MCQ
A constant torque acting on a uniform circular wheel changes its angular momentum from $A_0$ to $4 A_0$ in $4$ seconds. The magnitude of this torque is ...........
- ✓$\frac{3 A_0}{4}$
- B$A_0$
- C$4 A_0$
- D$12 A_0$
✓
Answer
Correct option: A.
$\frac{3 A_0}{4}$
a
(a)
(a)
$\tau=\frac{\Delta L}{\Delta t}$
$\tau=\frac{4 A_0-A_0}{4}=\frac{3 A_0}{4}$
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|
Column $-I$ Angle of projection |
Column $-II$ |
| $A.$ $\theta \, = \,{45^o}$ | $1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$ |
| $B.$ $\theta \, = \,{60^o}$ | $2.$ $\frac{{g{T^2}}}{R} = 8$ |
| $C.$ $\theta \, = \,{30^o}$ | $3.$ $\frac{R}{H} = 4\sqrt 3 $ |
| $D.$ $\theta \, = \,{\tan ^{ - 1}}\,4$ | $4.$ $\frac{R}{H} = 4$ |
$K_i :$ initial kinetic energy
$K_h :$ kinetic energy at the highest point