A fan initially rotating at 60 rad/sec clockwise accelerates and acquires an angular velocity of 110 rad/sec clockwise. If the angular acceleration of the fan is constant and it takes 2 sec to change its angular velocity, find the angular acceleration of the fan. A. 50 rad/sec2 in counterclockwise direction B. 25 rad/sec2 in counterclockwise direction C. 25 rad/sec2 in clockwise direction D. 50 rad/sec2 in clockwise direction

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The correct option is (C)

The equation of rotational motion for constant angular acceleration is given by

ω_f = ω_i + αt

⇒ α = \(\frac{\omega_f - \omega_i}{t}\)

Where,

ω_f = final angular velocity
ω_i = initial angular velocity
α = angular acceleration
t = time taken to reach final angular velocity from initial angular velocity.

Taking the clockwise direction as -ve and counterclockwise as +ve.

Given,

Initial angular velocity of the fan, ω_i = – 60 rad/s

final angular velocity of the fan, ω_f = – 110 rad/s

Time taken, t = 2 sec

Using the equation of rotational motion, angular acceleration (α) of the fan, is given by

α = \(\frac{(-110)-(-60)}{2}\) = – \(\frac{50}{2}\)

⇒ α = – 25 rad/sec²

Negative sign shows that the angular acceleration of the fan is in the clockwise direction.

Hence, the angular acceleration of the fan is 25 rad/sec² in clockwise direction.

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A fan initially rotating at 60 rad/sec clockwise accelerates and acquires an angular velocity of 110 rad/sec clockwise. If the angular acceleration of the fan is constant and it takes 2 sec to change its angular velocity, find the angular acceleration.

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2.The energy of an electron in an orbit in hydrogen atom is \( -3.4 \, \text{eV} \). Its angular momentum in the orbit will be:

3.The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

5.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

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