Question

$\left( A \right)$A ball takes $t$seconds to fall from height ${h_1}$and $2t$seconds to fall from height${h_2}$. Then $\dfrac{{{h_1}}}{{{h_2}}}$is equal to?
$\left( a \right)$ $0.5$
$\left( b \right)$ $0.25$
$\left( c \right)$2
$\left( d \right)$4
$\left( B \right)$The angle of contact for mercury glass combination is:
$\left( a \right)$ ${8^ \circ }$
$\left( b \right)$ ${48^ \circ }$
$\left( c \right)$ ${78^ \circ }$
$\left( d \right)$ ${160^ \circ }$

Answer 1

Equations of motion relate the displacement of an object with its speed, acceleration, and time. The motion of an object will follow many alternative methods. Here we'll target motion in an exceedingly line (one dimension). We will thus simply use positive and negative magnitudes of the displacement, speed, and acceleration, wherever negative values are within the other way to positive quantities.

Formula used: From the motion equation,
$\Rightarrow S = ut + \dfrac{1}{2}g{t^2}$
Where, $S$ is the distance or the height, $g$ is the acceleration due to gravity, and $t$ is the time.

Complete step by step solution:
$\left( a \right)$So for solving this problem we will use the equation of motion which we had mentioned in the formula.
$\Rightarrow S = ut + \dfrac{1}{2}g{t^2}$
Since initial velocity in both cases is zero because of the free fall.
Here, we will take the value of $g = 10m/{s^2}$
So now we will calculate the height,
$\Rightarrow {h_1} = \dfrac{1}{2} \times 10 \times {t^2}$
$\Rightarrow {h_1} = 5{t^2}$
Now we will calculate for ${h_2}$
$\Rightarrow {h_2} = \dfrac{1}{2} \times 10 \times {\left( {2t} \right)^2}$
$\Rightarrow {h_2} = 5 \times 4{t^2}$
$\Rightarrow {h_2} = 20{t^2}$
Now we will calculate the ratio $\dfrac{{{h_1}}}{{{h_2}}}$
$\Rightarrow \dfrac{{{h_1}}}{{{h_2}}} = \dfrac{{5{t^2}}}{{10{t^2}}}$
$\Rightarrow \dfrac{{{h_1}}}{{{h_2}}} = \dfrac{1}{4}$
$\Rightarrow \dfrac{{{h_1}}}{{{h_2}}} = 0.25$
Hence the ratio will be off $0.25$.

$\left( b \right)$. Suppose there is a glass shape tube and mercury is placed on it. Then due to the cohesive force of mercury which is greater as compared to resistive force between glass and mercury.
When a mercury drop meets the glass surface then a tangent will be made and the angle occurred between the glass and mercury. Since it’s a fact-based question and it is around ${138^ \circ }$, but we have to choose from the options and hence it will be ${160^ \circ }$. Therefore it is making an angle of ${160^ \circ }$.

Notes The angle between the tangent at the liquid surface at the purpose of contact and therefore the solid surface within the liquid is termed the angle of contact for a given try of solid and liquid. It’s portrayed by the value$\theta$.