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TEST SERIES (PHYSICS FST)

263827 The capacitors are uncharged in the begining . How much heat will be produced, when switch S is closed:

1

\frac{1}{2} {CV}^{2}

2

{Cl}^{2}

3

\frac{2}{3} {CV}^{2}

4

\frac{3}{2} {CV}^{2}

Explanation:

d
Net capacity $C_{Het} = 3 C$
$\begin{aligned} U = \frac{1}{2} c_{H} V^{2} \\ U = \frac{1}{2} (3 C) y^{2} \Rightarrow U = \frac{3}{2} C V^{2} \end{aligned}$

NCERT-XII-I-73

TEST SERIES (PHYSICS FST)

263828 A satellite revolve very near to the earth (radius $R_{e}$ ) then it's time - period is:

1

π \sqrt{\frac{2 R_{e}}{g}}

2

2 π \sqrt{\frac{2 R_{e}}{g}}

3

2 π \sqrt{\frac{R_{e}}{g}}

4

\frac{π}{2} \sqrt{\frac{R_{e}}{g}}

Explanation:

c
The time period at height h from earth's surface is .
$T = 2 π \sqrt{\frac{{(R_{e} + h)}^{3}}{g R_{e}^{2}}}$
for very пear $h \to 0$
so $T = 2 π \sqrt{\frac{R_{e}}{g}}$

NCERT-XI-I-137

TEST SERIES (PHYSICS FST)

263829 The sinusoidalA.C. current flows through a register of resistance $R$ . If the peak current is $I_{p}$ . then power dissipated is:

1

I_{p}^{2} R \cos θ

2

{\frac{1}{2} |}_{p}^{2} R

3

{\frac{4}{π} |}^{2} R

4

{\frac{1}{π^{2}} |}_{p}^{2} R

Explanation:

b
$I_{TmE} = \frac{I_{p}}{\sqrt{2}}$
power dissipated, $P = I_{mis}^{2} \times R = {\frac{1}{2} |}_{p}^{2} \times R$

NCERT-XII-I-180

TEST SERIES (PHYSICS FST)

263830 An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm . The size of the image is :

1 0.11 cm

2 0.50 cm

3 0.55 cm

4 0.60 m

Explanation:

c
$\begin{aligned} \frac{1}{0} & = \frac{f}{(f - u)} \\ \Rightarrow \frac{1}{5} & = \frac{- 10}{- 10 - (- 100)} \Rightarrow 1 = 0.55 cm \end{aligned}$

NCERT-XIIIII-225

TEST SERIES (PHYSICS FST)

263827 The capacitors are uncharged in the begining . How much heat will be produced, when switch S is closed:

1

\frac{1}{2} {CV}^{2}

2

{Cl}^{2}

3

\frac{2}{3} {CV}^{2}

4

\frac{3}{2} {CV}^{2}

Explanation:

d
Net capacity $C_{Het} = 3 C$
$\begin{aligned} U = \frac{1}{2} c_{H} V^{2} \\ U = \frac{1}{2} (3 C) y^{2} \Rightarrow U = \frac{3}{2} C V^{2} \end{aligned}$

NCERT-XII-I-73

TEST SERIES (PHYSICS FST)

263828 A satellite revolve very near to the earth (radius $R_{e}$ ) then it's time - period is:

1

π \sqrt{\frac{2 R_{e}}{g}}

2

2 π \sqrt{\frac{2 R_{e}}{g}}

3

2 π \sqrt{\frac{R_{e}}{g}}

4

\frac{π}{2} \sqrt{\frac{R_{e}}{g}}

Explanation:

c
The time period at height h from earth's surface is .
$T = 2 π \sqrt{\frac{{(R_{e} + h)}^{3}}{g R_{e}^{2}}}$
for very пear $h \to 0$
so $T = 2 π \sqrt{\frac{R_{e}}{g}}$

NCERT-XI-I-137

TEST SERIES (PHYSICS FST)

263829 The sinusoidalA.C. current flows through a register of resistance $R$ . If the peak current is $I_{p}$ . then power dissipated is:

1

I_{p}^{2} R \cos θ

2

{\frac{1}{2} |}_{p}^{2} R

3

{\frac{4}{π} |}^{2} R

4

{\frac{1}{π^{2}} |}_{p}^{2} R

Explanation:

b
$I_{TmE} = \frac{I_{p}}{\sqrt{2}}$
power dissipated, $P = I_{mis}^{2} \times R = {\frac{1}{2} |}_{p}^{2} \times R$

NCERT-XII-I-180

TEST SERIES (PHYSICS FST)

263830 An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm . The size of the image is :

1 0.11 cm

2 0.50 cm

3 0.55 cm

4 0.60 m

Explanation:

c
$\begin{aligned} \frac{1}{0} & = \frac{f}{(f - u)} \\ \Rightarrow \frac{1}{5} & = \frac{- 10}{- 10 - (- 100)} \Rightarrow 1 = 0.55 cm \end{aligned}$

NCERT-XIIIII-225

TEST SERIES (PHYSICS FST)

263827 The capacitors are uncharged in the begining . How much heat will be produced, when switch S is closed:

1

\frac{1}{2} {CV}^{2}

2

{Cl}^{2}

3

\frac{2}{3} {CV}^{2}

4

\frac{3}{2} {CV}^{2}

Explanation:

d
Net capacity $C_{Het} = 3 C$
$\begin{aligned} U = \frac{1}{2} c_{H} V^{2} \\ U = \frac{1}{2} (3 C) y^{2} \Rightarrow U = \frac{3}{2} C V^{2} \end{aligned}$

NCERT-XII-I-73

TEST SERIES (PHYSICS FST)

263828 A satellite revolve very near to the earth (radius $R_{e}$ ) then it's time - period is:

1

π \sqrt{\frac{2 R_{e}}{g}}

2

2 π \sqrt{\frac{2 R_{e}}{g}}

3

2 π \sqrt{\frac{R_{e}}{g}}

4

\frac{π}{2} \sqrt{\frac{R_{e}}{g}}

Explanation:

c
The time period at height h from earth's surface is .
$T = 2 π \sqrt{\frac{{(R_{e} + h)}^{3}}{g R_{e}^{2}}}$
for very пear $h \to 0$
so $T = 2 π \sqrt{\frac{R_{e}}{g}}$

NCERT-XI-I-137

TEST SERIES (PHYSICS FST)

263829 The sinusoidalA.C. current flows through a register of resistance $R$ . If the peak current is $I_{p}$ . then power dissipated is:

1

I_{p}^{2} R \cos θ

2

{\frac{1}{2} |}_{p}^{2} R

3

{\frac{4}{π} |}^{2} R

4

{\frac{1}{π^{2}} |}_{p}^{2} R

Explanation:

b
$I_{TmE} = \frac{I_{p}}{\sqrt{2}}$
power dissipated, $P = I_{mis}^{2} \times R = {\frac{1}{2} |}_{p}^{2} \times R$

NCERT-XII-I-180

TEST SERIES (PHYSICS FST)

263830 An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm . The size of the image is :

1 0.11 cm

2 0.50 cm

3 0.55 cm

4 0.60 m

Explanation:

c
$\begin{aligned} \frac{1}{0} & = \frac{f}{(f - u)} \\ \Rightarrow \frac{1}{5} & = \frac{- 10}{- 10 - (- 100)} \Rightarrow 1 = 0.55 cm \end{aligned}$

NCERT-XIIIII-225

TEST SERIES (PHYSICS FST)

263827 The capacitors are uncharged in the begining . How much heat will be produced, when switch S is closed:

1

\frac{1}{2} {CV}^{2}

2

{Cl}^{2}

3

\frac{2}{3} {CV}^{2}

4

\frac{3}{2} {CV}^{2}

Explanation:

d
Net capacity $C_{Het} = 3 C$
$\begin{aligned} U = \frac{1}{2} c_{H} V^{2} \\ U = \frac{1}{2} (3 C) y^{2} \Rightarrow U = \frac{3}{2} C V^{2} \end{aligned}$

NCERT-XII-I-73

TEST SERIES (PHYSICS FST)

263828 A satellite revolve very near to the earth (radius $R_{e}$ ) then it's time - period is:

1

π \sqrt{\frac{2 R_{e}}{g}}

2

2 π \sqrt{\frac{2 R_{e}}{g}}

3

2 π \sqrt{\frac{R_{e}}{g}}

4

\frac{π}{2} \sqrt{\frac{R_{e}}{g}}

Explanation:

c
The time period at height h from earth's surface is .
$T = 2 π \sqrt{\frac{{(R_{e} + h)}^{3}}{g R_{e}^{2}}}$
for very пear $h \to 0$
so $T = 2 π \sqrt{\frac{R_{e}}{g}}$

NCERT-XI-I-137

TEST SERIES (PHYSICS FST)

263829 The sinusoidalA.C. current flows through a register of resistance $R$ . If the peak current is $I_{p}$ . then power dissipated is:

1

I_{p}^{2} R \cos θ

2

{\frac{1}{2} |}_{p}^{2} R

3

{\frac{4}{π} |}^{2} R

4

{\frac{1}{π^{2}} |}_{p}^{2} R

Explanation:

b
$I_{TmE} = \frac{I_{p}}{\sqrt{2}}$
power dissipated, $P = I_{mis}^{2} \times R = {\frac{1}{2} |}_{p}^{2} \times R$

NCERT-XII-I-180

TEST SERIES (PHYSICS FST)

263830 An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm . The size of the image is :

1 0.11 cm

2 0.50 cm

3 0.55 cm

4 0.60 m

Explanation:

c
$\begin{aligned} \frac{1}{0} & = \frac{f}{(f - u)} \\ \Rightarrow \frac{1}{5} & = \frac{- 10}{- 10 - (- 100)} \Rightarrow 1 = 0.55 cm \end{aligned}$

NCERT-XIIIII-225

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