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TEST SERIES (PHYSICS FST)

263809 If the kinetic energy of the particle is increased to 16 times, the percentage change in the de Broglie wavelength of the particle is:

1

25 %

2

75 %

3

60 %

4

50 %

Explanation:

b
$\begin{aligned} λ = \frac{h}{\sqrt{2 k m}} \frac{Δ λ}{λ} \times 100 = \frac{\frac{1}{\sqrt{k_{t}}} - \frac{1}{\sqrt{k_{1}}}}{\frac{1}{\sqrt{k_{1}}}} \times 100 \\ - \frac{\frac{1}{\sqrt{16 k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 \end{aligned} = \frac{\frac{1}{4 \sqrt{k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 = \frac{- \frac{3}{4 \sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 1000$

NCERT-XII-II-285

TEST SERIES (PHYSICS FST)

263810 Ratio of magnetic fields at the centre of a current carrying coil of radius $R$ and at distance of $3 R$ on its axis is :

1

10 \sqrt{10}

2

20 \sqrt{10}

3

2 \sqrt{10}

4

\sqrt{10}

Explanation:

a
The desired ratio is
$\frac{\frac{μ_{0} i}{2 R}}{\frac{μ_{c} R^{2}}{2 {(R^{2} + 9 R^{2})}^{3 / 2}}} = 10 \sqrt{10}$

NCERT-XII-I-115

TEST SERIES (PHYSICS FST)

263811 A dia-magnetic material in a magnetic field moves:

1 Perpendicular to the field

2 From weakerto stronger parts

3 From stronger to weaker parts

4 In пone of the above directions

NCERT-XII-I-197

TEST SERIES (PHYSICS FST)

263812 A stone is thrown at $25 m / s$ at $53^{\circ}$ above the horizontal. At what time its velocity is at $45^{\circ}$ below the horizontal :

1 0.5 s

2 4 s

3 3.5 s

4 2.5 s

Explanation:

c

Horizontal component of velcoity, throughout the motion remain constant
$\begin{aligned} u \cos θ = v \cos α \\ 25 \cos 53^{\circ} = v \cos 45^{\circ} \\ \Rightarrow v = 25 \times \frac{3}{5} \cdot \sqrt{2} = 15 \sqrt{2} \\ \Rightarrow v_{y} = v \sin 45 = 15 \end{aligned}$
Now, using $V_{y} = u_{y} + a_{y} t$ in $y$ -direction
$\begin{aligned} - 15 = 25 \sin 53^{\circ} - g t \\ - 15 = 25 \times \frac{4}{5} - 10 t \\ t = 3.5 \sec . \end{aligned}$

NCERT-XII-I-39

TEST SERIES (PHYSICS FST)

263813 The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :

1

π / 4

2

π / 2

3

3 π / 4

4

π

Explanation:

b
$\frac{distance}{displacement} = \frac{π r}{2 r} = \frac{π r}{2 r} = \frac{π}{2}$

NCERT-XI-I-35

TEST SERIES (PHYSICS FST)

263809 If the kinetic energy of the particle is increased to 16 times, the percentage change in the de Broglie wavelength of the particle is:

1

25 %

2

75 %

3

60 %

4

50 %

Explanation:

b
$\begin{aligned} λ = \frac{h}{\sqrt{2 k m}} \frac{Δ λ}{λ} \times 100 = \frac{\frac{1}{\sqrt{k_{t}}} - \frac{1}{\sqrt{k_{1}}}}{\frac{1}{\sqrt{k_{1}}}} \times 100 \\ - \frac{\frac{1}{\sqrt{16 k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 \end{aligned} = \frac{\frac{1}{4 \sqrt{k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 = \frac{- \frac{3}{4 \sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 1000$

NCERT-XII-II-285

TEST SERIES (PHYSICS FST)

263810 Ratio of magnetic fields at the centre of a current carrying coil of radius $R$ and at distance of $3 R$ on its axis is :

1

10 \sqrt{10}

2

20 \sqrt{10}

3

2 \sqrt{10}

4

\sqrt{10}

Explanation:

a
The desired ratio is
$\frac{\frac{μ_{0} i}{2 R}}{\frac{μ_{c} R^{2}}{2 {(R^{2} + 9 R^{2})}^{3 / 2}}} = 10 \sqrt{10}$

NCERT-XII-I-115

TEST SERIES (PHYSICS FST)

263811 A dia-magnetic material in a magnetic field moves:

1 Perpendicular to the field

2 From weakerto stronger parts

3 From stronger to weaker parts

4 In пone of the above directions

NCERT-XII-I-197

TEST SERIES (PHYSICS FST)

263812 A stone is thrown at $25 m / s$ at $53^{\circ}$ above the horizontal. At what time its velocity is at $45^{\circ}$ below the horizontal :

1 0.5 s

2 4 s

3 3.5 s

4 2.5 s

Explanation:

c

Horizontal component of velcoity, throughout the motion remain constant
$\begin{aligned} u \cos θ = v \cos α \\ 25 \cos 53^{\circ} = v \cos 45^{\circ} \\ \Rightarrow v = 25 \times \frac{3}{5} \cdot \sqrt{2} = 15 \sqrt{2} \\ \Rightarrow v_{y} = v \sin 45 = 15 \end{aligned}$
Now, using $V_{y} = u_{y} + a_{y} t$ in $y$ -direction
$\begin{aligned} - 15 = 25 \sin 53^{\circ} - g t \\ - 15 = 25 \times \frac{4}{5} - 10 t \\ t = 3.5 \sec . \end{aligned}$

NCERT-XII-I-39

TEST SERIES (PHYSICS FST)

263813 The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :

1

π / 4

2

π / 2

3

3 π / 4

4

π

Explanation:

b
$\frac{distance}{displacement} = \frac{π r}{2 r} = \frac{π r}{2 r} = \frac{π}{2}$

NCERT-XI-I-35

TEST SERIES (PHYSICS FST)

263809 If the kinetic energy of the particle is increased to 16 times, the percentage change in the de Broglie wavelength of the particle is:

1

25 %

2

75 %

3

60 %

4

50 %

Explanation:

b
$\begin{aligned} λ = \frac{h}{\sqrt{2 k m}} \frac{Δ λ}{λ} \times 100 = \frac{\frac{1}{\sqrt{k_{t}}} - \frac{1}{\sqrt{k_{1}}}}{\frac{1}{\sqrt{k_{1}}}} \times 100 \\ - \frac{\frac{1}{\sqrt{16 k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 \end{aligned} = \frac{\frac{1}{4 \sqrt{k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 = \frac{- \frac{3}{4 \sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 1000$

NCERT-XII-II-285

TEST SERIES (PHYSICS FST)

263810 Ratio of magnetic fields at the centre of a current carrying coil of radius $R$ and at distance of $3 R$ on its axis is :

1

10 \sqrt{10}

2

20 \sqrt{10}

3

2 \sqrt{10}

4

\sqrt{10}

Explanation:

a
The desired ratio is
$\frac{\frac{μ_{0} i}{2 R}}{\frac{μ_{c} R^{2}}{2 {(R^{2} + 9 R^{2})}^{3 / 2}}} = 10 \sqrt{10}$

NCERT-XII-I-115

TEST SERIES (PHYSICS FST)

263811 A dia-magnetic material in a magnetic field moves:

1 Perpendicular to the field

2 From weakerto stronger parts

3 From stronger to weaker parts

4 In пone of the above directions

NCERT-XII-I-197

TEST SERIES (PHYSICS FST)

263812 A stone is thrown at $25 m / s$ at $53^{\circ}$ above the horizontal. At what time its velocity is at $45^{\circ}$ below the horizontal :

1 0.5 s

2 4 s

3 3.5 s

4 2.5 s

Explanation:

c

Horizontal component of velcoity, throughout the motion remain constant
$\begin{aligned} u \cos θ = v \cos α \\ 25 \cos 53^{\circ} = v \cos 45^{\circ} \\ \Rightarrow v = 25 \times \frac{3}{5} \cdot \sqrt{2} = 15 \sqrt{2} \\ \Rightarrow v_{y} = v \sin 45 = 15 \end{aligned}$
Now, using $V_{y} = u_{y} + a_{y} t$ in $y$ -direction
$\begin{aligned} - 15 = 25 \sin 53^{\circ} - g t \\ - 15 = 25 \times \frac{4}{5} - 10 t \\ t = 3.5 \sec . \end{aligned}$

NCERT-XII-I-39

TEST SERIES (PHYSICS FST)

263813 The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :

1

π / 4

2

π / 2

3

3 π / 4

4

π

Explanation:

b
$\frac{distance}{displacement} = \frac{π r}{2 r} = \frac{π r}{2 r} = \frac{π}{2}$

NCERT-XI-I-35

TEST SERIES (PHYSICS FST)

263809 If the kinetic energy of the particle is increased to 16 times, the percentage change in the de Broglie wavelength of the particle is:

1

25 %

2

75 %

3

60 %

4

50 %

Explanation:

b
$\begin{aligned} λ = \frac{h}{\sqrt{2 k m}} \frac{Δ λ}{λ} \times 100 = \frac{\frac{1}{\sqrt{k_{t}}} - \frac{1}{\sqrt{k_{1}}}}{\frac{1}{\sqrt{k_{1}}}} \times 100 \\ - \frac{\frac{1}{\sqrt{16 k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 \end{aligned} = \frac{\frac{1}{4 \sqrt{k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 = \frac{- \frac{3}{4 \sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 1000$

NCERT-XII-II-285

TEST SERIES (PHYSICS FST)

263810 Ratio of magnetic fields at the centre of a current carrying coil of radius $R$ and at distance of $3 R$ on its axis is :

1

10 \sqrt{10}

2

20 \sqrt{10}

3

2 \sqrt{10}

4

\sqrt{10}

Explanation:

a
The desired ratio is
$\frac{\frac{μ_{0} i}{2 R}}{\frac{μ_{c} R^{2}}{2 {(R^{2} + 9 R^{2})}^{3 / 2}}} = 10 \sqrt{10}$

NCERT-XII-I-115

TEST SERIES (PHYSICS FST)

263811 A dia-magnetic material in a magnetic field moves:

1 Perpendicular to the field

2 From weakerto stronger parts

3 From stronger to weaker parts

4 In пone of the above directions

NCERT-XII-I-197

TEST SERIES (PHYSICS FST)

263812 A stone is thrown at $25 m / s$ at $53^{\circ}$ above the horizontal. At what time its velocity is at $45^{\circ}$ below the horizontal :

1 0.5 s

2 4 s

3 3.5 s

4 2.5 s

Explanation:

c

Horizontal component of velcoity, throughout the motion remain constant
$\begin{aligned} u \cos θ = v \cos α \\ 25 \cos 53^{\circ} = v \cos 45^{\circ} \\ \Rightarrow v = 25 \times \frac{3}{5} \cdot \sqrt{2} = 15 \sqrt{2} \\ \Rightarrow v_{y} = v \sin 45 = 15 \end{aligned}$
Now, using $V_{y} = u_{y} + a_{y} t$ in $y$ -direction
$\begin{aligned} - 15 = 25 \sin 53^{\circ} - g t \\ - 15 = 25 \times \frac{4}{5} - 10 t \\ t = 3.5 \sec . \end{aligned}$

NCERT-XII-I-39

TEST SERIES (PHYSICS FST)

263813 The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :

1

π / 4

2

π / 2

3

3 π / 4

4

π

Explanation:

b
$\frac{distance}{displacement} = \frac{π r}{2 r} = \frac{π r}{2 r} = \frac{π}{2}$

NCERT-XI-I-35

TEST SERIES (PHYSICS FST)

263809 If the kinetic energy of the particle is increased to 16 times, the percentage change in the de Broglie wavelength of the particle is:

1

25 %

2

75 %

3

60 %

4

50 %

Explanation:

b
$\begin{aligned} λ = \frac{h}{\sqrt{2 k m}} \frac{Δ λ}{λ} \times 100 = \frac{\frac{1}{\sqrt{k_{t}}} - \frac{1}{\sqrt{k_{1}}}}{\frac{1}{\sqrt{k_{1}}}} \times 100 \\ - \frac{\frac{1}{\sqrt{16 k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 \end{aligned} = \frac{\frac{1}{4 \sqrt{k}} - \frac{1}{\sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 100 = \frac{- \frac{3}{4 \sqrt{k}}}{\frac{1}{\sqrt{k}}} \times 1000$

NCERT-XII-II-285

TEST SERIES (PHYSICS FST)

263810 Ratio of magnetic fields at the centre of a current carrying coil of radius $R$ and at distance of $3 R$ on its axis is :

1

10 \sqrt{10}

2

20 \sqrt{10}

3

2 \sqrt{10}

4

\sqrt{10}

Explanation:

a
The desired ratio is
$\frac{\frac{μ_{0} i}{2 R}}{\frac{μ_{c} R^{2}}{2 {(R^{2} + 9 R^{2})}^{3 / 2}}} = 10 \sqrt{10}$

NCERT-XII-I-115

TEST SERIES (PHYSICS FST)

263811 A dia-magnetic material in a magnetic field moves:

1 Perpendicular to the field

2 From weakerto stronger parts

3 From stronger to weaker parts

4 In пone of the above directions

NCERT-XII-I-197

TEST SERIES (PHYSICS FST)

263812 A stone is thrown at $25 m / s$ at $53^{\circ}$ above the horizontal. At what time its velocity is at $45^{\circ}$ below the horizontal :

1 0.5 s

2 4 s

3 3.5 s

4 2.5 s

Explanation:

c

Horizontal component of velcoity, throughout the motion remain constant
$\begin{aligned} u \cos θ = v \cos α \\ 25 \cos 53^{\circ} = v \cos 45^{\circ} \\ \Rightarrow v = 25 \times \frac{3}{5} \cdot \sqrt{2} = 15 \sqrt{2} \\ \Rightarrow v_{y} = v \sin 45 = 15 \end{aligned}$
Now, using $V_{y} = u_{y} + a_{y} t$ in $y$ -direction
$\begin{aligned} - 15 = 25 \sin 53^{\circ} - g t \\ - 15 = 25 \times \frac{4}{5} - 10 t \\ t = 3.5 \sec . \end{aligned}$

NCERT-XII-I-39

TEST SERIES (PHYSICS FST)

263813 The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :

1

π / 4

2

π / 2

3

3 π / 4

4

π

Explanation:

b
$\frac{distance}{displacement} = \frac{π r}{2 r} = \frac{π r}{2 r} = \frac{π}{2}$

NCERT-XI-I-35