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A plumb bob of mass 1 kg is hung from the ceiling of a train compartment. The train moves on an inclined plane with constant velocity. If the angle of incline is ${30^0}$. Find the angle made by the string with the normal to the ceiling. Also, find the tension in the string. $\left( {g = 10m{s^{ - 2}}} \right)$A) ${30^0},10N$B) ${10^0},30N$C) ${10^0},10N$D) ${30^0},30N$

Last updated date: 13th Jul 2024
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Hint: Constant velocity means rate of change of velocity with time is zero, i.e. $\frac{{dV}}{{dt}} = 0 \Rightarrow a = 0$. It means that on the bob there is only one acceleration which is acceleration due to gravity in vertical direction, no horizontal acceleration is present.

Complete step by step solution:
At constant velocity there is no acceleration on bob in horizontal direction, so the plumb bob will remain in vertical position.

From diagram we are able to see that angle $\theta$ will be equal to ${30^0}$, we can find it as-
$\angle OAB = {60^0}$
And $\theta + \angle OAB = {90^0}$
$\Rightarrow \theta + {60^0} = {90^0}$
$\Rightarrow \theta = {30^0}$
So, the angle between the plumb bob and normal of the ceiling is ${30^0}$.
In the diagram we can see that tension is the reaction force to the weight of the plumb bob in the vertical direction.
By equilibrium condition of plumb bob-
$T = mg$
$T = 1 \times 10kgm{s^{ - 2}}$
$\Rightarrow T = 10N$

Therefore, the option $\left( A \right)$ is the correct option.

Note: In above question, there is no any acceleration in vertical direction, but in vertical direction there is gravitational acceleration, so plumb bob is in vertical position while the train is moving. But if we assume another case in which horizontal acceleration is there then plumb bob will not remain in vertical position, it will definitely deviate from that position.