If the length of the sides BC, CA and AB of a △ABC are a, b and c respectively and AD
Question: If the length of the sides BC, CA and AB of a △ABC area,bandcrespectively and AD is the bisector of A then find the length of BD and DC. Solution: Let DC =x BD =axBy using angle bisector theore in △ABC, we have $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}$ $\Rightarrow \frac{c}{b}=\frac{a-x}{x}$ $\Rightarrow c x=a b-b x$ $\Rightarrow x(b+c)=a b$ $\Rightarrow x=\frac{a b}{(b+c)}$ Now, $a-x=a-\frac{a b}{b+c}$ $=\frac{a b+a c-a b}{b+c}$ $=\frac{a c}{(a+b)}$...
Read More →Draw all the isomers (geometrical and optical) of:
Question: Draw all the isomers (geometrical and optical) of: (i)[CoCl2(en)2]+ (ii)[Co(NH3)Cl(en)2]2+ (iii) [Co(NH3)2Cl2(en)]+ Solution: (i)[CoCl2(en)2]+ Geometrical isomerism Optical isomerism Since onlycisisomer is optically active, it shows optical isomerism. In total, three isomers are possible. (ii)[Co(NH3)Cl(en)2]2+ Geometrical isomerism Optical isomerism Since only cis isomer is optically active, it shows optical isomerism. (iii)[Co(NH3)2Cl2(en)]+...
Read More →The surface area of a solid metallic sphere is 616 cm2.
Question: The surface area of a solid metallic sphere is $616 \mathrm{~cm}^{2}$. It is melted and recast into a cone of height $28 \mathrm{~cm}$. Find the diameter of the base of the cone so formed (Use $\left.\pi=22 / 7\right)$. Solution: The surface area of the metallic sphere is 616 square cm. Let the radius of the metallic sphere isr. Therefore, we have $4 \pi r^{2}=616$ $\Rightarrow r^{2}=\frac{616 \times 7}{4 \times 22}$ $\Rightarrow r^{2}=7 \times 7$ $\Rightarrow r=7$ Therefore, the radiu...
Read More →Draw the structures of optical isomers of:
Question: Draw the structures of optical isomers of: (i)[Cr(C2O4)3]3 (ii)[PtCl2(en)2]2+ (iii)[Cr(NH3)2Cl2(en)]+ Solution: (i)[Cr(C2O4)3]3 (ii)[PtCl2(en)2]2+ (iii)[Cr(NH3)2Cl2(en)]+...
Read More →A man goes 12 m due south and then 35 m due west.
Question: A man goes 12 m due south and then 35 m due west. How far is he from the starting point? Solution: In right triangle SOWBy using Pythagoras theorem, we have $\mathrm{OW}^{2}=\mathrm{WS}^{2}+\mathrm{SO}^{2}$ $=35^{2}+12^{2}$ $=1225+144$ $=1369$ $\therefore \mathrm{OW}^{2}=1369$ $\Rightarrow \mathrm{OW}=37 \mathrm{~m}$ Hence, the man is 37 m away from the starting point....
Read More →How many geometrical isomers are possible in the following coordination entities?
Question: How many geometrical isomers are possible in the following coordination entities? (i)[Cr(C2O4)3]3(ii) [Co(NH3)3Cl3] Solution: (i)For [Cr(C2O4)3]3, no geometric isomer is possible as it is a bidentate ligand. (ii)[Co(NH3)3Cl3] Two geometrical isomers are possible....
Read More →List various types of isomerism possible for coordination compounds,
Question: List various types of isomerism possible for coordination compounds, giving an example of each. Solution: (a)Geometric isomerism: This type of isomerism is common in heteroleptic complexes. It arises due to the different possible geometric arrangements of the ligands. For example: (b)Optical isomerism: This type of isomerism arises in chiral molecules. Isomers are mirror images of each other and are non-superimposable. (c)Linkage isomerism:This type of isomerism is found in complexes t...
Read More →A well of diameter 3 m is dug 14 m deep
Question: A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment. Solution: The inner radius of the well is $\frac{3}{2} m$ and the height is $14 m$. Therefore, the volume of the Earth taken out of it is $V_{1}=\pi \times\left(\frac{3}{2}\right)^{2} \times 14 \mathrm{~m}^{3}$ The inner and outer radii of the embankment are $\frac{3}{2} m$ and $4+\frac{3}{2}=\frac{11}{2} m$ ...
Read More →Using IUPAC norms write the systematic names of the following:
Question: Using IUPAC norms write the systematic names of the following: (i)[Co(NH3)6]Cl3 (ii)[Pt(NH3)2Cl(NH2CH3)]Cl (iii)[Ti(H2O)6]3+ (iv)[Co(NH3)4Cl(NO2)]Cl (v)[Mn(H2O)6]2+ (vi)[NiCl4]2 (vii)[Ni(NH3)6]Cl2 (viii)[Co(en)3]3+ (ix)[Ni(CO)4] Solution: (i)Hexaamminecobalt(III) chloride (ii)Diamminechlorido(methylamine) platinum(II) chloride (iii)Hexaquatitanium(III) ion (iv)Tetraamminichloridonitrito-N-Cobalt(III) chloride (v)Hexaquamanganese(II) ion (vi)Tetrachloridonickelate(II) ion (vii)Hexaammin...
Read More →Each of the equal sides of an isosceles triangle is 25 cm.
Question: Each of the equal sides of an isosceles triangle is 25 cm. Find the length of its altitude if the base is 14 cm. Solution: We know that the altitude drawn from the vertex opposite to the non equal side bisects the non equal side.Suppose ABC is an isosceles triangle having equal sides AB and BC.So, the altitude drawn from the vertex will bisect the opposite side.Now, In right triangle ABDBy using Pythagoras theorem, we have $\mathrm{AB}^{2}=\mathrm{BD}^{2}+\mathrm{DA}^{2}$ $\Rightarrow ...
Read More →Using IUPAC norms write the formulas for the following:
Question: Using IUPAC norms write the formulas for the following: (i)Tetrahydroxozincate(II) (ii)Potassium tetrachloridopalladate(II) (iii)Diamminedichloridoplatinum(II) (iv)Potassium tetracyanonickelate(II) (v)Pentaamminenitrito-O-cobalt(III) (vi)Hexaamminecobalt(III) sulphate (vii)Potassium tri(oxalato)chromate(III) (viii)Hexaammineplatinum(IV) (ix)Tetrabromidocuprate(II) (x)Pentaamminenitrito-N-cobalt(III) Solution: (i)[Zn(OH)4]2 (ii)K2[PdCl4] (iii)[Pt(NH3)2Cl2] (iv)K2[Ni(CN)4] (v)[Co(ONO) (N...
Read More →A farmer runs a pipe of internal diameter 20 cm from
Question: A farmer runs a pipe of internal diameter 20 cm from the canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled? Solution: The internal radius of the pipe is $10 \mathrm{~cm}=0.1 \mathrm{~m}$. The water is flowing in the pipe at $3 \mathrm{~km} / \mathrm{hr}=3000 \mathrm{~m} / \mathrm{hr}$. Let the cylindrical tank will be filled in $t$ hours. Therefore, the lengt...
Read More →In a triangle BMP and CNR it is given that PB = 5 cm, MP = 6 cm, BM = 9 cm and NR = 9 cm.
Question: In a triangle BMP and CNR it is given that PB = 5 cm, MP = 6 cm, BM = 9 cm and NR = 9 cm. If △BMP △CNR, then find the perimeter of the △CNR Solution: When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.Here, △BMP △CNR $\therefore \frac{\mathrm{BM}}{\mathrm{CN}}=\frac{\mathrm{BP}}{\mathrm{CR}}=\frac{\mathrm{MP}}{\mathrm{NR}} \quad \ldots(1)$ Now, $\frac{\mathrm{BM}}{\mathrm{CN}}=\frac{\mathrm{MP}}{\mathrm{NR}} \quad[$ Using $(1)]$...
Read More →Specify the oxidation numbers of the metals in the following coordination entities:
Question: Specify the oxidation numbers of the metals in the following coordination entities: (i)[Co(H2O)(CN)(en)2]2+ (ii)[CoBr2(en)2]+ (iii)[PtCl4]2 (iv)K3[Fe(CN)6] (v)[Cr(NH3)3Cl3] Solution: (i) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{CN})(\mathrm{en})_{2}\right]^{2+}$ Let the oxidation number of Co bex. The charge on the complex is +2. (ii) Let the oxidation number of Pt bex. The charge on the complex is 2. x+ 4(1) = 2 x= + 2 (iv)...
Read More →Specify the oxidation numbers of the metals in the following coordination entities:
Question: Specify the oxidation numbers of the metals in the following coordination entities: (i)[Co(H2O)(CN)(en)2]2+ (ii)[CoBr2(en)2]+ (iii)[PtCl4]2 (iv)K3[Fe(CN)6] (v)[Cr(NH3)3Cl3] Solution: (i) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{CN})(\mathrm{en})_{2}\right]^{2+}$ Let the oxidation number of Co bex. The charge on the complex is +2. (ii) Let the oxidation number of Pt bex. The charge on the complex is 2. x+ 4(1) = 2 x= + 2 (iv)...
Read More →Water in a canal 1.5 m wide and 6 m deep
Question: Water in a canal 1.5 m wide and 6 m deep is flowing with a speed of 10km/hr. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired? Solution: The canal is 1.5 m wide and 6 m deep. The water is flowing in the canal at 10 km/hr. Hence, in 30 minutes, the length of the flowing standing water is $=10 \times \frac{30}{60} \mathrm{~km}$ $=5 \mathrm{~km}$ $=5000 \mathrm{~m}$ Therefore, the volume of the flowing water in 30 min is $V_{1}=5000 \times 1.5 \times 6 \ma...
Read More →In the given figure, MN∥BC and AM : MB = 1 : 2.
Question: In the given figure, MN∥BC and AM : MB = 1 : 2. Find $\frac{\operatorname{area}(\triangle \mathrm{AMN})}{\operatorname{area}(\triangle \mathrm{ABC})}$ Solution: We haveAM : MB = 1 : 2 $\Rightarrow \frac{\mathrm{MB}}{\mathrm{AM}}=\frac{2}{1}$ Adding 1 to both sides, we get $\Rightarrow \frac{\mathrm{MB}}{\mathrm{AM}}+1=\frac{2}{1}+1$ $\Rightarrow \frac{\mathrm{MB}+\mathrm{AM}}{\mathrm{AM}}=\frac{2+1}{1}$ $\Rightarrow \frac{\mathrm{AB}}{\mathrm{AM}}=\frac{3}{1}$ Now, In △AMN and △ABCAMN ...
Read More →A well with inner radius 4 m is dug 14 m deep.
Question: A well with inner radius 4 m is dug 14 m deep. Earth taken out of it has been spread evenly all around a width of 3 m it to form an embankment. Find the height of the embankment. Solution: The inner radius of the well is 4m and the height is 14m. Therefore, the volume of the Earth taken out of it is $V_{1}=\pi \times(4)^{2} \times 14 \mathrm{~m}^{3}$ The inner and outer radii of the embankment are 4m and 4+3=7m respectively. Let the height of the embankment beh. Therefore, the volume o...
Read More →What is meant by unidentate, didentate and ambidentate ligands?
Question: What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each. Solution: A ligand may contain one or more unshared pairs of electrons which are called the donor sites of ligands. Now, depending on the number of these donor sites, ligands can be classified as follows: (a) Unidentate ligands: Ligands with only one donor sites are called unidentate ligands. For e.g., $\ddot{\mathrm{N}} \mathrm{H}_{3}, \mathrm{Cl}^{-}$etc. (b)Didentate ligands:Ligands that have...
Read More →How many coins 1.75 cm in diameter and 2 mm
Question: How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm 10 cm 7 cm? Solution: The dimension of the cuboid is 11cm10cm7cm. Therefore, the volume of the cuboid is $V_{1}=11 \times 10 \times 7=770 \mathrm{~cm}^{3}$ The radius and thickness of each coin are $\frac{1.75}{2}=0.875 \mathrm{~cm}$ and $2 \mathrm{~mm}=0.2 \mathrm{~cm}$ respectively. Therefore, the volume of each coin is $V_{2}=\pi \times(0.875)^{2} \times 0.2 \mathrm{~cm}^{3}$ Since, the total vol...
Read More →Two traingles DEF and GHK are such that ∠D = 48∘and ∠H = 57∘ .
Question: Two traingles DEF and GHK are such that D = 48∘and H = 57∘ . If△DEF △GHK, then find the measure of F. Solution: If two traingle are similar then the corresponding angles of the two tringles are equal.Here, △DEF △GHKE = H = 57∘Now, In △DEFD + E + F = 180∘ (Angle sum property of triangle)⇒ F = 180∘ 48∘ 57∘ = 75∘...
Read More →Explain with two examples each of the following:
Question: Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic. Solution: (i)Coordination entity: A coordination entity is an electrically charged radical or species carrying a positive or negative charge. In a coordination entity, the central atom or ion is surrounded by a suitable number of neutral molecules or negative ions ( called ligands). For example: $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\...
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: Suppose ABCD is a rhombus.We know that the digonals of a rhombus perpendicularly bisect each other. AOB = 900, AO = 12 cm and BO = 5 cmNow, In right triangle AOBBy using Pythagoras theorem we haveAB2= AO2+ BO2= 122+ 52= 144 + 25= 169AB2= 169⇒ AB = 13 cmSince, all the sides of a rhombus are equal.Hence, AB = BC = CD = DA = 13 cm...
Read More →A right angled triangle whose sides are 3 cm, 4 cm and 5 cm
Question: A right angled triangle whose sides are 3 cm, 4 cm and 5 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two cones so formed. Also, find their curved surfaces. Solution: We consider the following figure as follows Let the angle B is right angle and the sides of the triangle are AB= 4cm, BC= 3cm,AC= 5cm. When the triangle is revolved about the side AB, then the base-radius, height and slant height of the produced cone becomes ...
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: We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides..Suppose ABC is an equilateral triangle having AB = BC = CA =a.Suppose AD is the altitude drawn from the vertex A to the side BC.So, It will bisects the side BC $\therefore \mathrm{DC}=\frac{1}{2} a$ Now, In right triangle ADCBy using Pythagoras the...
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