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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.
QuantitiesTheir numerical Values (in SI units)
a. Angular acceleration of smaller wheel1. \(\frac{5}{3}\)
b. Torque on the larger wheel2. \(\frac{100}{3}\)
c. Angular acceleration of larger wheel3. \(\frac{5}{2}\)

  1. A a - ii, b-iii, c - i
  2. B a-iii, b-i, c-ii
  3. C a-ii, b-i, c-iii
  4. D a - iii, b-ii, c-i
Verified Solution

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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