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A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take g= 10 m/s²)
- a)60°
- b)45°
- c)30°
- d)zero
Correct answer is option 'B'. Can you explain this answer?
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The plumb bob is always directed towards the center of the circular track due to the centripetal force acting on it. The force of gravity acts downward on the plumb bob, and the tension in the rod acts upward to balance the forces.
Let's consider the forces acting on the plumb bob in the vertical and horizontal directions. In the vertical direction, the tension in the rod balances the weight of the plumb bob:
$T \cos \theta = mg$
where T is the tension in the rod, $\theta$ is the angle made by the rod with the vertical, and m is the mass of the plumb bob.
In the horizontal direction, the centripetal force acting on the plumb bob is provided by the horizontal component of the tension in the rod:
$T \sin \theta = \frac{mv^2}{r}$
where v is the speed of the car and r is the radius of the circular track.
We can solve these two equations for $\theta$:
$\tan \theta = \frac{T \sin \theta}{T \cos \theta} = \frac{v^2}{rg}$
Substituting the given values, we get:
$\tan \theta = \frac{(10 \text{ m/s})^2}{(10 \text{ m}) (10 \text{ m/s}^2)} = 1$
Therefore, $\theta = \tan^{-1} 1 = \boxed{45^\circ}$.
The plumb bob is always directed towards the center of the circular track due to the centripetal force acting on it. The force of gravity acts downward on the plumb bob, and the tension in the rod acts upward to balance the forces.
Let's consider the forces acting on the plumb bob in the vertical and horizontal directions. In the vertical direction, the tension in the rod balances the weight of the plumb bob:
$T \cos \theta = mg$
where T is the tension in the rod, $\theta$ is the angle made by the rod with the vertical, and m is the mass of the plumb bob.
In the horizontal direction, the centripetal force acting on the plumb bob is provided by the horizontal component of the tension in the rod:
$T \sin \theta = \frac{mv^2}{r}$
where v is the speed of the car and r is the radius of the circular track.
We can solve these two equations for $\theta$:
$\tan \theta = \frac{T \sin \theta}{T \cos \theta} = \frac{v^2}{rg}$
Substituting the given values, we get:
$\tan \theta = \frac{(10 \text{ m/s})^2}{(10 \text{ m}) (10 \text{ m/s}^2)} = 1$
Therefore, $\theta = \tan^{-1} 1 = \boxed{45^\circ}$.
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A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take g= 10 m/s²)a)60°b)45°c)30°d)zeroCorrect answer is option 'B'. Can you explain this answer?
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A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take g= 10 m/s²)a)60°b)45°c)30°d)zeroCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take g= 10 m/s²)a)60°b)45°c)30°d)zeroCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take g= 10 m/s²)a)60°b)45°c)30°d)zeroCorrect answer is option 'B'. Can you explain this answer?.
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