Question

A smooth inclined plane of length$L$. inclining with a horizontal is inside a lift which is moving down with a retardation$a$. The time taken by the body to slide down the inclined plane from rest will be:
$(A)\sqrt {\dfrac{{2L}}{{\left( {g + a} \right)\sin \theta }}}$
$(B)\sqrt {\dfrac{{2L}}{{\left( {g - a} \right)\sin \theta }}}$
$(C)\sqrt {\dfrac{{2L}}{{a\sin \theta }}}$
$(D)\sqrt {\dfrac{{2L}}{{g\sin \theta }}}$

Answer 1

We know that change in acceleration is known as velocity. Acceleration with a negative sign is called Retardation. We are able to solve the given question with the help of the physical concept called retardation.

Formula used:
The pseudo force can be defined as the product of the mass and acceleration. That is,
$\Rightarrow F = ma$
Where $m$is the mass of the block and $a$is the acceleration.
the acceleration of the block relative to the plane $\Rightarrow {a_r} = \left( {g + a} \right)\sin \theta$
the equation of motion,
$\Rightarrow S = \dfrac{1}{2}a{t^2}$

Complete step by step answer:
To solve the given question first let us try to understand the term called Retardation. Retardation is nothing but an acceleration with a negative sign. The change in velocity of a body Is known as acceleration.
Remember that when a body moves along a straight line with constant retardation, the speed will be decreasing in moving away along the straight line.
In the question they have given that a smooth inclined plane of length L. inclining with a horizontal is inside a lift which is moving down with a retardation a.
We have to find the time taken by the body to slide down the inclined plane from rest.
Now let us try to answer the given question.
Moving down with the retardation means the lift is being accelerated upwards.
There will be some force that acts on the lift. So, let us consider that force as pseudo force, this pseudo force acts with respect to lift. The pseudo force can be defined as the product of the mass and acceleration. That is,
$\Rightarrow F = ma$
Where $m$is the mass of the block and $a$is the acceleration.
Consider $mg$as the downward force and it will be replaced by $m\left( {g + a} \right)$ .
With the help of the value, we can calculate the acceleration of the block relative to the plane. That is,
$\Rightarrow {a_r} = \left( {g + a} \right)\sin \theta$
We can use the equation of motion,
$\Rightarrow S = \dfrac{1}{2}a{t^2}$
We can substitute the values that are given in the question in the above equation. we get,
$\Rightarrow L = \dfrac{1}{2}{a_r}{t^2}$
We can simplify the given equation by taking the t into the left hand side. We get,
$\Rightarrow {t^2} = \dfrac{{2L}}{{{a_r}}}$
To remove the square term on the left hand side, we will take the square root on the right-hand side. That is
$\Rightarrow t = \sqrt {\dfrac{{2L}}{{{a_r}}}}$
We have the value of relative acceleration. We can substitute in the given equation we get,
$\Rightarrow t = \sqrt {\dfrac{{2L}}{{\left( {g + a} \right)\sin \theta }}}$
Therefore, the time taken by the body to slide down the inclined plane from rest will be,
$\sqrt {\dfrac{{2L}}{{\left( {g + a} \right)\sin \theta }}}$

Hence option (A) is the correct answer.

Note: Students should not get confused between acceleration and retardation. The S.I unit of the retardation is $m/{s^2}$. The equation of motion used here is known as the second equation of motion. We use $\Rightarrow S = ut + \dfrac{1}{2}a{t^2}$ this equation mostly and it is derived from the algebraic method.