IIT JEE , AIEEE Forum - Physics , Optics

zamuka

please bataao ki aise question kaise solve karte hain or is ka sol.n bhi do.....

a mirror faces -ve x axis.while a particle starts moving such that its image is formed in the miirrror. at a certain instant, velo. of particle is 3i+4j+5k and that of mirror is i-j+k. now match....

COLUMN A

1. magnitude of relative velocity (RV) of image w.r.t. mirror

2.mag of RV of image w.r.t. object

3. mag of RV of object w.r.t. mirror

4. absolute velocity of image w.r.t. ground

COLUMN B

1.root 42

2.root 45

3.root 16(=4)

4.root 43

zamuka

please solve the question!!!!!!!!!!!

arey yaar mujhe to initiate kaise karna yahi samajh nahi aa raha hai...

sujay_iit

seriously someone pls solve...was about to ask this only...
fiitjee ke full test mein aaya tha...palle hi nahi pada

zamuka

arey doesn't any 1 know how to solve this? plz koi to kar do.......

anandghegde

in column B is ther no option like root 23?

zamuka

NOPES!!!!!

but what formula/concept r u using??

anandghegde

edited!!!

zamuka

yaa u r correct.......

par ur 3rd optio isn't correct........

arey ye waala to easy tha(in fact yahi option mujhe aaya tha).......

object ki velo. relative to mirror to jo velo. vectors diye hain, unse hi nikal jaayegi.......

ur all other ans r correct

please explain the concept

anandghegde

wait wait type out the answers ill try to give the explanation.

anandghegde

type the answers given. I'll try to explain if i can

zamuka

ans r...

1-2

2-3

2-2

4-1

anandghegde

wait.....

anandghegde

$V_{om} = V_{o}-V_{m} = 3i+4j+5k -(i-j+k) =2i+5j+4k$

Hence velocity of object wrt mirror is $\sqrt{2^2+5^2+4^2}=\sqrt{45}$

Hence 3 is matched with 2.

now look...the x component of velocity of object wrt mirror is 2i. The object is moving perpendicularly wrt the mirror with a velocity of 2m/s. Hence the x component of velocity of image wrt mirror is -2i.The other two components of velocity of object and image are same..( see if you move leftwards in front of a mirror, your image will also move leftwards and not rightwards)

Hence the velocity of image wrt mirror is $V_{im} = -2i+5j+4k$

Hence the magnitude of rv of image wrt mirror is $\sqrt{2^2+5^2+4^2}=\sqrt{45}$

Hence 1 is matched with 2

$V_{io} =V_{im}-V_{om}= (-2i+5j+4k)-(2i+5j+4k) = -4i$

Hence 2 is matched with 3.

$V_{i} =V_{im}+V_{m}= (-2i+5j+4k)+(i-j+k) = -i+4j+5k$

Hence the magnitude of the velocity of img wrt ground is

$\sqrt{1^2+5^2+4^2}=\sqrt{42}$

Hence 4 is matched with 1

zamuka

great job maan!!!!!

thanks a lot!!

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