A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: [2025]

(Take acceleration due to gravity as 10 $m / s^{2}$ )

Question

A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: [2025]

(Take acceleration due to gravity as 10 $m / s^{2}$ )

Smart Achievers · Accepted Answer

(3)

$\frac{1}{2} m \times 100 + 0 = \frac{1}{2} m V_{B}^{2} + m g (R - \frac{R \sqrt{3}}{2})$

$100 = V_{B}^{2} + 2 g R (1 - \frac{\sqrt{3}}{2})$

$V_{B}^{2} = 100 - 20 (2 - \sqrt{3})$

$\Rightarrow V_{B}^{2} = 60 + 20 \sqrt{3}$

$K . E ._{B} = \frac{1}{2} m V_{B}^{2} = \frac{m}{2} (60 + 20 \sqrt{3})$ ...(i)

$\frac{1}{2} m (100) = \frac{1}{2} m V_{C}^{2} + m g (\frac{3 R}{2})$

$100 = V_{C}^{2} + 60$

$V_{C}^{2} = 40$

$K . E ._{C} = \frac{1}{2} m V_{C}^{2} = \frac{1}{2} m (40)$

$\Rightarrow \frac{K_{B}}{K_{C}} = {(\frac{V_{B}}{V_{C}})}^{2} = \frac{3 + \sqrt{3}}{2}$

Solution

Q.
A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: [2025]

(Take acceleration due to gravity as 10 $m / s^{2}$ )

1 $\frac{2 + \sqrt{3}}{3}$

2 $\frac{2 + \sqrt{2}}{3}$

3 $\frac{3 + \sqrt{3}}{2}$

4 $\frac{3 - \sqrt{2}}{2}$

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Solution

Q. A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: [2025] (Take acceleration due to gravity as 10 m/s2)

1 2+33

2 2+23

3 3+32

4 3–22