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TEST SERIES (PHYSICS FST)

266235 In the following figure, electric flux passing through the surface is:

1

\frac{q}{ε_{0}}

2

\frac{q}{2 ε_{a}}

3

\frac{q}{12 ε_{p}}

4

\frac{q}{6 ε_{α}}

Explanation:

c
total flux passing through 6 surfaces. $ϕ_{E} = \frac{q / 2}{ε_{0}}$
$∴$ Flux passing through one surface $= \frac{1}{12} \frac{q}{z_{0}}$

**NCERT-XII-I-29

TEST SERIES (PHYSICS FST)

266236 A stationary wave is represented by $Y = A \sin (100 t) \cos (0.01 x)$ , where $y$ and $A$ are in millimeters, $t$ is in seconds, and $x$ is in meteres, the velocity of super imposing waves is:

1

10^{4} m / s

2 Not derivable

3

1 m / s

4

10^{2} m / s

Explanation:

a
Compering with standard equation.
$\begin{aligned} y = A \sin [\frac{2 π}{T} t] \cos [\frac{2 π}{λ} x] \\ T = \frac{π}{50}, λ = 200 π \\ V = \frac{λ}{T} = 200 \times 50 = 10^{4} m / \sec \end{aligned}$

**NCERT-XI-II-290**

TEST SERIES (PHYSICS FST)

266237 In the figure shown below, the charge on the left plate of the $10_{μ} F$ capacitor is $- 30_{μ} C$ . The charge on the right plate of the $6 μ F$ capacitor is:

1

- 18 μ C

2

- 12 μ C

3

+ 12 μ C

4

+ 18 μ C

Explanation:

d
$6 μ F$ and $4 μ F$ are in parallel and total charge on this combination is $30 μ C$
$∴$ Charge on $6 μ F$ capacitor
$= \frac{6}{6 + 4} \times 30 = 18 μ C$
Since charge is asked on right plate therefore is $+ 18 μ C$ .

**NCERT-XII-I-72**

TEST SERIES (PHYSICS FST)

266238 In stationary wave all particles between two nodes pass through the mean position:

1 At different times with different velocities

2 At different times with the same velocity

3 At the same time with equal velocity

4 At the same time with different velocities

**NCERT-XI-II-291**

TEST SERIES (PHYSICS FST)

266235 In the following figure, electric flux passing through the surface is:

1

\frac{q}{ε_{0}}

2

\frac{q}{2 ε_{a}}

3

\frac{q}{12 ε_{p}}

4

\frac{q}{6 ε_{α}}

Explanation:

c
total flux passing through 6 surfaces. $ϕ_{E} = \frac{q / 2}{ε_{0}}$
$∴$ Flux passing through one surface $= \frac{1}{12} \frac{q}{z_{0}}$

**NCERT-XII-I-29

TEST SERIES (PHYSICS FST)

266236 A stationary wave is represented by $Y = A \sin (100 t) \cos (0.01 x)$ , where $y$ and $A$ are in millimeters, $t$ is in seconds, and $x$ is in meteres, the velocity of super imposing waves is:

1

10^{4} m / s

2 Not derivable

3

1 m / s

4

10^{2} m / s

Explanation:

a
Compering with standard equation.
$\begin{aligned} y = A \sin [\frac{2 π}{T} t] \cos [\frac{2 π}{λ} x] \\ T = \frac{π}{50}, λ = 200 π \\ V = \frac{λ}{T} = 200 \times 50 = 10^{4} m / \sec \end{aligned}$

**NCERT-XI-II-290**

TEST SERIES (PHYSICS FST)

266237 In the figure shown below, the charge on the left plate of the $10_{μ} F$ capacitor is $- 30_{μ} C$ . The charge on the right plate of the $6 μ F$ capacitor is:

1

- 18 μ C

2

- 12 μ C

3

+ 12 μ C

4

+ 18 μ C

Explanation:

d
$6 μ F$ and $4 μ F$ are in parallel and total charge on this combination is $30 μ C$
$∴$ Charge on $6 μ F$ capacitor
$= \frac{6}{6 + 4} \times 30 = 18 μ C$
Since charge is asked on right plate therefore is $+ 18 μ C$ .

**NCERT-XII-I-72**

TEST SERIES (PHYSICS FST)

266238 In stationary wave all particles between two nodes pass through the mean position:

1 At different times with different velocities

2 At different times with the same velocity

3 At the same time with equal velocity

4 At the same time with different velocities

**NCERT-XI-II-291**

TEST SERIES (PHYSICS FST)

266235 In the following figure, electric flux passing through the surface is:

1

\frac{q}{ε_{0}}

2

\frac{q}{2 ε_{a}}

3

\frac{q}{12 ε_{p}}

4

\frac{q}{6 ε_{α}}

Explanation:

c
total flux passing through 6 surfaces. $ϕ_{E} = \frac{q / 2}{ε_{0}}$
$∴$ Flux passing through one surface $= \frac{1}{12} \frac{q}{z_{0}}$

**NCERT-XII-I-29

TEST SERIES (PHYSICS FST)

266236 A stationary wave is represented by $Y = A \sin (100 t) \cos (0.01 x)$ , where $y$ and $A$ are in millimeters, $t$ is in seconds, and $x$ is in meteres, the velocity of super imposing waves is:

1

10^{4} m / s

2 Not derivable

3

1 m / s

4

10^{2} m / s

Explanation:

a
Compering with standard equation.
$\begin{aligned} y = A \sin [\frac{2 π}{T} t] \cos [\frac{2 π}{λ} x] \\ T = \frac{π}{50}, λ = 200 π \\ V = \frac{λ}{T} = 200 \times 50 = 10^{4} m / \sec \end{aligned}$

**NCERT-XI-II-290**

TEST SERIES (PHYSICS FST)

266237 In the figure shown below, the charge on the left plate of the $10_{μ} F$ capacitor is $- 30_{μ} C$ . The charge on the right plate of the $6 μ F$ capacitor is:

1

- 18 μ C

2

- 12 μ C

3

+ 12 μ C

4

+ 18 μ C

Explanation:

d
$6 μ F$ and $4 μ F$ are in parallel and total charge on this combination is $30 μ C$
$∴$ Charge on $6 μ F$ capacitor
$= \frac{6}{6 + 4} \times 30 = 18 μ C$
Since charge is asked on right plate therefore is $+ 18 μ C$ .

**NCERT-XII-I-72**

TEST SERIES (PHYSICS FST)

266238 In stationary wave all particles between two nodes pass through the mean position:

1 At different times with different velocities

2 At different times with the same velocity

3 At the same time with equal velocity

4 At the same time with different velocities

**NCERT-XI-II-291**

TEST SERIES (PHYSICS FST)

266235 In the following figure, electric flux passing through the surface is:

1

\frac{q}{ε_{0}}

2

\frac{q}{2 ε_{a}}

3

\frac{q}{12 ε_{p}}

4

\frac{q}{6 ε_{α}}

Explanation:

c
total flux passing through 6 surfaces. $ϕ_{E} = \frac{q / 2}{ε_{0}}$
$∴$ Flux passing through one surface $= \frac{1}{12} \frac{q}{z_{0}}$

**NCERT-XII-I-29

TEST SERIES (PHYSICS FST)

266236 A stationary wave is represented by $Y = A \sin (100 t) \cos (0.01 x)$ , where $y$ and $A$ are in millimeters, $t$ is in seconds, and $x$ is in meteres, the velocity of super imposing waves is:

1

10^{4} m / s

2 Not derivable

3

1 m / s

4

10^{2} m / s

Explanation:

a
Compering with standard equation.
$\begin{aligned} y = A \sin [\frac{2 π}{T} t] \cos [\frac{2 π}{λ} x] \\ T = \frac{π}{50}, λ = 200 π \\ V = \frac{λ}{T} = 200 \times 50 = 10^{4} m / \sec \end{aligned}$

**NCERT-XI-II-290**

TEST SERIES (PHYSICS FST)

266237 In the figure shown below, the charge on the left plate of the $10_{μ} F$ capacitor is $- 30_{μ} C$ . The charge on the right plate of the $6 μ F$ capacitor is:

1

- 18 μ C

2

- 12 μ C

3

+ 12 μ C

4

+ 18 μ C

Explanation:

d
$6 μ F$ and $4 μ F$ are in parallel and total charge on this combination is $30 μ C$
$∴$ Charge on $6 μ F$ capacitor
$= \frac{6}{6 + 4} \times 30 = 18 μ C$
Since charge is asked on right plate therefore is $+ 18 μ C$ .

**NCERT-XII-I-72**

TEST SERIES (PHYSICS FST)

266238 In stationary wave all particles between two nodes pass through the mean position:

1 At different times with different velocities

2 At different times with the same velocity

3 At the same time with equal velocity

4 At the same time with different velocities

**NCERT-XI-II-291**

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