4 -160-Masses -160-of -160-three -160-wires -160-of -160-same -160-material -160-is -160-on -160-ratio -160-1-2-3 -160-and -160-their -160-lengthsis -160-3-2-1the -160-resistance -160-will -160-be -160-in -160-ratio

Q4 #160;Density #160;=massvolumeresistance= #961;lAso #160;density #215;area #215;length=massA=massdensity #215;lAs #160;material #160;is #1 ...

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Byju's Answer

Standard XII

Physics

Carrier Concentration

4 #160;Mass...

Question

$4 M a s s e s o f t h r e e w i r e s o f s a m e m a t e r i a l i s o n r a t i o 1 : 2 : 3 a n d t h e i r l e n g t h s i s 3 : 2 : 1 t h e r e s i s \tan c e w i l l b e i n r a t i o$

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Solution

$Q 4 D e n s i t y = \frac{m a s s}{v o l u m e} r e s i s \tan c e = \frac{ρ l}{A} s o d e n s i t y \times a r e a \times l e n g t h = m a s s A = \frac{m a s s}{d e n s i t y \times l} A s m a t e r i a l i s s a m e t h a n d e n s i t y i s s a m e s o r e s i s \tan c e r a t i o i s \frac{l^{2}}{M} s o r a t i o b e c o m e s 27 : 6 : 3 R e g a r d s$

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4 Masses of three wires of same material is on ratio 1:2:3 and their lengthsis 3:2:1the resistance will be in ratio

Q4 Density =massvolumeresistance=ρlAso density×area×length=massA=massdensity×lAs material is same than density is same so resistance ratiois l2M so ratio becomes 27:6:3Regards

$4 M a s s e s o f t h r e e w i r e s o f s a m e m a t e r i a l i s o n r a t i o 1 : 2 : 3 a n d t h e i r l e n g t h s i s 3 : 2 : 1 t h e r e s i s \tan c e w i l l b e i n r a t i o$