Pseudo Force In Non-inertial Frames

Viewed from the accelerating carriage, a non-inertial frame, the passenger must invoke a pseudo force of magnitude $ma$ acting on the bob opposite to the carriage's acceleration. In this frame the bob is in equilibrium under three forces: tension along the thread, gravity $mg$ downward, and the pseudo force $ma$ horizontal. Resolving along and across the thread, the horizontal and vertical balances give $T\sin \theta =ma$ and $T\cos \theta =mg$, so dividing yields $\tan \theta =\frac{a}{g}=\frac{5}{10}=0.50$. The 0.25 value would need an acceleration of only 2.5 m/s^2. The 2.00 value inverts the ratio, using $g/a$ instead of $a/g$. The 0.75 value does not satisfy the force balance for the given numbers. An external observer standing on the ground would describe the same situation without any pseudo force, attributing the bob's tilt instead to the horizontal component of tension providing the real force $ma$ that accelerates the bob along with the carriage. Both descriptions give the identical angle, which illustrates that pseudo forces are a bookkeeping device for working comfortably inside accelerating frames. This matches the NCERT and JEE treatment of pseudo forces. As a check, a larger carriage acceleration would tilt the thread further from vertical, consistent with $\tan \theta$ rising with $a$.