A person with his hands in his pockets is skating on ice at the rate of 10 ms -1 and describes a circle of 50 m radius. What is his inclination to the vertical? (Given: g=10 ms -2 )

Question

A person with his hands in his pockets is skating on ice at the rate of 10 ms -1 and describes a circle of 50 m radius. What is his inclination to the vertical? (Given: g=10 ms -2 ) (A)  tan -1((1/2)) (B)  tan -1((1/5)) (C)  tan -1((3/5)) (D)  tan -1((1/10))

Tardigrade Admin · Accepted Answer

tan θ=(v2/R g)=((10 ms -1)2/(50 m )(10 ms -2))=(1/5) ⇒ θ= tan -1((1/5))

Q. A person with his hands in his pockets is skating on ice at the rate of $10 m s^{- 1}$ and describes a circle of $50 m$ radius. What is his inclination to the vertical? (Given: $g = 10 m s^{- 2})$

$tan^{- 1} (\frac{1}{2})$

$tan^{- 1} (\frac{1}{5})$

$tan^{- 1} (\frac{3}{5})$

$tan^{- 1} (\frac{1}{10})$

$tan θ = \frac{v ^{2}}{R g} = \frac{( 10 m s ^{- 1} ) ^{2}}{( 50 m ) ( 10 m s ^{- 2} )} = \frac{1}{5}$
$\Rightarrow θ = tan^{- 1} (\frac{1}{5})$

Q. A person with his hands in his pockets is skating on ice at the rate of 10ms−1 and describes a circle of 50m radius. What is his inclination to the vertical? (Given: g=10ms−2)

tan−1(21​)

tan−1(51​)

tan−1(53​)

tan−1(101​)