Physics Q23 – NEET-UG 2024

Home Physics Rotational MotionQuestion #441

A bob is whirled in a horizontal plane by means of a string with an initial speed of

ω

rpm. The tension in the string is

T

. If speed becomes

2 ω

while keeping the same radius, the tension in the string becomes :

Clue 1 of 3: Uniform Circular Motion

The tension in the string provides the necessary centripetal force for the bob to move in a horizontal circle.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify the Governing Principle✦ Active

For a bob whirled in a horizontal plane, the tension in the string provides the centripetal force required for circular motion. The formula for centripetal force is $F_{c} = m r ω^{2}$ , where $m$ is the mass of the bob, $r$ is the radius of the circular path, and $ω$ is the angular speed.

Step 2: Formulate Initial Tension○ Expand

Given that the initial angular speed is $ω$ and the tension is $T$ . Therefore, the initial tension can be written as:

T = m r ω^{2} (Equation 1)

💡 Teacher's Secret Hint

Remember that 'rpm' (revolutions per minute) is a unit of angular speed, and the formula uses angular speed directly.

Step 3: Calculate New Tension○ Expand

If the speed becomes $2 ω$ while keeping the same radius $r$ , the new tension $T^{'}$ will be:

T^{'} = m r (2 ω)^{2}

Simplifying this expression:

T^{'} = m r (4 ω^{2})

T^{'} = 4 (m r ω^{2})

From Equation 1, we know that $T = m r ω^{2}$ . Substituting this into the expression for $T^{'}$ :

T^{'} = 4 T

Thus, the new tension in the string becomes $4 T$ .

💡 Teacher's Secret Hint

Pay close attention to how the square of the angular speed affects the tension. Doubling the speed quadruples the tension.

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Physics Question 23 – NEET-UG 2024

Clue 1 of 3: Uniform Circular Motion

🧠 Full Solution Path