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Work, Energy and Power

268734 A body of mass $m$ is rotated in a vertical circle with help of light string such that velocity of body at a point is equal to critical velocity at that point. If $T_{1}, T_{2}$ be the tensions in the string when the body is crossing the highest and the lowest positions then the following relation is correct

1

T_{2} - T_{1} = 6 mg

2

T_{2} - T_{1} = 4 mg

3

T_{2}^{2} - T_{1}^{1} = 3 mg

4

T_{2}^{2} - T_{1}^{1} = 2 mg

Explanation:

$T_{1} = \frac{m v_{1}^{2}}{r} - m g, T_{2} = \frac{m v_{2}^{2}}{r} + m g$ $v_{2}^{2} - v_{1}^{2} = 4 g r, v_{1} = \sqrt{g r}$

Work, Energy and Power

268735 A vehicle is travelling with uniform speed along a concave road of radius of curvature $19.6 m$ . At lowest point of concave road if the normal reaction on the vehicle is three times its weight, the speed of vehicle is

1

4.9 m / s

2

9.8 m / s

3

14.7 m / s

4

19.6 m / s

Explanation:

Normal reaction, $N = m g + \frac{m v^{2}}{r}$ ;Given $N = 3 mg$

Work, Energy and Power

268736 A car is travelling along a flyover bridge which is a part of vertical circle of radius $10 m$ . At the highest point of it the normal reaction on the car is half of its weight, the speed of car is

1

7 m / s

2

10 m / s

3

14 m / s

4

20 m / s

Explanation:

Normal reaction, $N = m g - \frac{m v^{2}}{r}$
Given that $N = \frac{m g}{2}$

Work, Energy and Power

268737 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it has to leave the hemisphere without sliding down its surface $(g = 9.8 {m s}^{- 2})$ is

1

\sqrt{9.8} m / s

2

\sqrt{4.9} m / s

3

\sqrt{1.96} m / s

4

\sqrt{3.92} m / s

Explanation:

$V_{min} = \sqrt{r g}$

Work, Energy and Power

268788 A simple pendulum of length ' $l$ ' carries a bob of mass ' $m$ '. If the breaking strength of the string is $2 mg$ , the maximum angular amplitude from the vertical can be

1

0^{\circ}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

If $θ$ is angular amplitude, $T = m g (3 - 2 \cos θ)$

Work, Energy and Power

268734 A body of mass $m$ is rotated in a vertical circle with help of light string such that velocity of body at a point is equal to critical velocity at that point. If $T_{1}, T_{2}$ be the tensions in the string when the body is crossing the highest and the lowest positions then the following relation is correct

1

T_{2} - T_{1} = 6 mg

2

T_{2} - T_{1} = 4 mg

3

T_{2}^{2} - T_{1}^{1} = 3 mg

4

T_{2}^{2} - T_{1}^{1} = 2 mg

Explanation:

$T_{1} = \frac{m v_{1}^{2}}{r} - m g, T_{2} = \frac{m v_{2}^{2}}{r} + m g$ $v_{2}^{2} - v_{1}^{2} = 4 g r, v_{1} = \sqrt{g r}$

Work, Energy and Power

268735 A vehicle is travelling with uniform speed along a concave road of radius of curvature $19.6 m$ . At lowest point of concave road if the normal reaction on the vehicle is three times its weight, the speed of vehicle is

1

4.9 m / s

2

9.8 m / s

3

14.7 m / s

4

19.6 m / s

Explanation:

Normal reaction, $N = m g + \frac{m v^{2}}{r}$ ;Given $N = 3 mg$

Work, Energy and Power

268736 A car is travelling along a flyover bridge which is a part of vertical circle of radius $10 m$ . At the highest point of it the normal reaction on the car is half of its weight, the speed of car is

1

7 m / s

2

10 m / s

3

14 m / s

4

20 m / s

Explanation:

Normal reaction, $N = m g - \frac{m v^{2}}{r}$
Given that $N = \frac{m g}{2}$

Work, Energy and Power

268737 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it has to leave the hemisphere without sliding down its surface $(g = 9.8 {m s}^{- 2})$ is

1

\sqrt{9.8} m / s

2

\sqrt{4.9} m / s

3

\sqrt{1.96} m / s

4

\sqrt{3.92} m / s

Explanation:

$V_{min} = \sqrt{r g}$

Work, Energy and Power

268788 A simple pendulum of length ' $l$ ' carries a bob of mass ' $m$ '. If the breaking strength of the string is $2 mg$ , the maximum angular amplitude from the vertical can be

1

0^{\circ}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

If $θ$ is angular amplitude, $T = m g (3 - 2 \cos θ)$

Work, Energy and Power

268734 A body of mass $m$ is rotated in a vertical circle with help of light string such that velocity of body at a point is equal to critical velocity at that point. If $T_{1}, T_{2}$ be the tensions in the string when the body is crossing the highest and the lowest positions then the following relation is correct

1

T_{2} - T_{1} = 6 mg

2

T_{2} - T_{1} = 4 mg

3

T_{2}^{2} - T_{1}^{1} = 3 mg

4

T_{2}^{2} - T_{1}^{1} = 2 mg

Explanation:

$T_{1} = \frac{m v_{1}^{2}}{r} - m g, T_{2} = \frac{m v_{2}^{2}}{r} + m g$ $v_{2}^{2} - v_{1}^{2} = 4 g r, v_{1} = \sqrt{g r}$

Work, Energy and Power

268735 A vehicle is travelling with uniform speed along a concave road of radius of curvature $19.6 m$ . At lowest point of concave road if the normal reaction on the vehicle is three times its weight, the speed of vehicle is

1

4.9 m / s

2

9.8 m / s

3

14.7 m / s

4

19.6 m / s

Explanation:

Normal reaction, $N = m g + \frac{m v^{2}}{r}$ ;Given $N = 3 mg$

Work, Energy and Power

268736 A car is travelling along a flyover bridge which is a part of vertical circle of radius $10 m$ . At the highest point of it the normal reaction on the car is half of its weight, the speed of car is

1

7 m / s

2

10 m / s

3

14 m / s

4

20 m / s

Explanation:

Normal reaction, $N = m g - \frac{m v^{2}}{r}$
Given that $N = \frac{m g}{2}$

Work, Energy and Power

268737 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it has to leave the hemisphere without sliding down its surface $(g = 9.8 {m s}^{- 2})$ is

1

\sqrt{9.8} m / s

2

\sqrt{4.9} m / s

3

\sqrt{1.96} m / s

4

\sqrt{3.92} m / s

Explanation:

$V_{min} = \sqrt{r g}$

Work, Energy and Power

268788 A simple pendulum of length ' $l$ ' carries a bob of mass ' $m$ '. If the breaking strength of the string is $2 mg$ , the maximum angular amplitude from the vertical can be

1

0^{\circ}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

If $θ$ is angular amplitude, $T = m g (3 - 2 \cos θ)$

Work, Energy and Power

268734 A body of mass $m$ is rotated in a vertical circle with help of light string such that velocity of body at a point is equal to critical velocity at that point. If $T_{1}, T_{2}$ be the tensions in the string when the body is crossing the highest and the lowest positions then the following relation is correct

1

T_{2} - T_{1} = 6 mg

2

T_{2} - T_{1} = 4 mg

3

T_{2}^{2} - T_{1}^{1} = 3 mg

4

T_{2}^{2} - T_{1}^{1} = 2 mg

Explanation:

$T_{1} = \frac{m v_{1}^{2}}{r} - m g, T_{2} = \frac{m v_{2}^{2}}{r} + m g$ $v_{2}^{2} - v_{1}^{2} = 4 g r, v_{1} = \sqrt{g r}$

Work, Energy and Power

268735 A vehicle is travelling with uniform speed along a concave road of radius of curvature $19.6 m$ . At lowest point of concave road if the normal reaction on the vehicle is three times its weight, the speed of vehicle is

1

4.9 m / s

2

9.8 m / s

3

14.7 m / s

4

19.6 m / s

Explanation:

Normal reaction, $N = m g + \frac{m v^{2}}{r}$ ;Given $N = 3 mg$

Work, Energy and Power

268736 A car is travelling along a flyover bridge which is a part of vertical circle of radius $10 m$ . At the highest point of it the normal reaction on the car is half of its weight, the speed of car is

1

7 m / s

2

10 m / s

3

14 m / s

4

20 m / s

Explanation:

Normal reaction, $N = m g - \frac{m v^{2}}{r}$
Given that $N = \frac{m g}{2}$

Work, Energy and Power

268737 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it has to leave the hemisphere without sliding down its surface $(g = 9.8 {m s}^{- 2})$ is

1

\sqrt{9.8} m / s

2

\sqrt{4.9} m / s

3

\sqrt{1.96} m / s

4

\sqrt{3.92} m / s

Explanation:

$V_{min} = \sqrt{r g}$

Work, Energy and Power

268788 A simple pendulum of length ' $l$ ' carries a bob of mass ' $m$ '. If the breaking strength of the string is $2 mg$ , the maximum angular amplitude from the vertical can be

1

0^{\circ}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

If $θ$ is angular amplitude, $T = m g (3 - 2 \cos θ)$

Work, Energy and Power

268734 A body of mass $m$ is rotated in a vertical circle with help of light string such that velocity of body at a point is equal to critical velocity at that point. If $T_{1}, T_{2}$ be the tensions in the string when the body is crossing the highest and the lowest positions then the following relation is correct

1

T_{2} - T_{1} = 6 mg

2

T_{2} - T_{1} = 4 mg

3

T_{2}^{2} - T_{1}^{1} = 3 mg

4

T_{2}^{2} - T_{1}^{1} = 2 mg

Explanation:

$T_{1} = \frac{m v_{1}^{2}}{r} - m g, T_{2} = \frac{m v_{2}^{2}}{r} + m g$ $v_{2}^{2} - v_{1}^{2} = 4 g r, v_{1} = \sqrt{g r}$

Work, Energy and Power

268735 A vehicle is travelling with uniform speed along a concave road of radius of curvature $19.6 m$ . At lowest point of concave road if the normal reaction on the vehicle is three times its weight, the speed of vehicle is

1

4.9 m / s

2

9.8 m / s

3

14.7 m / s

4

19.6 m / s

Explanation:

Normal reaction, $N = m g + \frac{m v^{2}}{r}$ ;Given $N = 3 mg$

Work, Energy and Power

268736 A car is travelling along a flyover bridge which is a part of vertical circle of radius $10 m$ . At the highest point of it the normal reaction on the car is half of its weight, the speed of car is

1

7 m / s

2

10 m / s

3

14 m / s

4

20 m / s

Explanation:

Normal reaction, $N = m g - \frac{m v^{2}}{r}$
Given that $N = \frac{m g}{2}$

Work, Energy and Power

268737 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it has to leave the hemisphere without sliding down its surface $(g = 9.8 {m s}^{- 2})$ is

1

\sqrt{9.8} m / s

2

\sqrt{4.9} m / s

3

\sqrt{1.96} m / s

4

\sqrt{3.92} m / s

Explanation:

$V_{min} = \sqrt{r g}$

Work, Energy and Power

268788 A simple pendulum of length ' $l$ ' carries a bob of mass ' $m$ '. If the breaking strength of the string is $2 mg$ , the maximum angular amplitude from the vertical can be

1

0^{\circ}

2

30^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

If $θ$ is angular amplitude, $T = m g (3 - 2 \cos θ)$

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