QBManager

PHXI06:WORK ENERGY AND POWER

355727 A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $θ$ at which the speed of the bob is half of that at $A$ , satisfies
supporting img

1

θ = \frac{π}{4}

2

\frac{π}{4} < θ < \frac{π}{2}

3

\frac{π}{2} < θ < \frac{3 π}{4}

4

\frac{3 π}{4} < θ < π

Explanation:

$v = \sqrt{5 g ℓ}$
$v_{final} = \frac{v}{2} = \frac{\sqrt{5 g ℓ}}{2} = \sqrt{\frac{5}{4} g ℓ} = \sqrt{u^{2} - 2 g h}$
$\sqrt{\frac{5 g ℓ}{4}} = \sqrt{u^{2} - 2 g ℓ (1 - \cos θ)}$
$\frac{5 g ℓ}{4} - 5 g ℓ = - 2 g ℓ (1 - \cos θ)$
$\frac{15}{4} = 2 (1 - \cos θ) \Rightarrow \cos θ = - \frac{7}{4} \Rightarrow \frac{3 π}{4} < θ < π$ .

PHXI06:WORK ENERGY AND POWER

355728 A mass $m$ starting from $A$ reaches $B$ of a frictionless track. On reaching $B$ , it pushes the track with a force equal to $x$ times its weight, then the applicable relation is
supporting img

1

h = \frac{(x + 5)}{2} r

2

h = \frac{x}{2} r

3

h = r

4

h = (\frac{x + 1}{2}) r

Explanation:

From conservation of mechanical energy
$K . E_{B} = P . E_{A} - P . E_{B}$
$\frac{1}{2} m v^{2} = m g h - m g 2 r = m g (h - 2 r)$
$v^{2} = 2 g (h - 2 r) (1)$
Given that
$\frac{m v^{2}}{r} = x m g + m g$
$\Rightarrow v^{2} = (x + 1) r g (2)$
From Eqs (1) and (2), we get
$2 g (h - 2 r) = (x + 1) g r$
or $2 g h = (x + 1) g r + 4 g r = (x + 5) g r$
$h = (\frac{x + 5}{2}) r$

PHXI06:WORK ENERGY AND POWER

355729 A particle is given an initial speed $u$ inside a smooth spherical shell of radius $R = 2 m$ so that it just completes a circle. Acceleration of the particle when it is in vertical circle is $(g = 10 m / s^{2}, \sqrt{10} = 3.16)$
supporting img

1

31.6 m / s^{2}

2

11.25 m / s^{2}

3

27.3 m / s^{2}

4

42.7 m / s^{2}

Explanation:

Vertical velocity at bottom,
$u^{2} = 5 g R$
$∴ v^{2} = u^{2} - 2 g R = 5 g R - 2 g R = 3 g R$
Tangential acceleration at bottom is $a_{t} = g$
(downwards)
Centripetal acceleration at bottom is
$a_{c} = \frac{v^{2}}{R} = 3 g$
$∴ a = \sqrt{a_{c}^{2} + a_{1}^{2}} = \sqrt{g^{2} + 9 g^{2}}$
$= g \sqrt{10} = 10 \sqrt{10}$
$∴ a = 31.6 m / s^{2}$

PHXI06:WORK ENERGY AND POWER

355730 A stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n r e v / min$ . The total tension in the string when the stone is at the lowest point is

1

m g

2

m (g + π n r^{2})

3

m (g + n r)

4

m (g + \frac{π^{2} n^{2} r}{900})

Explanation:

Tension at lowest point is given by
$T = \frac{m v^{2}}{r} + m g = m r ω^{2} + m g$
supporting img

$= m r {(\frac{2 π n}{60})}^{2} + m g = m (\frac{π^{2} n^{2} r}{900} + g)$

PHXI06:WORK ENERGY AND POWER

355731 A sphere is suspended by a thread of length $l$ . What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension?

1

g l

2

\sqrt{2 g l}

3

2 g l

4

\sqrt{g l}

Explanation:

Kinetic energy given to a sphere at lowest point $=$ potential energy at the height of suspensions
$\begin{matrix} \Rightarrow \frac{1}{2} m v^{2} = m g l \\ ∴ v = \sqrt{2 g l} \end{matrix}$

PHXI06:WORK ENERGY AND POWER

355727 A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $θ$ at which the speed of the bob is half of that at $A$ , satisfies
supporting img

1

θ = \frac{π}{4}

2

\frac{π}{4} < θ < \frac{π}{2}

3

\frac{π}{2} < θ < \frac{3 π}{4}

4

\frac{3 π}{4} < θ < π

Explanation:

$v = \sqrt{5 g ℓ}$
$v_{final} = \frac{v}{2} = \frac{\sqrt{5 g ℓ}}{2} = \sqrt{\frac{5}{4} g ℓ} = \sqrt{u^{2} - 2 g h}$
$\sqrt{\frac{5 g ℓ}{4}} = \sqrt{u^{2} - 2 g ℓ (1 - \cos θ)}$
$\frac{5 g ℓ}{4} - 5 g ℓ = - 2 g ℓ (1 - \cos θ)$
$\frac{15}{4} = 2 (1 - \cos θ) \Rightarrow \cos θ = - \frac{7}{4} \Rightarrow \frac{3 π}{4} < θ < π$ .

PHXI06:WORK ENERGY AND POWER

355728 A mass $m$ starting from $A$ reaches $B$ of a frictionless track. On reaching $B$ , it pushes the track with a force equal to $x$ times its weight, then the applicable relation is
supporting img

1

h = \frac{(x + 5)}{2} r

2

h = \frac{x}{2} r

3

h = r

4

h = (\frac{x + 1}{2}) r

Explanation:

From conservation of mechanical energy
$K . E_{B} = P . E_{A} - P . E_{B}$
$\frac{1}{2} m v^{2} = m g h - m g 2 r = m g (h - 2 r)$
$v^{2} = 2 g (h - 2 r) (1)$
Given that
$\frac{m v^{2}}{r} = x m g + m g$
$\Rightarrow v^{2} = (x + 1) r g (2)$
From Eqs (1) and (2), we get
$2 g (h - 2 r) = (x + 1) g r$
or $2 g h = (x + 1) g r + 4 g r = (x + 5) g r$
$h = (\frac{x + 5}{2}) r$

PHXI06:WORK ENERGY AND POWER

355729 A particle is given an initial speed $u$ inside a smooth spherical shell of radius $R = 2 m$ so that it just completes a circle. Acceleration of the particle when it is in vertical circle is $(g = 10 m / s^{2}, \sqrt{10} = 3.16)$
supporting img

1

31.6 m / s^{2}

2

11.25 m / s^{2}

3

27.3 m / s^{2}

4

42.7 m / s^{2}

Explanation:

Vertical velocity at bottom,
$u^{2} = 5 g R$
$∴ v^{2} = u^{2} - 2 g R = 5 g R - 2 g R = 3 g R$
Tangential acceleration at bottom is $a_{t} = g$
(downwards)
Centripetal acceleration at bottom is
$a_{c} = \frac{v^{2}}{R} = 3 g$
$∴ a = \sqrt{a_{c}^{2} + a_{1}^{2}} = \sqrt{g^{2} + 9 g^{2}}$
$= g \sqrt{10} = 10 \sqrt{10}$
$∴ a = 31.6 m / s^{2}$

PHXI06:WORK ENERGY AND POWER

355730 A stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n r e v / min$ . The total tension in the string when the stone is at the lowest point is

1

m g

2

m (g + π n r^{2})

3

m (g + n r)

4

m (g + \frac{π^{2} n^{2} r}{900})

Explanation:

Tension at lowest point is given by
$T = \frac{m v^{2}}{r} + m g = m r ω^{2} + m g$
supporting img

$= m r {(\frac{2 π n}{60})}^{2} + m g = m (\frac{π^{2} n^{2} r}{900} + g)$

PHXI06:WORK ENERGY AND POWER

355731 A sphere is suspended by a thread of length $l$ . What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension?

1

g l

2

\sqrt{2 g l}

3

2 g l

4

\sqrt{g l}

Explanation:

Kinetic energy given to a sphere at lowest point $=$ potential energy at the height of suspensions
$\begin{matrix} \Rightarrow \frac{1}{2} m v^{2} = m g l \\ ∴ v = \sqrt{2 g l} \end{matrix}$

PHXI06:WORK ENERGY AND POWER

355727 A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $θ$ at which the speed of the bob is half of that at $A$ , satisfies
supporting img

1

θ = \frac{π}{4}

2

\frac{π}{4} < θ < \frac{π}{2}

3

\frac{π}{2} < θ < \frac{3 π}{4}

4

\frac{3 π}{4} < θ < π

Explanation:

$v = \sqrt{5 g ℓ}$
$v_{final} = \frac{v}{2} = \frac{\sqrt{5 g ℓ}}{2} = \sqrt{\frac{5}{4} g ℓ} = \sqrt{u^{2} - 2 g h}$
$\sqrt{\frac{5 g ℓ}{4}} = \sqrt{u^{2} - 2 g ℓ (1 - \cos θ)}$
$\frac{5 g ℓ}{4} - 5 g ℓ = - 2 g ℓ (1 - \cos θ)$
$\frac{15}{4} = 2 (1 - \cos θ) \Rightarrow \cos θ = - \frac{7}{4} \Rightarrow \frac{3 π}{4} < θ < π$ .

PHXI06:WORK ENERGY AND POWER

355728 A mass $m$ starting from $A$ reaches $B$ of a frictionless track. On reaching $B$ , it pushes the track with a force equal to $x$ times its weight, then the applicable relation is
supporting img

1

h = \frac{(x + 5)}{2} r

2

h = \frac{x}{2} r

3

h = r

4

h = (\frac{x + 1}{2}) r

Explanation:

From conservation of mechanical energy
$K . E_{B} = P . E_{A} - P . E_{B}$
$\frac{1}{2} m v^{2} = m g h - m g 2 r = m g (h - 2 r)$
$v^{2} = 2 g (h - 2 r) (1)$
Given that
$\frac{m v^{2}}{r} = x m g + m g$
$\Rightarrow v^{2} = (x + 1) r g (2)$
From Eqs (1) and (2), we get
$2 g (h - 2 r) = (x + 1) g r$
or $2 g h = (x + 1) g r + 4 g r = (x + 5) g r$
$h = (\frac{x + 5}{2}) r$

PHXI06:WORK ENERGY AND POWER

355729 A particle is given an initial speed $u$ inside a smooth spherical shell of radius $R = 2 m$ so that it just completes a circle. Acceleration of the particle when it is in vertical circle is $(g = 10 m / s^{2}, \sqrt{10} = 3.16)$
supporting img

1

31.6 m / s^{2}

2

11.25 m / s^{2}

3

27.3 m / s^{2}

4

42.7 m / s^{2}

Explanation:

Vertical velocity at bottom,
$u^{2} = 5 g R$
$∴ v^{2} = u^{2} - 2 g R = 5 g R - 2 g R = 3 g R$
Tangential acceleration at bottom is $a_{t} = g$
(downwards)
Centripetal acceleration at bottom is
$a_{c} = \frac{v^{2}}{R} = 3 g$
$∴ a = \sqrt{a_{c}^{2} + a_{1}^{2}} = \sqrt{g^{2} + 9 g^{2}}$
$= g \sqrt{10} = 10 \sqrt{10}$
$∴ a = 31.6 m / s^{2}$

PHXI06:WORK ENERGY AND POWER

355730 A stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n r e v / min$ . The total tension in the string when the stone is at the lowest point is

1

m g

2

m (g + π n r^{2})

3

m (g + n r)

4

m (g + \frac{π^{2} n^{2} r}{900})

Explanation:

Tension at lowest point is given by
$T = \frac{m v^{2}}{r} + m g = m r ω^{2} + m g$
supporting img

$= m r {(\frac{2 π n}{60})}^{2} + m g = m (\frac{π^{2} n^{2} r}{900} + g)$

PHXI06:WORK ENERGY AND POWER

355731 A sphere is suspended by a thread of length $l$ . What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension?

1

g l

2

\sqrt{2 g l}

3

2 g l

4

\sqrt{g l}

Explanation:

Kinetic energy given to a sphere at lowest point $=$ potential energy at the height of suspensions
$\begin{matrix} \Rightarrow \frac{1}{2} m v^{2} = m g l \\ ∴ v = \sqrt{2 g l} \end{matrix}$

PHXI06:WORK ENERGY AND POWER

355727 A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $θ$ at which the speed of the bob is half of that at $A$ , satisfies
supporting img

1

θ = \frac{π}{4}

2

\frac{π}{4} < θ < \frac{π}{2}

3

\frac{π}{2} < θ < \frac{3 π}{4}

4

\frac{3 π}{4} < θ < π

Explanation:

$v = \sqrt{5 g ℓ}$
$v_{final} = \frac{v}{2} = \frac{\sqrt{5 g ℓ}}{2} = \sqrt{\frac{5}{4} g ℓ} = \sqrt{u^{2} - 2 g h}$
$\sqrt{\frac{5 g ℓ}{4}} = \sqrt{u^{2} - 2 g ℓ (1 - \cos θ)}$
$\frac{5 g ℓ}{4} - 5 g ℓ = - 2 g ℓ (1 - \cos θ)$
$\frac{15}{4} = 2 (1 - \cos θ) \Rightarrow \cos θ = - \frac{7}{4} \Rightarrow \frac{3 π}{4} < θ < π$ .

PHXI06:WORK ENERGY AND POWER

355728 A mass $m$ starting from $A$ reaches $B$ of a frictionless track. On reaching $B$ , it pushes the track with a force equal to $x$ times its weight, then the applicable relation is
supporting img

1

h = \frac{(x + 5)}{2} r

2

h = \frac{x}{2} r

3

h = r

4

h = (\frac{x + 1}{2}) r

Explanation:

From conservation of mechanical energy
$K . E_{B} = P . E_{A} - P . E_{B}$
$\frac{1}{2} m v^{2} = m g h - m g 2 r = m g (h - 2 r)$
$v^{2} = 2 g (h - 2 r) (1)$
Given that
$\frac{m v^{2}}{r} = x m g + m g$
$\Rightarrow v^{2} = (x + 1) r g (2)$
From Eqs (1) and (2), we get
$2 g (h - 2 r) = (x + 1) g r$
or $2 g h = (x + 1) g r + 4 g r = (x + 5) g r$
$h = (\frac{x + 5}{2}) r$

PHXI06:WORK ENERGY AND POWER

355729 A particle is given an initial speed $u$ inside a smooth spherical shell of radius $R = 2 m$ so that it just completes a circle. Acceleration of the particle when it is in vertical circle is $(g = 10 m / s^{2}, \sqrt{10} = 3.16)$
supporting img

1

31.6 m / s^{2}

2

11.25 m / s^{2}

3

27.3 m / s^{2}

4

42.7 m / s^{2}

Explanation:

Vertical velocity at bottom,
$u^{2} = 5 g R$
$∴ v^{2} = u^{2} - 2 g R = 5 g R - 2 g R = 3 g R$
Tangential acceleration at bottom is $a_{t} = g$
(downwards)
Centripetal acceleration at bottom is
$a_{c} = \frac{v^{2}}{R} = 3 g$
$∴ a = \sqrt{a_{c}^{2} + a_{1}^{2}} = \sqrt{g^{2} + 9 g^{2}}$
$= g \sqrt{10} = 10 \sqrt{10}$
$∴ a = 31.6 m / s^{2}$

PHXI06:WORK ENERGY AND POWER

355730 A stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n r e v / min$ . The total tension in the string when the stone is at the lowest point is

1

m g

2

m (g + π n r^{2})

3

m (g + n r)

4

m (g + \frac{π^{2} n^{2} r}{900})

Explanation:

Tension at lowest point is given by
$T = \frac{m v^{2}}{r} + m g = m r ω^{2} + m g$
supporting img

$= m r {(\frac{2 π n}{60})}^{2} + m g = m (\frac{π^{2} n^{2} r}{900} + g)$

PHXI06:WORK ENERGY AND POWER

355731 A sphere is suspended by a thread of length $l$ . What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension?

1

g l

2

\sqrt{2 g l}

3

2 g l

4

\sqrt{g l}

Explanation:

Kinetic energy given to a sphere at lowest point $=$ potential energy at the height of suspensions
$\begin{matrix} \Rightarrow \frac{1}{2} m v^{2} = m g l \\ ∴ v = \sqrt{2 g l} \end{matrix}$

PHXI06:WORK ENERGY AND POWER

355727 A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $θ$ at which the speed of the bob is half of that at $A$ , satisfies
supporting img

1

θ = \frac{π}{4}

2

\frac{π}{4} < θ < \frac{π}{2}

3

\frac{π}{2} < θ < \frac{3 π}{4}

4

\frac{3 π}{4} < θ < π

Explanation:

$v = \sqrt{5 g ℓ}$
$v_{final} = \frac{v}{2} = \frac{\sqrt{5 g ℓ}}{2} = \sqrt{\frac{5}{4} g ℓ} = \sqrt{u^{2} - 2 g h}$
$\sqrt{\frac{5 g ℓ}{4}} = \sqrt{u^{2} - 2 g ℓ (1 - \cos θ)}$
$\frac{5 g ℓ}{4} - 5 g ℓ = - 2 g ℓ (1 - \cos θ)$
$\frac{15}{4} = 2 (1 - \cos θ) \Rightarrow \cos θ = - \frac{7}{4} \Rightarrow \frac{3 π}{4} < θ < π$ .

PHXI06:WORK ENERGY AND POWER

355728 A mass $m$ starting from $A$ reaches $B$ of a frictionless track. On reaching $B$ , it pushes the track with a force equal to $x$ times its weight, then the applicable relation is
supporting img

1

h = \frac{(x + 5)}{2} r

2

h = \frac{x}{2} r

3

h = r

4

h = (\frac{x + 1}{2}) r

Explanation:

From conservation of mechanical energy
$K . E_{B} = P . E_{A} - P . E_{B}$
$\frac{1}{2} m v^{2} = m g h - m g 2 r = m g (h - 2 r)$
$v^{2} = 2 g (h - 2 r) (1)$
Given that
$\frac{m v^{2}}{r} = x m g + m g$
$\Rightarrow v^{2} = (x + 1) r g (2)$
From Eqs (1) and (2), we get
$2 g (h - 2 r) = (x + 1) g r$
or $2 g h = (x + 1) g r + 4 g r = (x + 5) g r$
$h = (\frac{x + 5}{2}) r$

PHXI06:WORK ENERGY AND POWER

355729 A particle is given an initial speed $u$ inside a smooth spherical shell of radius $R = 2 m$ so that it just completes a circle. Acceleration of the particle when it is in vertical circle is $(g = 10 m / s^{2}, \sqrt{10} = 3.16)$
supporting img

1

31.6 m / s^{2}

2

11.25 m / s^{2}

3

27.3 m / s^{2}

4

42.7 m / s^{2}

Explanation:

Vertical velocity at bottom,
$u^{2} = 5 g R$
$∴ v^{2} = u^{2} - 2 g R = 5 g R - 2 g R = 3 g R$
Tangential acceleration at bottom is $a_{t} = g$
(downwards)
Centripetal acceleration at bottom is
$a_{c} = \frac{v^{2}}{R} = 3 g$
$∴ a = \sqrt{a_{c}^{2} + a_{1}^{2}} = \sqrt{g^{2} + 9 g^{2}}$
$= g \sqrt{10} = 10 \sqrt{10}$
$∴ a = 31.6 m / s^{2}$

PHXI06:WORK ENERGY AND POWER

355730 A stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n r e v / min$ . The total tension in the string when the stone is at the lowest point is

1

m g

2

m (g + π n r^{2})

3

m (g + n r)

4

m (g + \frac{π^{2} n^{2} r}{900})

Explanation:

Tension at lowest point is given by
$T = \frac{m v^{2}}{r} + m g = m r ω^{2} + m g$
supporting img

$= m r {(\frac{2 π n}{60})}^{2} + m g = m (\frac{π^{2} n^{2} r}{900} + g)$

PHXI06:WORK ENERGY AND POWER

355731 A sphere is suspended by a thread of length $l$ . What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension?

1

g l

2

\sqrt{2 g l}

3

2 g l

4

\sqrt{g l}

Explanation:

Kinetic energy given to a sphere at lowest point $=$ potential energy at the height of suspensions
$\begin{matrix} \Rightarrow \frac{1}{2} m v^{2} = m g l \\ ∴ v = \sqrt{2 g l} \end{matrix}$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: