KCET · Physics · Rotational Motion
Two fly wheels are connected by a non-slipping belt as shown in the figure. \(I_1=4 \mathrm{~kg} \mathrm{~m}^2, r_1=20 \mathrm{~cm}\), \(\mathrm{I}_2=20 \mathrm{kgm}^2\) and \(\mathrm{r}_2=30 \mathrm{~cm}\). A torque of 10 Nm is applied on the smaller wheel. Then match the entries of column I with appropriate entries of column II.
| Quantities | Their numerical Values (in SI units) |
| a. Angular acceleration of smaller wheel | 1. \(\frac{5}{3}\) |
| b. Torque on the larger wheel | 2. \(\frac{100}{3}\) |
| c. Angular acceleration of larger wheel | 3. \(\frac{5}{2}\) |
- A a - ii, b-iii, c - i
- B a-iii, b-i, c-ii
- C a-ii, b-i, c-iii
- D a - iii, b-ii, c-i
Answer & Solution
Correct Answer
(D) a - iii, b-ii, c-i
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}\mathrm{T}_1=\mathrm{I}_1 \alpha_1 & \alpha_2 \mathrm{H}_2=\alpha_1 \mathrm{H}_1 \\ 10=\alpha_1 \times 4 & \alpha_2 \times 0.3=\frac{5}{2} \times 0.2 \\ \therefore \alpha_1=\frac{5}{2} \mathrm{SI} & \alpha_2=\frac{5}{3} \mathrm{SI} \\ \therefore \mathrm{T}_2=\mathrm{I}_2 \alpha_2=20 \times \frac{5}{3}=\frac{100}{3} \mathrm{SI}\end{array}\)
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