A right circular cylinder and aright circular cone have equal bases and equal heights.
Question: A right circular cylinder and aright circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them. Solution: In cylinder radius =r height =h Surface area $=2 \times r \times h \times \pi$ In cone radius =r height =h Slant height $=\sqrt{h^{2}+r^{2}}$ Surface area $=\pi \times r \times \sqrt{h^{2}+r^{2}}$ $\frac{\text { As CSA of cylinder }}{\text { CSA of cone }}=\frac{8}...
Read More →Arrange the following in increasing order of their basic strength:
Question: Arrange the following in increasing order of their basic strength: (i)C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2and (C2H5)2NH (ii)C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (iii)CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. Solution: (i)Considering the inductive effect of alkyl groups, NH3, C2H5NH2,and (C2H5)2NH can be arranged in the increasing order of their basic strengths as: $\mathrm{NH}_{3}\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}...
Read More →In the given figure, XY ∥ AC and XY divides ∆ABC into two regions,
Question: In the given figure, $X Y \| A C$ and $X Y$ divides $\triangle A B C$ into two regions, equal in area. Show that $\frac{A X}{A B}=\frac{(2-\sqrt{2})}{2}$. Solution: In $\triangle A B C$ and $\triangle B X Y$, we have: $\angle B=\angle B$ $\angle B X Y=\angle B A C \quad$ (Corresponding angles) Thus, $\triangle A B C \sim \triangle B X Y \quad($ AA criterion $)$ $\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle B X Y)}=\frac{A B^{2}}{B X^{2}}=\frac{A B^{2...
Read More →A heat engine is involved with exchange of heat
Question: A heat engine is involved with exchange of heat of $1915 \mathrm{~J}$, $-40 \mathrm{~J},+125 \mathrm{~J}$ and $-Q \mathrm{~J}$, during one cycle achieving an efficiency of $50.0 \%$. The value of $Q$ is :$640 \mathrm{~J}$$40 \mathrm{~J}$$980 \mathrm{~J}$$400 \mathrm{~J}$Correct Option: , 3 Solution: (3) Efficiency, $\eta=\frac{\text { Work done }}{\text { Heat absorbed }}=\frac{W}{\Sigma Q}$ $=\frac{Q_{1}+Q_{2}+Q_{3}+Q_{4}}{Q_{1}+Q_{3}}=0.5$ Here, $Q_{1}=1915 \mathrm{~J}, Q_{2}=-40 \ma...
Read More →The surface area of a sphere is the same as the
Question: The surface area of a sphere is the same as the curved surface area of a cone having the radius of the base as 120 cm and height 160 cm. Find the radius of the sphere. Solution: Lateral height of cone $l=\sqrt{(120)^{2}+(160)^{2}}$ $=\sqrt{14400+25600}$ $=\sqrt{40000}$ $=200$ Surface area of sphere = surface area of cone $4 \pi \mathrm{r}_{1}{ }^{2}=\pi \mathrm{rl}$ $\mathrm{r}_{1}{ }^{2}=\frac{r l}{4}$ $\mathrm{r}_{1}^{2}=\frac{120 \times 200}{4}$ $\mathrm{r}_{1}^{2}=6000$ Radius of s...
Read More →How will you convert?
Question: How will you convert? (i)Benzene into aniline (ii)Benzene into N, N-dimethylaniline (iii)Cl(CH2)4Cl into hexan-1, 6-diamine? Solution: (i) (ii) (iii)...
Read More →Write structures of different isomeric amines corresponding to the molecular formula,
Question: (i)Write structures of different isomeric amines corresponding to the molecular formula, C4H11N (ii)Write IUPAC names of all the isomers. (iii)What type of isomerism is exhibited by different pairs of amines? Solution: (i), (ii)The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, C4H11N are given below: (a)CH3-CH2-CH2-CH2-NH2 Butanamine (10) (b) Butan-2-amine (10) (c) 2-Methylpropanamine (10) (d) 2-Methylpropan-2-amine (10) (e)CH3-CH...
Read More →In the given figure, ∆ABC and ∆DBC have the same base BC.
Question: In the given figure, $\triangle A B C$ and $\triangle D B C$ have the same base $B C$. If $A D$ and $B C$ intersect at $O$, prove that $\frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D B C)}=\frac{A O}{D O}$. Solution: Construction : Draw $\mathrm{AX} \perp \mathrm{CO}$ and DY $\perp \mathrm{BO}$. As, $\frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{DBC})}=\frac{1 / 2 \times \mathrm{AX} \times \mathrm{BC}}{1 / 2 \times \mathrm{DY} \...
Read More →A solid is composed of a cylinder with hemispherical ends.
Question: A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. Solution: Height of the cylinder = height of entire solid - height of sphere 1 - height of sphere 2= 108 - 18 - 18 = 108 36 = 72 cm r= 18 cm C.S.A. of cylinder $=2 \pi \mathrm{rh}$ $=2 \pi \times 18 \times 72$ $=8138.88 \mathrm{~cm}^{2}$ C.S.A. of 2 hemispheres = surface ar...
Read More →Classify the following amines as primary, secondary or tertiary:
Question: Classify the following amines as primary, secondary or tertiary: (i) (ii) (iii)(C2H5)2CHNH2 (iv)(C2H5)2NH Solution: Primary:(i)and(iii) Secondary:(iv) Tertiary:(ii)...
Read More →An engine takes in 5 mole of air
Question: An engine takes in 5 mole of air at $20^{\circ} \mathrm{C}$ and 1 atm, and compresses it adiabaticaly to $1 / 10^{\text {th }}$ of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be $X \mathrm{~kJ}$. The value of $X$ to the nearest integer is_______ Solution: $(46)$ For adiabatic process, $T V^{\gamma-1}=$ constant or, $T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$ $T_{1}=20^{\c...
Read More →The height of a solid cylinder is 15 cm
Question: The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm and height 4 cm are cut off. Find the volume of the remaining solid. Solution: The height of cylinderh= 15 cm Radius of cylinder $r=\frac{7}{2}$ The volume of cylinder $=\pi r^{2} h$ $=\pi \times\left(\frac{7}{2}\right)^{2} \times 15 \mathrm{~cm}^{2}$ $=183.75 \pi$ The radius of conical holes = 3 cm Height of conical holes = 4 cm. The volume of conical holes $=\frac...
Read More →Prove that the ratio of the perimeters of two similar triangles is the same as the
Question: Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides. Solution: Let the two triangles be ABC and PQR.We have: $\triangle A B C \sim \triangle P Q R$ Here,BC =a, AC =band AB =cPQ =r, PR =qand QR =pWe have to prove: $\frac{a}{p}=\frac{b}{q}=\frac{c}{r}=\frac{a+b+c}{p+q+r}$ $\triangle A B C \sim \triangle P Q R$; therefore, their corresponding sides will be proportional. $\Rightarrow \frac{a}{p}=\frac{b}{q}=\frac{c}{r}=k$ (s...
Read More →Lead spheres of diameter 6 cm are dropped into a cylindrical
Question: Lead spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18 cm and water rises by 40 cm. find the number of lead spheres dropped in the water. Solution: Radius of sphere $=\frac{6}{2}=3 \mathrm{~cm}$ Volume of lead sphere $=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \pi(3)^{3}$ $=\frac{4}{3} \pi \times 27$ $=36 \pi \mathrm{cm}^{3}$ Letnbe the no. of spheres are fully submerged. Radius of cylinder beake...
Read More →A solid sphere of radius 'r' is melted and recast into
Question: A solid sphere of radius 'r' is melted and recast into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm, its height 24 cm and thickness 2 cm, find the value of 'r'. Solution: Volume of sphere $=\frac{4}{3} \pi r^{3}$ ......(i) Since, The sphere is recast in to a hollow cylinder of uniform thickness 2 cm. The externalradius of hollow cylinderr1= 4 cm The internal radius of hollow cylinder $r_{2}=4-2=2 \mathrm{~cm}$ and height,h= 24 cm Cl...
Read More →Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.
Question: Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long. Solution: LetABCD be the rhombus with diagonals AC and BD intersecting each other at O.We know that the diagonals of a rhombus bisect each other at right angles.If AC = 24 cm and BD =10 cm,AO = 12 cm and BO = 5 cmApplying Pythagoras theorem in right-angled triangle AOB, we get: $A B^{2}=A O^{2}+B O^{2}=12^{2}+5^{2}=144+25=169$ $A B=13 \mathrm{~cm}$ Hence, the length of each side of the given rhombus is ...
Read More →Find the weight of a hollow sphere of metal
Question: Find the weight of a hollow sphere of metal having internal and external diameters as $20 \mathrm{~cm}$ and $22 \mathrm{~cm}$, respectively if $1 \mathrm{~m}^{3}$ of metal weighs $21 \mathrm{~g}$. Solution: External radius of hollow sphere, $r_{1}=\frac{22}{2}=11 \mathrm{~cm}$ Internal radius of hollow sphere, $r_{2}=\frac{20}{2}=10 \mathrm{~cm}$ The volume of hollow sphere $=\frac{4}{3} \pi\left(r_{1}^{3}-r_{2}^{2}\right)$ $=\frac{4}{3} \pi\left(11^{3}-10^{3}\right)$ $=\frac{4}{3} \pi...
Read More →In an equilateral triangle with side a, prove that area
Question: In an equilateral triangle with side $a$, prove that area $=\frac{\sqrt{3}}{4} a^{2} .$ Solution: Let ABC be the equilateral triangle with each side equal toa.Let AD be the altitude from A, meeting BC at D.Therefore, D is the midpoint of BC.Let AD beh.Applying Pythagoras theorem in right-angled triangle ABD, we have: $A B^{2}=A D^{2}+B D^{2}$ $\Rightarrow a^{2}=h^{2}+\left(\frac{a}{2}\right)^{2}$ $\Rightarrow h^{2}=a^{2}-\frac{a^{2}}{4}=\frac{3}{4} a^{2}$ $\Rightarrow h=\frac{\sqrt{3}}...
Read More →Although phenoxide ion has more number of resonating structures than carboxylate ion,
Question: Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why? Solution: Resonance structures of phenoxide ion are: It can be observed from the resonance structures of phenoxide ion that in II, III and IV, less electronegative carbon atoms carry a negative charge. Therefore, these three structures contribute negligibly towards the resonance stability of the phenoxide ion. Hence, these structures can be eliminat...
Read More →An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen.
Question: An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound. Solution: % of carbon = 69.77 % % of hydrogen = 11.63 % % of oxygen = {100 (69.77 + 11.63)}% = 18.6 % Thus, the ratio of the number of ca...
Read More →The largest sphere is carved out of a cube of side 10.5 cm.
Question: The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere. Solution: The side of cubea= 10.5 cm. Since, a largest sphere is curved out of that cube i.e.,radius of sphere, $r=\frac{a}{2}$ $r=\frac{10.5}{2} \mathrm{~cm}$ $r=5.25 \mathrm{~cm}$ The volume of sphere $=\frac{4}{3} \pi(5.25)^{3}$ $=22 \times 0.75 \times 1.75 \times 21$ $=606.375 \mathrm{~cm}^{3}$...
Read More →Prove that the internal bisector of an angle of a triangle divides
Question: Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Solution: Let the triangle be $\mathrm{ABC}$ with $\mathrm{AD}$ as the bisector of $\angle A$ which meets $\mathrm{BC}$ at $\mathrm{D}$. We have to prove: $\frac{B D}{D C}=\frac{A B}{A C}$ Draw CE∥DA, meeting BA produced at E.CE∥DATherefore, $\angle 2=\angle 3 \quad$ (Alternate angles) and $\angle 1=\angle 4 \quad$ (Corresponding angles) But, $...
Read More →Give plausible explanation for each of the following:
Question: Give plausible explanation for each of the following: (i)Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6 trimethylcyclohexanone does not. (ii)There are two NH2groups in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii)During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed. Solution: (i)Cyclohexanones form cyanohydrins accor...
Read More →A hollow sphere of internal and external diameters 4 and 8 cm
Question: A hollow sphere of internal and external diameters 4 and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone. Solution: Internal radius of hemisphere $r_{1}=\frac{4}{2}=2 \mathrm{~cm}$ External radius of hemisphere $r_{2}=\frac{8}{2}=4 \mathrm{~cm}$ Volume of hollow sphere $=\frac{4}{3} \pi\left(r_{2}^{3}-r_{1}^{3}\right)$ $=\frac{4}{3} \pi(64-8)$ $=\frac{4}{3} \pi \times 56$ $=\frac{224}{3} \pi \mathrm{cm}^{2}$ Since, The hemisphere melted into a...
Read More →Complete each synthesis by giving missing starting material, reagent or products
Question: Complete each synthesis by giving missing starting material, reagent or products (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) Solution: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)...
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