F1 and F2 at θ1
Fnet1=P2+Q2+P2+Q2+2(P2+Q2)cosθ1
Fnet2=P2+Q2+P2+Q2+2(P2+Q2)cosθ2
If Fnet1=3(P2+Q2)=2(P2+Q2)+2(P2+Q2)cosθ1
⇒cosθ1=2(P2+Q2)P2+Q2
⇒θ1=60∘
Fnet2=2(P2+Q2)=2(P2+Q2)+2(P2+Q2)cosθ2
⇒cosθ2=0⇒θ2=90∘
Statement-I : Two forces ( P+ Q) and ( P- Q) where Pperp Q, when act at an angle θ _1 each other, the magnitude of their resultant is √3(P^2+Q^2),…
JEE Main 2021 — Physics Mechanics
Statement-I : Two forces (P+Q) and (P−Q) where P⊥Q, when act at an angle θ1 each other, the magnitude of their resultant is 3(P2+Q2), when they act at an angle θ2, the magnitude of their resultant becomes 2(P2+Q2). This is possible only when θ1<θ2.
Statement-II : In the situation given above.
θ1=60∘ and θ2=90∘
In the light of the above statement, choose the most appropriate answer from the options given below :
Held on 31 Aug 2021 · Verified 1 Jul 2026.
Options
- A
Statement I is false but Statement II is true.
- B
Both Statement I and Statement II are true.
- C
Both Statement I and Statement II are false.
- D
Statement I is true but Statement II is false.
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Mechanics in other years
More JEE Main Mechanics questions
If $\epsilon, E$ and $t$ represent the free space permittivity, electric field and time respectively, then the unit of $\frac{\epsilon E}{t}$ will be :
When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height $h$. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by $1 \%$ each, then the height of the liquid in the tube will change by $\%$.
The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s. The magnitude of acceleration at point $(1, 2, 4)$ is _______ m/s$^2$.
Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^{2}$. The value of $x$ is $\_\_\_\_$. 
A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm. If $(15000-x) \mu \mathrm{J}$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is $\_\_\_\_$. ($\pi=22 / 7$)
Other Physics topics for JEE Main
Keep practicing Mechanics
Work through every JEE Main Mechanics PYQ, year by year.