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TEST SERIES (PHYSICS FST)

263943 A body of mass 5 kg is suspended by a spring balance on an inclined plane as shown in figure. The spring balance measure:

1 50 N

2 25 N

3 500 N

4 10 N SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

b
force applied on spring balance
$\begin{aligned} = m g \sin θ \\ 5.10 \times \sin 30^{\circ} \\ = 25 N \end{aligned}$
SECTION-B

**NCERT-XI-I-59**

TEST SERIES (PHYSICS FST)

263944 A constant torque of $1000 N - m$ , turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis perpendicular to it through the centre. Angular velocity of the wheel after 3 s will be:

1

15 rad / s

2

10 rad / s

3

5 rad / s

4

1 rad / s

TEST SERIES (PHYSICS FST)

263945 A body of mass $m$ rises to height $h =$ R 15 from the earth's surface, where $R$ is earth's radius. If $g$ is acceleration due to gravity at earth's surface, the increase in potential energy is:

1 mgh

2

\frac{4}{5} m g h

3

\frac{5}{6} m g h

4

\frac{6}{7} m g h

Explanation:

c
From $ω = ω_{0} + a t = 0 + 5 \times 3 = 15 rad / s$
$a = \frac{τ}{I} = \frac{1000}{200} = 5 Rad / \sec^{2}$
[NCERT-XI-I-134]
$Δ u = \frac{m g h}{1 + h / R}$
Substituting $R = 5 h$
we get $Δ U = \frac{m g h}{1 + 1 / 5} = \frac{5}{6} mgh$

TEST SERIES (PHYSICS FST)

263947 Match of the column $A$ and $B$ : |A|B| |-----|-----| |i. Magnetisation|a. $\frac{\overset{―}{M}}{\overset{―}{H}}$ | |ii. Magnetic field in material|b. $\frac{magnetic moment}{volume}$ | |iii. Magnetic susceptibility|c. $\frac{B}{μ_{0} H}$ | |iv. Relative magnetic permeability |d. ${\underset{―}{H}}_{0} (\overset{―}{H} + \overset{―}{M})$ |

1 i-b, ii-d, iii-c, iv-a

2 i-b, ii-d, iii-a, iv-c

3 i-b, ii-a, iii-d, iv-c

4 i-a, ii-b, iii-d, iv-c

**NCERT-XII-I-150**

TEST SERIES (PHYSICS FST)

263948 The position of a particle at time ' $t$ ' is given by the relation $x (t) = (\frac{v_{0}}{α}) (1 - c^{- α 1})$ , where $v_{0}$ is a constant and $α > 0$ . The dimensions of $v_{0}$ and ' $u$ ' are res pectively

1

M^{0} L^{1} T^{- 1}

and

T^{- 1}

2

M^{0} L^{1} T^{0}

and

T^{- 1}

3

M^{0} L^{1} T^{- 1}

and

L T^{- 2}

4

M^{0} L^{1} T^{- 1}

and T

TEST SERIES (PHYSICS FST)

263943 A body of mass 5 kg is suspended by a spring balance on an inclined plane as shown in figure. The spring balance measure:

1 50 N

2 25 N

3 500 N

4 10 N SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

b
force applied on spring balance
$\begin{aligned} = m g \sin θ \\ 5.10 \times \sin 30^{\circ} \\ = 25 N \end{aligned}$
SECTION-B

**NCERT-XI-I-59**

TEST SERIES (PHYSICS FST)

263944 A constant torque of $1000 N - m$ , turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis perpendicular to it through the centre. Angular velocity of the wheel after 3 s will be:

1

15 rad / s

2

10 rad / s

3

5 rad / s

4

1 rad / s

TEST SERIES (PHYSICS FST)

263945 A body of mass $m$ rises to height $h =$ R 15 from the earth's surface, where $R$ is earth's radius. If $g$ is acceleration due to gravity at earth's surface, the increase in potential energy is:

1 mgh

2

\frac{4}{5} m g h

3

\frac{5}{6} m g h

4

\frac{6}{7} m g h

Explanation:

c
From $ω = ω_{0} + a t = 0 + 5 \times 3 = 15 rad / s$
$a = \frac{τ}{I} = \frac{1000}{200} = 5 Rad / \sec^{2}$
[NCERT-XI-I-134]
$Δ u = \frac{m g h}{1 + h / R}$
Substituting $R = 5 h$
we get $Δ U = \frac{m g h}{1 + 1 / 5} = \frac{5}{6} mgh$

TEST SERIES (PHYSICS FST)

263947 Match of the column $A$ and $B$ : |A|B| |-----|-----| |i. Magnetisation|a. $\frac{\overset{―}{M}}{\overset{―}{H}}$ | |ii. Magnetic field in material|b. $\frac{magnetic moment}{volume}$ | |iii. Magnetic susceptibility|c. $\frac{B}{μ_{0} H}$ | |iv. Relative magnetic permeability |d. ${\underset{―}{H}}_{0} (\overset{―}{H} + \overset{―}{M})$ |

1 i-b, ii-d, iii-c, iv-a

2 i-b, ii-d, iii-a, iv-c

3 i-b, ii-a, iii-d, iv-c

4 i-a, ii-b, iii-d, iv-c

**NCERT-XII-I-150**

TEST SERIES (PHYSICS FST)

263948 The position of a particle at time ' $t$ ' is given by the relation $x (t) = (\frac{v_{0}}{α}) (1 - c^{- α 1})$ , where $v_{0}$ is a constant and $α > 0$ . The dimensions of $v_{0}$ and ' $u$ ' are res pectively

1

M^{0} L^{1} T^{- 1}

and

T^{- 1}

2

M^{0} L^{1} T^{0}

and

T^{- 1}

3

M^{0} L^{1} T^{- 1}

and

L T^{- 2}

4

M^{0} L^{1} T^{- 1}

and T

TEST SERIES (PHYSICS FST)

263943 A body of mass 5 kg is suspended by a spring balance on an inclined plane as shown in figure. The spring balance measure:

1 50 N

2 25 N

3 500 N

4 10 N SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

b
force applied on spring balance
$\begin{aligned} = m g \sin θ \\ 5.10 \times \sin 30^{\circ} \\ = 25 N \end{aligned}$
SECTION-B

**NCERT-XI-I-59**

TEST SERIES (PHYSICS FST)

263944 A constant torque of $1000 N - m$ , turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis perpendicular to it through the centre. Angular velocity of the wheel after 3 s will be:

1

15 rad / s

2

10 rad / s

3

5 rad / s

4

1 rad / s

TEST SERIES (PHYSICS FST)

263945 A body of mass $m$ rises to height $h =$ R 15 from the earth's surface, where $R$ is earth's radius. If $g$ is acceleration due to gravity at earth's surface, the increase in potential energy is:

1 mgh

2

\frac{4}{5} m g h

3

\frac{5}{6} m g h

4

\frac{6}{7} m g h

Explanation:

c
From $ω = ω_{0} + a t = 0 + 5 \times 3 = 15 rad / s$
$a = \frac{τ}{I} = \frac{1000}{200} = 5 Rad / \sec^{2}$
[NCERT-XI-I-134]
$Δ u = \frac{m g h}{1 + h / R}$
Substituting $R = 5 h$
we get $Δ U = \frac{m g h}{1 + 1 / 5} = \frac{5}{6} mgh$

TEST SERIES (PHYSICS FST)

263947 Match of the column $A$ and $B$ : |A|B| |-----|-----| |i. Magnetisation|a. $\frac{\overset{―}{M}}{\overset{―}{H}}$ | |ii. Magnetic field in material|b. $\frac{magnetic moment}{volume}$ | |iii. Magnetic susceptibility|c. $\frac{B}{μ_{0} H}$ | |iv. Relative magnetic permeability |d. ${\underset{―}{H}}_{0} (\overset{―}{H} + \overset{―}{M})$ |

1 i-b, ii-d, iii-c, iv-a

2 i-b, ii-d, iii-a, iv-c

3 i-b, ii-a, iii-d, iv-c

4 i-a, ii-b, iii-d, iv-c

**NCERT-XII-I-150**

TEST SERIES (PHYSICS FST)

263948 The position of a particle at time ' $t$ ' is given by the relation $x (t) = (\frac{v_{0}}{α}) (1 - c^{- α 1})$ , where $v_{0}$ is a constant and $α > 0$ . The dimensions of $v_{0}$ and ' $u$ ' are res pectively

1

M^{0} L^{1} T^{- 1}

and

T^{- 1}

2

M^{0} L^{1} T^{0}

and

T^{- 1}

3

M^{0} L^{1} T^{- 1}

and

L T^{- 2}

4

M^{0} L^{1} T^{- 1}

and T

TEST SERIES (PHYSICS FST)

263943 A body of mass 5 kg is suspended by a spring balance on an inclined plane as shown in figure. The spring balance measure:

1 50 N

2 25 N

3 500 N

4 10 N SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

b
force applied on spring balance
$\begin{aligned} = m g \sin θ \\ 5.10 \times \sin 30^{\circ} \\ = 25 N \end{aligned}$
SECTION-B

**NCERT-XI-I-59**

TEST SERIES (PHYSICS FST)

263944 A constant torque of $1000 N - m$ , turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis perpendicular to it through the centre. Angular velocity of the wheel after 3 s will be:

1

15 rad / s

2

10 rad / s

3

5 rad / s

4

1 rad / s

TEST SERIES (PHYSICS FST)

263945 A body of mass $m$ rises to height $h =$ R 15 from the earth's surface, where $R$ is earth's radius. If $g$ is acceleration due to gravity at earth's surface, the increase in potential energy is:

1 mgh

2

\frac{4}{5} m g h

3

\frac{5}{6} m g h

4

\frac{6}{7} m g h

Explanation:

c
From $ω = ω_{0} + a t = 0 + 5 \times 3 = 15 rad / s$
$a = \frac{τ}{I} = \frac{1000}{200} = 5 Rad / \sec^{2}$
[NCERT-XI-I-134]
$Δ u = \frac{m g h}{1 + h / R}$
Substituting $R = 5 h$
we get $Δ U = \frac{m g h}{1 + 1 / 5} = \frac{5}{6} mgh$

TEST SERIES (PHYSICS FST)

263947 Match of the column $A$ and $B$ : |A|B| |-----|-----| |i. Magnetisation|a. $\frac{\overset{―}{M}}{\overset{―}{H}}$ | |ii. Magnetic field in material|b. $\frac{magnetic moment}{volume}$ | |iii. Magnetic susceptibility|c. $\frac{B}{μ_{0} H}$ | |iv. Relative magnetic permeability |d. ${\underset{―}{H}}_{0} (\overset{―}{H} + \overset{―}{M})$ |

1 i-b, ii-d, iii-c, iv-a

2 i-b, ii-d, iii-a, iv-c

3 i-b, ii-a, iii-d, iv-c

4 i-a, ii-b, iii-d, iv-c

**NCERT-XII-I-150**

TEST SERIES (PHYSICS FST)

263948 The position of a particle at time ' $t$ ' is given by the relation $x (t) = (\frac{v_{0}}{α}) (1 - c^{- α 1})$ , where $v_{0}$ is a constant and $α > 0$ . The dimensions of $v_{0}$ and ' $u$ ' are res pectively

1

M^{0} L^{1} T^{- 1}

and

T^{- 1}

2

M^{0} L^{1} T^{0}

and

T^{- 1}

3

M^{0} L^{1} T^{- 1}

and

L T^{- 2}

4

M^{0} L^{1} T^{- 1}

and T

TEST SERIES (PHYSICS FST)

263943 A body of mass 5 kg is suspended by a spring balance on an inclined plane as shown in figure. The spring balance measure:

1 50 N

2 25 N

3 500 N

4 10 N SECTION - B This section will have 15 questions. Candidate can choose to attempt any 10 questions out of these 15 questions. In case if candidate attempts more than 10 questions, first 10 attempted questions will be considered for marking.

Explanation:

b
force applied on spring balance
$\begin{aligned} = m g \sin θ \\ 5.10 \times \sin 30^{\circ} \\ = 25 N \end{aligned}$
SECTION-B

**NCERT-XI-I-59**

TEST SERIES (PHYSICS FST)

263944 A constant torque of $1000 N - m$ , turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis perpendicular to it through the centre. Angular velocity of the wheel after 3 s will be:

1

15 rad / s

2

10 rad / s

3

5 rad / s

4

1 rad / s

TEST SERIES (PHYSICS FST)

263945 A body of mass $m$ rises to height $h =$ R 15 from the earth's surface, where $R$ is earth's radius. If $g$ is acceleration due to gravity at earth's surface, the increase in potential energy is:

1 mgh

2

\frac{4}{5} m g h

3

\frac{5}{6} m g h

4

\frac{6}{7} m g h

Explanation:

c
From $ω = ω_{0} + a t = 0 + 5 \times 3 = 15 rad / s$
$a = \frac{τ}{I} = \frac{1000}{200} = 5 Rad / \sec^{2}$
[NCERT-XI-I-134]
$Δ u = \frac{m g h}{1 + h / R}$
Substituting $R = 5 h$
we get $Δ U = \frac{m g h}{1 + 1 / 5} = \frac{5}{6} mgh$

TEST SERIES (PHYSICS FST)

263947 Match of the column $A$ and $B$ : |A|B| |-----|-----| |i. Magnetisation|a. $\frac{\overset{―}{M}}{\overset{―}{H}}$ | |ii. Magnetic field in material|b. $\frac{magnetic moment}{volume}$ | |iii. Magnetic susceptibility|c. $\frac{B}{μ_{0} H}$ | |iv. Relative magnetic permeability |d. ${\underset{―}{H}}_{0} (\overset{―}{H} + \overset{―}{M})$ |

1 i-b, ii-d, iii-c, iv-a

2 i-b, ii-d, iii-a, iv-c

3 i-b, ii-a, iii-d, iv-c

4 i-a, ii-b, iii-d, iv-c

**NCERT-XII-I-150**

TEST SERIES (PHYSICS FST)

263948 The position of a particle at time ' $t$ ' is given by the relation $x (t) = (\frac{v_{0}}{α}) (1 - c^{- α 1})$ , where $v_{0}$ is a constant and $α > 0$ . The dimensions of $v_{0}$ and ' $u$ ' are res pectively

1

M^{0} L^{1} T^{- 1}

and

T^{- 1}

2

M^{0} L^{1} T^{0}

and

T^{- 1}

3

M^{0} L^{1} T^{- 1}

and

L T^{- 2}

4

M^{0} L^{1} T^{- 1}

and T

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