Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor…

Solution & Explanation

Related Formula

h = ut + (1)/(2)gt²

Core Logic

Since drops fall at regular intervals, find the total time of fall for the first drop and divide it by the number of intervals to get the time gap between successive drops.

Step 1: Total Time of Fall

Time taken by the first drop to reach the ground ( $h = 5 ~m$ $h = 5 \mathrm{~m}$):

t = \sqrt((2h)/(g)) = \sqrt((2 \times 5)/(10)) = 1 ~sec

Step 2: Time Interval Between Drops

The first drop strikes the floor when the sixth drop begins to fall. There are exactly 5 intervals between the 1st and 6th drop. Time between each drop $Δ t = 1 ~sec5 = 0.2 ~sec$ .

Step 3: Distance of the Fourth Drop

The 4th drop has been falling for 2 intervals (since drops 6, 5, 4 mean the 4th drop was released $2 \times 0.2 ~s$ $2 \times 0.2 \mathrm{~s}$ after the 6th drop started... wait, no. The 4th drop was released before the 5th and 6th). Time of fall for the $4th$ $4^{\text{th}}$ drop is: It was released 2 intervals after the 1st drop, so it has been falling for $1.0 - 2(0.2) = 0.6 ~s$ $1.0 - 2(0.2) = 0.6 \mathrm{~s}$? Wait, the PDF solution says: "Time of fall for $4th$ $4^{\text{th}}$ drop is $1 - 0.6 = 0.4 ~sec$ $1 - 0.6 = 0.4 \mathrm{~sec}$" Let's trace: 6th drop (0s fall), 5th drop (0.2s fall), 4th drop (0.4s fall). Distance fallen by $4th$ $4^{\text{th}}$ drop:

h' = (1)/(2) g t² = (1)/(2) \times 10 \times (0.4)² = 0.8 ~m

Step 4: Height from Ground

Height from ground = Total height - Distance fallen

H = 5 - 0.8 = 4.2 ~m

Pattern Recognition

In 'falling drops' problems, if N drops are in the air, there are (N-1) time intervals. The $kth$ $k^{\text{th}}$ drop from the top has fallen for $(k-1)$ $(k-1)$ intervals.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

More Motion in a Straight Line Previous-Year Questions — Page 8

Q44 jee_main_2024_31_jan_evening Vector Algebra

If two vectors

A

and

B

having equal magnitude

R

are inclined at an angle

θ

, then

A. $| A - B| = \sqrt(2)R ((θ)/(2))$
B. $| A + B| = 2R ((θ)/(2))$
C. $| A + B| = 2R ((θ)/(2))$
D. $| A - B| = 2R ((θ)/(2))$

Solution

Related Formula

The magnitude of the resultant vector is given by:

| Rᵣₑₛ| = \sqrt(A² + B² + 2AB θ)

Core Logic

Let $| A| = | B| = R$ $|\vec{A}| = |\vec{B}| = R$. Then for vector addition:

| A + B| = \sqrt(R² + R² + 2R² θ)

Step 1: Simplify Addition Form

| A + B| = \sqrt(2R² (1 + θ))

Using the trigonometric identity $1 + θ = 2 ² ((θ)/(2))$ :

| A + B| = \sqrt(2R² \times 2 ² ((θ)/(2))) = 2R ((θ)/(2))

Step 2: Cross-check Subtraction Form

For subtraction:

| A - B| = \sqrt(R² + R² - 2R² θ)

| A - B| = \sqrt(2R² (1 - θ)) = \sqrt(2R² \times 2 ² ((θ)/(2))) = 2R ((θ)/(2))

Checking options, only $| A + B| = 2R ((θ)/(2))$ is correctly paired in the choice list.

Pattern Recognition

Standard geometry shortcut: Addition of two equal vectors yields a cosine half-angle dependency. Subtraction yields a sine half-angle dependency. $(+ \to )$ $(+ \to \cos)$, $(- \to )$ $(- \to \sin)$.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q jee_main_2024_31_jan_morning Projectile Motion

A body starts falling freely from height

H

hits an inclined plane in its path at height

h

. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of

(H)/(h)

for which the body will take the maximum time to reach the ground is

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

t = \sqrt((2d)/(g))

Core Logic

Projectile Motion diagram for Q55 - JEE Main 2024 Morning

The body falls freely from height $H$ $H$ to height $h$ $h$. The distance fallen is $(H - h)$ $(H - h)$. Time taken to fall this distance:

t₁ = \sqrt((2(H - h))/(g))

After elastic impact, the vertical velocity becomes zero (the entire velocity is directed horizontally). From height $h$ $h$, it now acts as a horizontal projectile. The time taken to reach the ground vertically from height $h$ $h$ is:

t₂ = \sqrt((2h)/(g))

Total time of flight $T = t₁ + t₂$ $T = t_1 + t_2$:

T = \sqrt((2(H - h))/(g)) + \sqrt((2h)/(g))

Step 2: Maximizing Time

To find the maximum time $T$ $T$, differentiate $T$ $T$ with respect to $h$ $h$ and equate to zero:

(dT)/(dh) = \sqrt((2)/(g)) ( -12\sqrt(H - h) + 12\sqrt(h) ) = 0

12\sqrt(h) = 12\sqrt(H - h)

\sqrt(H - h) = \sqrt(h)

Squaring both sides: $H - h = h$ $H - h = h$ $H = 2h$ $H = 2h$

(H)/(h) = 2

Chapter Mix

Class 11 Physics: Kinematics

Q33 jee_main_2024_31_jan_morning Differentiation In Kinematics

The relation between time '

t

' and distance '

x

' is

t = α x² + β x

, where

α

and

β

are constants. The relation between acceleration (

a

) and velocity (

v

) is:

A. $a = -2α v³$ $a = -2\alpha v^3$
B. $a = -5α v⁵$ $a = -5\alpha v^5$
C. $a = -3α v²$ $a = -3\alpha v^2$
D. $a = -4α v⁴$ $a = -4\alpha v^4$

Solution

Related Formula

v = (dx)/(dt)

a = (dv)/(dt) = v(dv)/(dx)

Step 1: Differentiate with respect to time

Given the equation:

t = α x² + β x

Differentiating with respect to time $t$ $t$:

(dt)/(dt) = (d)/(dt)(α x² + β x)

1 = 2α x (dx)/(dt) + β (dx)/(dt)

1 = (2α x + β) v

v = (2α x + β)⁻¹

Step 2: Calculate Acceleration

Now, acceleration $a = (dv)/(dt)$ $a = \frac{dv}{dt}$. Differentiating $v$ $v$ with respect to time $t$ $t$:

a = (d)/(dt) [ (2α x + β)⁻¹ ]

a = -1(2α x + β)⁻² \cdot (d)/(dt)(2α x + β)

a = -(2α x + β)⁻² \cdot (2α) (dx)/(dt)

Substitute $v$ $v$ and $(2α x + β)⁻² = v²$ $(2\alpha x + \beta)^{-2} = v^2$:

a = -(v²) \cdot (2α) \cdot v

a = -2α v³

Pattern Recognition

Standard kinematic shortcut: Whenever $t = Ax² + Bx$ $t = Ax^2 + Bx$, $v = (2Ax+B)⁻¹$ $v = (2Ax+B)^{-1}$ and $a = -2A v³$ $a = -2A v^3$. Memorizing this directly saves derivation time during the exam.

Chapter Mix

Class 11 Physics: Kinematics

Solution & Explanation

Related Formula

Core Logic

Step 1: Total Time of Fall

Step 2: Time Interval Between Drops

Step 3: Distance of the Fourth Drop

Step 4: Height from Ground

Pattern Recognition

Chapter Mix

Reference Study Guides

More Motion in a Straight Line Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Simplify Addition Form

Step 2: Cross-check Subtraction Form

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Maximizing Time

Chapter Mix

Solution

Related Formula

Step 1: Differentiate with respect to time

Step 2: Calculate Acceleration

Pattern Recognition

Chapter Mix

More Motion in a Straight Line Questions — jee_main_2026_28_january_morning

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