
Balls are dropped from the roof of a tower at a fixed interval of time. At the moment when the \[{9^{th}}\] ball reaches the ground \[{n^{th}}\] ball is \[3/{4^{th}}\] height of the tower. The value of n is
A. 13
B. 7
C. 6
D. 5
Answer
161.1k+ views
Hint: Free fall is when an object is dropped from above the earth’s surface and it falls under the force of gravity. This motion will have the effect of acceleration due to gravity. Free fall does not depend on the mass of the falling object and only depends on the height and time taken for the fall.
Formula used :
The height of the object from which it falls,
\[h = \dfrac{1}{2}g{t^2}\]
Where, g – acceleration due to gravity and t – time taken for the free fall.
Complete step by step solution:
When an object moves only with the force due to gravity, then the object is said to be in free fall. Since, the object is acted upon by acceleration due to gravity, the motion will be downwards. During free fall the initial velocity is always zero.
In this case, 9 balls are dropped from the tower at fixed time interval, so the time taken for a ball to fall down is,
\[t = \sqrt {\dfrac{{2h}}{g}} \]
When the \[{9^{th}}\] ball reaches the ground, the \[{n^{th}}\] ball reaches \[3/{4^{th}}\] height of the tower. So, only 8 balls have completely dropped to the ground.
So, the time interval for each ball is,
\[\dfrac{t}{8} = \dfrac{1}{8}\sqrt {\dfrac{{2h}}{g}} \]
Time taken for the \[{n^{th}}\] ball is,
\[{t_n} = (n - 1)\dfrac{1}{8}\sqrt {\dfrac{{2h}}{g}} \]
For, the \[{n^{th}}\] ball the height is \[\dfrac{{3h}}{4}\], from the top the height can be written as \[\dfrac{h}{4}\].
So,
\[\dfrac{h}{4} = \dfrac{1}{2}g{t^2}\]
Substituting t in above equation,
\[\dfrac{h}{4} = \dfrac{1}{2}g{(n - 1)^2}\dfrac{1}{{{{(8)}^2}}}\left( {\dfrac{{2h}}{g}} \right) \\ \]
\[\Rightarrow \dfrac{1}{4} = {(n - 1)^2}\dfrac{1}{{{{(8)}^2}}} \\ \]
\[\Rightarrow \dfrac{{8{\rm{x8}}}}{4} = {(n - 1)^2} \\ \]
\[\Rightarrow {n^2} - 2n + 1 = 16 \]
\[\Rightarrow {n^2} - 2n - 15 = 0\]
Solving the quadratic equation, we get two roots as
\[n = 5,n = - 3\]
Since the number of balls cannot be negative, n = 5 is the answer.
So, the correct answer is option D.
Note : A state of free fall known as weightlessness occurs when the inertial force created by orbital flight or other scenarios that negate gravity completely. It is felt since there is no longer a sense of weight. When no forces from touch are occurring on the bodies, it occurs (human body).
Formula used :
The height of the object from which it falls,
\[h = \dfrac{1}{2}g{t^2}\]
Where, g – acceleration due to gravity and t – time taken for the free fall.
Complete step by step solution:
When an object moves only with the force due to gravity, then the object is said to be in free fall. Since, the object is acted upon by acceleration due to gravity, the motion will be downwards. During free fall the initial velocity is always zero.
In this case, 9 balls are dropped from the tower at fixed time interval, so the time taken for a ball to fall down is,
\[t = \sqrt {\dfrac{{2h}}{g}} \]
When the \[{9^{th}}\] ball reaches the ground, the \[{n^{th}}\] ball reaches \[3/{4^{th}}\] height of the tower. So, only 8 balls have completely dropped to the ground.
So, the time interval for each ball is,
\[\dfrac{t}{8} = \dfrac{1}{8}\sqrt {\dfrac{{2h}}{g}} \]
Time taken for the \[{n^{th}}\] ball is,
\[{t_n} = (n - 1)\dfrac{1}{8}\sqrt {\dfrac{{2h}}{g}} \]
For, the \[{n^{th}}\] ball the height is \[\dfrac{{3h}}{4}\], from the top the height can be written as \[\dfrac{h}{4}\].
So,
\[\dfrac{h}{4} = \dfrac{1}{2}g{t^2}\]
Substituting t in above equation,
\[\dfrac{h}{4} = \dfrac{1}{2}g{(n - 1)^2}\dfrac{1}{{{{(8)}^2}}}\left( {\dfrac{{2h}}{g}} \right) \\ \]
\[\Rightarrow \dfrac{1}{4} = {(n - 1)^2}\dfrac{1}{{{{(8)}^2}}} \\ \]
\[\Rightarrow \dfrac{{8{\rm{x8}}}}{4} = {(n - 1)^2} \\ \]
\[\Rightarrow {n^2} - 2n + 1 = 16 \]
\[\Rightarrow {n^2} - 2n - 15 = 0\]
Solving the quadratic equation, we get two roots as
\[n = 5,n = - 3\]
Since the number of balls cannot be negative, n = 5 is the answer.
So, the correct answer is option D.
Note : A state of free fall known as weightlessness occurs when the inertial force created by orbital flight or other scenarios that negate gravity completely. It is felt since there is no longer a sense of weight. When no forces from touch are occurring on the bodies, it occurs (human body).
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
