1sum1 plz help me out in di.s..
hey frnds.... facing a bit of confusion in the concept dat..... when the length of the rod of resisitivity Ï is doubled, what will be the change in the resistance (initially length = l and area of cross-section = A)
on doubling the length of the rod, will the area or the radius of cross-section become halved... plz explain....
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8 Answers
106volume is constant (so when length is doubled area is halved)
1hw can u be sure abt it dat area is halved.... can' t it be dat radius is halved.....
1see the quantity of the wire remains the same..
so its Volume (V) has to remain constant..
and V = length * area of cross section
so inverse relation between length and area of cross section
hope now u got it cleared.
1apply,
A1L1=A2L2 (AS VOLUME REMAINS CONSTANT)
WHERE A1 IS INITIAL AREA OF CROSS SECTION AND L1 IS INITIAL LENGTH OF ROD.
A2 IS NEW AREA OF CROSS SECTION AND L2 IS NEW LENTH OF ROD.
ACC. TO QUESTION ----
A1L1=A2.2L1
A2A1=12 ............................EQ.n 1
SO , NEW AREA OF CROSS-SECTION BECOMES HALF OF INITIAL AREA AND NEW RADIUS BECOMES 1√2 TIMES OF INITIAL RADIUS..
NOW,
RINTIAL = p L1A1
RNEW = p L2A2 (BECAUSE AS NATURE OF MATERIAL REMAINS SAME RESISTIVITY REMAINS SAME)
RNEW= p 2L1A2
RINITIAL RNEW =12 A2A1
RINITIAL RNEW= 12.12 (FROM EQ.n 1)
RINITIAL RNEW=14
RNEW=4.RINITIAL
SO , NEW RESISTANCE BECOMES 4 TIMES OF INITIAL RESISTANCE BY DOUBLING THE LENGTH OF ROD.......
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1i 've understood da concept....
par ye last post mein nihal kya saabit karna chaah raha hai