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Water drops fall at regular intervals from a roof. At an instant when a drop is about to leave the roof, the separations between 3 successive drops below the roof are in the ratio
A. $1:2:3$
B. $1:4:9$
C. $1:3:5$
D. $1:5:13$
Answer
459.9k+ views
Hint:Here, as the drops are dropped in successive separations of time, assume that the drops are falling at an interval of 1 second. Now, using the second equation of motion for linear kinematics, we find the height of the three drops. Additionally, in the equation, as the water drops are dropped, their initial velocity is considered as zero.
Complete step by step answer:
Here, the drops are falling at successive intervals. Thus, at the 0th second, none of the drop falls, while at the 1st second, the first drop falls, in the 2nd second, the second drop falls and at the 3rd second, the third drop falls. Let these times be t1,t2,t,3 respectively and the distance travelled by these drops be d1, d2, d3 respectively. Now, we know the kinematic equation for a free fall as follows:
\[h = \dfrac{1}{2}g{t^2}\]
Where h is the height of the free fall, g is the acceleration due to gravity and t is the time travelled by an object during the free fall. Thus here, for the first drop, the distance travelled would be
\[
{d_1} = \dfrac{1}{2}g{t_1}^2 = \dfrac{1}{2}(10){(3)^2} \\
\Rightarrow {d_1} = 45m \\
\]
The distance travelled by the second drop would be;
\[
{d_2} = \dfrac{1}{2}g{t_2}^2 = \dfrac{1}{2}(10){(2)^2} \\
\Rightarrow {d_2} = 20m \\
\]
The distance travelled by the third drop is:
\[
{d_3} = \dfrac{1}{2}g{t_3}^2 = \dfrac{1}{2}(10)(1) \\
\therefore {d_3} = 5m
\]
Thus, the ratio of the separations is given by \[{d_3}:{d_2} - {d_3}:{d_1} - {d_2}\] which is equal $5:15:25$. Thus the ratio of separations is $1:3:5$.
Thus, option C is the correct answer.
Note:Here, instead of 1 second, we can consider any timer interval such as 2 seconds, 3 seconds or so on. We don’t consider time intervals in terms of $t-1$, t and $t+1$ as the calculations get extremely tedious and the ratio of the succession of the distances would be in terms of the variable t and we won’t be able to obtain numerical values of the ratio.
Complete step by step answer:
Here, the drops are falling at successive intervals. Thus, at the 0th second, none of the drop falls, while at the 1st second, the first drop falls, in the 2nd second, the second drop falls and at the 3rd second, the third drop falls. Let these times be t1,t2,t,3 respectively and the distance travelled by these drops be d1, d2, d3 respectively. Now, we know the kinematic equation for a free fall as follows:
\[h = \dfrac{1}{2}g{t^2}\]
Where h is the height of the free fall, g is the acceleration due to gravity and t is the time travelled by an object during the free fall. Thus here, for the first drop, the distance travelled would be
\[
{d_1} = \dfrac{1}{2}g{t_1}^2 = \dfrac{1}{2}(10){(3)^2} \\
\Rightarrow {d_1} = 45m \\
\]
The distance travelled by the second drop would be;
\[
{d_2} = \dfrac{1}{2}g{t_2}^2 = \dfrac{1}{2}(10){(2)^2} \\
\Rightarrow {d_2} = 20m \\
\]
The distance travelled by the third drop is:
\[
{d_3} = \dfrac{1}{2}g{t_3}^2 = \dfrac{1}{2}(10)(1) \\
\therefore {d_3} = 5m
\]
Thus, the ratio of the separations is given by \[{d_3}:{d_2} - {d_3}:{d_1} - {d_2}\] which is equal $5:15:25$. Thus the ratio of separations is $1:3:5$.
Thus, option C is the correct answer.
Note:Here, instead of 1 second, we can consider any timer interval such as 2 seconds, 3 seconds or so on. We don’t consider time intervals in terms of $t-1$, t and $t+1$ as the calculations get extremely tedious and the ratio of the succession of the distances would be in terms of the variable t and we won’t be able to obtain numerical values of the ratio.
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