Tag: maths

Questions Related to maths

The surface area of a solid sphere is always greater than the surface area of a hemisphere for the same value of radius.

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. True

  2. False

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Correct Option: A
Explanation:
Let radius of solid sphere=radius of hemisphere$=r$
Then, S.A of solid sphere$=4\pi { r }^{ 2 }$
S.A of hemisphere$=2\pi { r }^{ 2 }$
$\therefore $S.A. of solid sphere$>$ S.A of hemisphere
Hence statement is true

A hemispherical bowl has inner radius $5cm$ and outer radius $6cm$. What will be the volume of solid enclosed between the two hemispheres? (Correct upto 2 decimal places)

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $904.78 \ {cm}^{3}$

  2. $523.60 \ {cm}^{3}$

  3. $381.18 \ {cm}^{3}$

  4. $190.59 \ {cm}^{3}$

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Correct Option: D
Explanation:

The volume of hemispherical shell$= \dfrac{2}{3}\pi*R^{3}-\dfrac{2}{3}\pi*r^{3}$
where R and r are the outer and inner radius of the hemisphere
On solving the equation we get Volume$= 190.59 \ {cm}^{3}$

The surface area of a solid spherical ball of diameter $10\ cm$ is equal to :

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $25\pi\ {cm}^2$

  2. $50\pi\ {cm}^2$

  3. $100\pi\ {cm}^2$

  4. $200\pi\ {cm}^2$

Show answer
Correct Option: C
Explanation:

Given: Diameter of the sphere $= 10\ cm$
Hence, Radius ($r$) of the sphere will be $5\ cm$

We know that, 
Surface area of the sphere is $4\pi r^2$
Therefore, Area will be $4\pi (5)^2 = 100\pi\ {cm}^2$

What will be the Inner and Outer radius of a spherical shell of inner surface area $452.39\ {cm}^2$ and outer surface area $804.25\ {cm}^2$ ? (Surface areas are accurate upto 2 decimal places)

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $6 \ cm, 7 \ cm$

  2. $7 \ cm, 8 \ cm$

  3. $6 \ cm, 8 \ cm$

  4. $7 \ cm, 9\ cm$

Show answer
Correct Option: C
Explanation:
Inner surface area$=4\pi { \left( inner\quad radius \right)  }^{ 2 }$
$\Rightarrow 4\pi { \left( { r } _{ 1 } \right)  }^{ 2 }=452.39cm^{2}\Rightarrow { r } _{ 1 }^{ 2 }=\cfrac { 452.39\times 7 }{ 4\times 22 } =35.99$
$\Rightarrow { r } _{ 2 }=\sqrt { 35.99 } =5.99\approx 6cm$
Outer surface area$=4\pi { \left( outer\quad radius \right)  }^{ 2 }$
$\Rightarrow 4\pi { \left( { r } _{ 2 } \right)  }^{ 2 }=804.25㎠\Rightarrow { r } _{ 2 }^{ 2 }=\cfrac { 804.25\times 7 }{ 4\times 22 } =63.97$
$\Rightarrow { r } _{ 2 }=\sqrt { 63.97 } =7.99\approx 8cm$

The value of radius for which the numerical value of total surface area of a sphere and the volume of sphere are equal, will be:(Consider the units of volume and surface area as ${cm}^3\  \text{and}\ {cm}^2$)

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $1cm$

  2. $2cm$

  3. $3cm$

  4. $4cm$

Show answer
Correct Option: C
Explanation:
Let radius of sphere be $'r'㎝$, then
TSA of sphere=volume of sphere
$\Rightarrow 4\pi { r }^{ 2 }=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }\Rightarrow r=3cm$

The increase in the total surface area of a sphere of Radius ${R}$ when it is cut to make two hemispheres of same Radius will be equal to:

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $5\ \pi{R}^2$

  2. $4\ \pi{R}^2$

  3. $3\ \pi{R}^2$

  4. $2\ \pi{R}^2$

Show answer
Correct Option: D
Explanation:
Total surface area of sphere$=4\pi { R }^{ 2 }$
TSA of hemisphere=CSA of hemisphere+CSA of circle
$=2\pi { R }^{ 2 }+\pi { R }^{ 2 }=3\pi { R }^{ 2 }$
$\therefore $TSA of two hemisphere$=2\times 3\pi { R }^{ 2 }=6\pi { R }^{ 2 }$
Therefore, increase in TSA$=6\pi { R }^{ 2 }-4\pi { R }^{ 2 }=2\pi { R }^{ 2 }$

The height of a cylinder is  $14 { cm }$  and its  $ { CSA }$  is  $264 { cm } ^ { 2 },$  then cylinder is.........${ cm }^{ { 3 } }$

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $183$

  2. $396$

  3. $896$

  4. $968$

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Correct Option: A

The points with position vectors $60\hat{i}+3\hat{j}$, $40\hat{i}-8\hat{j}$, $a\hat{i}-52\hat{j}$  are collinear if

direction cosines and direction ratios three dimensional geometry maths

  1. $a=-40$

  2. $a=40$

  3. $a=20$

  4. $None\ of\ these$

Show answer
Correct Option: A
Explanation:

suppose ${60i + 3j}$ , ${40i - 8j}$ and ${ai - 52j}$ is the three position of vector $A,B,C$


$\begin{array}{l} \overrightarrow { AB } =\left( { 40i-8j } \right) -\left( { 60i+3j } \right)  \ \overrightarrow { AB } =-20i-11j \ \overrightarrow { BC } =\left( { ai-52j } \right) -\left( { 40i-8j } \right)  \ \overrightarrow { BC } =\left( { a-40 } \right) i-44j \ \left( { a-40 } \right) i-44j=m\left( { -20i-11j } \right)  \ \left( { a-40 } \right) i-44j=-20im-11jm \ -44=-11m \ m=\frac { { -44 } }{ { -11 } }  \ m=4 \ a-40=-20m \ a-40=-20\left( 4 \right)  \ a=-80+40 \ a=-40 \end{array}$

 The points with position vectors $\vec {a}=\hat {i}-2\hat {j}+3\hat {k}, \vec {b}=2\hat {i}+3\hat {j}-4\hat {k}$ & $-7\hat {j}+10\hat {k}$ are collinear.

direction cosines and direction ratios three dimensional geometry maths

  1. True

  2. False

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Correct Option: A

The points $i + j + k, \, i + 2j, \, 2i+2j+k,\, 2i+3j+2k$ are

direction cosines and direction ratios three dimensional geometry maths

  1. collinear

  2. coplanar but not collinear

  3. non-coplanar

  4. none

Show answer
Correct Option: C
Explanation:

$\begin{matrix} A& B& C& D\i+j+k, &i+2j, &2i+2j+k,&2i+3j+2k \end{matrix}$
$\overline{AC} = (2-1)i + (2-1)j + k-k$
$=i+j$
$\overline{AB} = o + j - \overline{k} = j - \overline{k}$
$\overline{AD} = i + 2j + k$
$\begin{vmatrix} 1&1&0 \0&2 &1\end{vmatrix} = 1(1+2)-1(0+1)$
$=3-1 = 2 \neq 0$
Non coplanar.