# CBSE Class 10th Math 13 – Surface Areas and Volumes MCQs

#### A playground has an area of 158400 sq. m. If a roller of length 3.5 m takes 6000 revolutions for levelling it, then the diameter of the roller is ____.

Area of the ground = Curved surface area of roller × Number of revolutions D = diameter of the roller h = height of the roller = 3.5 m Given, Number of revolutions = 6000 Area of the play ground = 158400 m^{2} Curved surface area of the roller = 158400/6000 = 26.4 m^{2} Curved surface area of roller = πdh ⇒26.4 = 22/7 × d × 3.5 ⇒ d = 2.4m Hence, the diameter of the roller is 2.4 m.

#### A well of diameter 4 m is dug to a depth of 21 m and the earth collected is spread evenly on a platform 80 m × 60 m. The raise in the level of the platform is ____.

Volume of the earth = Volume of the platform r= Radius of well h= Depth of well l= length of the platform b= breadth of the platform x= height of the platform ⇒π r^{2}h=lbx ⇒227 × 4 × 21 = 80 × 60 ×x ⇒x = 2644800 =0.055 m =5.5 cm. Hence, the rise in the level of the platform is 5.5 cm

#### A sphere of radius 3.5 cm is melted and made into small cones of the same radius and height 1cm. The number of such cones can be made is ____.

Number of cones formed=Volume of sphere/Volume of cone ⇒ (4/3πr^{3})/ (1/3πr^{2}h) = 4r = 4 × 3.5 Hence, the number of cones formed is 14.

#### Two cubes of edges 5 cm are kept side to side. The surface area of a cuboid so formed is ____.

When two cubes of edges5 cm are kept side to side, then the length of the cuboid so formed is 10 cm. breadth = height = 5 cm Surface area of a cuboid =2(lb+bh+hl) =2(10 × 5 + 5 × 5 + 5 × 10) =2(50 + 25 + 50) =250 cm^{2} Hence, thesurface area of the cuboid is 250 cm^{2}.

#### A right circular cone, which is sliced through by a plane parallel to its base, when the smaller conical portion is removed, the resulting solid is called _____.

A right circular cone, which is sliced through by a plane parallel to its base, when the smaller conical portion is removed, the resulting solid is called frustum of a right circular cone.

#### A cube of a certain edge is immersed in water contained in a cylinder. There will be no change in the water level.

Whenever a solid is immersed in water contained in a vessel, water equal in volume to the solid immersed, will get displaced. Consequently, there will be a rise in the water level. Applying this to the given problem, there will be change in the water level.

#### A solid cone of 6 cm diameter and a height of 21 cm is melted and recast into a cylinder of diameter 4 cm. The height of the cylinder is ____.

Let h be the height of the cone and H be the height of the cylinder Let r be the radius of the cone and R be the radius of the cylinder Volume of the cone = Volume of the cylinder 13 πr^{2}h = πR^{2}H ⇒ 13 × 9 × 21 = 4H ⇒H = 15.75 cm. Hence, height of the cylinder is 15.75 cm

#### A metallic hemisphere of radius r is melted and recast into a cone of the same radius. If 'H' is the height of the cone, then the value of H: r is ____.

Volume of the cone = Volume of the hemisphere 1/3 πr^{2}H = 2/3 πr^{3} H = 2r ⇒ H/r = 2/1 Hence, the value of H : r = 2 : 1

#### A cylinder is melted and recast into a cone of the same base shall have 3 times its earlier height.

Volume of cylinder = Volume of cone πr^{2}h = 1/3πr^{2}h_{1} ⇒ h = 1/3h_{1 } ⇒ h_{1}=3h Hence, if a cylinder is melted and recast into a cone of the same base shall have 3 times its earlier height.