Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

If $T_1, T_2, T_3, T_4$ and $T_5$ represent the tension in the string of a simple pendulum when the bob is at the left extreme, right extreme, mean, any intermediate left and any intermediate right positions, respectively. Then, which of the following relations are correct?

(A) $T_1 = T_2$ (B) $T_3 > T_2$ (C) $T_4 > T_3$ (D) $T_3 = T_4$ (E) $T_5 > T_2$

Choose the most appropriate answer from the options given below:

(A), (B) and (C) only

(B), (C) and (D) only

(A), (B) and (E) only

(C), (D) and (E) only

Step-by-Step Solution

Tension Formula: The tension $T$ in the string of a simple pendulum of length $L$ and mass $m$ at an angular position $\theta$ is given by $T = mg\cos\theta + \frac{mv^2}{L}$ , where $v$ is the velocity of the bob.
Analysis of Positions:

Mean Position ( $T_3$ ): At the mean position ( $eta = 0^{\circ}$ ), velocity is maximum ( $v_{max}$ ) and $\cos 0^{\circ} = 1$ . Thus, tension is maximum: $T_3 = T_{max} = mg + \frac{mv_{max}^2}{L}$ .
Extreme Positions ( $T_1, T_2$ ): At the extreme positions ( $eta = eta_{max}$ ), velocity is zero ( $v=0$ ). Thus, tension is minimum: $T_1 = T_2 = mg\cos\theta_{max}$ . Since the motion is symmetric, $T_1 = T_2$ . (Statement A is Correct).
Intermediate Positions ( $T_4, T_5$ ): At intermediate positions ( $0 < eta < eta_{max}$ ), the velocity $v$ is non-zero (but less than $v_{max}$ ) and $\cos\theta < 1$ . The tension lies between the maximum and minimum values ( $T_{ext} < T_{int} < T_{mean}$ ).

Evaluating Inequalities:

$T_3$ vs $T_2$ : Since $T_3$ is maximum and $T_2$ is minimum, $T_3 > T_2$ . (Statement B is Correct).
$T_4$ vs $T_3$ : $T_4$ is intermediate and $T_3$ is maximum. Thus, $T_4 < T_3$ . The statement $T_4 > T_3$ is Incorrect (Statement C is False).
$T_5$ vs $T_2$ : $T_5$ is intermediate tension and $T_2$ is minimum tension. Thus, $T_5 > T_2$ . (Statement E is Correct).

Conclusion: The correct relations are (A), (B), and (E).

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