therefore \frac{a r(\triangle A O B)}{a r(\triangle C O D)}=...
Read More →An organic compound (A)
Question: An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene.Write equations for the reactions involved. Solution: An organic compound A with molecular formula C8H16O2gives a carboxylic acid (B) and an alcohol (C) on hydrolysis with dilute sulphuric acid. Thus, compound A must be an ester. Further, alcohol C gives acid B ...
Read More →∆ABC ∼ ∆DEF such that ar(∆ABC) = 64 cm2 and ar(∆DEF) = 169 cm2.
Question: ∆ABC ∆DEFsuch that ar(∆ABC) = 64 cm2and ar(∆DEF) = 169 cm2. IfBC= 4 cm, findEF. Solution: $\because \triangle A B C \sim \triangle D E F$ $\therefore \frac{a r(\triangle A B C)}{a r(\triangle D E F)}=\frac{B C^{2}}{E F^{2}}$ $\Rightarrow \frac{64}{169}=\frac{4^{2}}{E F^{2}}$ $\Rightarrow E F^{2}=\frac{16 \times 169}{64}$ $\Rightarrow E F=\frac{4 \times 13}{8}=6.5 \mathrm{~cm}$...
Read More →An organic compound with the molecular formula
Question: An organic compound with the molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollens reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound. Solution: It is given that the compound (with molecular formula C9H10O) forms 2, 4-DNP derivative and reduces Tollens reagent. Therefore, the given compound must be an aldehyde. Again, the compound undergoes cannizzaro reaction and on oxidation gives 1, 2-benzened...
Read More →A reversible heat engine converts one- fourth of the heat
Question: A reversible heat engine converts one- fourth of the heat input into work. When the temperature of the sink is reduced by $52 \mathrm{~K}$, its efficiency is doubled. The temperature in Kelvin of the source will be Solution: (208) $\because \mathrm{n}=\frac{\mathrm{W}}{\mathrm{Q}_{\mathrm{in}}}=\frac{1}{4}$ $\frac{1}{4}=1-\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{T_{1}}{T_{2}}=\frac{3}{4}$ When the temperature of the sink is reduced by $52 \mathrm{k}$ then its efficiency is double...
Read More →Write structural formulas and names of four possible aldol condensation products from propanal and butanal.
Question: Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile. Solution: (i)Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile. (ii)Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile. (iii)Taking one molecule each of propanal and butanal in which propanal acts as a...
Read More →Find the length of the altitude of an equilateral triangle of side 2a cm.
Question: Find the length of the altitude of an equilateral triangle of side 2acm. Solution: Let the triangle be ABC with AD as its altitude. Then, D is the midpoint of BC.In right-angled triangle ABD, we have: $A B^{2}=A D^{2}+D B^{2}$ $\Rightarrow A D^{2}=A B^{2}-D B^{2}=4 a^{2}-a^{2} \quad\left(\because B D=\frac{1}{2} B C\right)$ $=3 a^{2}$ $A D=\sqrt{3} a$ Hence, the length of the altitude of an equilateral triangle of side $2 a \mathrm{~cm}$ is $\sqrt{3} a \mathrm{~cm}$....
Read More →A ladder 10 m long reaches the window of a house 8 m above the ground.
Question: A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall. Solution: Let the ladder be AB and BC be the height of the window from the ground. We have:AB = 10 m and BC = 8 mApplying Pythagoras theorem in right-angled triangle ACB, we have: $A B^{2}=A C^{2}+B C^{2}$ $\Rightarrow A C^{2}=A B^{2}-B C^{2}=10^{2}-8^{2}=100-64=36$ $\Rightarrow A C=6 \mathrm{~m}$ Hence, the foot of the ladder is 6 m away from th...
Read More →How will you convert ethanal into the following compounds?
Question: How will you convert ethanal into the following compounds? (i)Butane-1, 3-diol (ii)But-2-enal (iii)But-2-enoic acid Solution: (i)On treatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane-1, 3-diol on reduction. (ii)On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal. (iii)When treated with Tollens reagent, But-2-enal produced in the above reaction produces but-2-enoic acid ....
Read More →Which of the following compounds would undergo aldol condensation,
Question: Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i)Methanal(ii)2-Methylpentanal (iii)Benzaldehyde(iv)Benzophenone (v)Cyclohexanone(vi)1-Phenylpropanone (vii)Phenylacetaldehyde(viii)Butan-1-ol (ix)2, 2-Dimethylbutanal Solution: Aldehydes and ketones having at least one -hydrogen undergo aldol condensation. The compounds (ii) 2me...
Read More →In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm
Question: In the given figure,DE∥BCsuch thatAD=xcm,DB= (3x+ 4) cm,AE= (x+ 3) cm andEC= (3x+ 19) cm. Find the value of x. Solution: $\because D E \| B C$ $\therefore \frac{A D}{D B}=\frac{A E}{E C} \quad$ (Basic proportionality theorem) $\frac{x}{3 x+4}=\frac{x+3}{3 x+19}$ $\Rightarrow x(3 x+19)=(x+3)(3 x+4)$ $\Rightarrow 3 x^{2}+19 x=3 x^{2}+4 x+9 x+12$ $\Rightarrow 19 x-13 x=12$ $\Rightarrow 6 x=12$ $\Rightarrow x=2$...
Read More →The diameter of a copper sphere is 18 cm.
Question: The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter. Solution: The radius of copper sphere, $R=\frac{18}{2}=9 \mathrm{~cm}$ The volume of sphere $=\frac{4}{3} \pi \times(9)^{3}$ $=\frac{4}{3} \pi \times 729$ $=972 \pi \mathrm{cm}^{2}$ Since, The sphere is melted and drawn into a long circular wire of length 108 m = 10800 cm Letrbe the radius of wire, Clearly...
Read More →Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
Question: Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i)PhMgBr and then H3O+ (ii)Tollens reagent (iii)Semicarbazide and weak acid (iv)Excess ethanol and acid (v)Zinc amalgam and dilute hydrochloric acid Solution: (i) (ii) (iii) (iv) (v)...
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