Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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
Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

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)

  1. $904.78 \ {cm}^{3}$
  2. $523.60 \ {cm}^{3}$
  3. $381.18 \ {cm}^{3}$
  4. $190.59 \ {cm}^{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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$= 190.59 \ {cm}^{3}$

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

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

  1. $25\pi\ {cm}^2$
  2. $50\pi\ {cm}^2$
  3. $100\pi\ {cm}^2$
  4. $200\pi\ {cm}^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $4\pi (5)^2 = 100\pi\ {cm}^2$

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

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)

  1. $6 \ cm, 7 \ cm$
  2. $7 \ cm, 8 \ cm$
  3. $6 \ cm, 8 \ cm$
  4. $7 \ cm, 9\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$=4\pi { \left( outer\quad radius \right)  }^{ 2 }$
$\Rightarrow 4\pi { \left( { r } _{ 2 } \right)  }^{ 2 }=804.25㎠\Rightarrow { r } _{ 2 }^{ 2 }=\cfrac { 804.25\times 7 }{ 4\times 22 } =63.97$
$\Rightarrow { r } _{ 2 }=\sqrt { 63.97 } =7.99\approx 8cm$
Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

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 ${cm}^3\  \text{and}\ {cm}^2$)

  1. $1cm$
  2. $2cm$
  3. $3cm$
  4. $4cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let radius of sphere be $'r'㎝$, then
TSA of sphere=volume of sphere
$\Rightarrow 4\pi { r }^{ 2 }=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }\Rightarrow r=3cm$
Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

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:

  1. $5\ \pi{R}^2$
  2. $4\ \pi{R}^2$
  3. $3\ \pi{R}^2$
  4. $2\ \pi{R}^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
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$=6\pi { R }^{ 2 }-4\pi { R }^{ 2 }=2\pi { R }^{ 2 }$
Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. $a=-40$
  2. $a=40$
  3. $a=20$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

Multiple choice direction cosines and direction ratios three dimensional geometry maths

 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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Three points with position vectors a, b, and c are collinear if (b-a) is a scalar multiple of (c-a). Given a = i - 2j + 3k, b = 2i + 3j - 4k, and c = 0i - 7j + 10k: b-a = i + 5j - 7k; c-a = -i - 5j + 7k. Since c-a = -1(b-a), the vectors are collinear.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. collinear

  2. coplanar but not collinear

  3. non-coplanar

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
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.