Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal?
Question: Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal? (i) (ii) Solution: (i) (ii)...
Read More →Each side of a box made of metal sheet in cubic shape is ' a ' at room temperature 'T',
Question: Each side of a box made of metal sheet in cubic shape is ' $a$ ' at room temperature 'T', the coefficient of linear expansion of the metal sheet is ' $\alpha^{\prime}$. The metal sheet is heated uniformly, by a small temperature $\Delta \mathrm{T}$, so that its new temperature is $\mathrm{T}+\Delta \mathrm{T}$. Calculate the increase in the volume of the metal box-$\frac{4}{3} \pi a^{3} \alpha \Delta T$$4 \pi \mathrm{a}^{3} \alpha \Delta \mathrm{T}$$3 a^{3} \alpha \Delta T$$4 \mathrm{a...
Read More →Name the following compounds according to IUPAC system.
Question: Name the following compounds according to IUPAC system. (i) (ii) (iii) (iv) (v) Solution: (i)3-Chloromethyl-2-isopropylpentan-1-ol (ii)2, 5-Dimethylhexane-1, 3-diol (iii)3-Bromocyclohexanol (iv)Hex-1-en-3-ol (v)2-Bromo-3-methylbut-2-en-1-ol...
Read More →If ∆ABC ∼ ∆EDF and ∆ABC is not similar to ∆DEF,
Question: If ∆ABC ∆EDFand ∆ABCis not similar to ∆DEF, then which of the following is not true?(a)BCEF=ACFD(b)ABEF=ACDE(c)BCDE=ABEF(d)BCDE=ABFD Solution: (c)BCDE=ABEF∆ABC ∆EDFTherefore, $\frac{A B}{D E}=\frac{A C}{E F}=\frac{B C}{D F}$ $\Rightarrow B C \cdot D E \neq A B \cdot E F$...
Read More →Identify allylic alcohols in the above examples.
Question: Identify allylic alcohols in the above examples. Solution: The alcohols given in (ii) and (vi) are allylic alcohols....
Read More →Two identical metal wires of
Question: Two identical metal wires of thermal conductivities $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively are connected in series.The effective thermal conductivity of the combination is:$\frac{2 \mathrm{~K}_{1} \mathrm{~K}_{2}}{\mathrm{~K}_{1}+\mathrm{K}_{2}}$$\frac{K_{1}+K_{2}}{2 K_{1} K_{2}}$$\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{\mathrm{~K}_{1} \mathrm{~K}_{2}}$$\frac{\mathrm{K}_{1} \mathrm{~K}_{2}}{\mathrm{~K}_{1}+\mathrm{K}_{2}}$Correct Option: 1 Solution: \ $\mathrm{R}_{\text {eff }...
Read More →Classify the following as primary, secondary and tertiary alcohols:
Question: Classify the following as primary, secondary and tertiary alcohols: (i) (ii) (iii) (iv) (v) (vi) Solution: Primary alcohol (i), (ii), (iii) Secondary alcohol (iv), (v) Tertiary alcohol (vi)...
Read More →A reservoir in the form of the frustum of a right
Question: A reservoir in the form of the frustum of a right circular cone contains 44 107litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir. (Take: = 22/7) Solution: Let the depth of the frustum cone like reservoir ishm. The radii of the top and bottom circles of the frustum cone like reservoir arer1=100m andr2=50m respectively. The volume of t...
Read More →What happens when
Question: What happens when (i)n-butyl chloride is treated with alcoholic KOH, (ii)bromobenzene is treated with Mg in the presence of dry ether, (iii)chlorobenzene is subjected to hydrolysis, (iv)ethyl chloride is treated with aqueous KOH, (v)methyl bromide is treated with sodium in the presence of dry ether, (vi)methyl chloride is treated with KCN. Solution: (i)When nbutyl chloride is treated with alcoholic KOH, the formation of butlene takes place. This reaction is a dehydrohalogenation reacti...
Read More →For an ideal heat engine,
Question: For an ideal heat engine, the temperature of the source is $127^{\circ} \mathrm{C}$. In order to have $60 \%$ efficiency the temperature of the sink should be ${ }^{\circ} \mathrm{C}$. (Round off to the Nearest Integer) (write modulus or absolute value of the temperature) Solution: $(-113)$ $\mathrm{n}=0.60=1=\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ $\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}=0.4 \Rightarrow \mathrm{T}_{\mathrm{L}}}=0.4 \times 400$ $=160 \ma...
Read More →Solve the following
Question: Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b).Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. Solution: There are two primary alkyl halides having the formula, C4H9Br. They aren bulyl bromide and...
Read More →A tent consists of a frustum of a cone capped by a cone.
Question: A tent consists of a frustum of a cone capped by a cone. If the radii of the ends of the frustum be 13 m and 7 m , the height of the frustum be 8 m and the slant height of the conical cap be 12 m, find the canvas required for the tent. (Take : = 22/7) Solution: The height of the frustum cone ish= 8 m. The radii of the end circles of the frustum arer1= 13m andr2=7m. The slant height of the frustum cone is $l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$ $=\sqrt{(13-7)^{2}+8^{2}}$ $=\sqrt{1...
Read More →In ∆DEF and ∆PQR, it is given that ∠D = ∠Q and ∠R = ∠E,
Question: In ∆DEFand ∆PQR,it is given that D= Qand R= E, then which of the following is not true? (a) $\frac{E F}{P R}=\frac{D F}{P Q}$ (b) $\frac{D E}{P Q}=\frac{E F}{R P}$ (c) $\frac{D E}{Q R}=\frac{D F}{P Q}$ (d) $\frac{E F}{R P}=\frac{D E}{Q R}$ Solution: (b) $\frac{D E}{P Q}=\frac{E F}{R P}$ In ∆DEFand ∆PQR,we have: $\angle D=\angle Q$ and $\angle R=\angle E$ Applying $A A$ similarity theorem, we conclude that $\triangle D E F \sim \triangle Q R P$. Hence, $\frac{D E}{Q R}=\frac{D F}{Q P}=\...
Read More →A bimetallic strip consists of metals A and B.
Question: A bimetallic strip consists of metals $\mathrm{A}$ and $\mathrm{B}$. It is mounted rigidly as shown. The metal A has higher coefficient of expansion compared to that of metal B. When the bimetallic strip is placed in a cold both, it will : Bend towards the rightNot bend but shrinkNeither bend nor shrinkBend towards the leftCorrect Option: , 4 Solution: (4) $\alpha_{\mathrm{A}}\alpha_{\mathrm{B}}$ Length of both strips will decrease $\Delta \mathrm{L}_{\mathrm{A}}\Delta \mathrm{L}_{\mat...
Read More →The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH,
Question: The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain. Solution: In an aqueous solution, KOH almost completely ionizes to give OHions. OHion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol. On the other hand, an alcoholic solution of KOH contains alkoxide (RO) ion, which is a strong base. Thus, it can abstract a hydrogen fro...
Read More →The radii of the circular bases of a frustum of a right circular cone are 12 cm
Question: The radii of the circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 12 cm. Find the total surface area and the volume of the frustum. Solution: The height of the frustum cone ish= 12 cm. The radii of the bottom and top circles arer1= 12cm andr2= 3cm respectively. The slant height of the frustum cone is $l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$ $=\sqrt{(12-3)^{2}+12^{2}}$ $=\sqrt{225}$ $=15 \mathrm{~cm}$ The total surface area of the frustum co...
Read More →In ∆ABC and ∆DEF, it is given that
Question: In $\triangle A B C$ and $\triangle D E F$, it is given that $\frac{A B}{D E}=\frac{B C}{F D}$, then (a) B= E(b) A= D(c) B= D(d) F= F Solution: (c) $\angle B=\angle D$ Disclaimer: In the question, the ratio should be $\frac{A B}{D E}=\frac{B C}{F D}=\frac{A C}{E F}$. We can write it as: $\frac{A B}{E D}=\frac{B C}{D F}=\frac{A C}{F E}$ Therefore, $\triangle A B C \sim E D F$ Hence, the corresponding angles, i. e., $\angle B$ and $\angle D$, will be equal. i. e., $\angle B=\angle D$...
Read More →How the following conversions can be carried out?
Question: How the following conversions can be carried out? (i)Propene to propan-1-ol (ii)Ethanol to but-1-yne (iii)1-Bromopropane to 2-bromopropane (iv)Toluene to benzyl alcohol (v)Benzene to 4-bromonitrobenzene (vi)Benzyl alcohol to 2-phenylethanoic acid (vii)Ethanol to propanenitrile (viii)Aniline to chlorobenzene (ix)2-Chlorobutane to 3, 4-dimethylhexane (x)2-Methyl-1-propene to 2-chloro-2-methylpropane (xi)Ethyl chloride to propanoic acid (xii)But-1-ene to n-butyliodide (xiii)2-Chloropropan...
Read More →If the radii of the circular ends of a bucket 24 cm
Question: If the radii of the circular ends of a bucket 24 cm high are 5 cm and 15 cm respectively, find the surface area of the bucket. Solution: The height of the conical bucket ish= 24 cm. The radii of the bottom and top circles arer1= 15cm andr2= 5cm respectively. The slant height of the bucket is $l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$ $=\sqrt{(15-5)^{2}+24^{2}}$ $=\sqrt{676}$ $=26 \mathrm{~cm}$ The curved surface area of the bucket is $=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{2}^...
Read More →Solve this
Question: In $\triangle A B C, A B=6 \sqrt{3} \mathrm{~cm}, A C=12 \mathrm{~cm}$ and $B C=6 \mathrm{~cm}$. Then $\angle B$ is (a) 45o(b) 60o(c) 90o(d) 120o Solution: $\mathrm{AB}=6 \sqrt{3} \mathrm{~cm}$ $\Rightarrow \mathrm{AB}^{2}=108 \mathrm{~cm}^{2}$ $\mathrm{AC}=12 \mathrm{~cm}$ $\Rightarrow \mathrm{AC}^{2}=144 \mathrm{~cm}^{2}$ $\mathrm{BC}=6 \mathrm{~cm}$ $\Rightarrow \mathrm{BC}^{2}=36 \mathrm{~cm}$ $\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$ Since, the square of the lon...
Read More →The height of a cone is 20 cm.
Question: The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If its volume be 1/125 of the volume of the original cone, determine at what height above the base the section is made. Solution: We have the following situation as shown in the figure Let VAB be a cone of heighth1=VO1=20cm. Then from the symmetric triangles VO1A and VOA1, we have $\frac{\mathrm{VO}_{1}}{\mathrm{VO}}=\frac{\mathrm{O}_{1} \mathrm{~A}}{\mathrm{OA}_{1}}$ $\Rightarrow \frac...
Read More →p-Dichlorobenzene has higher m.p.
Question: p-Dichlorobenzene has higher m.p. and lower solubility than those ofo- and m-isomers. Discuss. Solution: p-Dichlorobenzene is more symmetrical thano-andm-isomers. For this reason, it fits more closely thano-andm-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice ofp-dichlorobenzene. As a result,p-dichlorobenzene has a higher melting point and lower solubility thano-andm-isomers....
Read More →In the given figure ∠BAC = 90° and AD ⊥ BC. Then,
Question: In the given figure BAC= 90 andADBC. Then, (a)BCCD=BC2(b)ABAC=BC2(c)BDCD=AD2(d)ABAC=AD2 Solution: (c)BDCD=AD2 In $\triangle B D A$ and $\triangle A D C$, we have : $\angle B D A=\angle A D C=90^{\circ}$ $\angle A B D=90^{\circ}-\angle D A B$ $=90^{\circ}-\left(90^{\circ}-\angle D A C\right)$ $=90^{\circ}-90^{\circ}+\angle D A C$ $=\angle D A C$ Applying $A A$ similarity theorem, we conclude that $\triangle B D A \sim \triangle A D C$. $\Rightarrow \frac{B D}{A D}=\frac{A D}{C D}$ $\Rig...
Read More →Solve the following
Question: Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH? Solution: Hydrolysis by aqueous $\mathrm{KOH}$ proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous $\mathrm{KOH}$. Now, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}$ forms $1^{\circ}$-carbocation, while $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHClC}_{6} \mathrm{H}_{5}$ forms $2^{\circ}$-carbocation, which is more stable t...
Read More →If the radii of the circular ends of a conical bucket
Question: If the radii of the circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm, find the capacity of the bucket. (Use = 22/7). Solution: The height of the conical bucket ish= 45 cm. The radii of the bottom and top circles arer1= 28cm andr2=7cm respectively. The volume/capacity of the conical bucket is $V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$ $=\frac{1}{3} \pi\left(28^{2}+28 \times 7+7^{2}\right) \times 45$ $=\frac{1}{3} \times \frac{22}{7} \...
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