If the volume of a right circular cone of height $9 \mathrm{~cm}$ is $48 \pi \mathrm{cm}^{3}$,
[question] Question. If the volume of a right circular cone of height $9 \mathrm{~cm}$ is $48 \pi \mathrm{cm}^{3}$, find the diameter of its base. [/question] [solution] Solution: Height (h) of cone = 9 cm Let the radius of the cone be r. Volume of cone $=48 \pi \mathrm{cm}^{3}$ $\Rightarrow \frac{1}{3} \pi r^{2} h=48 \pi \mathrm{cm}^{3}$ $\Rightarrow r^{2}=16 \mathrm{~cm}^{2}$ $\Rightarrow r=4 \mathrm{~cm}$ Diameter of base $=2 r=8 \mathrm{~cm}$ [/solution]...
Read More →Give reasons for the following observation:
[question] Question. Give reasons for the following observation: The smell of hot sizzling food reaches you several meters away, but to get the smell from cold food you have to go close. [/question] [solution] Solution: The smell of hot sizzling food reaches severed meters away, as the particles of hot food have more kinetic energy and hence the rate of diffusion is more than the particles of cold food. [/solution]...
Read More →The height of a cone is 15 cm.
[question] Question. The height of a cone is $15 \mathrm{~cm}$. If its volume is $1570 \mathrm{~cm}^{3}$, find the diameter of its base. [Use $\left.\pi=3.14\right]$ [solution] Solution: Height (h) of cone = 15 cm Let the radius of the cone be r. Volume of cone $=1570 \mathrm{~cm}^{3}$ $\frac{1}{3} \pi r^{2} h=1570 \mathrm{~cm}^{3}$ $\Rightarrow\left(\frac{1}{3} \times 3.14 \times r^{2} \times 15\right) \mathrm{cm}=1570 \mathrm{~cm}^{3}$ $\Rightarrow r^{2}=100 \mathrm{~cm}^{2}$ ⇒ r = 10 cm There...
Read More →A solution of a substance ' $X$ ' is used for white washing.
[question] Question. A solution of a substance ' $X$ ' is used for white washing. (i) Name the substance ' $X$ ' and write its formula. (ii) Write the reaction of the substance 'X' with water. [/question] [solution] solution: (i) The substance ' $\mathrm{X}$ ' is calcium oxide. Its chemical formula is $\mathrm{CaO}$. (ii) Calcium oxide reacts vigorously with water to form calcium hydroxide (slaked lime). [/solution]...
Read More →Find the capacity in litres of a conical vessel with
[question] Question. Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Radius (r) of cone = 7 cm Slant height (l) of cone = 25 cm Height $(h)$ of cone $=\sqrt{l^{2}-r^{2}}$ $=\left(\sqrt{25^{2}-7^{2}}\right) \mathrm{cm}$ $=24 \mathrm{~cm}$ Volume of cone $=\frac{1}{3} \pi r^{2} h$ $=\left(\frac{1}{3} \times \frac{22}{7} \times(7)...
Read More →Which of the following are matter?
[question] Question. Which of the following are matter? Chair, air, love, smell, hate, almonds, thought, cold, cold-drink, smell of perfume. [/question] [solution] Solution: Chair, air, almonds, and cold-drink. [/solution]...
Read More →Write a balanced chemical equation with state symbols for the following reactions.
[question] Question. Write a balanced chemical equation with state symbols for the following reactions. (i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride. (ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water. [/question] [solution] solution: (i) $\mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{SO}_{4}$ (aq) $\rightarro...
Read More →Write the balanced equation for the following chemical reactions.
[question] Question. Write the balanced equation for the following chemical reactions. (i) Hydrogen + Chlorine Hydrogen chloride (ii) Barium chloride + Aluminium sulphate Barium sulphate + Aluminium chloride (iii) Sodium + Water Sodium hydroxide + Hydrogen [/question] [solution] solution: (i) $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})$ (ii) $3 \mathrm{BaCl}_{2}$ (aq) $+\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq}) \righta...
Read More →Why should a magnesium ribbon be cleaned before it is burnt in air?
[question] Question. Why should a magnesium ribbon be cleaned before it is burnt in air? [/question] [solution] Solution: Magnesium is an extremely reactive metal. When stored, it reacts with oxygen to form a layer of magnesium oxide on its surface. This layer of magnesium oxide is quite stable and prevents further reaction of magnesium with oxygen. The magnesium ribbon is cleaned by sand paper for removing this layer so that the underlying metal can be exposed to air. [/solution]...
Read More →Find the volume of the right circular cone with
[question] Question. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Radius (r) of cone = 6 cm Height $(h)$ of cone $=7 \mathrm{~cm}$ $=\left[\frac{1}{3} \times \frac{22}{7} \times(6)^{2} \times 7\right] \mathrm{cm}^{3}$ $=(12 \times 22) \mathrm{cm}^{3}$ $=264 \mathrm{~cm}^{3}$ Therefore, the volume of the cone is $264 \mathrm{~cm}^{3}$. (ii) R...
Read More →Convert the following temperature to the Celsius scale
[question] Question Convert the following temperature to the Celsius scale - (i) 293 K (ii) 470 K [/question] [solution] Solution (i) 293 - 273 = 20°C (ii) 470-273 = 197°C [/solution]...
Read More →The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres.
[question] Question. The capacity of a closed cylindrical vessel of height $1 \mathrm{~m}$ is $15.4$ litres. How many square metres of metal sheet would be needed to make it? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Let the radius of the circular end be r. Height (h) of cylindrical vessel = 1 m Volume of cylindrical vessel $=15.4$ litres $=0.0154 \mathrm{~m}^{3}$ $\pi r^{2} h=0.0154 \mathrm{~m}^{3}$ $\left(\frac{22}{7} \times r^{2} \times 1\right) \...
Read More →It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep.
[question] Question. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel $10 \mathrm{~m}$ deep. If the cost of painting is at the rate of Rs 20 per $\mathrm{m}^{2}$, find (i) Inner curved surface area of the vessel (ii) Radius of the base (iii) Capacity of the vessel Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Rs 20 is the cost of painting $1 \mathrm{~m}^{2}$ area. Rs 2200 is the cost of painting $=\left(\frac{1}{20} \times 2200\right) \math...
Read More →If the lateral surface of a cylinder is $94.2 \mathrm{~cm}^{2}$
[question] Question. If the lateral surface of a cylinder is $94.2 \mathrm{~cm}^{2}$ and its height is $5 \mathrm{~cm}$, then find (i) radius of its base (ii) its volume. [Use $\pi=3.14]$ [/question] [solution] Solution: (i) Height $(h)$ of cylinder $=5 \mathrm{~cm}$ Let radius of cylinder be r. CSA of cylinder $=94.2 \mathrm{~cm}^{2}$ $2 \pi r h=94.2 \mathrm{~cm}^{2}$ $(2 \times 3.14 \times r \times 5) \mathrm{cm}=94.2 \mathrm{~cm}^{2}$ $r=3 \mathrm{~cm}$ (ii) Volume of cylinder $=\pi r^{2} h$ ...
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