JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 18)
A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the
figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25
rotations per second in 5s, is close to :
_10th_April_Evening_Slot_en_18_1.png)
7.9 × 10–6 Nm
4.0 × 10–6 Nm
2.0 × 10–5 Nm
1.6 × 10–5 Nm
Explanation
m = 5 × 10–3 kg, r = 10–2 m
$$\omega $$ = 25 × 2$$\pi $$ rad/5
= 50 $$\omega $$ rad/sec
$$\omega = {\tau \over I}t$$
$$\tau = {{I\omega } \over t} = {{5m{r^2}} \over 4} \times {\omega \over t}$$
$$ = {{5 \times 5 \times {{10}^{ - 3}} \times {{10}^{ - 4}} \times 50\pi } \over {4 \times 5}}$$
$$ = {{25\pi } \over 4} \times {10^{ - 6}} = 2 \times {10^{ - 5}}$$ Nm
$$\omega $$ = 25 × 2$$\pi $$ rad/5
= 50 $$\omega $$ rad/sec
$$\omega = {\tau \over I}t$$
$$\tau = {{I\omega } \over t} = {{5m{r^2}} \over 4} \times {\omega \over t}$$
$$ = {{5 \times 5 \times {{10}^{ - 3}} \times {{10}^{ - 4}} \times 50\pi } \over {4 \times 5}}$$
$$ = {{25\pi } \over 4} \times {10^{ - 6}} = 2 \times {10^{ - 5}}$$ Nm
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