NCERT Exemplar Class 10 Maths Chapter 12 Exercise 12.2 Solutions

Verdict: False. The total surface area is 4π r², not 6π r².

Concept used. Two hemispheres joined base-to-base make a sphere. The curved surface of one hemisphere is 2π r², and the joined flat bases vanish.

1. Curved surface of each hemisphere: [] 2π r² each.
2. The two flat circular bases are stuck together, so they are no longer on the outside.
3. Add the two curved surfaces only: [] 2π r²+2π r²=4π r².

False: the combination is a sphere of surface area 4π r², so the claimed 6π r² is wrong.

Verdict: False. The correct total surface area is 4π rh+2π r².

Concept used. Stacking two identical cylinders makes one tall cylinder of height 2h. Its surface is 2π R H+2π R² with R=r and H=2h.

1. The combined solid is a single cylinder of radius r and height 2h.
2. Curved surface: [] 2π r(2h)=4π rh.
3. Two circular ends (top and bottom): [] 2×π r²=2π r².
4. Add the parts: [] total =4π rh+2π r².

False: the surface area is 4π rh+2π r², not 4π rh+4π r².

Verdict: False. The correct total surface area is π r[√r²+h²+2h+r].

Concept used. Total surface = CSA of cone + CSA of cylinder + base of cylinder. The cone's slant is l=√r²+h²; the join between cone base and cylinder top is hidden.

1. Curved surface of the cone: [] π r l=π r√r²+h².
2. Curved surface of the cylinder: [] 2π rh.
3. Only the bottom circle of the cylinder is exposed (its top is covered by the cone): [] π r².
4. Add the exposed parts: [] π r√r²+h²+2π rh+π r²=π r[√r²+h²+2h+r].

False: the correct area is π r[√r²+h²+2h+r], with a single r inside, not 3r.

Verdict: False. The volume of the ball is 16π a³.

Concept used. A ball that exactly fits a cube touches all six faces, so its diameter equals the side a, giving radius a2. Volume =43π r³.

1. The ball touches opposite faces, so its diameter is the side: [] diameter =a, hence r=a2.
2. Apply the sphere volume: [] 43π r³=43π(a2)³.
3. Cube the radius: [] 43π·a³8=4π a³24=π a³6.

False: the true volume is π a³6, eight times smaller than the claimed 43π a³.

Verdict: False. The correct formula has a +r₁r₂ term: V=13π h[r₁²+r₂²+r₁r₂].

Concept used. The standard frustum-volume formula carries a plus sign before the product r₁r₂, not a minus.

1. Recall the correct frustum volume: [] V=13π h(r₁²+r₂²+r₁r₂).
2. The given statement uses -r₁r₂, which changes the value.
3. Check with equal radii r₁=r₂=r (frustum becomes a cylinder): [] correct: 13π h(3r²)=π r²h (a cylinder, right); [] given: 13π h(r²) (one-third of a cylinder, wrong).

False: the sign before r₁r₂ must be +, giving V=13π h(r₁²+r₂²+r₁r₂).

Verdict: True. The capacity is indeed π r²3(3h-2r).

Concept used. The hemisphere pushes into the vessel, so it removes space. Capacity = volume of cylinder - volume of hemisphere.

1. Volume of the cylinder (radius r, height h): [] π r² h.
2. Volume of the hemisphere bulging in (radius r): [] 23π r³.
3. Subtract, since the hemisphere eats into the holding space: [] capacity =π r²h-23π r³.
4. Take out π r²3: [] capacity =π r²3(3h-2r).

True: capacity =π r²3(3h-2r), matching the figure.

Verdict: False. The curved-surface formula π l(r₁+r₂) is correct, but the slant height must be l=√h²+(r₁-r₂)², with a minus.

Concept used. The slant height of a frustum is the hypotenuse of a right triangle whose legs are the vertical height h and the difference of radii r₁-r₂.

1. Drop a perpendicular from the smaller rim to the larger rim. The horizontal leg is the difference of radii: [] horizontal leg =r₁-r₂.
2. The vertical leg is the height h.
3. By Pythagoras, the slant height is: [] l=√h²+(r₁-r₂)².
4. So the statement's (r₁+r₂)² inside the root is wrong; the CSA expression π l(r₁+r₂) itself is fine.

False: it should be l=√h²+(r₁-r₂)², the difference of radii, not the sum.

Verdict: True. The sheet used is exactly CSA of frustum + area of circular base + CSA of cylinder.

Concept used. ``Metallic sheet used'' means every face you have to cut from the sheet: the open frustum has no lids, the bottom is a flat circle, and the cylindrical base is a tube.

1. The bucket is open at the top, so the frustum contributes only its curved face: CSA of frustum.
2. The base of the bucket is a flat circle of metal: area of circular base.
3. The hollow cylindrical stand contributes only its curved wall: CSA of cylinder.
4. Add these three, which is precisely the stated expression.

True: sheet = CSA(frustum) + base circle + CSA(cylinder).