Two walls and the ceiling of a room meet at right angles at point P. A fly is in the air, 1 meter from one wall, 8 meters from the other wall and 9 meters from the point P. How many meters is the fly from the ceiling?
[1] 14
[2] 15
[3] 4
[4] None of these
$1 ^ { 2 } + 8 ^ { 2 } + z ^ { 2 } = 9 ^ { 2 }$ so $z = 4$
Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with the adjacent marble(s). Also, each marble is in contact all around the funnel wall. The smallest marble has a radius of 8 mm. The largest marble has a radius of 18 mm. What is the radius (in mm) of the middle marble?
[1] 10
[2] 11
[3] 12
[4] 15
Triangles shown in the figure are similar by AA property. Hence
$\frac { r _ { 1 } - r _ { 2 } } { r _ { 1 } + r _ { 2 } } = \frac { r _ { 2 } - r _ { 3 } } { r _ { 2 } + r _ { 3 } } = r _ { 2 } ^ { 2 } = r _ { 1 } \cdot r _ { 3 }$
Hence $r _ { 2 }$ is geometric mean of $r _ { 1 }$ and $r _ { 3 } .$ And hence $r _ { 1 } , r _ { 2 }$ and $r _ { 3 }$ in G.P.
Similarly radius of these five marbles are in G.P. Therefore $r _ { 2 } ^ { 2 } = 8 \times 18 \Rightarrow r _ { 2 } = 12 \mathrm { mm }$
Three identical cones with base radius r are placed on their base so that each is touching the other two. The radius of the circle drawn through their vertices is
[1] smaller than r
[2] equal to r
[3] larger than r
[4] depends on the height of the cones
The centres of the bases of the cones from a triangle of side 2$\mathrm { r. }$ . The circumcircle of the circle will be identical to a circle drawn through the vertices of the cones and thus, it will have a radius of $\frac { 2 } { \sqrt { 3 } }$ times $r$ . which is greater than $\mathrm { r }$ .
Glass spheres (which submerge fully in water) of radius 7 mm are dropped into a cylindrical vessel containing some water. The diameter of the vessel is 14 cm. Find how many spheres have been dropped in it if the water level rises by 3.5 cm?
[1] 375
[2] 300
[3] 37
[4] Data insufficient
Volume of water displaced = $\pi r ^ { 2 } h = \left( \pi \times ( 7 ) ^ { 2 } \times 3.5 \right) \mathrm { cm } ^ { 3 }$
Volume of glass spheres $= \frac { \text { Volume of water displaced } } { \text { Volume of a glass sphere } }$$= \frac { \pi \times 7 \times 7 \times 3.5 } { \frac { 4 } { 3 } \times \pi \times \frac { 7 } { 10 } \times \frac { 7 } { 10 } } = 375 .$
A regular hexagonal prism has its perimeter of bases as 144 cm and height 120 cm. It contains water upto a height of 110 cm. A spherical ball of diameter of 35 cm is dropped into it due to which the level of water rises and some water flows out. Find the volume of water that flows out.
[1] 9000 cc
[2] 7500 cc
[3] 11000 cc
[4] Cannot be determined
Side of hexagonal base $= \frac { 144 } { 6 } = 24 \mathrm { cm }$
Area of hexagonal base $= \frac { 3 \sqrt { 3 } } { 2 } \times$ side $^ { 2 } = \frac { 3 \sqrt { 3 } } { 2 } \times 24 \times 24 = 864 \sqrt { 3 }$
Volume of sphere with diameter $35 \mathrm { cm } = \frac { 4 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 35 } { 2 } \right) ^ { 3 } = 22458 \mathrm { cc }$
volume of empty space in prism $= 864 \sqrt { 3 } \times 120 - 864 \sqrt { 3 } \times 110$
$= 864 \sqrt { 3 } \times 10 \approx 14965 \mathrm { cc }$
Therefore $\quad$ Volume of water that flows out
$= 22458 - 14965 = 7493 \mathrm { cc } \approx 7500 \mathrm { cc }$
A solid cone of metal of radius r and height h cm is melted into a cube of length a. If h + r = 10 cms, and the volume of the cube formed is maximum then the radius of the cone in cms is
[1] $\frac { 5 } { 3 }$
[2] $\frac { 20 } { 3 }$
[3] $\frac { 10 } { 3 }$
[4] 5
For total surface area of cube to be maximum the volume of the cone has to be maximum i.e., $\frac { 1 } { 3 } \pi r ^ { 2 }$ has to be maximum provided $h + r = 10 .$ Going by options it is clear that radius will be $\frac { 20 } { 3 }$ and height will be $10 - \frac { 20 } { 3 } \cdot$ Ans. $( 4 )$
Find the edge of the cube that can be inscribed in a cone with base diameter 12 cm and height 8 cm.
[1] 4.8
[2] $\frac { 24 } { 2 \sqrt { 2 } }$
[3] $\frac { 24 } { 2 \sqrt { 2 } + 3 }$
[4] None of These
Let a be the edge of the cube $\mathrm { QS } = \mathrm { a } \sqrt { 2 }$. Now $\Delta \mathrm { AQS } \approx \Delta \mathrm { ABC }$ as $\mathrm { QS }$ is parallel to $\mathrm { BC }$, so by AAA symmetry. $\frac { \mathrm { AO } } { \mathrm { AK } } = \frac { Q \mathrm { S } } { \mathrm { BC } }$ so $\frac { 8 - \mathrm { a } } { 8 } = \frac { \mathrm { a } \sqrt { 2 } } { 12 }$.
$a = \frac { 24 } { 2 \sqrt { 2 } + 3 }$
Water flows at the rate of 10 metres/minute from a cylindrical pipe 5 mm in diameter. How long will it take to fill up a conical vessel whose diameter at the base is 40 cms and depth 24 cms?
[1] 55 min
[2] 52 min
[3] 51 min
[4] 48 min
Volume of conical vessel $= \frac { 1 } { 3 } \pi ( 20 ) ^ { 2 } \times 24 = 3200 \pi$.
Volume flows out from cylindrical pipe in $1 \mathrm { min. } = \pi \frac { 2.5 } { 10 } \times \frac { 2.5 } { 10 } \times 1000 = 62.5 \pi$.
Therefore, time taken to fill conical vessel $= \frac { 3200 \pi } { 62.5 \pi }$ = 51 min 12 sec
All five faces of a regular pyramid with a square base are found to be of the same area. The height of the pyramid is 3 cm. The total area of all its surfaces is
[1] 10 sq. cm
[2] 12 sq.cm
[3] 15 sq. cm
[4] 20 sq. cm
Height of the triangular face with base length 'a' $= \sqrt { \frac { a ^ { 2 } } { 4 } + 3 ^ { 2 } }$
Where a = edge length of a square base
Therefore Area of one triangular base $= \frac { 1 } { 2 } \times a \times \sqrt { \frac { a ^ { 2 } } { 4 } + 3 ^ { 2 } }$
By problem, $= \frac { 1 } { 2 } \times \mathrm { a } \times \sqrt { \frac { \mathrm { a } ^ { 2 } } { 4 } + 3 ^ { 2 } } = \mathrm { a } ^ { 2 } \Rightarrow 15 \mathrm { a } ^ { 2 } = 36$.
Therefore, sum of all faces $\Rightarrow 5 \mathrm { a } ^ { 2 } = 12 \mathrm { sq } . \mathrm { cm }$
A piece of string is wound uniformly around a triangular prism ABCDEF, of an equilateral base. The thread starts from the corner A at the base and ends at the corner D of the top, as shown in the figure. If there are total 14 rounds of string and the side of the base and the height of the prism are 2 cm and 13 cm respectively, then find the length of the string.
[1] 85 cm
[2] 84 cm
[3] 87 cm
[4] 97 cm
The prism is rolled out and opened into a rectangle ABCD. The length of the string will be equal to the length of the hypotenuse of the following right angled triangle whose base will be 2 × 3 × 14 and the length is 13 cm.
$= \sqrt { 84 ^ { 2 } + 13 ^ { 2 } } = 85 \mathrm { cm }$
A circular hall, surmounted by a hemispherical roof, contains 5236$\mathrm { m } ^ { 3 }$ of air and the internal diameter of the room is equal to the height of the highest point of the roof above the ground. Find the height of the room (Take $\pi = 3.1416 )$ .
[1] 10 m
[2] 20 m
[3] 12 m
[4] 15 m
Let the radius of the circular floor be r metres
Therefore, volume of the cylindrical portion is $\pi r ^ { 2 } \cdot r = \pi r ^ { 3 } m ^ { 3 }$
Volume of the hemispherical portion is $\frac { 2 } { 3 } \pi r ^ { 3 } m ^ { 3 }$
Therefore, volume of air in the hall $= \pi r ^ { 3 } + \frac { 2 } { 3 } \pi r ^ { 3 } = \frac { 5 } { 3 } \pi r ^ { 3 } m ^ { 3 }$
Given $\frac { 5 } { 3 } \pi r ^ { 3 } = 5236$ or $r ^ { 3 } = \frac { 5236 } { 3.1416 } \times \frac { 3 } { 5 } = 1000 \mathrm { m } ^ { 3 } ,$ nearly
Or, $ r = 10 \mathrm { m } .$ So the height of the room is $2 r = 20$ metres
The material within a hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted and cast into the form of a right circular cone of base diameter 8 cm. Find the height of the cone.
[1] 28 cm
[2] 15 cm
[3] 20 cm
[4] 14 cm
The volume of the material of the hollow sphere $= \frac { 4 } { 3 } \pi \left[ 4 ^ { 3 } - 2 ^ { 3 } \right] = \frac { 4 \pi } { 3 } \times 56$
The radius of the base of the cone cast $= 4 \mathrm { cm }$
If h cm is the height of the cone, $\frac { 1 } { 3 } \pi 4 ^ { 2 } \mathrm { h } = \frac { 4 } { 3 } \pi \times 56$
i.e., $\mathrm { h } = \frac { 4 \times 56 } { 16 } = 14 \mathrm { cm } ,$ is the height of the cone.
If a cube of maximum possible volume is cut off from a solid sphere of diameter d, then the volume of the remaining (waste) material of the sphere would be equal to
[1] $\frac { d ^ { 3 } } { 3 } \left( \pi - \frac { d } { 2 } \right)$
[2] $\frac { d ^ { 3 } } { 3 } \left( \frac { \pi } { 2 } - \frac { 1 } { \sqrt { 3 } } \right)$
[3] $\frac { \mathrm { d } ^ { 2 } } { 4 } ( \sqrt { 2 } - \pi )$
[4] None of these
The diagonal of cube will be equal to the diameter of sphere.
Therefore, volume of sphere $= \frac { 4 } { 3 } \pi \left( \frac { \mathrm { d } } { 2 } \right) ^ { 3 } = \frac { \pi \mathrm { d } ^ { 3 } } { 6 }$
and each side of cube $= \mathrm { a } = \frac { \mathrm { d } } { \sqrt { 3 } }$
Volume of cube $= a ^ { 3 } = \frac { d ^ { 3 } } { 3 \sqrt { 3 } }$
Remaining volume $= \frac { \pi \mathrm { d } ^ { 3 } } { 6 } - \frac { \mathrm { d } ^ { 3 } } { 3 \sqrt { 3 } } = \frac { \mathrm { d } ^ { 3 } } { 3 } \left( \frac { \pi } { 2 } - \frac { 1 } { \sqrt { 3 } } \right)$
In a bullet the gun powder is to be filled up inside the metallic enclosure. The metallic enclosure is made up of a cylindrical base and conical top with the base of radius 5 cm. The ratio of height of cylinder and cone is 3 : 2. A cylindrical hole is drilled through the metal solid with height two-third the height of metal solid. What should be the radius of the hole, so that the volume of the hole (in which gun powder is to be filled up) is one-third the volume of metal solid after drilling?
[1] $\sqrt { \frac { 88 } { 5 } } \mathrm { cm }$
[2] $\sqrt { \frac { 55 } { 8 } } \mathrm { cm }$
[3] $\frac { 55 } { 8 } \mathrm { cm }$
[4] 33$\pi \mathrm { cm }$
Volume of the whole body
$\mathrm { V } _ { 1 } = \frac { 1 } { 3 } \pi r _ { 1 } ^ { 2 } \mathrm { h } _ { 1 } + \pi \mathrm { r } _ { 1 } ^ { 2 } \mathrm { h } _ { 2 }$
But $\frac { \mathrm { h } _ { 1 } } { \mathrm { h } _ { 2 } } = \frac { 2 } { 3 }$
So, $ \mathrm { V } _ { 1 } = \pi \mathrm { r } _ { 1 } ^ { 2 } \frac { 11 \mathrm { h } _ { 1 } } { 6 }$
and $\mathrm { h } _ { 3 } = \frac { 2 } { 3 } \left( \mathrm { h } _ { 1 } + \mathrm { h } _ { 2 } \right) = \frac { 5 \mathrm { h } _ { 1 } } { 3 }$
Hence, volume of the hole $\left( \mathrm { V } _ { 2 } \right) = \pi \mathrm { r } _ { 2 } ^ { 2 } \mathrm { h } _ { 3 }= \frac { 5 } { 3 } \pi \mathrm { r } _ { 2 } ^ { 2 } \mathrm { h } _ { 1 }$
But it is given that $V _ { 2 } = \frac { V _ { 1 } - V _ { 2 } } { 3 }$
Therefore, $ V _ { 1 } = 4 V _ { 2 }$
$\Rightarrow 4 \times \frac { 5 } { 3 } \pi r _ { 2 } ^ { 2 } h _ { 1 } = \pi r _ { 1 } ^ { 2 } \times \frac { 11 } { 6 } h _ { 1 } \Rightarrow r _ { 2 } = \sqrt { \frac { 55 } { 8 } } \mathrm { cm }$
The radius of a cone is $\sqrt { 2 }$ times the height of the cone. A cube of maximum possible volume is cut from the same cone. What is the ratio of the volume of the cone to the volume of the cube?
[1] 2.25$\pi$
[2] 3.18$\pi$
[3] 2.35
[4] can’t be determined
Let the each side of cube be a, then
$\mathrm { CD } = \sqrt { 2 \mathrm { a } }$
Therefore $\mathrm { CQ } = \frac { \mathrm { a } } { \sqrt { 2 } }$
Let the radius of cone be r and height be h, then
$\mathrm { r } = \mathrm { h } \sqrt { 2 }$
Therefore in $\Delta \mathrm { APO }$ and $\Delta \mathrm { CQO }$ (Similar triangles)
$\frac { \mathrm { AP } } { \mathrm { PO } } = \frac { \mathrm { CQ } } { \mathrm { OQ } } = \frac { \mathrm { r } } { \mathrm { h } } = \frac { \frac { \mathrm { a } } { \sqrt { 2 } } } { ( \mathrm { h } - \mathrm { a } ) }$
$\Rightarrow \frac { \frac { \mathrm { a } } { \sqrt { 2 } } } { ( \mathrm { h } - \mathrm { a } ) } = \sqrt { 2 } \Rightarrow \mathrm { a } = 2 ( \mathrm { h } - \mathrm { a } ) \Rightarrow \mathrm { h } = \frac { 3 \mathrm { a } } { 2 } \quad \mathrm { r } = \frac { 3 \mathrm { a } } { 2 } \times \sqrt { 2 }$ and $h = \frac { 3 a } { 2 }$
Therefore Volume of cone $= \frac { 1 } { 3 } \pi \times \left( \frac { 3 \mathrm { a } \sqrt { 2 } } { 2 } \right) ^ { 2 } \times \frac { 3 \mathrm { a } } { 2 } = \frac { 9 } { 4 } \mathrm { a } ^ { 3 } \pi$
and Volume of cube $= \mathrm { a } ^ { 3 }$
Or, required ratio $= \frac { \frac { 9 } { 4 } \pi \mathrm { a } ^ { 3 } } { \mathrm { a } ^ { 3 } } = \frac { 9 } { 4 } \pi = 2.25 \pi $