Q.30. A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In th

Q: What is the correct answer for JEE Main Physics Q30?

The correct answer is B. Sol. Initial angular velocity $\omega_0 = 0$. Uniform angular acceleration $\alpha$. For uniform angular acceleration, the angular displacement is given by $\theta = \omega_0 t + (1/2) \alpha t^2$. In the first 2 s: $\theta_1 = 0 \cdot 2 + (1/2) \alpha (2)^2 = (1/2) \alpha \cdot 4 = 2\alpha$. In the next 2 s (i.e., from t=2s to t=4s): The total angle rotated in 4s is $\theta_{total} = 0 \cdot 4 +

Q: Q.30. A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle . In the next 2 s it rotates through an angle . The ratio…

The correct answer is B. Sol. Initial angular velocity $\omega_0 = 0$. Uniform angular acceleration $\alpha$. For uniform angular acceleration, the angular displacement is given by $\theta = \omega_0 t + (1/2) \alpha t^2$. In the first 2 s: $\theta_1 = 0 \cdot 2 + (1/2) \alpha (2)^2 = (1/2) \alpha \cdot 4 = 2\alpha$. In the next 2 s (i.e., from t=2s to t=4s): The total angle rotated in 4s is $\theta_{total} = 0 \cdot 4 +

Question

Q.30. A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle . In the next 2 s it rotates through an angle . The ratio  is ________.

FirstInTest · Accepted Answer

Sol. Initial angular velocity $\omega_0 = 0$. Uniform angular acceleration $\alpha$. For uniform angular acceleration, the angular displacement is given by $	heta = \omega_0 t + (1/2) \alpha t^2$.
In the first 2 s: $	heta_1 = 0 \cdot 2 + (1/2) \alpha (2)^2 = (1/2) \alpha \cdot 4 = 2\alpha$.
In the next 2 s (i.e., from t=2s to t=4s): The total angle rotated in 4s is $	heta_{total} = 0 \cdot 4 + (1/2) \alpha (4)^2 = (1/2) \alpha \cdot 16 = 8\alpha$.
So, $	heta_2 = 	heta_{total} - 	heta_1 = 8\alpha - 2\alpha = 6\alpha$.
The ratio $\frac{	heta_2}{	heta_1} = \frac{6\alpha}{2\alpha} = 3$.
Alternatively, using the formula for displacement in nth interval: $	heta_n = \omega_0 + \alpha(n - 1/2)$.
For the first 2s (n=1, considering 2s as a single interval): $	heta_1 = (1/2) \alpha (2)^2 = 2\alpha$.
For the next 2s (which is the second 2s interval, so from t=2 to t=4, this is the 2nd interval of 2s duration). Let's use the standard approach for consecutive intervals.
Angle in first 2 sec: $	heta_1 = (1/2) \alpha (2)^2 = 2\alpha$.
Angle in next 2 sec: This is the angle rotated from $t=2s$ to $t=4s$. The total angle rotated in $4s$ is $	heta_{4s} = (1/2) \alpha (4)^2 = 8\alpha$.
So, $	heta_2 = 	heta_{4s} - 	heta_1 = 8\alpha - 2\alpha = 6\alpha$.
Ratio $\frac{	heta_2}{	heta_1} = \frac{6\alpha}{2\alpha} = 3$.

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