A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to `2 : 1`. What will be the ratio of their nuclear size (nuclear radius)?
A. `2^(1/3) :1`
B. `1 : 3^(1/2)`
C. `3^(1/2) : 1`
D. `10^(9) K`
Correct Answer - A
We know, self-inductance of a solenoid,
`L=(mu_(0)N^(2)A)/(l)=(mu_(0)N^(2)(pir^(2)))/(l)`
So, `(L_(1))/(L_(2))=((N_(1))/(N_(2)))^(2)((r_(1))/(r_(2)))^(2)((l_(2))/(l_(1)))=(1)/(4)xx(2)/(1)=(1)/(2)`
Correct Answer - D
When an `alpha` particle of the mass m moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no less of...
Correct Answer - A
By conservation of momentum, `m_(1)v_(1)=m_(2)v_(2)`
`implies " "(v_(1))/(v_(2))=8/1=(m_(2))/(m_(1))" "...(i)`
Also, from `r propA^(1//3),`
`therefore" "(r_(1))/(r_(2))=((A_(1))/(A_(2)))^(1//3)=((1)/(8))^(1//3)=1/2`
Correct Answer - B
Momentum remains constant,
`K=(p^(2))/(2m)or Kprop(1)/(m)implies(K_(n))/(K_(alpha))=(m_(alpha))/(m_(n))=(4)/(206)=(2)/(103)`
`therefore` Kinetic energy of the residual nucleus
`K_(n)=((2)/(103))K_(alpha)=((2)/(103))E`
Correct Answer - `({26!}^(4))/({13!}^(4)xx52!)`
We have to find the probability that out of 26 cards drawn at random from the pack of cards, 13 will be red and 13 black. The...
`because P= =(V^(2))/(R) therefore R=(V^(2))/(P)=((200)^(2))/(100)=400Omega` Resistance of half piece `=(400)/(2)=200Omega`
Resistance of pieces connected in parallel `=(200)/(2)=100Omega`
Energy liberated/sec ond `=P=(V^(2))/(R)=(200xx200)/(100)=400W`
Power `=(V^(2))/(R)` where R denotes the resistance of the coil
`therefore R=(V^(2))/(P)=(V^(2))/(1000)Omega`
Resistance of each part `=(R)/(2)=(V^(2))/(2000)Omega`
`therefore` power of each part `=(V^(2)xx2000)/(V^(2))` =2000W
`therefore` combined wattage `=2xx200=4000` W