QBManager

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361263 A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ punctured with a pin, the thread.
supporting img

1 Becomes concave towards

A

2 Becomes convex towards

A

3 Remains in the initial position

4 Either (1) or (2) depending on the size of

A

w.r.t.

B

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361264 The surface tension of a liquid is $5 N m^{- 1}$ . If a thin film formed on a loop of area $0.02 m^{- 2}$ then its surface energy will be

1

5 \times 10^{- 2} J

2

2.5 \times 10^{- 2} J

3

3 \times 10^{- 1} J

4

2 \times 10^{- 1} J

Explanation:

Effective area $= 2 \times 0.02 m^{2} = 0.04 m^{2}$
Surface energy $= 5 N m^{- 1} \times 0.04 m^{2} = 2 \times 10^{- 1} J$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361265 A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become

1

\frac{1^{th}}{100}

2

{\frac{1}{10}}^{th}

3 100 times

4 10 times

Explanation:

Let radius of big drop is $R$ and radius of small drop is $r$ . Here, volume remain constant.
Final volume $=$ Initial volume
So, $\frac{4}{3} π R^{3} = 1000 \cdot \frac{4}{3} π r^{3}$
$R^{3} = 10^{3} r^{3} \Rightarrow R = 10 r$
Initial surface energy, $U_{i} = 1000 \times T \times$ Area
$= 1000 \times T \times 4 π r^{2}$
Final surface energy,
$U_{f} = T \times 4 π R^{2} = T \times 4 π (10 r)^{2}$
$= 100 \times T \times 4 π r^{2}$
$\frac{U_{f}}{U_{i}} = \frac{100 \times 4 π r^{2}}{1000 \times 4 π r^{2}} \Rightarrow U_{f} = \frac{1}{10} U_{i}$
Thus, surface energy will become $1 / 10^{th}$ times.

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361266 A large number of liquid drops each of radius ' $a$ ' coalesce to form a single spherical drop of radius ' $b$ '. The energy released in the process is converted into kinetic energy of the big drop formed. The speed of big drop is (The density of liquid is $ρ$ and surface tension is $T$ )

1

\sqrt{\frac{6 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

2

\sqrt{\frac{4 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

3

\sqrt{\frac{8 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

4

\sqrt{\frac{5 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

Explanation:

Let $N$ be the number of drops each having radius ' $a$ '. When they coalesce the total volume remains constant
$\begin{aligned} N \frac{4}{3} π a^{3} = \frac{4}{3} π b^{3} \\ b = a N^{1 / 3} \end{aligned}$
The change in surface area is
$Δ A = N 4 π a^{2} - 4 π b^{2}$
$Δ A = 4 π [N a^{2} - a^{2} N^{2 / 3}] = 4 π a^{2} [N - N^{2 / 3}]$
The change in K.E $= T Δ A = 4 π a^{2} T [N - N^{2 / 3}]$
$\begin{aligned} \frac{1}{2} (ρ \frac{4}{3} π b^{3}) V^{2} = \frac{1}{2} (N ρ \frac{4}{3} π a^{3}) V^{2} \\ = 4 π a^{2} T [N - N^{2 / 3}] \\ \Rightarrow V = \sqrt{\frac{6 T}{a ρ} [1 - \frac{1}{N^{2 / 3}}]} = \sqrt{\frac{6 T}{ρ a} [1 - \frac{a}{b}]} \\ \Rightarrow V = \sqrt{\frac{6 T}{ρ} [\frac{1}{a} - \frac{1}{b}]} \\ (b = a N^{1 / 3}) \end{aligned}$

JEE - 2014

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361263 A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ punctured with a pin, the thread.
supporting img

1 Becomes concave towards

A

2 Becomes convex towards

A

3 Remains in the initial position

4 Either (1) or (2) depending on the size of

A

w.r.t.

B

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361264 The surface tension of a liquid is $5 N m^{- 1}$ . If a thin film formed on a loop of area $0.02 m^{- 2}$ then its surface energy will be

1

5 \times 10^{- 2} J

2

2.5 \times 10^{- 2} J

3

3 \times 10^{- 1} J

4

2 \times 10^{- 1} J

Explanation:

Effective area $= 2 \times 0.02 m^{2} = 0.04 m^{2}$
Surface energy $= 5 N m^{- 1} \times 0.04 m^{2} = 2 \times 10^{- 1} J$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361265 A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become

1

\frac{1^{th}}{100}

2

{\frac{1}{10}}^{th}

3 100 times

4 10 times

Explanation:

Let radius of big drop is $R$ and radius of small drop is $r$ . Here, volume remain constant.
Final volume $=$ Initial volume
So, $\frac{4}{3} π R^{3} = 1000 \cdot \frac{4}{3} π r^{3}$
$R^{3} = 10^{3} r^{3} \Rightarrow R = 10 r$
Initial surface energy, $U_{i} = 1000 \times T \times$ Area
$= 1000 \times T \times 4 π r^{2}$
Final surface energy,
$U_{f} = T \times 4 π R^{2} = T \times 4 π (10 r)^{2}$
$= 100 \times T \times 4 π r^{2}$
$\frac{U_{f}}{U_{i}} = \frac{100 \times 4 π r^{2}}{1000 \times 4 π r^{2}} \Rightarrow U_{f} = \frac{1}{10} U_{i}$
Thus, surface energy will become $1 / 10^{th}$ times.

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361266 A large number of liquid drops each of radius ' $a$ ' coalesce to form a single spherical drop of radius ' $b$ '. The energy released in the process is converted into kinetic energy of the big drop formed. The speed of big drop is (The density of liquid is $ρ$ and surface tension is $T$ )

1

\sqrt{\frac{6 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

2

\sqrt{\frac{4 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

3

\sqrt{\frac{8 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

4

\sqrt{\frac{5 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

Explanation:

Let $N$ be the number of drops each having radius ' $a$ '. When they coalesce the total volume remains constant
$\begin{aligned} N \frac{4}{3} π a^{3} = \frac{4}{3} π b^{3} \\ b = a N^{1 / 3} \end{aligned}$
The change in surface area is
$Δ A = N 4 π a^{2} - 4 π b^{2}$
$Δ A = 4 π [N a^{2} - a^{2} N^{2 / 3}] = 4 π a^{2} [N - N^{2 / 3}]$
The change in K.E $= T Δ A = 4 π a^{2} T [N - N^{2 / 3}]$
$\begin{aligned} \frac{1}{2} (ρ \frac{4}{3} π b^{3}) V^{2} = \frac{1}{2} (N ρ \frac{4}{3} π a^{3}) V^{2} \\ = 4 π a^{2} T [N - N^{2 / 3}] \\ \Rightarrow V = \sqrt{\frac{6 T}{a ρ} [1 - \frac{1}{N^{2 / 3}}]} = \sqrt{\frac{6 T}{ρ a} [1 - \frac{a}{b}]} \\ \Rightarrow V = \sqrt{\frac{6 T}{ρ} [\frac{1}{a} - \frac{1}{b}]} \\ (b = a N^{1 / 3}) \end{aligned}$

JEE - 2014

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361263 A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ punctured with a pin, the thread.
supporting img

1 Becomes concave towards

A

2 Becomes convex towards

A

3 Remains in the initial position

4 Either (1) or (2) depending on the size of

A

w.r.t.

B

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361264 The surface tension of a liquid is $5 N m^{- 1}$ . If a thin film formed on a loop of area $0.02 m^{- 2}$ then its surface energy will be

1

5 \times 10^{- 2} J

2

2.5 \times 10^{- 2} J

3

3 \times 10^{- 1} J

4

2 \times 10^{- 1} J

Explanation:

Effective area $= 2 \times 0.02 m^{2} = 0.04 m^{2}$
Surface energy $= 5 N m^{- 1} \times 0.04 m^{2} = 2 \times 10^{- 1} J$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361265 A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become

1

\frac{1^{th}}{100}

2

{\frac{1}{10}}^{th}

3 100 times

4 10 times

Explanation:

Let radius of big drop is $R$ and radius of small drop is $r$ . Here, volume remain constant.
Final volume $=$ Initial volume
So, $\frac{4}{3} π R^{3} = 1000 \cdot \frac{4}{3} π r^{3}$
$R^{3} = 10^{3} r^{3} \Rightarrow R = 10 r$
Initial surface energy, $U_{i} = 1000 \times T \times$ Area
$= 1000 \times T \times 4 π r^{2}$
Final surface energy,
$U_{f} = T \times 4 π R^{2} = T \times 4 π (10 r)^{2}$
$= 100 \times T \times 4 π r^{2}$
$\frac{U_{f}}{U_{i}} = \frac{100 \times 4 π r^{2}}{1000 \times 4 π r^{2}} \Rightarrow U_{f} = \frac{1}{10} U_{i}$
Thus, surface energy will become $1 / 10^{th}$ times.

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361266 A large number of liquid drops each of radius ' $a$ ' coalesce to form a single spherical drop of radius ' $b$ '. The energy released in the process is converted into kinetic energy of the big drop formed. The speed of big drop is (The density of liquid is $ρ$ and surface tension is $T$ )

1

\sqrt{\frac{6 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

2

\sqrt{\frac{4 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

3

\sqrt{\frac{8 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

4

\sqrt{\frac{5 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

Explanation:

Let $N$ be the number of drops each having radius ' $a$ '. When they coalesce the total volume remains constant
$\begin{aligned} N \frac{4}{3} π a^{3} = \frac{4}{3} π b^{3} \\ b = a N^{1 / 3} \end{aligned}$
The change in surface area is
$Δ A = N 4 π a^{2} - 4 π b^{2}$
$Δ A = 4 π [N a^{2} - a^{2} N^{2 / 3}] = 4 π a^{2} [N - N^{2 / 3}]$
The change in K.E $= T Δ A = 4 π a^{2} T [N - N^{2 / 3}]$
$\begin{aligned} \frac{1}{2} (ρ \frac{4}{3} π b^{3}) V^{2} = \frac{1}{2} (N ρ \frac{4}{3} π a^{3}) V^{2} \\ = 4 π a^{2} T [N - N^{2 / 3}] \\ \Rightarrow V = \sqrt{\frac{6 T}{a ρ} [1 - \frac{1}{N^{2 / 3}}]} = \sqrt{\frac{6 T}{ρ a} [1 - \frac{a}{b}]} \\ \Rightarrow V = \sqrt{\frac{6 T}{ρ} [\frac{1}{a} - \frac{1}{b}]} \\ (b = a N^{1 / 3}) \end{aligned}$

JEE - 2014

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361263 A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ punctured with a pin, the thread.
supporting img

1 Becomes concave towards

A

2 Becomes convex towards

A

3 Remains in the initial position

4 Either (1) or (2) depending on the size of

A

w.r.t.

B

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361264 The surface tension of a liquid is $5 N m^{- 1}$ . If a thin film formed on a loop of area $0.02 m^{- 2}$ then its surface energy will be

1

5 \times 10^{- 2} J

2

2.5 \times 10^{- 2} J

3

3 \times 10^{- 1} J

4

2 \times 10^{- 1} J

Explanation:

Effective area $= 2 \times 0.02 m^{2} = 0.04 m^{2}$
Surface energy $= 5 N m^{- 1} \times 0.04 m^{2} = 2 \times 10^{- 1} J$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361265 A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become

1

\frac{1^{th}}{100}

2

{\frac{1}{10}}^{th}

3 100 times

4 10 times

Explanation:

Let radius of big drop is $R$ and radius of small drop is $r$ . Here, volume remain constant.
Final volume $=$ Initial volume
So, $\frac{4}{3} π R^{3} = 1000 \cdot \frac{4}{3} π r^{3}$
$R^{3} = 10^{3} r^{3} \Rightarrow R = 10 r$
Initial surface energy, $U_{i} = 1000 \times T \times$ Area
$= 1000 \times T \times 4 π r^{2}$
Final surface energy,
$U_{f} = T \times 4 π R^{2} = T \times 4 π (10 r)^{2}$
$= 100 \times T \times 4 π r^{2}$
$\frac{U_{f}}{U_{i}} = \frac{100 \times 4 π r^{2}}{1000 \times 4 π r^{2}} \Rightarrow U_{f} = \frac{1}{10} U_{i}$
Thus, surface energy will become $1 / 10^{th}$ times.

JEE - 2024

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361266 A large number of liquid drops each of radius ' $a$ ' coalesce to form a single spherical drop of radius ' $b$ '. The energy released in the process is converted into kinetic energy of the big drop formed. The speed of big drop is (The density of liquid is $ρ$ and surface tension is $T$ )

1

\sqrt{\frac{6 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

2

\sqrt{\frac{4 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

3

\sqrt{\frac{8 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

4

\sqrt{\frac{5 T}{ρ} [\frac{1}{a} - \frac{1}{b}]}

Explanation:

Let $N$ be the number of drops each having radius ' $a$ '. When they coalesce the total volume remains constant
$\begin{aligned} N \frac{4}{3} π a^{3} = \frac{4}{3} π b^{3} \\ b = a N^{1 / 3} \end{aligned}$
The change in surface area is
$Δ A = N 4 π a^{2} - 4 π b^{2}$
$Δ A = 4 π [N a^{2} - a^{2} N^{2 / 3}] = 4 π a^{2} [N - N^{2 / 3}]$
The change in K.E $= T Δ A = 4 π a^{2} T [N - N^{2 / 3}]$
$\begin{aligned} \frac{1}{2} (ρ \frac{4}{3} π b^{3}) V^{2} = \frac{1}{2} (N ρ \frac{4}{3} π a^{3}) V^{2} \\ = 4 π a^{2} T [N - N^{2 / 3}] \\ \Rightarrow V = \sqrt{\frac{6 T}{a ρ} [1 - \frac{1}{N^{2 / 3}}]} = \sqrt{\frac{6 T}{ρ a} [1 - \frac{a}{b}]} \\ \Rightarrow V = \sqrt{\frac{6 T}{ρ} [\frac{1}{a} - \frac{1}{b}]} \\ (b = a N^{1 / 3}) \end{aligned}$

JEE - 2014

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: