JEE Main Physics Class 11 Motion in a Straight Line and Plane Part 2 Questions

Given, force vectors P⊥Q, i.e. θ=90°
Let resultant of P + Q = x
and resultant of P − Q = y
∴x2=P2+Q2+2‌PQ‌cos‌90°=P2+Q2 ...(i)
y2=P2+Q2−2‌PQ‌cos‌90°=P2+Q2 ...(ii)
When θ1 is the angle between (P + Q) and (P − Q),
then, their resultant is given by
|x+y|=√3(P2+Q2)
∴√x2+y2+2xy‌cos‌θ1=√3P2+3Q2 ...(iii)
Substituting the values of x2 and y2 in Eq. (iii), we get 2P2+2Q2+(2P2+2Q2)‌cos‌θ1=3P2+3Q2
⇒cos‌θ1=

P2+Q2

2(P2+Q2)

=

1

2

∴θ1=60°
When θ2 is the angle between (P + Q) and (P − Q), then the magnitude of their resultant is given by
|x+y|=√2(P2+Q2)
x2+y2+2xy‌cos‌θ2=2P2+2Q2 ...(iv)
Substituting the values, we get
P2+Q2+P2+Q2+2(P2+Q2)‌cos‌θ2 =2P2+2Q2
cos θ2=0
⇒θ2=90°
So, θ1=60° and θ2=90° and θ1<θ2
Hence, both statement I and statement II are true.