An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V, respectively. If a resistnace is now joined in parallel with the voltmeter

Options

A. `(4^(1//2))g`
B. `(4^(1//3))g`
C. `(4^(1//2) - 1)g`
D. `(4^(1//3) - 1)g`

Answer

D

Explanation

`P' = 2P_(0) = 2rhoAv_(0)^(3)`
Let the new velocity of air pushed down be v
`rhoAv_(0)^(3) = 2rhoAv_(0)^(3)`
`v = 2^(1//3)v_(0)`
New thrust `F = rhoAv^(2)`
`= rhoA2^(2//3)v_(0)^(2)`
`= (mg)4^(1//3) " " [mg = rhoAv_(0)^(2)]`
`:.` Acceleration `= a = (F_("net"))/(m) = (F_("new") - F_(0))/(m)`
`= (mg4^(1//3) - mg)/(m) = (4^(1//3) - 1)`