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Work, Energy and Power

268800 A canon shell fired breaks into two equal parts at its highest point. One part retraces the path to the canon with kinetic energy $E_{1}$ and the kinetic energy of the second part is $E_{2}$ . Relation between $E_{1}$ and $E_{2}$ is
$(2014$ E)

1

E_{2} = E_{1}

2

E_{2} = 4 E_{1}

3

E_{2} = 9 E_{1}

4

E_{2} = 15 E_{1}

Explanation:

At highest point according to L.C.M, $m u \cos θ = - \frac{m}{2} u \cos θ + \frac{m}{2} v$
$E_{1} = \frac{1}{2} ◻ \frac{m}{2} \exists (u \cos θ)^{2}, E_{2} = \frac{1}{2} ∵ \frac{m}{2} θ v^{2}$

Work, Energy and Power

268801 A body of $200 g$ begins to fall from a height where its potential energy is $80 J$ . Its velocity at a point where its kinetic and potential energies are equal(in $m / s) (2014 M)$

1

10 \sqrt{8}

2 4

3 400

4 20

Explanation:

T.E at top $= T . E$ at given point $; P . E =$ P.E $E^{1} + K^{K} E^{1}$
$80 = 2 K . E^{1} (∵ P . E^{1} = K . E^{1})$

Work, Energy and Power

268802 The work done by a force $\bar{F} = 2 \bar{i} - \bar{j} - \bar{k}$ in moving an object from origin to a point whose position vector is $\bar{r} = 3 \bar{i} + 2 \bar{j} - 5 \bar{k} (2013$ E)

1 1 unit

2 9 units

3 13 units

4 60 units

Explanation:

$W = \vec{F} \cdot \vec{S} = \vec{F} \cdot (\vec{r_{2}} - \vec{r_{1}})$

Work, Energy and Power

268803 A ball at rest is dropped from a height of $12 m$ . If it looses $25 %$ of its kinetic energy on striking the ground and bounces back to a height ' $h$ '. the value of ' $h$ ' is equal to ( $2013 E$ )

1

3 m

2

6 m

3

9 m

4

12 m

Explanation:

percentage loss of $PE = \frac{h_{1} - h_{2}}{h_{1}} \times 100 %$

Work, Energy and Power

268804 A block of mass $2.9 kg$ is suspended from a string of length $50 cm$ and is at rest. Another block of mass $100 g$ , which is moving with a speed of $150 m / s$ strikes and sticks to the first block. Subsequently when the string makes an angle of $60^{\circ}$ . With the vertical, the tension in the string will be $(g = 10 {m s}^{- 2})$ ( $2013 E$ )

1

140 N

2

135 N

3

125 N

4

90 N

Explanation:

$v_{common} = \frac{m u}{M + m}$ and $T = m g \cos θ + \frac{m v^{2}}{r}$ Where $v = \sqrt{v_{bottom}^{2} - 2 g r (1 - \cos θ)}$

Work, Energy and Power

268800 A canon shell fired breaks into two equal parts at its highest point. One part retraces the path to the canon with kinetic energy $E_{1}$ and the kinetic energy of the second part is $E_{2}$ . Relation between $E_{1}$ and $E_{2}$ is
$(2014$ E)

1

E_{2} = E_{1}

2

E_{2} = 4 E_{1}

3

E_{2} = 9 E_{1}

4

E_{2} = 15 E_{1}

Explanation:

At highest point according to L.C.M, $m u \cos θ = - \frac{m}{2} u \cos θ + \frac{m}{2} v$
$E_{1} = \frac{1}{2} ◻ \frac{m}{2} \exists (u \cos θ)^{2}, E_{2} = \frac{1}{2} ∵ \frac{m}{2} θ v^{2}$

Work, Energy and Power

268801 A body of $200 g$ begins to fall from a height where its potential energy is $80 J$ . Its velocity at a point where its kinetic and potential energies are equal(in $m / s) (2014 M)$

1

10 \sqrt{8}

2 4

3 400

4 20

Explanation:

T.E at top $= T . E$ at given point $; P . E =$ P.E $E^{1} + K^{K} E^{1}$
$80 = 2 K . E^{1} (∵ P . E^{1} = K . E^{1})$

Work, Energy and Power

268802 The work done by a force $\bar{F} = 2 \bar{i} - \bar{j} - \bar{k}$ in moving an object from origin to a point whose position vector is $\bar{r} = 3 \bar{i} + 2 \bar{j} - 5 \bar{k} (2013$ E)

1 1 unit

2 9 units

3 13 units

4 60 units

Explanation:

$W = \vec{F} \cdot \vec{S} = \vec{F} \cdot (\vec{r_{2}} - \vec{r_{1}})$

Work, Energy and Power

268803 A ball at rest is dropped from a height of $12 m$ . If it looses $25 %$ of its kinetic energy on striking the ground and bounces back to a height ' $h$ '. the value of ' $h$ ' is equal to ( $2013 E$ )

1

3 m

2

6 m

3

9 m

4

12 m

Explanation:

percentage loss of $PE = \frac{h_{1} - h_{2}}{h_{1}} \times 100 %$

Work, Energy and Power

268804 A block of mass $2.9 kg$ is suspended from a string of length $50 cm$ and is at rest. Another block of mass $100 g$ , which is moving with a speed of $150 m / s$ strikes and sticks to the first block. Subsequently when the string makes an angle of $60^{\circ}$ . With the vertical, the tension in the string will be $(g = 10 {m s}^{- 2})$ ( $2013 E$ )

1

140 N

2

135 N

3

125 N

4

90 N

Explanation:

$v_{common} = \frac{m u}{M + m}$ and $T = m g \cos θ + \frac{m v^{2}}{r}$ Where $v = \sqrt{v_{bottom}^{2} - 2 g r (1 - \cos θ)}$

Work, Energy and Power

268800 A canon shell fired breaks into two equal parts at its highest point. One part retraces the path to the canon with kinetic energy $E_{1}$ and the kinetic energy of the second part is $E_{2}$ . Relation between $E_{1}$ and $E_{2}$ is
$(2014$ E)

1

E_{2} = E_{1}

2

E_{2} = 4 E_{1}

3

E_{2} = 9 E_{1}

4

E_{2} = 15 E_{1}

Explanation:

At highest point according to L.C.M, $m u \cos θ = - \frac{m}{2} u \cos θ + \frac{m}{2} v$
$E_{1} = \frac{1}{2} ◻ \frac{m}{2} \exists (u \cos θ)^{2}, E_{2} = \frac{1}{2} ∵ \frac{m}{2} θ v^{2}$

Work, Energy and Power

268801 A body of $200 g$ begins to fall from a height where its potential energy is $80 J$ . Its velocity at a point where its kinetic and potential energies are equal(in $m / s) (2014 M)$

1

10 \sqrt{8}

2 4

3 400

4 20

Explanation:

T.E at top $= T . E$ at given point $; P . E =$ P.E $E^{1} + K^{K} E^{1}$
$80 = 2 K . E^{1} (∵ P . E^{1} = K . E^{1})$

Work, Energy and Power

268802 The work done by a force $\bar{F} = 2 \bar{i} - \bar{j} - \bar{k}$ in moving an object from origin to a point whose position vector is $\bar{r} = 3 \bar{i} + 2 \bar{j} - 5 \bar{k} (2013$ E)

1 1 unit

2 9 units

3 13 units

4 60 units

Explanation:

$W = \vec{F} \cdot \vec{S} = \vec{F} \cdot (\vec{r_{2}} - \vec{r_{1}})$

Work, Energy and Power

268803 A ball at rest is dropped from a height of $12 m$ . If it looses $25 %$ of its kinetic energy on striking the ground and bounces back to a height ' $h$ '. the value of ' $h$ ' is equal to ( $2013 E$ )

1

3 m

2

6 m

3

9 m

4

12 m

Explanation:

percentage loss of $PE = \frac{h_{1} - h_{2}}{h_{1}} \times 100 %$

Work, Energy and Power

268804 A block of mass $2.9 kg$ is suspended from a string of length $50 cm$ and is at rest. Another block of mass $100 g$ , which is moving with a speed of $150 m / s$ strikes and sticks to the first block. Subsequently when the string makes an angle of $60^{\circ}$ . With the vertical, the tension in the string will be $(g = 10 {m s}^{- 2})$ ( $2013 E$ )

1

140 N

2

135 N

3

125 N

4

90 N

Explanation:

$v_{common} = \frac{m u}{M + m}$ and $T = m g \cos θ + \frac{m v^{2}}{r}$ Where $v = \sqrt{v_{bottom}^{2} - 2 g r (1 - \cos θ)}$

Work, Energy and Power

268800 A canon shell fired breaks into two equal parts at its highest point. One part retraces the path to the canon with kinetic energy $E_{1}$ and the kinetic energy of the second part is $E_{2}$ . Relation between $E_{1}$ and $E_{2}$ is
$(2014$ E)

1

E_{2} = E_{1}

2

E_{2} = 4 E_{1}

3

E_{2} = 9 E_{1}

4

E_{2} = 15 E_{1}

Explanation:

At highest point according to L.C.M, $m u \cos θ = - \frac{m}{2} u \cos θ + \frac{m}{2} v$
$E_{1} = \frac{1}{2} ◻ \frac{m}{2} \exists (u \cos θ)^{2}, E_{2} = \frac{1}{2} ∵ \frac{m}{2} θ v^{2}$

Work, Energy and Power

268801 A body of $200 g$ begins to fall from a height where its potential energy is $80 J$ . Its velocity at a point where its kinetic and potential energies are equal(in $m / s) (2014 M)$

1

10 \sqrt{8}

2 4

3 400

4 20

Explanation:

T.E at top $= T . E$ at given point $; P . E =$ P.E $E^{1} + K^{K} E^{1}$
$80 = 2 K . E^{1} (∵ P . E^{1} = K . E^{1})$

Work, Energy and Power

268802 The work done by a force $\bar{F} = 2 \bar{i} - \bar{j} - \bar{k}$ in moving an object from origin to a point whose position vector is $\bar{r} = 3 \bar{i} + 2 \bar{j} - 5 \bar{k} (2013$ E)

1 1 unit

2 9 units

3 13 units

4 60 units

Explanation:

$W = \vec{F} \cdot \vec{S} = \vec{F} \cdot (\vec{r_{2}} - \vec{r_{1}})$

Work, Energy and Power

268803 A ball at rest is dropped from a height of $12 m$ . If it looses $25 %$ of its kinetic energy on striking the ground and bounces back to a height ' $h$ '. the value of ' $h$ ' is equal to ( $2013 E$ )

1

3 m

2

6 m

3

9 m

4

12 m

Explanation:

percentage loss of $PE = \frac{h_{1} - h_{2}}{h_{1}} \times 100 %$

Work, Energy and Power

268804 A block of mass $2.9 kg$ is suspended from a string of length $50 cm$ and is at rest. Another block of mass $100 g$ , which is moving with a speed of $150 m / s$ strikes and sticks to the first block. Subsequently when the string makes an angle of $60^{\circ}$ . With the vertical, the tension in the string will be $(g = 10 {m s}^{- 2})$ ( $2013 E$ )

1

140 N

2

135 N

3

125 N

4

90 N

Explanation:

$v_{common} = \frac{m u}{M + m}$ and $T = m g \cos θ + \frac{m v^{2}}{r}$ Where $v = \sqrt{v_{bottom}^{2} - 2 g r (1 - \cos θ)}$

Work, Energy and Power

268800 A canon shell fired breaks into two equal parts at its highest point. One part retraces the path to the canon with kinetic energy $E_{1}$ and the kinetic energy of the second part is $E_{2}$ . Relation between $E_{1}$ and $E_{2}$ is
$(2014$ E)

1

E_{2} = E_{1}

2

E_{2} = 4 E_{1}

3

E_{2} = 9 E_{1}

4

E_{2} = 15 E_{1}

Explanation:

At highest point according to L.C.M, $m u \cos θ = - \frac{m}{2} u \cos θ + \frac{m}{2} v$
$E_{1} = \frac{1}{2} ◻ \frac{m}{2} \exists (u \cos θ)^{2}, E_{2} = \frac{1}{2} ∵ \frac{m}{2} θ v^{2}$

Work, Energy and Power

268801 A body of $200 g$ begins to fall from a height where its potential energy is $80 J$ . Its velocity at a point where its kinetic and potential energies are equal(in $m / s) (2014 M)$

1

10 \sqrt{8}

2 4

3 400

4 20

Explanation:

T.E at top $= T . E$ at given point $; P . E =$ P.E $E^{1} + K^{K} E^{1}$
$80 = 2 K . E^{1} (∵ P . E^{1} = K . E^{1})$

Work, Energy and Power

268802 The work done by a force $\bar{F} = 2 \bar{i} - \bar{j} - \bar{k}$ in moving an object from origin to a point whose position vector is $\bar{r} = 3 \bar{i} + 2 \bar{j} - 5 \bar{k} (2013$ E)

1 1 unit

2 9 units

3 13 units

4 60 units

Explanation:

$W = \vec{F} \cdot \vec{S} = \vec{F} \cdot (\vec{r_{2}} - \vec{r_{1}})$

Work, Energy and Power

268803 A ball at rest is dropped from a height of $12 m$ . If it looses $25 %$ of its kinetic energy on striking the ground and bounces back to a height ' $h$ '. the value of ' $h$ ' is equal to ( $2013 E$ )

1

3 m

2

6 m

3

9 m

4

12 m

Explanation:

percentage loss of $PE = \frac{h_{1} - h_{2}}{h_{1}} \times 100 %$

Work, Energy and Power

268804 A block of mass $2.9 kg$ is suspended from a string of length $50 cm$ and is at rest. Another block of mass $100 g$ , which is moving with a speed of $150 m / s$ strikes and sticks to the first block. Subsequently when the string makes an angle of $60^{\circ}$ . With the vertical, the tension in the string will be $(g = 10 {m s}^{- 2})$ ( $2013 E$ )

1

140 N

2

135 N

3

125 N

4

90 N

Explanation:

$v_{common} = \frac{m u}{M + m}$ and $T = m g \cos θ + \frac{m v^{2}}{r}$ Where $v = \sqrt{v_{bottom}^{2} - 2 g r (1 - \cos θ)}$

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