NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2

Add volumes; keep π in symbol form. The whole question turns on reading two instructions correctly.

• Volumes add: joined solids lose no volume, so the answer is the cone plus the hemisphere with nothing subtracted.
• Stop at π: ``in terms of π'' means leave the answer as π cm³, never as a decimal like 3.14.
• Light algebra: with r=1 you have r²=r³=1, so the cone gives 13π, the hemisphere gives 23π, and the thirds add to a whole π almost by inspection.
• Quick check: both parts share the radius and the cone height equals it, so the squat solid landing on a tidy π cm³ signals the numbers behaved.

π cm³, the sum of 13π (cone) and 23π (hemisphere).

Keep everything over a denominator of 7. Working in sevenths keeps the arithmetic exact from start to finish.

• Common denominator: because the value of pi used here is twenty-two over seven, every volume term naturally carries a seven in the denominator. If you write the cylinder and the two cones over that same seven, you can add the numerators in one stroke and divide once at the end, with no decimals to round midway and no small errors creeping in.
• No wall thickness: the phrase about the outer and inner dimensions being nearly the same is telling you to treat the aluminium sheet as having no thickness at all, so the air inside fills exactly the full solid volume and no subtraction for the wall is needed.
• Double one cone: the two cones share the same radius and the same height, so the safest book-keeping is to work out one cone and simply double it, rather than inventing a single tall cone of combined height.
• Find the cylinder height: the given length runs from one cone tip to the other, so you must subtract both cone lengths to get the straight cylinder in the middle. Forgetting to remove both cones is the single most common error on this problem, and it makes the cylinder volume far too large.
• Ratio check: a cone and a cylinder of equal radius and height stand in volumes of one to three. The cones here are shorter than the cylinder, so together they add only a small fraction of the total, and a cone share larger than the cylinder would mean a height had been swapped somewhere.
• Whole-number finish: the final fraction divides exactly to a clean whole number, so a leftover decimal would flag a slip, most likely in squaring the radius or in handling the denominator.

66 cm³, from 3967 (cylinder) plus 667 (two cones).

Find one piece, scale up, then take the percentage. Work in that order so the numbers stay manageable.

• One real sum: the volume of a single gulab jamun, about 25.05 cm³, is the only calculation. The rest is multiplying by 45 and then by 0.30.
• Caps make a sphere: treat the two ends as one sphere of radius 1.4 cm, giving the 43π r³ term in one shot instead of 2×23π r³.
• Trim the cylinder: the cylinder length is 5-2.8=2.2 cm after removing the two rounded ends, and missing that subtraction is the usual error.
• Round late: ``about'' allows 338 cm³, but keep three decimals in the working and round only at the last line so the figure stays reliable.

≈ 338 cm³ of syrup in 45 pieces (about 25.05 cm³ each).

Compute one cone, scale to four, subtract once. The cuboid is easy, so the care goes into the four small cones.

• One cone: with r=0.5 the square r²=0.25 is small, and 227× 0.25× 1.4=1.1, so one cone is about 0.3667 cm³ and four remove only about 1.47 cm³.
• Near-equality check: the depressions are tiny next to the block, so 523.53 cm³ sits just below 525. A number far below 525 means a cone was overstated, perhaps by using the radius where the depth belongs.
• Round late: carry four decimals in the cone volume and round only the final answer, since trimming 0.3667 to 0.37 early drifts the total once you multiply by four.

523.53 cm³ of wood (525 cuboid minus 1.47 in four cones).

Cancel π before you reach for a calculator. Keeping π in symbol form to the end is faster and avoids rounding error.

• Pi cancels: both the overflow and one shot carry a factor of π, so it drops out in the division and the count is a pure number.
• Clean divide: the overflow 50π3 divided by one shot π6 has its threes and sixes simplify to exactly 100.
• Whole number expected: you cannot drop part of a lead shot, so a decimal answer means a volume was set up wrongly.
• Exact cube: 0.5³ = 0.125 = 18, so 43π·18 = π6. Spotting this keeps the fractions exact instead of slipping into decimals.

100 lead shots, found by dividing the overflow 50π3 by one shot π6.

Factor π once, convert grams to kilograms at the end. A single multiply and a clean unit are all this question rewards.

• One multiply: keep π outside the bracket while you add 31680 and 3840 to get 35520, then multiply by 3.14 just once rather than scaling each cylinder.
• Grams to kilograms: the density is per cm³, so the mass arrives in grams. Dividing 892262.4 g by 1000 gives the natural 892.26 kg.
• Always label: a bare ``892262.4'' with no gram or kilogram tag loses a mark, so carry the unit through.
• Dominance check: the tall lower cylinder gives 31680π of the 35520π, about nine-tenths. An upper cylinder larger than the lower one means a radius was swapped.

≈ 892.26 kg (i.e. 892262.4 g) for a volume of 35520π cm³.

Keep π in symbol form through the subtraction. Symbol-form π keeps every step clean until the very last line.

• Equal parts: the cone and the hemisphere here come to exactly the same volume, because the cone height happens to be twice the radius, which makes its formula collapse to the hemisphere formula. Spotting that coincidence lets you write both volumes down at once instead of grinding through each.
• Clean subtraction: keeping pi as a symbol, you add the cone and hemisphere together and take the sum away from the cylinder, which leaves a single tidy multiple of pi before you have done any decimal work at all.
• Convert last: only at the very end do you substitute the numerical value of pi to reach the figure in cubic centimetres. A thousand cubic centimetres make one litre and a million make one cubic metre, so the same answer reads as roughly eleven hundred litres or a little over one cubic metre.
• Proportion check: the solid fills almost half of the cylinder, so a little over half the water stays behind. A rough mental check of just under half is a quick way to catch a gross error before you reach the final figure.
• Read the wording: the phrase about touching the bottom is confirming that the solid is fully submerged, so all of its volume displaces water. A solid only partly under the surface would displace just the part below the water line, and the working would change.

360000π cm³ ≈ 1131428.57 cm³ of water remains.

Compute the true value, then judge the claim. A verification question is answered only when you find the real volume and state a verdict.

• Cube decides it: the sphere dominates, so the step (4.25)³ = 76.765625 settles the answer. Rush this cube and the comparison goes wrong.
• Add the parts: the small neck 25.12 cm³ plus the sphere 321.39 cm³ gives about 346.51 cm³.
• State the verdict: since 346.51 differs from 345, the child is wrong, though only slightly, the gap coming partly from her rounding while measuring water.
• Hold two decimals: keys quote 346.51 cm³, so rounding the sphere term early to 321 could mislead you into calling 345 correct. Precision is the whole point here.

True volume ≈ 346.51 cm³; the child's 345 cm³ is not correct.