If the KE of the particle is increased 16 times the percentage change in deBroglie wavelength of the particle is?

Question

dls · Answer

$$\LARGE \lambda=\frac{h}{mv}=\frac{h}{\sqrt{2m KE}}=\frac{h}{\sqrt{2m (KE+16KE)}}$$

anonymous · Answer

!? why did you add? :O

dls · Answer

increased 16 times o.o

dls · Answer

first it was x..
after increasing 16 times..
16x+x=17x :"D

anonymous · Answer

its said increased 16 times..  do you know the meaning of "times" ?!

anonymous · Answer

5 times 2 is not 5+2....  :P

dls · Answer

so ..
$$\Huge \lambda=\frac{h}{4 \sqrt{2 KE m}}$$

dls · Answer

so it becomes 1/4 times..
% decrease: 25%
so new one is 75% i got this answer earlier too but its wrong..answer isn't that..
@Mashy

dls · Answer

@Mashy

anonymous · Answer

it decreases 75%.. 

the new one is 1/4times.. means the new is one 25 %

dls · Answer

yes i got the same 25% wahi but ans is 60%!!

dls · Answer

(verified ans)

dls · Answer

that is what confused me :(

anonymous · Answer

thats preposterous

dls · Answer

A resistance of 20 ohms is connected to a source of ac rated 110V,50Hz.
Find the time taken by the current to change from its max to rms value?? @Mashy

anonymous · Answer

just use the the current equation..

dls · Answer

how??

anonymous · Answer

whats the instantaneous current equation!?

dls · Answer

$$\LARGE i=i_0 \sin (\omega t+\phi)$$

anonymous · Answer

its pure resistive.. you can calculate i_0 and omega.. right?!