∣P∣=2F,∣Q∣=3F
∣R∣=P2+Q2+2PQcosθ
=4F2+9F2+12F2cosθ=F13+12cosθ
If Q is doubled
2∣R∣=4F2+36F2+24F2cosθ=F40+24cosθ
4(13+12cosθ)=40+24cosθ
12=−24cosθ
cosθ=−21
θ=120∘
Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle θ is:
Held on 10 Jan 2019 · Verified 1 Jul 2026.
120∘
60∘
30∘
90∘
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
A particle of mass 2 kg is projected vertically upward with a speed of 30 m/s. The maximum height reached by the particle is (g = 10 m/s²):
Two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1:x$. The value of $x$ is:
In an experiment the values of two spring constants were measured as $k_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m}$ and $k_{2}=(20 \pm 0.3) \mathrm{N} / \mathrm{m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _____. ($\pi = 22/7$)
Work through every JEE Main Mechanics PYQ, year by year.