Explain briefly the different modes of heat transfer ?
Answers
MODES OF HEAT TRANSFER : Tʜᴇʀᴇ ᴀʀᴇ ᴛʜʀᴇᴇ ᴍᴏᴅᴇs ᴏғ ʜᴇᴀᴛ ᴛʀᴀɴsғᴇʀ ɴᴀᴍᴇʟʏ ᴄᴏɴᴅᴜᴄᴛɪᴏɴ, ᴄᴏɴᴠᴇᴄᴛɪᴏɴ ᴀɴᴅ ʀᴀᴅɪᴀᴛɪᴏɴ. Cᴏɴᴅᴜᴄᴛɪᴏɴ : Cᴏɴᴅᴜᴄᴛɪᴏɴ ʀᴇғᴇʀs ᴛᴏ ᴛʜᴇ ʜᴇᴀᴛ ᴛʀᴀɴsғᴇʀ ᴛʜᴀᴛ ᴏᴄᴄᴜʀs ᴀᴄʀᴏss ᴛʜᴇ ᴍᴇᴅɪᴜᴍ. Mᴇᴅɪᴜᴍ ᴄᴀɴ ʙᴇ sᴏʟɪᴅ ᴏʀ ᴀ ғʟᴜɪᴅ
Hᴏᴘᴇ ɪᴛ ʜᴇʟᴘs....
Mᴀʀᴋ ᴍᴇ ᴀs ʙʀᴀɪɴʟɪᴇsᴛ ❤❤
Answer:
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Given :-
The quadratic equation (a-b)x²+(b-c)x+(c-a)=0 are equal real roots.
To prove :-
2a=b+c
Theory :-
For a Quadratic equation of the form
ax²+bx+c= 0 , the expression b²-4ac is called the discriminant.
Nature of roots:-
The roots of a quadratic equation can be of three types.
If D>0, the equation has two distinct real roots.
If D=0, the equation has two equal real roots.
If D<0, the equation has no real roots.
Solution :
We have , (a-b)x²+(b-c)x+(c-a)=0
On comparing with the standard form of Quadratic equation ax²+bx+c= 0.
Here ,
a= (a-b)
b= (b-c)
and c = (c-a)
When equation have equal ro ots then,
Discriminant = 0
\sf{\implies ({b-c}) ^{2} - 4(a-b)(c-a)=0 }⟹(b−c)
2
−4(a−b)(c−a)=0
\sf{\implies {b}^{2} + {c}^{2} - 2cb - 4ac + 4 {a}^{2} + 4bc - 4ab = 0 }⟹b
2
+c
2
−2cb−4ac+4a
2
+4bc−4ab=0
\sf{\implies {b}^{2} + {c}^{2} + 4 {a}^{2} + 4bc - 4ac - 4 ab = 0 }⟹b
2
+c
2
+4a
2
+4bc−4ac−4ab=0
\sf{\implies {b}^{2} + {c}^{2} +( {-2a})^{2}+2bc + 2c(-2a) + 2(-2a)b = 0}⟹b
2
+c
2
+(−2a)
2
+2bc+2c(−2a)+2(−2a)b=0
\sf{\implies {(b+c-2a)}^{2} = 0 }⟹(b+c−2a)
2
=0
\sf{\implies b + c -2a = 0 }⟹b+c−2a=0
Hence proved ✔
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