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Page 1
Edurev123
4. Sphere and its Properties
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =??
and passing through the ci zle
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =??
(2009 : 12 Marks)
Solution:
Approach: General equation of sphere through any circle is used. The parameter san be
found by the Centre satisfying equation of plane.
General equation of a sphere passing through the circle is
?? +???? =0
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3
? 3?? +37=6
? ?? =
-31
3
? Required sphere is
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a
circle of radius 1 and find the equation of the sphere which has this circle as great
circle.
(2010: 12 Marks)
Solution:
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3
Page 2
Edurev123
4. Sphere and its Properties
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =??
and passing through the ci zle
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =??
(2009 : 12 Marks)
Solution:
Approach: General equation of sphere through any circle is used. The parameter san be
found by the Centre satisfying equation of plane.
General equation of a sphere passing through the circle is
?? +???? =0
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3
? 3?? +37=6
? ?? =
-31
3
? Required sphere is
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a
circle of radius 1 and find the equation of the sphere which has this circle as great
circle.
(2010: 12 Marks)
Solution:
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3
Centre of given circle =(
1
2
,-
1
2
,0)
Radius =
v
1
4
,
1
4
,2=
v
2+
1
2
=
v
5
2
Let ?? be the centre of this circle.
Distance of plane from centre
|
1
2
-
1
2
-3|
v1
2
+1
2
+2
2
=
3
v6
? Radius of circle with ???? as radius
v
5
2
-
9
6
=
v
2
2
=1
Equation of sphere with circle as great circle.
?? 2
+?? 2
+?? 2
-?? +?? -2+?? (?? +?? -2?? -3)=0
? ?? 2
+?? 2
+?? 2
-?? +?? -2+???? +???? -2???? -?? 3=0
? ?? 2
+?? 2
+?? 2
+(1-?? )?? +(1+?? )?? -2???? -3?? +2=0
Page 3
Edurev123
4. Sphere and its Properties
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =??
and passing through the ci zle
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =??
(2009 : 12 Marks)
Solution:
Approach: General equation of sphere through any circle is used. The parameter san be
found by the Centre satisfying equation of plane.
General equation of a sphere passing through the circle is
?? +???? =0
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3
? 3?? +37=6
? ?? =
-31
3
? Required sphere is
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a
circle of radius 1 and find the equation of the sphere which has this circle as great
circle.
(2010: 12 Marks)
Solution:
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3
Centre of given circle =(
1
2
,-
1
2
,0)
Radius =
v
1
4
,
1
4
,2=
v
2+
1
2
=
v
5
2
Let ?? be the centre of this circle.
Distance of plane from centre
|
1
2
-
1
2
-3|
v1
2
+1
2
+2
2
=
3
v6
? Radius of circle with ???? as radius
v
5
2
-
9
6
=
v
2
2
=1
Equation of sphere with circle as great circle.
?? 2
+?? 2
+?? 2
-?? +?? -2+?? (?? +?? -2?? -3)=0
? ?? 2
+?? 2
+?? 2
-?? +?? -2+???? +???? -2???? -?? 3=0
? ?? 2
+?? 2
+?? 2
+(1-?? )?? +(1+?? )?? -2???? -3?? +2=0
Radius of the sphere =1
? (
1-?? 2
)
2
+(
1+?? 2
)
2
+?? 2
+3?? +2=1
2
?
1+?? 2
+2?? 4
+
1+?? 2
-2?? 4
+?? 2
+3?? +2=1
?
3?? 2
2
+
1
2
+3?? =-1
?
3?? 2
2
+3?? +
3
2
=0
? ?? 2
+2?? +1=0
? (?? +1)
2
=0
? ?? =-1
Using this value of ?? , the equation of sphere is
?? 2
+?? 2
+?? 2
-?? +?? -2-1(?? +?? -2?? -3)=0
? ?? 2
+?? 2
+?? 2
-?? +?? -?? -?? -?? +2?? +3=0
? ?? 2
+?? 2
+?? 2
-3?? +2?? +1=0
4.3 Show that the equation of the sphere which touches the sphere
?? (?? ?? +?? ?? +?? ?? )+???? ?? -???? ?? -?? ?? =??
at the point (?? ,?? ,-?? ) and passes through the point (-?? ,?? ,?? ) is
?? ?? +?? ?? +?? ?? +?? ?? -?? ?? +?? =??
(2011 : 10 Marks)
Solution:
The equation of the given sphere is
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? =0 (??)
The equation of the tangent plane to the sphere (i) at the point (1,2,-2) is
4(?? ·1+?? ·2+?? (-2))+5·(?? +1)-
25
2
(?? +2)-(?? -2)=0
Or 18?? -9?? -18?? +14=0 (???? )
? The equation of the sphere which touches the sphere (i) at (1,2,-2) is
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? +?? (18?? -9?? -18?? +14)=0 (?????? )
If (iii) passes through (-1,0,1) then,
4(1+0+0)+10(-1)-0-0+?? (-18-0-0+14)=0
Page 4
Edurev123
4. Sphere and its Properties
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =??
and passing through the ci zle
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =??
(2009 : 12 Marks)
Solution:
Approach: General equation of sphere through any circle is used. The parameter san be
found by the Centre satisfying equation of plane.
General equation of a sphere passing through the circle is
?? +???? =0
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3
? 3?? +37=6
? ?? =
-31
3
? Required sphere is
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a
circle of radius 1 and find the equation of the sphere which has this circle as great
circle.
(2010: 12 Marks)
Solution:
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3
Centre of given circle =(
1
2
,-
1
2
,0)
Radius =
v
1
4
,
1
4
,2=
v
2+
1
2
=
v
5
2
Let ?? be the centre of this circle.
Distance of plane from centre
|
1
2
-
1
2
-3|
v1
2
+1
2
+2
2
=
3
v6
? Radius of circle with ???? as radius
v
5
2
-
9
6
=
v
2
2
=1
Equation of sphere with circle as great circle.
?? 2
+?? 2
+?? 2
-?? +?? -2+?? (?? +?? -2?? -3)=0
? ?? 2
+?? 2
+?? 2
-?? +?? -2+???? +???? -2???? -?? 3=0
? ?? 2
+?? 2
+?? 2
+(1-?? )?? +(1+?? )?? -2???? -3?? +2=0
Radius of the sphere =1
? (
1-?? 2
)
2
+(
1+?? 2
)
2
+?? 2
+3?? +2=1
2
?
1+?? 2
+2?? 4
+
1+?? 2
-2?? 4
+?? 2
+3?? +2=1
?
3?? 2
2
+
1
2
+3?? =-1
?
3?? 2
2
+3?? +
3
2
=0
? ?? 2
+2?? +1=0
? (?? +1)
2
=0
? ?? =-1
Using this value of ?? , the equation of sphere is
?? 2
+?? 2
+?? 2
-?? +?? -2-1(?? +?? -2?? -3)=0
? ?? 2
+?? 2
+?? 2
-?? +?? -?? -?? -?? +2?? +3=0
? ?? 2
+?? 2
+?? 2
-3?? +2?? +1=0
4.3 Show that the equation of the sphere which touches the sphere
?? (?? ?? +?? ?? +?? ?? )+???? ?? -???? ?? -?? ?? =??
at the point (?? ,?? ,-?? ) and passes through the point (-?? ,?? ,?? ) is
?? ?? +?? ?? +?? ?? +?? ?? -?? ?? +?? =??
(2011 : 10 Marks)
Solution:
The equation of the given sphere is
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? =0 (??)
The equation of the tangent plane to the sphere (i) at the point (1,2,-2) is
4(?? ·1+?? ·2+?? (-2))+5·(?? +1)-
25
2
(?? +2)-(?? -2)=0
Or 18?? -9?? -18?? +14=0 (???? )
? The equation of the sphere which touches the sphere (i) at (1,2,-2) is
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? +?? (18?? -9?? -18?? +14)=0 (?????? )
If (iii) passes through (-1,0,1) then,
4(1+0+0)+10(-1)-0-0+?? (-18-0-0+14)=0
4-10+?? (-4) =0
-6 =4?? ??? =-
3
2
Put ?? =-
3
2
in (iii),
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? -
3
2
(18?? -9?? -18?? +14)=0
? 8(?? 2
+?? 2
+?? 2
)+20?? -50?? -4?? -54?? +27?? +54?? -42=0
?8(?? 2
+?? 2
+?? 2
)-34?? -23?? +50?? -42=0, which is the required equation of the
sphere.
4.4 Show that three mutually perpendicular tangent lines can be drawn to the
sphere ?? ?? +?? ?? +?? ?? =?? ?? from any point on the sphere ?? (?? ?? +?? ?? +?? ?? )=?? ?? ?? .
(2013 : 15 Marks)
Solution:
Let (?? ,?? ,?? ) be any point. Equation of enveloping cone from this point to sphere
?? 2
+?? 2
+?? 2
=1
2
(??)
is ?? ?? 1
=?? 1
2
?(?? 2
+?? 2
+?? 2
-?? 2
)(?? 2
+?? 2
+?? 2
-?? 2
)=(???? +???? +???? -?? 2
)
2
This cone will have three mutually perpendicular generators if coefficient of ?? 2
+
coefficient of ?? 2
+ coefficient of ?? 2
=0.
i.e., ?? +?? +?? =0
? (?? 2
+?? 2
-?? 2
)+(?? 2
+?? 2
-?? 2
)+(?? 2
+?? 2
-?? 2
)=0
? 2(?? 2
+?? 2
+?? 2
)=0
Since this is also the condition that three tangent lines from (?? ,?? ,?? ) to sphere are
mutually perpendicular, so locus of (?? ,?? ,?? ) is
2(?? 2
+?? 2
+?? 2
)=3?? 2
4.5 Find the co-ordinates of the points on the sphere ?? ?? +?? ?? +?? ?? -?? ?? +?? ?? =?? ,
the tangent planes at which are paraliel to the plane ?? ?? -?? +?? ?? =?? .
(2014 : 10 Marks)
Solution:
Let, the equation of planes ?? 1
and ?? 2
parallel to
Page 5
Edurev123
4. Sphere and its Properties
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =??
and passing through the ci zle
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =??
(2009 : 12 Marks)
Solution:
Approach: General equation of sphere through any circle is used. The parameter san be
found by the Centre satisfying equation of plane.
General equation of a sphere passing through the circle is
?? +???? =0
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3
? 3?? +37=6
? ?? =
-31
3
? Required sphere is
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a
circle of radius 1 and find the equation of the sphere which has this circle as great
circle.
(2010: 12 Marks)
Solution:
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3
Centre of given circle =(
1
2
,-
1
2
,0)
Radius =
v
1
4
,
1
4
,2=
v
2+
1
2
=
v
5
2
Let ?? be the centre of this circle.
Distance of plane from centre
|
1
2
-
1
2
-3|
v1
2
+1
2
+2
2
=
3
v6
? Radius of circle with ???? as radius
v
5
2
-
9
6
=
v
2
2
=1
Equation of sphere with circle as great circle.
?? 2
+?? 2
+?? 2
-?? +?? -2+?? (?? +?? -2?? -3)=0
? ?? 2
+?? 2
+?? 2
-?? +?? -2+???? +???? -2???? -?? 3=0
? ?? 2
+?? 2
+?? 2
+(1-?? )?? +(1+?? )?? -2???? -3?? +2=0
Radius of the sphere =1
? (
1-?? 2
)
2
+(
1+?? 2
)
2
+?? 2
+3?? +2=1
2
?
1+?? 2
+2?? 4
+
1+?? 2
-2?? 4
+?? 2
+3?? +2=1
?
3?? 2
2
+
1
2
+3?? =-1
?
3?? 2
2
+3?? +
3
2
=0
? ?? 2
+2?? +1=0
? (?? +1)
2
=0
? ?? =-1
Using this value of ?? , the equation of sphere is
?? 2
+?? 2
+?? 2
-?? +?? -2-1(?? +?? -2?? -3)=0
? ?? 2
+?? 2
+?? 2
-?? +?? -?? -?? -?? +2?? +3=0
? ?? 2
+?? 2
+?? 2
-3?? +2?? +1=0
4.3 Show that the equation of the sphere which touches the sphere
?? (?? ?? +?? ?? +?? ?? )+???? ?? -???? ?? -?? ?? =??
at the point (?? ,?? ,-?? ) and passes through the point (-?? ,?? ,?? ) is
?? ?? +?? ?? +?? ?? +?? ?? -?? ?? +?? =??
(2011 : 10 Marks)
Solution:
The equation of the given sphere is
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? =0 (??)
The equation of the tangent plane to the sphere (i) at the point (1,2,-2) is
4(?? ·1+?? ·2+?? (-2))+5·(?? +1)-
25
2
(?? +2)-(?? -2)=0
Or 18?? -9?? -18?? +14=0 (???? )
? The equation of the sphere which touches the sphere (i) at (1,2,-2) is
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? +?? (18?? -9?? -18?? +14)=0 (?????? )
If (iii) passes through (-1,0,1) then,
4(1+0+0)+10(-1)-0-0+?? (-18-0-0+14)=0
4-10+?? (-4) =0
-6 =4?? ??? =-
3
2
Put ?? =-
3
2
in (iii),
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? -
3
2
(18?? -9?? -18?? +14)=0
? 8(?? 2
+?? 2
+?? 2
)+20?? -50?? -4?? -54?? +27?? +54?? -42=0
?8(?? 2
+?? 2
+?? 2
)-34?? -23?? +50?? -42=0, which is the required equation of the
sphere.
4.4 Show that three mutually perpendicular tangent lines can be drawn to the
sphere ?? ?? +?? ?? +?? ?? =?? ?? from any point on the sphere ?? (?? ?? +?? ?? +?? ?? )=?? ?? ?? .
(2013 : 15 Marks)
Solution:
Let (?? ,?? ,?? ) be any point. Equation of enveloping cone from this point to sphere
?? 2
+?? 2
+?? 2
=1
2
(??)
is ?? ?? 1
=?? 1
2
?(?? 2
+?? 2
+?? 2
-?? 2
)(?? 2
+?? 2
+?? 2
-?? 2
)=(???? +???? +???? -?? 2
)
2
This cone will have three mutually perpendicular generators if coefficient of ?? 2
+
coefficient of ?? 2
+ coefficient of ?? 2
=0.
i.e., ?? +?? +?? =0
? (?? 2
+?? 2
-?? 2
)+(?? 2
+?? 2
-?? 2
)+(?? 2
+?? 2
-?? 2
)=0
? 2(?? 2
+?? 2
+?? 2
)=0
Since this is also the condition that three tangent lines from (?? ,?? ,?? ) to sphere are
mutually perpendicular, so locus of (?? ,?? ,?? ) is
2(?? 2
+?? 2
+?? 2
)=3?? 2
4.5 Find the co-ordinates of the points on the sphere ?? ?? +?? ?? +?? ?? -?? ?? +?? ?? =?? ,
the tangent planes at which are paraliel to the plane ?? ?? -?? +?? ?? =?? .
(2014 : 10 Marks)
Solution:
Let, the equation of planes ?? 1
and ?? 2
parallel to
2?? -?? +2?? =1
2?? -?? +2?? +?? =0
2?? -?? +2?? +?? =0
Now, of
be tangent to sphere length perpendicular to ?? 1
and ?? 2
= radius of sphere.
|
2(2)-1(-1)+2(0)+?? v4+1+4
|=3
So,
?? =14,-4
Now, to find points of constant of tangent plane, ?? 1
,?? 2
and sphere (point ?? and ?? in
diagram) equation of line ' ?? 1
' normal to tangent plane and passing through centre
(2,-1,0) is
?? -2
2
=
?? +1
-1
=
?? -0
2
=??
For ?? and ?? ??? =±??
???? ,
?? -2
2/3
=
?? +1
-1/3
=
?? 2/3
=±3
or (?? ,?? ,?? )˜(4,-2,2),(0,0,-2)
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FAQs on Sphere and its Properties - Mathematics Optional Notes for UPSC
| 1. What are the properties of a sphere? |  |
Ans. A sphere is a three-dimensional geometric shape that is perfectly round in shape. It has properties such as a constant diameter, a constant radius from the center to any point on the surface, and all points on the surface equidistant from the center.
| 2. How is the volume of a sphere calculated? | |
Ans. The volume of a sphere can be calculated using the formula V = (4/3)πr^3, where "r" represents the radius of the sphere and π is a constant approximately equal to 3.14159.
| 3. What is the surface area of a sphere formula? | |
Ans. The surface area of a sphere can be calculated using the formula A = 4πr^2, where "r" represents the radius of the sphere and π is a constant approximately equal to 3.14159.
| 4. How are spheres used in real life applications? | |
Ans. Spheres are used in various real-life applications such as in sports equipment like soccer balls, basketballs, and baseballs. They are also used in architectural design for creating domes and in packaging for designing round containers.
| 5. Can a sphere have different shapes or dimensions? | |
Ans. No, a sphere is a specific geometric shape that is perfectly round and symmetrical. It cannot have different shapes or dimensions as its properties are defined by its constant radius and equidistant points from the center.
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Dec 22, 2024 Last updated |
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Additional Information about Sphere and its Properties for UPSC Preparation
Sphere and its Properties Free PDF Download
The Sphere and its Properties is an invaluable resource that delves deep into the core of the UPSC exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Sphere and its Properties now and kickstart your journey towards success in the UPSC exam.
Importance of Sphere and its Properties
The importance of Sphere and its Properties cannot be overstated, especially for UPSC aspirants.
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It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Sphere and its Properties Notes
Sphere and its Properties Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Sphere and its Properties.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
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Sphere and its Properties UPSC Questions
The "Sphere and its Properties UPSC Questions" guide is a valuable resource for all aspiring students preparing for the
UPSC exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Sphere and its Properties on the App
Students of UPSC can study Sphere and its Properties alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Sphere and its Properties,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Sphere and its Properties is prepared as per the latest UPSC syllabus.
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