In the given figure, D is the midpoint of side BC and AE
Question: In the given figure,Dis the midpoint of sideBCandAEBC. IfBC=a,AC=b,AB=c,ED=x,AD=pandAE=h, prove that (i) $b^{2}=p^{2}+a x+\frac{a^{2}}{4}$ (ii) $c^{2}=p^{2}-a x+\frac{a^{2}}{4}$ (iii) $\left(b^{2}+c^{2}\right)=2 p^{2}+\frac{1}{2} a^{2}$ (iv) $\left(b^{2}-c^{2}\right)=2 a x$ Solution: (i)In right-angled triangle AEC, applying Pythagoras theorem, we have: $A C^{2}=A E^{2}+E C^{2}$ $\Rightarrow b^{2}=h^{2}+\left(x+\frac{a}{2}\right)^{2}=h^{2}+x^{2}+\frac{a^{2}}{4}+a x \ldots(i)$ In right ...
Read More →A hemispherical bowl of internal radius 9 cm is full of liquid.
Question: A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped small bottles each of diameter 3 cm and height 4 cm. How many bottles are necessary to empty the bowl? Solution: The internal radius of the hemispherical bowl is 9cm. Therefore, the volume of the water in the hemispherical bowl is $V=\frac{2}{3} \pi \times(9)^{3} \mathrm{~cm}^{3}$ The water in the hemispherical bowl is required to transfer into the cylindrical bottles each...
Read More →Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Question: Compare the chemistry of actinoids with that of the lanthanoids with specialreference to: (i)electronic configuration(iii)oxidation state (ii)atomic and ionic sizes and(iv)chemical reactivity. Solution: (i)Electronic configuration The general electronic configuration for lanthanoids is [Xe]544f0-145d0-16s2and that for actinoids is [Rn]865f1-146d0-17s2. Unlike 4forbitals, 5forbitals are not deeply buried and participate in bonding to a greater extent. (ii)Oxidation states The principal ...
Read More →A rectangular tank 15 m long and 11 m broad is required
Question: A rectangular tank 15 m long and 11 m broad is required to receive entire liquid contents from a fully cylindrical tank of internal diameter 21 m and length 5 m. Find the least height of the tank that will serve the purpose. Solution: Suppose height of the rectangular tank is equal toh. Length of the tank = 15 m Breadth of the tank = 11 m Further, length of cylindrical tank = 5 m Radius of cylindrical tank $=\frac{21}{2} \mathrm{~m}$ To find out the least height of the tank, equate the...
Read More →A conical flask is full of water.
Question: A conical flask is full of water. The flask has base-radiusrand heighth. The water is poured into a cylindrical flask of base-radiusmr. Find the height of water in the cylindrical flask. Solution: The base-radius and height of the conical flask arerandhrespectively. Let, the slant height of the conical flask isl. Therefore, the volume of the water in the conical flask is $V=\frac{1}{3} \times \pi \times r^{2} \times h$ The water in the conical flask is poured into a cylindrical flask o...
Read More →Compare the stability of +2 oxidation state for the elements of
Question: Compare the stability of +2 oxidation state for the elements of the first transition series. Solution: From the abovetable, it is evident that the maximum number of oxidation states is shown by Mn, varying from +2 to +7. The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons. The relative stability of the +2 oxidation state increases on moving fro...
Read More →Solve this
Question: In the given figure, ACB = 900and CD AB. Prove that $\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}$ Solution: Given: ACB = 900and CD AB To Prove: $\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}$ Proof: In $\triangle \mathrm{ACB}$ and $\triangle \mathrm{CDB}$ $\angle \mathrm{ACB}=\angle \mathrm{CDB}=90^{\circ}$ (Given) $\angle \mathrm{ABC}=\angle \mathrm{CBD}$ (Common) By AA similarity-criterion $\triangle \mathrm{ACB} \sim \triangle \m...
Read More →Rain water, which falls on a flat rectangular surface
Question: Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen? Solution: The fallen rains are in the form of a cuboid of height 1 cm, length 6 m = 600 cm and breadth 4 m = 400 cm. Therefore, the volume of the fallen rains is $V=600 \times 400 \times 1=240000 \mathrm{~cm}^{3}$ The fallen rains are transferred...
Read More →A cylindrical bucket, 32 cm high and 18 cm of radius of the base,
Question: A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap Solution: The height and radius of the cylindrical bucket are $h=32 \mathrm{~cm}$ and $r=18 \mathrm{~cm}$ respectively. Therefore, the volume of the cylindrical bucket is $V=\pi r^{2} h$ $=\frac{22}{7} \times(18)^{2} \times 32$ The bucket ...
Read More →Predict which of the following will be coloured in aqueous solution?
Question: Predict which of the following will be coloured in aqueous solution?Ti3+,V3+, Cu+,Sc3+,Mn2+, Fe3+andCo2+. Give reasons for each. Solution: Only the ions that haveelectrons ind-orbital and in which d-d transition is possible will be coloured. The ions in whichd-orbitals are empty or completely filled will be colourless as no d-d transitionis possible in those configurations. From the above table, it can be easily observed that onlySc3+has an emptyd-orbital and Cu+has completely filled d...
Read More →Find the volume of the largest right circular
Question: Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm. Solution: We have the following figure The length of each side of the cube is 9 cm. We have to find the volume of the largest right circular cone contained in the cube. The diameter of the base circle is same as the length of the side of the cube. Thus, the diameter of the base circle of the right circular cone is 9 cm. Therefore, the radius of the base of the right circular cone is $r=...
Read More →In ∆ABC, D is the midpoint of BC and AE ⊥ BC. If AC > AB, show that
Question: In ∆ABC, Dis the midpoint ofBCandAEBC. IfACAB, show that $A B^{2}=A D^{2}-B C \cdot D E+\frac{1}{4} B C^{2}$ Solution: In right-angled triangle AEB, applying Pythagoras theorem, we have: $A B^{2}=A E^{2}+E B^{2} \quad \ldots$ (i) In right-angled triangle AED, applying Pythagoras theorem, we have: $A D^{2}=A E^{2}+E D^{2}$ $\Rightarrow A E^{2}=A D^{2}-E D^{2}$ ....(ii) Therefore, $A B^{2}=A D^{2}-E D^{2}+E B^{2} \quad($ from $(\mathrm{i})$ and $(\mathrm{ii}))$ $A B^{2}=A D^{2}-E D^{2}+(...
Read More →Solve the following
Question: For $\mathrm{M}^{2+} / \mathrm{M}$ and $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ systems, the $\mathrm{E}^{\ominus}$ values for some metals are as follows: Cr2+/Cr 0.9V Cr3/Cr2+0.4 V Mn2+/Mn 1.2V Mn3+/Mn2++1.5 V Fe2+/Fe0.4V Fe3+/Fe2++0.8 V Use this data to comment upon: (i)The stability of Fe3+in acid solution as compared to that of Cr3+or Mn3+and (ii)The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal. Solution: (i) The ${ }^{\The...
Read More →A well of diameter 2 m is dug 14 m deep.
Question: A well of diameter 2 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to form an embankment of height 40 cm. Find the width of the embankment. Solution: Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is $r=\frac{2}{2}=1$ The well is $14 \mathrm{~m}$ deep. Thus, the height of the solid right circular cylinder is $h=14 \mathrm{~m}$. Therefore, the volume of the solid right circular cylinder is $V_{1}=\...
Read More →Describe the preparation of potassium permanganate.
Question: Describe the preparation of potassium permanganate. How does the acidifiedpermanganate solution react with (i) iron(II) ions (ii) SO2and (iii) oxalic acid? Write the ionic equations for the reactions. Solution: Potassium permanganate can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3or KClO4, to give K2MnO4 The green mass can be extracted with water and then oxidized either electrolytically ...
Read More →A 16 m deep well with diameter 3.5 m is dug up and the earth from
Question: A 16 m deep well with diameter 3.5 m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7 m. Find the height of the platform. Solution: Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is $r=\frac{3.5}{2}=1.75 \mathrm{~m}$ The well is 16m deep. Thus, the height of the solid right circular cylinder ism. Therefore, the volume of the solid right circular cylinder is $V_{1}=\pi r^{2} h$ $=\frac{22}{7} \time...
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 = 24 cm and BD = 10 cm) meeting at O.We know that the diagonals of a rhombus bisect each other at right angles.Applying Pythagoras theorem in right-angled triangle AOB, we get: $A B^{2}=A O^{2}+B O^{2}=12^{2}+5^{2}$ $A B^{2}=144+25=169$ $A B=\sqrt{169}=13 \mathrm{~cm}$ Hence, the length of each side of the rhombus is 13 cm....
Read More →Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
Question: Describe the oxidising action of potassium dichromate and write the ionicequations for its reaction with: (i)iodide(ii)iron(II) solution and(iii)H2S Solution: $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ acts as a very strong oxidising agent in the acidic medium. $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+4 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+4 \mathrm{H}_{2} \mathrm{O}+3[\mathrm{O}]$ $\mat...
Read More →A path 2 m wide surrounds a circular pond of diameter 40 m.
Question: A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic metres of gravel are required to grave the path to a depth of 20 cm? Solution: Diameter of the circular pond is given = 40 m So, the radius of this pond is 20 m There is a path surrounding the pond. We are given the thickness of this path as 2 m We have to grave this path with gravel. The depth of the path is also given 20 cm=0.2 m This circular path can be viewed as a hollow cylinder of thickness 0.2 m and dept...
Read More →Find the length of a diagonal of a rectangle whose adjacent sides are 30 cm and 16 cm.
Question: Find the length of a diagonal of a rectangle whose adjacent sides are 30 cm and 16 cm. Solution: LetABCD be the rectangle with diagonals AC and BD meeting at O.According to the question:AB = CD = 30 cm and BC = AD = 16 cm Applying Pythagoras theorem in right-angled triangle ABC, we get: $A C^{2}=A B^{2}+B C^{2}=30^{2}+16^{2}=900+256=1156$ $A C=\sqrt{1156}=34 \mathrm{~cm}$ Diagonals of a rectangle are equal.Therefore, AC = BD = 34 cm...
Read More →25 circular plates, each of radius 10.5 cm and thickness 1.6 cm,
Question: 25 circular plates, each of radius 10.5 cm and thickness 1.6 cm, are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed. Solution: We have 25 circular plates, each with radius = 10.5 cm and thickness = 1.6 cm These plates are stacked on top of one another. So, the total height of the arrangement becomes $=1.6 \times 25=40 \mathrm{~cm}$ Volume of this arrangement $=\pi r^{2} h=\pi(10.5)^{2} \times 40=13860 ...
Read More →Find the height of an equilateral triangle of side 12 cm.
Question: Find the height of an equilateral triangle of side 12 cm. Solution: Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then, D will be the midpoint of BC.Applying Pythagoras theorem in right-angled triangle ABD, we get: $A B^{2}=A D^{2}+B D^{2}$ $\Rightarrow A D^{2}=12^{2}-6^{2} \quad\left(\because B D=\frac{1}{2} B C=6\right)$ $\Rightarrow A D^{2}=144-36=108$ $\Rightarrow A D=\sqrt{108}=6 \sqrt{3} \mathrm{~cm}$ Hence, the height of the given triangle is...
Read More →50 circular plates each of diameter 14 cm and thickness 0.5 cm
Question: 50 circular plates each of diameter 14 cm and thickness 0.5 cm are placed one above the other to form a right circular cylinder. Find its total surface area. Solution: We have 50 circular plates, each with diameter = 14 cm That is, radius = 7 cm and thickness = 0.5 cm These plates are stacked on top of one another. So, the total thickness $=0.5 \times 50 \mathrm{~cm}=25 \mathrm{~cm}$ This is clearly a cylindrical arrangement.We know, Total surface area of a cylinder $=2 \pi r h+2 \pi r...
Read More →Describe the preparation of potassium dichromate from iron chromite ore.
Question: Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate? Solution: Potassium dichromate is prepared from chromite ore $\left(\mathrm{FeCr}_{2} \mathrm{O}_{4}\right)$ in the following steps. Step (1):Preparation of sodium chromate $4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+16 \mathrm{NaOH}+7 \mathrm{O}_{2} \longrightarrow 8 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{Fe}_{2} \mathrm{O}_{3}+8 \mathrm{H}_{2...
Read More →A cylindrical vessel having diameter equal to its height
Question: A cylindrical vessel having diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and height 21 cm which are filled completely. Find the diameter of the cylindrical vessel. Solution: A cylindrical vessel whose height is equal to its diameter is given. It is filled with water. We know that the volume of a cylinder $=\pi r^{2} h$ In this particular case, Height is equal to the diameter, that is $h=2 r$, The volume of cyli...
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