Saved Bookmarks
| 1. |
The massses of the three wires of copper are in the ratio 1 : 3 : 5. And their lengths are in th ratio 5 : 3 : 1. the ratio of their electrical resistance isA. `25 : 1 : 125`B. `1 : 125 : 25`C. `125 : 1 : 25`D. `125 : 25 : 1` |
|
Answer» Given , Ratio of masses = `1 : 3 : 5` As density will remain same Ratio of volumes = `1 : 3 : 5 " " …. (i)` and Ratio of length `5 : 3 : 1 " " …. (ii)` For ratio of resistance `R_(1) : R_(2) : R_(3) = (rho l_(1))/(A_(1)) : (rho l_(2))/(A_(2) ) : (rho l_(3))/(A_(3))` Since , resistivity remain same ` = (l_(1))/(A_(1)) : (l_(2))/(A_(2)) : (l_(3))/(A_(3)) = (l_(1)^(2))/(A_(1) l_(1)) : (l_(2)^(2))/(A_(2) l_(2)) : (l_(3)^(2))/(A_(3) l_(3))` From Eqs . (i) and (ii) `R_(1) : R_(2) : R_(3) = ([5]^(2))/(1) : ([3]^(2))/(3) : ([1]^(2))/(5) = 25 : 3 : (1)/(5) = 125 : 15 : 1` |
|