thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas.
Question: thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If $V_{2}=2 V_{1}$ then the ratio of temperature $T_{2} / T_{1}$ is: $\frac{1}{\sqrt{2}}$$\frac{1}{2}$2$\sqrt{2}$Correct Option: , 4 Solution: From $\mathrm{p}-\mathrm{v}$ diagram, Given $\mathrm{Pv}^{1 / 2}=$ constant $\ldots$ (i) We know that $\mathrm{PV}=\mathrm{nRT}$ $P \propto\left(\frac{T}{v}\right)$ Put in equation (i) $\left(\frac{T}{v}\right)(v)^{1 / 2}=$ constant $T \propto v^{1 / 2}$ $\frac{T...
Read More →thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas.
Question: thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If $V_{2}=2 V_{1}$ then the ratio of temperature $T_{2} / T_{1}$ is: $\frac{1}{\sqrt{2}}$$\frac{1}{2}$2$\sqrt{2}$Correct Option: , 4 Solution: From $\mathrm{p}-\mathrm{v}$ diagram, Given $\mathrm{Pv}^{1 / 2}=$ constant $\ldots$ (i) We know that $\mathrm{PV}=\mathrm{nRT}$ $P \propto\left(\frac{T}{v}\right)$ Put in equation (i) $\left(\frac{T}{v}\right)(v)^{1 / 2}=$ constant $T \propto v^{1 / 2}$ $\frac{T...
Read More →If ∆ABC ∼ ∆DEF such that 2 AB = DE and BC = 6 cm, find EF.
Question: If ∆ABC ∆DEFsuch that 2AB=DEandBC= 6 cm, findEF. Solution: $\because \triangle A B C \sim \triangle D E F$ $\therefore \frac{A B}{D E}=\frac{B C}{E F}$ $\Rightarrow \frac{1}{2}=\frac{6}{E F}$ $\Rightarrow E F=12 \mathrm{~cm}$...
Read More →Draw structures of the following derivatives.
Question: Draw structures of the following derivatives. (i)The 2,4-dinitrophenylhydrazone of benzaldehyde (ii)Cyclopropanone oxime (iii)Acetaldehydedimethylacetal (iv)The semicarbazone of cyclobutanone (v)The ethylene ketal of hexan-3-one (vi)The methyl hemiacetal of formaldehyde Solution: (i) (ii) (iii) (iv) (v) (vi)...
Read More →A solid metallic sphere of diameter 28 cm
Question: A solid metallic sphere of diameter $28 \mathrm{~cm}$ is melted and recast into a number of smaller cones, each of diameter $4 \frac{2}{3} \mathrm{~cm}$ and height $3 \mathrm{~cm}$. Find the number of cones so formed. Solution: The radius of solid metallic sphere, $R=\frac{28}{2}=14 \mathrm{~cm}$ The volume of sphere $=\frac{4}{3} \pi R^{3}$ $=\frac{4}{3} \times \pi \times(14)^{3}$ $=\frac{4}{3} \pi \times 14 \times 14 \times 14$ $=\frac{10976 \pi}{3} \mathrm{~cm}^{3}$ Given, the spher...
Read More →Write the IUPAC names of the following ketones and aldehydes.
Question: Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. (i)CH3CO(CH2)4CH3 (ii)CH3CH2CHBrCH2CH(CH3)CHO (iii)CH3(CH2)5CHO (iv)Ph-CH=CH-CHO (v) (vi)PhCOPh Solution: (i)CH3CO(CH2)4CH3 IUPAC name:Heptan-2-one Common name:Methyl n-propyl ketone (ii)CH3CH2CHBrCH2CH(CH3)CHO IUPAC name:4-Bromo-2-methylhaxanal Common name:(-Bromo--methyl-caproaldehyde) (iii)CH3(CH2)5CHO IUPAC name:Heptanal (iv)Ph-CH=CH-CHO IUPAC name:3-phenylprop-2-enal Common nam...
Read More →The areas of two similar triangles are 25 cm2 and 36 cm2 respectively.
Question: The areas of two similar triangles are 25 cm2and 36 cm2respectively. If the altitude of the first triangle is 3.5 cm, then the corresponding altitude of the other triangle is(a) 5.6 cm(b) 6.3 cm(c) 4.2 cm(d) 7 cm Solution: (c)We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.Lethbe the altitude of the other triangle.Therefore, $\frac{25}{36}=\frac{(3.5)^{2}}{h^{2}}$ $\Rightarrow h^{2}=\frac{(3.5)^{2} \times 36}{25}$ $...
Read More →Draw the structures of the following compounds.
Question: Draw the structures of the following compounds. (i)3-Methylbutanal(ii)p-Nitropropiophenone (iii)p-Methylbenzaldehyde(iv)4-Methylpent-3-en-2-one (v)4-Chloropentan-2-one(vi)3-Bromo-4-phenylpentanoic acid (vii)p,p-Dihydroxybenzophenone(viii)Hex-2-en-4-ynoic acid Solution: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)...
Read More →Two poles of height 6 m and 11 m stand vertically upright on a plane ground.
Question: Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their feet is 12 m, then the distance between their tops is(a) 12 m(b) 13 m(c) 14 m(d) 15 m Solution: (b) 13 m Let the poles be AB and CD.It is given that:AB = 6 m and CD = 11 mLet AC be 12 m.Draw a perpendicular from B on CD, meeting CD at E.Then,BE = 12 mWe have to find BD.Applying Pythagoras theorem in right-angled triangle BED, we have: $B D^{2}=B E^{2}+E D^{2}$ $=12^{2}+5^{2} \qu...
Read More →Name the following compounds according to IUPAC system of nomenclature:
Question: Name the following compounds according to IUPAC system of nomenclature: (i)CH3CH(CH3)CH2CH2CHO (ii)CH3CH2COCH(C2H5)CH2CH2Cl (iii)CH3CH=CHCHO (iv)CH3COCH2COCH3 (v)CH3CH(CH3)CH2C(CH3)2COCH3 (vi)(CH3)3CCH2COOH (vii)OHCC6H4CHO-p Solution: (i)4-methylpentanal (ii)6-Chloro-4-ethylhexan-3-one (iii)But-2-en-1-al (iv)Pentane-2,4-dione (v)3,3,5-Trimethylhexan-2-one (vi)3,3-Dimethylbutanoic acid (vii)Benzene-1,4-dicarbaldehyde...
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{14...
Read More →In the given figure, DE ∥ BC. If DE = 5 cm, BC = 8 cm
Question: In the given figure,DE∥BC. IfDE= 5 cm,BC= 8 cm andAD= 3.5 cm, thenAB= ? (a) 5.6 cm(b) 4.8 cm(c) 5.2 cm(d) 6.4 cm Solution: (a) 5.6 cm∵ DE∥ BC $\therefore \frac{A D}{A B}=\frac{A E}{A C}=\frac{D E}{B C} \quad$ (Thales' theorem) $\Rightarrow \frac{3.5}{A B}=\frac{5}{8}$ $\Rightarrow A B=\frac{3.5 \times 8}{5}=5.6 \mathrm{~cm}$...
Read More →What is meant by the following terms?
Question: What is meant by the following terms? Give an example of the reaction in each case. (i)Cyanohydrin(ii)Acetal (iii)Semicarbazone(iv)Aldol (v)Hemiacetal(vi)Oxime (vii)Ketal(vii)Imine (ix)2,4-DNP-derivative(x)Schiffs base Solution: (i)Cyanohydrin: Cyanohydrins are organic compounds having the formula RRC(OH)CN, where R and R can be alkyl or aryl groups. Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess sodium cyanide (NaCN) as a catalyst to field cyanohydri...
Read More →A diatomic gas,
Question: A diatomic gas, having $C_{p}=\frac{7}{2} R$ and $C_{v}=\frac{5}{2} R$, is heated at constant pressure. The ratio $\mathrm{dU}: \mathrm{dQ}: \mathrm{dW}$03:07:02$05: 07: 02$$05: 07: 03$03:05:02Correct Option: , 2 Solution: (2) $C_{p}=\frac{7}{2} R$ $C_{v}=\frac{5}{2} R$ $d U=n C_{\mathrm{v}} d T$ $d Q=n C_{p} d T$ $d W=n R d T$ $d U: d Q: d W$ $C_{v}: C_{p}: R$ $\frac{5}{2} R: \frac{7}{2} R: R$ 5: $7: 2$...
Read More →In a cylindrical vessel of diameter 24 cm,
Question: In a cylindrical vessel of diameter 24 cm, filled up with sufficient quantity of water, a solid spherical ball of radius 6 cm is completely immersed. Find the increase in height of water level. Solution: Radius of spherical ballr= 6 cm, radius of cylindrical vesselr1= 12 cm Since, the ball completely immersed into the vessel, the water level is increased. Let the height of increased level. Therefore, The volume of increase water level = volume of ball ' $144 x=\frac{4}{3} \times 216$ $...
Read More →∆ABC ∼ ∆DEF and their perimeters are 32 cm and 24 cm
Question: ∆ABC ∆DEFand their perimeters are 32 cm and 24 cm respectively. IfAB= 10 cm, thenDE=?(a) 8 cm(b) 7.5 cm(c) 15 cm (d) $5 \sqrt{3} \mathrm{~cm}$ Solution: (b) 7.5 cm $\because \Delta A B C \sim \triangle D E F$ $\therefore \frac{\text { Perimeter }(\triangle \mathrm{ABC})}{\text { Perimeter }(\triangle \mathrm{DEF})}=\frac{\mathrm{AB}}{\mathrm{DE}}$ $\Rightarrow \frac{32}{24}=\frac{10}{D E}$ $\Rightarrow D E=\frac{10 \times 24}{32}=7.5 \mathrm{~cm}$...
Read More →Which acid of each pair shown here would you expect to be stronger?
Question: Which acid of each pair shown here would you expect to be stronger? (i)CH3CO2H or CH2FCO2H (ii)CH2FCO2H or CH2ClCO2H (iii)CH2FCH2CH2CO2H or CH3CHFCH2CO2H (iv) Solution: (i) The +I effect of CH3group increases the electron density on the OH bond. Therefore, release of proton becomes difficult. On the other hand, the I effect of F decreases the electron density on the OH bond. Therefore, proton can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H. (ii) F has stronger I...
Read More →A hemispherical bowl of internal radius 15 cm contains a liquid.
Question: A hemispherical bowl of internal radius 15 cm contains a liquid. The liquid is to be filled into cylindrical-shaped bottles of diameter 5 cm and height 6 cm. How many bottles are necessary to empty the bowl? Solution: Internal radius of hemispherical bowlr= 15 cm The volume of bowl $=\frac{2}{3} \pi r^{3}=$ volume of liquid $=\frac{2}{3} \pi(15)^{3}$ Volume of liquid $=2250 \pi \mathrm{cm}^{3}$ Since, the liquid filled into the cylindrical shaped bottles of radius $\frac{5}{2} \mathrm{...
Read More →Show how each of the following compounds can be converted to benzoic acid.
Question: Show how each of the following compounds can be converted to benzoic acid. (i)Ethylbenzene(ii)Acetophenone (iii)Bromobenzene(iv)Phenylethene (Styrene) Solution: (i) (ii) (iii) (iv)...
Read More →Give the IUPAC names of the following compounds:
Question: Give the IUPAC names of the following compounds: (i)PhCH2CH2COOH(ii)(CH3)2C=CHCOOH (iii)(iv) Solution: (i)3-Phenylpropanoic acid (ii)3-Methylbut-2-enoic acid (iii)2-Methylcyclopentanecarboxylic acid (iv)2,4,6-Trinitrobenzoic acid...
Read More →Find the depth of a cylindrical tank of radius 28 m,
Question: Find the depth of a cylindrical tank of radius 28 m, if its capacity is equal to that of a rectangular tank of size 28 m 16 m 11 m. Solution: Letxbe the depth of cylindrical tank. The radius of tank r = 28 m. Since, The volume of cylindrical tank = volume of rectangular tank $\pi r^{2} x=28 \times 16 \times 11$ $\Rightarrow \frac{22}{7} \times 28 \times 28 \times \mathrm{x}=28 \times 16 \times 11$ $=\frac{16}{8}$ $x=2 \mathrm{~m}$ Thus, the depth of cylindrical tank = 2 m....
Read More →Predict the products of the following reactions:
Question: Predict the products of the following reactions: (i) (ii) (iii) (iv) Solution: (i) (ii) (iii) (iv)...
Read More →Match List I with List II
Question: Match List I with List II Choose the correct answer from the options given below -$(a)-(i i),(b)-(i v),(c)-(i i i),(d)-(i)$$(a)-(i i),(b)-(i i i),(c)-(i v),(d)-(i)$$(a)-(i),(b)-(i i i),(c)-(i i),(d)-(i v)$$(a)-(i i i),(b)-(i i),(c)-(i),(d)-(i v)$Correct Option: , 2 Solution: (2) $(\mathrm{a}) \rightarrow(\mathrm{ii}),(\mathrm{b}) \rightarrow(\mathrm{iii}),(\mathrm{c}) \rightarrow(\mathrm{iv}),(\mathrm{d}) \rightarrow(\mathrm{i})$ By theory In isothermal process, temperature is constant...
Read More →Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
Question: Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i)Ethanal, Propanal, Propanone, Butanone. (ii)Benzaldehyde,p-Tolualdehyde,p-Nitrobenzaldehyde, Acetophenone. Hint:Consider steric effect and electronic effect. Solution: (i) The +I effect of the alkyl group increases in the order: Ethanal Propanal Propanone Butanone The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances...
Read More →A spherical ball of iron has been melted and made into smaller balls.
Question: A spherical ball of iron has been melted and made into smaller balls. If the radius of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made? Solution: Let radius of spherical ball =r Then radius of smaller spherical ball $=\frac{r}{4}$ Letnbe the no. of balls made by big spherical ball. Clearly, Volume of big spherical balls =n volume of one smaller ball $r^{3}=n \times \frac{r^{3}}{64}$ $n=64$ Hence, the no. of balls = 64...
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