Motion in a Plane — NEET UG practice

1. Angular speed of a hand is the angle it sweeps per unit time, and it completes one full revolution ($2\pi$ radians) in the time it takes to go around the clock once.

For the second hand, it completes one revolution in $60$ seconds:

$\omega_1=\frac{2\pi}{60}$

Since the angular distance covered in $1$ minute ($60$ s) is $x_1=\omega_1\times60$:

$x_1=2\pi$

2. For the minute hand, it completes one revolution in $3600$ seconds (one hour):

$\omega_2=\frac{2\pi}{3600}$

So the angular distance in $1$ minute is:

$x_2=\omega_2\times60=\frac{2\pi}{60}$

3. For the hour hand, it completes one revolution in $3600\times12$ seconds (12 hours):

$\omega_3=\frac{2\pi}{3600\times12}$

So the angular distance in $1$ minute is:

$x_3=\omega_3\times60=\frac{2\pi}{720}$

4. Now check which ratio stays the same for all three hands. Dividing $\omega$ by $x$ for each hand:

$\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=\frac{1}{60}=k$

Hence, the answer is A.

1. When a bob is whirled in a horizontal circle, the tension in the string acts as the centripetal force that keeps the bob moving in the circle.

$T=\frac{mv^2}{r}$

2. Given that the radius $r$ and mass $m$ stay the same, and the tension is quadrupled, we can write the new tension as:

$T_2=4T_1$

3. Writing both tensions using the centripetal force formula:

$T_1=\frac{mv_1^2}{r},\qquad T_2=\frac{mv_2^2}{r}$

Substituting $T_2=4T_1$:

$4\left(\frac{mv_1^2}{r}\right)=\frac{mv_2^2}{r}$

4. Since $m$ and $r$ are common to both sides, they cancel out, giving:

$4v_1^2=v_2^2$

5. Taking the square root of both sides:

$v_2=2v_1$

Given $v_1=10$ rpm, therefore:

$v_2=2\times10=20\ \text{rpm}$

Hence, the answer is A (20 rpm).

1. For a projectile launched with speed $u$ at angle $\theta$ to the horizontal, only the vertical component of velocity decides how high it goes, and the maximum height is given by:

$H_{max}=\frac{u^2\sin^2\theta}{2g}$

2. Substituting the given values $u=280\ \text{m/s}$, $\sin30^\circ=0.5$, and $g=9.8\ \text{m/s}^2$:

$H_{max}=\frac{(280)^2(0.5)^2}{2(9.8)}$

3. Simplifying step by step:

$H_{max}=\frac{78400\times0.25}{19.6}$

$H_{max}=1000\ \text{m}$

Hence, the answer is B (1000 m).