All questions

• Complete the following chart (body cavity, germ layer, phylum: Porifera, Aschelminthes, Arthropoda). Completed chart: Body cavityGerm layerPhylum AbsentDiploblasticPorifera AbsentTriploblasticPlatyhelminthes PseudocoelomTriploblasticAschelmi... Animal Classification 1 answer
• Complete the following chart (type, character, example: Cyclostomata, gill respiration, Amphibia, whale, poikilotherms). Completed chart: Type (class)CharacterExample CyclostomataJawless mouth provided with sucker; skin without scalesPetromyzon (lamprey), Myxin... Animal Classification 1 answer
• The ratio of circumference and area of a circle is 2:7. Find its circumference. (A) 14π (B) 7/π (C) 7π (D) 14/π Answer: (A) 14π 2π r/π r²=2/r=2/7⇒ r=7 Circumference =2π×7=14π . Mensuration 1 answer
• If measure of an arc of a circle is 160° and its length is 44 cm, find the circumference of the circle. (A) 66 cm (B) 44 cm (C) 160 cm (D) 99 cm Answer: (D) 99 cm Arc length =160/360× circumference =44 circumference=44×360/160=99\ cm Mensuration 1 answer
• Find the perimeter of a sector of a circle if its measure is 90° and radius is 7 cm. (A) 44 cm (B) 25 cm (C) 36 cm (D) 56 cm Answer: (B) 25 cm Arc length =90/360×2×22/7×7=11 cm. Perimeter =2r+l=14+11=25 cm. Mensuration 1 answer
• Find the curved surface area of a cone of radius 7 cm and height 24 cm. (A) 440 cm² (B) 550 cm² (C) 330 cm² (D) 110 cm² Answer: (B) 550 cm² l=√(24²+7²)=√(625)=25 cm. CSA =π rl=22/7×7×25=550 cm². Mensuration 1 answer
• The curved surface area of a cylinder is 440 cm² and its radius is 5 cm. Find its height. (A) 44/π cm (B) 22π cm (C) 44π cm (D) 22/π cm Answer: (A) 44/π cm 2π rh=440⇒2π×5× h=440 h=440/10π=44/π\ cm Mensuration 1 answer
• A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone. (A) 15 cm (B) 10 cm (C) 18 cm (D) 5 cm Answer: (A) 15 cm Volumes are equal: 1/3π r²H=π r²×5 , so H=15 cm. Mensuration 1 answer
• Find the volume of a cube of side 0.01 cm. (A) 1 cm³ (B) 0.001 cm³ (C) 0.0001 cm³ (D) 0.000001 cm³ Answer: (D) 0.000001 cm³ V=(0.01)³=(10⁻²)³=10⁻⁶=0.000001 cm³. Mensuration 1 answer
• Find the side of a cube of volume 1 m³. (A) 1 cm (B) 10 cm (C) 100 cm (D) 1000 cm Answer: (C) 100 cm a³=1 m³ gives a=1 m =100 cm. Mensuration 1 answer
• A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub? (π = 22/7) Answer: 20350 cm³ = 20.35 litres V=1/3π h(r₁²+r₂²+r₁r₂) =1/3×22/7×21×(400+225+300) =22×925=20350\ cm³ Mensuration 1 answer
• Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube? Answer: 7830 balls Outer radius R = 30 cm, inner radius r = 30 − 2 = 28 cm, length h = 90 cm. V_tube=π(R²-r²)h=π(900-784)×90 =10440π\ cm³ Vo... Mensuration 1 answer
• A metal parallelopiped of measures 16 cm × 11 cm × 10 cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively? Answer: 2800 coins (π = 22/7) Volume of metal =16×11×10=1760 cm³. Each coin is a cylinder: r = 1 cm, h = 2 mm = 0.2 cm. V_coin=22/7×1²×0.2=4... Mensuration 1 answer
• The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of Rs. 10 per sq.m. Answer: Rs. 6336 r = 60 cm, h = 84 cm. Area covered in one rotation = curved surface area: 2π rh=2×22/7×60×84=31680\ cm² =3.168 m². In 200 r... Mensuration 1 answer
• The diameter and thickness of a hollow metal sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm³. Find the outer surface area and mass of the sphere. Answer: Outer surface area = 452.16 cm²; mass ≈ 3383 g (π = 3.14) Outer radius R = 6 cm; thickness 0.01 m = 1 cm, so inner radius r = 5 cm.... Mensuration 1 answer
• A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into a shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone? Answer: 2640 cm² Volume of sand =π r²h=22/7×14²×20=12320 cm³. For the cone, 1/3×base area×14=12320 . Base area=12320×3/14=2640\ cm² Mensuration 1 answer
• The radius of a metallic sphere is 9 cm. It was melted to make a wire of diameter 4 mm. Find the length of the wire. Answer: 243 m Volume of sphere =4/3π(9)³=972π cm³. Wire radius = 2 mm = 0.2 cm. Volume of wire =π(0.2)²L=0.04π L . 0.04π L=972π⇒ L=24300\ cm... Mensuration 1 answer
• The area of a sector of a circle of 6 cm radius is 15π sq.cm. Find the measure of the arc and length of the arc corresponding to the sector. Answer: Measure of arc = 150°, length of arc = 5π cm θ/360×π×36=15π⇒θ=15×360/36=150^° Area =l× r/2⇒15π=l×6/2⇒ l=5π cm. Mensuration 1 answer
• Seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion (minor segment AB). (π = 3.14, √3 = 1.73) Answer: ≈ 39.31 cm² Let PM ⊥ AB, PM = 4, PA = 8. In right ΔPMA, cos∠ APM=4/8=1/2 , so ∠ APM=60^° and ∠ APB=120^° . AM=√(64-16)=4sqrt3 , so A... Mensuration 1 answer
• Square ABCD is inscribed in the sector A-PCQ. The radius of sector C-BXD is 20 cm. Complete the activity to find the area of the shaded region. (π = 3.14) Answer: Total shaded area = 86 + 228 = 314 cm² Side of square = radius of sector C-BXD = 20 cm. Area of square =20²=400 cm². Shaded region i... Mensuration 1 answer
• Balance the following equation stepwise: H2S2O7(l) + H2O(l) → H2SO4(l) Balanced equation: H₂S₂O₇(l) + H₂O(l) → 2H₂SO₄(l) Step 1: Count atoms in the skeletal equation. ElementReactant sideProduct side H2 + 2 = 42... Chemical Reactions and Equations 1 answer
• Balance the following equation stepwise: SO2(g) + H2S(aq) → S(s) + H2O(l) Balanced equation: SO₂(g) + 2H₂S(aq) → 3S(s) + 2H₂O(l) Step 1: Count atoms in the skeletal equation. Left: S = 2, O = 2, H = 2. Right: S = 1... Chemical Reactions and Equations 1 answer
• Balance the following equation stepwise: Ag(s) + HCl(aq) → AgCl + H2 Balanced equation: 2Ag(s) + 2HCl(aq) → 2AgCldownarrow + H₂uparrow Step 1: Count atoms. Left: Ag = 1, H = 1, Cl = 1. Right: Ag = 1, Cl = 1, H... Chemical Reactions and Equations 1 answer
• Balance the following equation stepwise: NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) Balanced equation: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l) Step 1: Count atoms. Left: Na = 1, O = 1 + 4 = 5, H = 1 + 2 = 3, S = 1. Righ... Chemical Reactions and Equations 1 answer