Question

In the figure shown a particle of mass m is tied to a string and the other end of string is tied to the side wall of a carriage, which is accelerating horizontally with constant acceleration a. Assuming the string remains taut and the particle is not in contact with the base of carriage. Then:
(Particle is at rest with respect to carriage)

• A. The angle $\theta$ that string makes with the horizontal is $ta{n}^{-1}\left(\frac{g}{a}\right)$
• B. The angle $\theta$ that string makes with the horizontal is $ta{n}^{-1}\left(\frac{a}{g}\right)$
• C. Tension in the string is $m\sqrt{g^2+a^2}$
• D. Tension in the string is $m\sqrt{g^2-a^2}$

Answer 1

Answer: (A), (C)

The angle $\theta$ that string makes with the horizontal is $ta{n}^{-1}\left(\frac{g}{a}\right)$
Tension in the string is $m\sqrt{g^2+a^2}$

Answer is A and C.
From the diagram,
Weight of the particle (mg) will act vertically downwards and inertial force due to acceleration will be ma acting opposite to motion of carriage.
Therefore,
$\frac{mg}{ma}=tan\theta$
$\theta =ta{n}^{-1}\frac{g}{a}$
Tension T will be the resultant of mg and ma
$T=m\sqrt{g^2+a^2}$