
As an inclined plane is made slowly horizontal by reducing the value of angle $\theta $ with horizontal, the component of weight parallel to the plane of a block resting on the inclined plane:
A) Decreases.
B) remains the same.
C) increases.
D) increases if the plane is smooth.
Answer
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Hint:The hilly areas have roads having some angle of inclination if the angle of inclination is more, then it is difficult to ride on it the any vehicle and if the angle of inclination is more, then it is difficult to ride any vehicle as the parallel component to the weight on the inclined plane will not support it.
Formula used:The formula of the weight component parallel to the inclined is given by
$W = mg\sin \theta - \mu mg\cos \theta $
Where m is the mass g is the acceleration due to gravity $\mu $ is the coefficient of friction and $\theta $ is the angle of inclination.
Complete step by step answer:
It is given in the problem that an inclined plane is made slowly horizontal by reducing the value of angle $\theta $ with horizontal and we need to find that the component of weight parallel to the plane to the plane of a block resting on the inclined plane will increase or decrease.
The free body diagram for this condition will be,
From the above free body diagram we get,
Weight component which is parallel to the plane is given by,
$W = mg\sin \theta - \mu mg\cos \theta $
Here if the value of $\theta $ decreases then the value of $\sin \theta $ decreases and the value of $\cos \theta $ increases.
Since the first term in component decreases and the second component increases also there is negative sign in between the two signs therefore the value of the component which is parallel to the inclined plane will decrease with decrease in angle of inclination $\theta $.
Note:The force which acts on the body which is parallel to the inclined plane depends on the angle of inclination if the angle $\theta $ is increased then the value of the component parallel to the inclined plane also increases and if the angle of inclination decreases then the value of component of parallel to the inclined parallel decreases.
Formula used:The formula of the weight component parallel to the inclined is given by
$W = mg\sin \theta - \mu mg\cos \theta $
Where m is the mass g is the acceleration due to gravity $\mu $ is the coefficient of friction and $\theta $ is the angle of inclination.
Complete step by step answer:
It is given in the problem that an inclined plane is made slowly horizontal by reducing the value of angle $\theta $ with horizontal and we need to find that the component of weight parallel to the plane to the plane of a block resting on the inclined plane will increase or decrease.
The free body diagram for this condition will be,
From the above free body diagram we get,
Weight component which is parallel to the plane is given by,
$W = mg\sin \theta - \mu mg\cos \theta $
Here if the value of $\theta $ decreases then the value of $\sin \theta $ decreases and the value of $\cos \theta $ increases.
Since the first term in component decreases and the second component increases also there is negative sign in between the two signs therefore the value of the component which is parallel to the inclined plane will decrease with decrease in angle of inclination $\theta $.
Note:The force which acts on the body which is parallel to the inclined plane depends on the angle of inclination if the angle $\theta $ is increased then the value of the component parallel to the inclined plane also increases and if the angle of inclination decreases then the value of component of parallel to the inclined parallel decreases.
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