Suggest a method to liquefy atmospheric gases?
[question] Question Suggest a method to liquefy atmospheric gases? [/question] [solution] Solution The atmospheric gases are taken in a cylinder with piston fitted on it. By cooling and applying pressure on them, the gases can be liquefied. [solution]...
Read More →For any substance, why does the temperature remain constant during the change of state?
[question] Question For any substance, why does the temperature remain constant during the change of state? [/question] [solution] Solution During the change of state of any matter heat is supplied to the substance. The molecules of this matter use heat to overcome the force of attraction between the particles, at this period of time, temperature remains constant. This extra heat is acquired by the molecules in the form of hidden heat called latent heat to change from one state of matter to the ...
Read More →A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24 cm.
[question] Question. A joker's cap is in the form of right circular cone of base radius $7 \mathrm{~cm}$ and height $24 \mathrm{~cm}$. Find the area of the sheet required to make 10 such caps. [ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of conical cap $=7 \mathrm{~cm}$ Height $(h)$ of conical cap $=24 \mathrm{~cm}$ Slant height ( $l$ ) of conical cap $=\sqrt{r^{2}+h^{2}}$ $=\left[\sqrt{(7)^{2}+(24)^{2}}\right] \mathrm{cm}=(\sqrt{625}) \mathrm{cm}=25 \ma...
Read More →A conical tent is 10 m high and the radius of its base is 24 m. Find
[question] Question. A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent (ii) cost of the canvas required to make the tent, if the cost of $1 \mathrm{~m}^{2}$ canvas is Rs 70 . $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Let ABC be a conical tent. Height (h) of conical tent = 10 m Radius (r) of conical tent = 24 m Let the slant height of the tent be l. In $\triangle \mathrm{ABO}$ $\mathrm{AB}^{2}=\mathr...
Read More →Curved surface area of a cone is $308 \mathrm{~cm}^{2}$ and its slant height is $14 \mathrm{~cm}$.
[question] Question. Curved surface area of a cone is $308 \mathrm{~cm}^{2}$ and its slant height is $14 \mathrm{~cm}$. Find (i) radius of the base and (ii) total surface area of the cone. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Slant height (l) of cone $=14 \mathrm{~cm}$ Let the radius of the circular end of the cone be $r$. We know, CSA of cone $=\pi r$ $(308) \mathrm{cm}^{2}=\left(\frac{22}{7} \times r \times 14\right) \mathrm{cm}$ $\Rightar...
Read More →Find the total surface area of a cone,
[question] Question. Find the total surface area of a cone, if its slant height is $21 \mathrm{~m}$ and diameter of its base is $24 \mathrm{~m}$. [ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of the base of cone $=\left(\frac{24}{2}\right) \mathrm{m}=12 \mathrm{~m}$ Slant height (l) of cone $=21 \mathrm{~m}$ Total surface area of cone $=\pi r(r+l)$ $=\left[\frac{22}{7} \times 12 \times(12+21)\right] \mathrm{m}^{2}$ $=\left(\frac{22}{7} \times 12 \times 33...
Read More →Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm.
[question] Question. Diameter of the base of a cone is $10.5 \mathrm{~cm}$ and its slant height is $10 \mathrm{~cm}$. Find its curved surface area. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of the base of cone $=\left(\frac{10.5}{2}\right) \mathrm{cm}=5.25 \mathrm{~cm}$ Slant height (l) of cone $=10 \mathrm{~cm}$ CSA of cone $=\pi r l$ $=\left(\frac{22}{7} \times 5.25 \times 10\right) \mathrm{cm}^{2}=(22 \times 0.75 \times 10) \mathrm{cm...
Read More →Find
[question] Question. Find (i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high. (ii) How much steel was actually used, if $\frac{1}{12}$ of the steel actually used was wasted in making the tank. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Height (h) of cylindrical tank = 4.5 m Radius $(r)$ of the circular end of cylindrical tank $=\left(\frac{4.2}{2}\right) \mathrm{m}=2.1 \mathrm{~m...
Read More →In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm.
[question] Question. In a hot water heating system, there is a cylindrical pipe of length $28 \mathrm{~m}$ and diameter $5 \mathrm{~cm}$. Find the total radiating surface in the system. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m Radius (r) of circular end of pipe = = 2.5 cm = 0.025 m CSA of cylindrical pipe $=2 \pi r h$ $=\left(2 \times \frac{22}{7} \times 0.025 \times 28\right) \mathr...
Read More →The inner diameter of a circular well is 3.5 m.
[question] Question. The inner diameter of a circular well is $3.5 \mathrm{~m}$. It is $10 \mathrm{~m}$ deep. Find (i) Its inner curved surface area, (ii) The cost of plastering this curved surface at the rate of Rs 40 per $\mathrm{m}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Inner radius $(r)$ of circular well $=\left(\frac{3.5}{2}\right) \mathrm{m}=1.75 \mathrm{~m}$ Depth $(h)$ of circular well $=10 \mathrm{~m}$ Inner curved surface area $=2 \p...
Read More →Curved surface area of a right circular cylinder is $4.4 \mathrm{~m}^{2}$
[question] Question. Curved surface area of a right circular cylinder is $4.4 \mathrm{~m}^{2}$. If the radius of the base of the cylinder is $0.7 \mathrm{~m}$, find its height. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Let the height of the circular cylinder be h. Radius $(r)$ of the base of cylinder $=0.7 \mathrm{~m}$ CSA of cylinder $=4.4 \mathrm{~m}^{2}$ $2 \pi r h=4.4 \mathrm{~m}^{2}$ $\left(2 \times \frac{22}{7} \times 0.7 \times h\right) \mathr...
Read More →A cylindrical pillar is $50 \mathrm{~cm}$ in diameter and $3.5 \mathrm{~m}$ in height.
[question] Question. A cylindrical pillar is $50 \mathrm{~cm}$ in diameter and $3.5 \mathrm{~m}$ in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. $12.50$ per $\mathrm{m}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Height $(h)$ cylindrical pillar $=3.5 \mathrm{~m}$ Radius $(r)$ of the circular end of pillar $=\frac{50}{2}=25 \mathrm{~cm}$ $=0.25 \mathrm{~m}$ CSA of pillar $=2 \pi r h$ $=\left(2 \times \frac{22...
Read More →A metal pipe is 77 cm long.
[question] Question. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. (i) Inner curved surface area, (ii) Outer curved surface area (iii) Total surface area. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Inner radius $\left(r_{1}\right)$ of cylindrical pipe $=\left(\frac{4}{2}\right) \mathrm{cm}=2 \mathrm{~cm}$ Outer radius $\left(r_{2}\right)$ of cylindrical pipe $=\left(\frac{4.4}{2}\right) \ma...
Read More →It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet.
[question] Question. It is required to make a closed cylindrical tank of height $1 \mathrm{~m}$ and base diameter $140 \mathrm{~cm}$ from a metal sheet. How many square meters of the sheet are required for the same? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Height (h) of cylindrical tank = 1 m Base radius $(r)$ of cylindrical $\operatorname{tank}=\left(\frac{140}{2}\right) \mathrm{cm}=70 \mathrm{~cm}=0.7 \mathrm{~m}$ Area of sheet required $=$ Total ...
Read More →A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape.
[question] Question. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. (i) What is the area of the glass? (ii) How much of tape is needed for all the 12 edges? [/question] [solution] Solution: (i) Length $(l)$ of green house $=30 \mathrm{~cm}$ Breadth $(b)$ of green house $=25 \mathrm{~cm}$ Height $(h)$ of green house $=25 \mathrm{~cm}$ Total surface area of green house $=2[1 b+1 h+b h]$ $=...
Read More →A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long,
[question] Question. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high. (i) Which box has the greater lateral surface area and by how much? (ii) Which box has the smaller total surface area and by how much? [/question] [solution] Solution: (i) Edge of cube $=10 \mathrm{~cm}$ Length $(l)$ of box $=12.5 \mathrm{~cm}$ Breadth $(b)$ of box $=10 \mathrm{~cm}$ Height $(h)$ of box $=8 \mathrm{~cm}$ Lateral surface area of cubical box $=4(\text { edge }...
Read More →The paint in a certain container is sufficient to paint an area equal to 9.375 m2.
[question] Question. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container? [/question] [solution] Solution: Total surface area of one brick = 2(lb + bh + lh) $=[2(22.5 \times 10+10 \times 7.5+22.5 \times 7.5)] \mathrm{cm}^{2}$ $=2(225+75+168.75) \mathrm{cm}^{2}$ $=(2 \times 468.75) \mathrm{cm}^{2}$ $=9375 \mathrm{~cm}^{2}$ Let n bricks can be painted out by the paint of the c...
Read More →The floor of a rectangular hall has a perimeter 250 m.
[question] Question. The floor of a rectangular hall has a perimeter 250 m. If the cost of panting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. [/question] [solution] Solution: Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively. Area of four walls $=2 / h+2 b h$ $=2(I+b) h$ Perimeter of the floor of hall $=2(I+b)$ $=250 \mathrm{~m}$ $\therefore$ Area of four walls $=2(I+b) h=250 h \mathrm{~m}^{2}$ Cost of painting per...
Read More →The length, breadth and height of a room are 5 m, 4 m and 3 m respectively.
[question] Question. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m2. [/question] [solution] Solution: It is given that Length (l) of room = 5 m Breadth (b) of room = 4 m Height (h) of room = 3 m It can be observed that four walls and the ceiling of the room are to be white-washed. The floor of the room is not to be white-washed. Area to be white-washed = Area of walls +...
Read More →A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, i
[question] Question. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs 20. [/question] [solution] Solution: It is given that, length (l) of box = 1.5 m Breadth (b) of box = 1.25 m Depth (h) of box = 0.65 m (i) Box is to be open at top. Area of sheet required $=2 h+2 b h+l b$...
Read More →A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m.
[question] Question. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. [/question] [solution] Solution: Draw a line BE parallel to AD and draw a perpendicular BF on CD. It can be observed that ABED is a parallelogram. BE = AD = 13 m ED = AB = 10 m EC = 25 − ED = 15 m For $\triangle B E C$, Semi-perimeter, $s=\frac{(13+14+15) \mathrm{m}}{2}=21 \mathrm{~m}$ By Heron’s formula, Area of triangle $=\sqr...
Read More →A floral design on a floor is made up of 16 tiles which are triangular,
[question] Question. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see the given figure). Find the cost of polishing the tiles at the rate of 50p per cm2. [/question] [solution] Solution: It can be observed that Semi-perimeter of each triangular-shaped tile, $s=\frac{(35+28+9) \mathrm{cm}}{2}=36 \mathrm{~cm}$ By Heron’s formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of each tile $=[\sqrt{36(36-35)(36-28)(...
Read More →A kite in the shape of a square with a diagonal 32 cm
[question] Question. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangles of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it? [/question] [solution] Solution: We know that Area of square $=\frac{1}{2}$ (diagonal) $^{2}$ Area of the given $k i t e=\frac{1}{2}(32 \mathrm{~cm})^{2}=512 \mathrm{~cm}^{2}$ Area of $1^{\text {st }}$ shade $=$ Area of $2^{\text {nd }}$ shade $...
Read More →What is the physical state of water at :
[question] Question What is the physical state of water at : (a) 250°C (b) 100°C [/question] [solution] Solution (a) 250°C = gas (b) 100°C liquid as well as gas [/solution]...
Read More →An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure),
[question] Question. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella? [/question] [solution] Solution: For each triangular piece, Semi-perimeter, $s=\frac{(20+50+50) \mathrm{cm}}{2}=60 \mathrm{~cm}$ By Heron's formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of each triangular piece $=[\sqrt{60(60-50)(60-50)(60-20)}] \mat...
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