Tag: surface area and volume of sphere

Questions Related to surface area and volume of sphere

From a solid sphere of radius $R$, a concentric solid sphere of radius $\dfrac{R}{2}$ is removed. The total surface area increases by

  1. $0\%$

  2. $25\%$

  3. $50\%$

  4. $75\%$


Correct Option: B
Explanation:
Solution:- (B) $25 \%$
Initial area of sphere $ \left( {A} _{1} \right) = 4 \pi {R}^{2}$
New area of sphere $\left( {A} _{2} \right) = 4 \pi {R}^{2} + 4 \pi {\left( \cfrac{R}{2} \right)}^{2} = 5 \pi {R}^{2}$
$\therefore$ Increase in area $= \cfrac{{A} _{2} - {A} _{1}}{{A} _{1}} \times 100$
$\Rightarrow$ Increase in area $= \cfrac{5 \pi {R}^{2} - 4 \pi {R}^{2}}{4 \pi {R}^{2}} \times 100 = 25 \%$
Hence the area will be increased by $25 \%$.

The volume of triangular prism whose adjacent sides are $\bar a,\ \bar b,\ \bar c$ each of magnitude $4$ units and each is inclined at an angel $\dfrac {\pi}{3}$ with other two is  

  1. $16\sqrt {2}$

  2. $16\sqrt {3}$

  3. $8\sqrt {3}$

  4. $8\sqrt {2}$


Correct Option: B

If angle of minimum deviation through an equilateral prism is $40^o$, angle of incidence (being equal to angle of emergence) would be

  1. $50^o$

  2. $60^o$

  3. $40^o$

  4. none of these


Correct Option: A
Explanation:
For minimum deviation $ i=ei=e $
$a+δm=i+ea+δm=i+e $
$a+δm=2ia+δm=2i $
$60^o+40^o=2i$
$i=50^o$

If a regular square pyramid has a base of side $8 cm$ and height of $30 cm$, then its volume is

  1. $120 cm^3.$

  2. $240 cm^3.$

  3. $640 cm^3.$

  4. $900 cm^3.$


Correct Option: C
Explanation:
Given that:
Side $a=8\ cm$, Height $h=30\ cm$
As we know that
Volume of regular square pyramid 
$\Rightarrow a^2\dfrac{h}{3}$
$\Rightarrow 8^2\dfrac{30}{3}$
$\Rightarrow 64\times 10$
$\Rightarrow 640\ cm^3$
This is the required solution.

A square pyramid can contain $16\ m^3$ of water. The height of the pyramid is $3\ m$. Calculate the length of base of the square pyramid.

  1. $16\ m$

  2. $12\ m$

  3. $4\ m$

  4. $2\ m$


Correct Option: C
Explanation:

Volume of square pyramid  $=16{ m }^{ 3 }$

$\Rightarrow \quad \dfrac { 1 }{ 3 } \times { a }^{ 2 }\times h=16{ m }^{ 3 }\Rightarrow \dfrac { 1 }{ 3 } \times { a }^{ 2 }\times 3=16\Rightarrow a=\sqrt { 16 } =4cm$
$\therefore $  Length of base $= 4m$

$VPQRS$ is rectangle based pyramid where $PQ = 30\ cm, QR = 20\ cm$ and volume is $2000\ {cm}^3$, then height (in cm) is

  1. $20$

  2. $40$

  3. $10$

  4. $30$


Correct Option: C
Explanation:

Given : Length of base$(l)=30\ cm$, width of base$(h)=20\ cm$, Volume of pyramid$=2000\ cm^3$

Let $h$ be the height of the pyramid
We know that, volume of pyramid $=\dfrac{l\times w\times h}{3}$
$\implies 2000\ cm^3 = \dfrac{30 cm\times 20 cm\times h}{3}$
$\implies h=\dfrac{2000\times 3}{30\times 20} cm=10 cm$
Hence, height is $10 cm$.

State whether true or false : 

If $s$ is the perimeter of the base of a prism, $n$ is the number of sides of the base, $S$ is the total length of the edges and $h$ is the height, then $S=nh+2s$.

  1. True

  2. False


Correct Option: A
Explanation:

True. Total length of Prism $= S$
Length of all the heights $= nh$
Perimeter of base and head $= 2s$
$\therefore S = nh + 2s$

The base of right prism is a triangle whose perimeter is 28 cm and the inradius of the triangle is 4 cm. If the volume of the prism is 366 cc, then its height is 

  1. 6.54 cm

  2. 8 cm

  3. 4 cm

  4. None of these


Correct Option: A
Explanation:

Perimeter of triangle is $2{s}=28\ cm\implies s=14$

Inradius of triangle $r=4\implies \Delta=r.s=56$
Volume of prism $=366$ cc
$\implies $ Area of triangle $\times $ height $=366$
Height $=\dfrac{366}{56}=6.54$

For a prism, $A = 60^o$, $\mu = \sqrt{\dfrac{7}{3}}$, then the minimum possible angle of incidence, so that the light ray is refracted from the second surface is

  1. $30^o$

  2. $60^o$

  3. $90^o$

  4. $40^o$


Correct Option: A

The slant height of a right pyramid having square base of side $10cm$ and vertical height $15cm$ is

  1. $5\sqrt{10}cm$

  2. $6\sqrt{10}cm$

  3. $7\sqrt{10}cm$

  4. $8\sqrt{10}cm$


Correct Option: A