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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Physics
Chapter
Laws of Motion

BITSAT 2010 · Physics · Laws of Motion

Two blocks are connected over a massless pulley as shown in fig....

Two blocks are connected over a massless pulley as shown in fig. The mass of block AA is 10kg10 \mathrm{ kg} and the coefficient of kinetic Two blocks are connected over a massless pulley as shown in fig. The mass of block A is 10 kg and the coefficient of
friction is 0.2. Block A slides down the incline at constant speed. The mass of block B\mathrm{B} in kg\mathrm{kg} is:

Options

  1. A3.5
  2. B3.3
  3. C3
  4. D2.5

Correct answer

B. 3.3

Step-by-step solution

Considering the equilibrium of A, we get 10a=10 gsin⁡30∘−T−μN where N=10 gcos⁡30∘\begin{array}{l}10 \mathrm{a}=10 \mathrm{~g} \sin30^{{}^{\circ}}-\mathrm{T}-\mu \mathrm{N}\\ \text{ where } \mathrm{N}=10 \mathrm{~g} \cos30^{{}^{\circ}} \end{array} Two blocks are connected over a massless pulley as shown in fig. The mass of block A is 10 kg and the coefficient of
∴10a=102 g−T−\therefore 10 \mathrm{a}=\frac{10}{2} \mathrm{~g}-\mathrm{T}-\mu \times 10  gcos⁡30∘buta=0\mathrm{~g} \cos 30^{{}^{\circ}} but \mathrm{a}=0, \mathrm{~T}=\mathrm{m}_{\mathrm{B}} \mathrm{g}0=5 \mathrm{~g}-mBg\mathrm{m}_{\mathrm{B}} \mathrm{g}-0.232×10×g\frac{0.2 \sqrt{3}}{2} \times 10 \times \mathrm{g}\Rightarrow mB=3.268≈3.3 kg\mathrm{m}_{\mathrm{B}}=3.268 \approx 3.3 \mathrm{~kg}
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