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Question 1: Two cubes each of volume 64 cm$^3$ are joined end to end. The surface area of the resulting cuboid is:
Solution: (C) 160 cm$^2$
Step 1: Volume of cube $= a^3 = 64 \Rightarrow a = 4$ cm.
Step 2: Dimensions of cuboid: $L = 4+4=8$, $B=4$, $H=4$.
Step 3: TSA $= 2(LB+BH+HL) = 2(32+16+32) = 2(80) = 160$ cm$^2$. -
Question 2: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. The volume of the solid is:
Solution: (A) $\pi$ cm$^3$
Step 1: Radius $r=1$, Height of cone $h=1$.
Step 2: Volume $= \text{Vol(Cone)} + \text{Vol(Hemisphere)} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$.
Step 3: $= \frac{1}{3}\pi(1)(1) + \frac{2}{3}\pi(1) = \frac{1}{3}\pi + \frac{2}{3}\pi = \pi$ cm$^3$. -
Question 3: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. The height of the cylinder is:
Solution: (A) 2.74 cm
Step 1: Volume of Sphere = Volume of Cylinder. $\frac{4}{3}\pi R^3 = \pi r^2 h$.
Step 2: $\frac{4}{3}(4.2)^3 = (6)^2 h \Rightarrow h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 36}$.
Step 3: $h = \frac{296.352}{108} \approx 2.744$ cm. -
Question 4: From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. The volume of the remaining solid is:
Solution: (A) $240\pi$ cm$^3$
Step 1: Remaining Volume = Vol(Cylinder) - Vol(Cone).
Step 2: $= \pi r^2 h - \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^2 h$.
Step 3: $= \frac{2}{3}\pi (6)^2 (10) = \frac{2}{3}\pi (360) = 240\pi$ cm$^3$. -
Question 5: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Its surface area is:
Solution: (B) 220 mm$^2$
Step 1: Radius $r = 2.5$ mm. Height of cylinder $h = 14 - 2(2.5) = 9$ mm.
Step 2: TSA = CSA(Cylinder) + 2 $\times$ CSA(Hemisphere) $= 2\pi rh + 4\pi r^2 = 2\pi r(h+2r)$.
Step 3: $= 2 \times \frac{22}{7} \times 2.5 (9 + 5) = \frac{110}{7} \times 14 = 220$ mm$^2$. -
Question 6: A shuttlecock used for playing badminton has the shape of the combination of:
Solution: (D) Frustum of a cone and a hemisphere
Reason: The cork base is a hemisphere and the feathers form a frustum of a cone. -
Question 7: The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm is:
Solution: (C) 5
Step 1: Vol(Cylinder) $= \pi (2)^2 (45) = 180\pi$.
Step 2: Vol(Sphere) $= \frac{4}{3}\pi (3)^3 = 36\pi$.
Step 3: Number $= \frac{180\pi}{36\pi} = 5$. -
Question 8: The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is:
Solution: (D) 19.4 cm$^3$
Step 1: For largest cone, diameter = edge = 4.2 cm $\Rightarrow r = 2.1$ cm. Height $h = 4.2$ cm.
Step 2: Volume $= \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (2.1)^2 \times 4.2$.
Step 3: $= \frac{1}{3} \times 22 \times 0.63 \times 4.2 = 19.404$ cm$^3$. -
Question 9: A metallic solid sphere of radius 9 cm is melted to form a solid cylinder of radius 9 cm. The height of the cylinder is:
Solution: (A) 12 cm
Step 1: $\frac{4}{3}\pi R^3 = \pi R^2 h$ (since radii are same).
Step 2: $\frac{4}{3} R = h \Rightarrow h = \frac{4}{3}(9) = 12$ cm. -
Question 10: If the surface area of a sphere is $144\pi$ cm$^2$, then its volume is:
Solution: (A) $288\pi$ cm$^3$
Step 1: $4\pi r^2 = 144\pi \Rightarrow r^2 = 36 \Rightarrow r = 6$ cm.
Step 2: Volume $= \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi$ cm$^3$.